QUESTION:
Came across this particular question on a forum with a multitude of
different answers and their opinion of it, I would like to hear your input
on it. Here is the question: How many days will it take a spaceship to
acceleration to the speed of (3.0x10^{8} m/s) with the acceleration g ? how
far will it travel during this interval and what fraction of a light year is
the answer to the previous question?

ANSWER:
(This sounded like a homework problem to me.) What forum?

REPLY:
Reddit .

ANSWER:
Now I know it is homework because this section of AskReddit
has a title cheatatmathhomework ! And if you can
read you see that AskThePhysicist.com does not do homework
so I should just trash this question. However, this question
was submitted to AskReddit a year ago so the present
questioner is presumably interested in the answer since it
is marked UNSOLVED . But,
the problem is not unsolved because the one comment points
the reader to a site where the one-dimensional, constant
acceleration kinematics is discussed; this is the
appropriate way to help someone do their homework, not
working it out to the final answer. The solution to the
problem is a simple application of these equations. So here
is my answer:

You have not accurately conveyed the problem. Most important, the final speed is not 3x10^{8} m/s but rather 1% of that number. But that is a small enough fraction of the speed of light that you can do the problem classically, not relativistically
(which would be much more difficult). The problem is not unsolved since the single answer points you to the appropriate equations to use, Specifically,
x ≈½gt ^{2} and v≈gt . Here,
g =9.8 m/s^{2} , v =3x10^{6}
m/s. Do the algebra to solve for the two unknowns, convert
t from seconds to days and convert x from
meters to light years.

ADDED NOTE: I worked out the answers and
found that the problem, as stated, does not have the correct
answer for the distance traveled. The correct answer is
4.6x10^{11} m=4.9x10^{-5} ly.

QUESTION:
Hello, I'm not a student or anything I was watching a TED Talk video and it inspired a curiosity question. He was talking about antimatter accelerators and transverse seeing the universe at various speeds of light, and he said that if the ship could approach 99.99 percent the speed of light then we could reach the edge of the viewable Universe in 30 years relative to the spacecraft. However 13.9 billion years would have passed here on Earth.

My question is—how would a radio signal behave if this ship pointed a transmitter at the earth when it started this journey and broadcast a constant radio signal as it grew closer and closer to light speed traveling further away?

ANSWER:
Well, you got your numbers wrong somehow. 99.99% of c is not
really that big for such a long distance. If you go at that
speed, the distance (as seen by the ship) would be 13.9x10^{9} √(1-0.9999^{2} )=2x10^{8}
ly. Since the speed of the ship is 0.9999 ly/yr, the time
necessary would be 2x10^{8} /0.9999≈2x10^{8}
yr, not 30. I have seen this kind of estimate before and the
situation being described was probably that the ship has a
constant acceleration (as seen by the ship) equal to the
acceleration of gravity (about 10 m/s^{2} )
essentially forever; so the speed (as seen from earth) gets
to be much larger than 0.9999c . You can see the
calculation of the speed of accelerating objects as seen
from an inertial frame in an
earlier answer . If you want a really detailed discussion
of getting to the edge of the universe in a few decades, see
this post by
Dave Goldberg ; his number is more like 99.9999999999999999998% of
c !

But your question does
not depend on the details of how far you went and how long
it took you to get there. The radio signals being received
on earth will be Doppler shifted to lower frequency, the
shift getting larger as the speed gets larger. This is the
same idea as the Doppler shift of a siren going away from
you getting changed to a lower pitch. The expression for the
shift in frequency is f _{observed} =f _{source} √[(1-(v /c ))/(1+(v /c ))]=√[(1-0.9999/(1+0.9999)]=0.0007f _{source} ,
a much lower frequency.

QUESTION:
Theoretically, if I look through a telescope at a mirror 2 light years away, the light bouncing back would give me a instant view of my location from 4 years ago, right?
With that, if I was in a ship that safely accelerated to the speed of light, say 1 g acceleration, so that I would reach the speed of light after approximately 1 year, traveling 1/2 of a light year, then turning around and decelerating at 1 g for one year to come to a complete stop, then doing the same to get back to my original location, thereby traveling 4 years total, why would I change 4 years in age, but my original location (and final destination) be more than 4 years older?

ANSWER:
Yes, the mirror would give you the view four years earlier.
However, your understanding of the accelerating spaceship is
very wrong. It is a pretty difficult thing to understand, so
I suggest that you read an
earlier answer
as well as the earlier answer it links to. When you have
done that, you should be able to follow my answer to your
question.

First a qualitative discussion. You decide that you want
your spaceship to have a constant acceleration of g
as measured by you, captain of the ship. You therefore
arrange to have your engines always exert a force on the
ship of F=mg where m is the mass of the
ship. So you will always see your ship having the desired
acceleration. But, as I have emphasized in many earlier
answers, acceleration you measure is not the same as
acceleration someone on earth measures. In Newtonian
physics, everyone measures the same acceleration. This is
really a glorified
twin paradox problem and I am not going to calculate how
long the round trip will be for you (but I will calculate
how long it takes in earth time). What I can tell you is that you
will take longer than 4 years to return to earth and the
time elapsed on earth will be longer than your time. Your
desire to accelerate "to
the speed of light" is impossible, as you will see, so
forget that; with your scenario the fastest you will ever go
is about 86% of c .

What I want to focus on is what someone on
earth will measure regarding your motion. This is mostly
worked out in the earlier answers I have referred you to, in
particular v /c =(gt /c )/√[1+(gt /c )^{2} ]
and a /g =[1+(gt /c )^{2} ]^{-3/2} ;
these are plotted in the graph above with the red and blue
curves. Things to notice are that after 1 year you have only
acquired a speed of about 0.7c and your
acceleration has dropped to about 0.35g . What was
not derived earlier was your position as a function of time;
integrating the velocity equation, I find gx /c ^{2} =√[1+(gt /c )^{2} ]-1,
shown by the black curve. Things to notice are that it
starts out looking like a parbola as it would if you had a
constant acceleration of g , but eventually becomes
a straight line as you approach the speed of light which you
never reach. Reading off the graph, the time it takes you to
get halfway, 1 ly, is about 1.7 years (as measured from
earth); your speed will be about 0.86c and your acceleration
(again as measured from earth) will be about 0.13g .
This will be ¼ of your tota l trip, 6.8 years. The time you measure will be somewhere
between 4 and 6.8 years.

Incidentally, the axes
may be shown to be numerically equal to years and light
years if you approximate g ≈10 m/s^{2} .

ADDED COMMENT: I see that I have misread
the question. For some reason the questioner reverses his thrust at ½
ly out so he only goes to 1 ly before turning the ship back
to earth. From the graph you can see that the time to get to
this position is about 1.1 years and the speed there is
about 0.75c . Thus the total elapsed time on earth
before your return would be 4.4 years in this scenario.
Light would take 2 years to make this trip. Again, I have
not calculated the time on your clock but it would be
between 2 and 4.4 years.

QUESTION:
I understand that the motion of a light source doesn't add to or subtract from the speed, or change the direction, of the light that it emits. So, if I was a passenger on a spaceship, and I shone a flashlight from one side of the ship to the other, perpendicular to the direction of travel, the light beam wouldn’t strike the wall exactly where the flashlight was pointed, but rather a little towards the rear, as the opposite wall would have moved forward some small distance in the time it took the photons to cross the ship. And if the ship was traveling at, say, half the speed of light, the light beam would leave the flashlight at a noticeable angle. Is this correct?

ANSWER:
If the spaceship were moving with constant velocity, the
light beam would go straight across the ship. As viewed from
outside, the light would have a component of its velocity
along the direction of motion equal to the speed of the
ship, so it would still hit exactly opposite of your
position which also moves with speed v). The speed of the
light would still be c. If the width of the ship is W, you
would see the light take a time W /c but
the outside observer would see it take a time (W /c )/ √[1-(v /c )^{2} ].
If the spaceship were accelerating you would see the light
hit somewhere backward (forward) of where you aimed it
depending if you were speeding up (slowing down).

FOLLOWUP QUESTION:
Your answer confirmed what I thought was the case for three-dimensional objects propelled mechanically through space. For example, an arrow shot from a bow across the spaceship would hit the bullseye or miss, according to whether the ship's velocity was constant or accelerating/decelerating.
However, I thought electromagnetic radiation, such as light, was a different animal. What led to my original question was reading that the speed of light is a constant which can't be exceeded, so I assumed it couldn't take on additional velocity from the motion of its frame of reference. When you say "the light would have a component of its velocity along the direction of motion equal to the speed of the ship" it sounds like you're saying the light is leaving the flashlight at speed c, and then the ship's speed of, say, 10 km/sec is being added to make the effective speed c + 10 km/sec (viewed from outside) which I thought was impossible.
My mental image of light radiating is of energy being "conducted" (analogous to electricity conducted by a copper wire) at a constant speed through an intrinsically motionless medium (space), not of a physical object being propelled.

ANSWER:
First a minor point. The arrow, speed u across the
ship which moves with speed v , will have a speed u'= √(u ^{2} +v ^{2} ),
not u+v as you suggest. However, as you correctly
surmise, this cannot be true for light since, in your
example, you could not have the light with speed √(c ^{2} +10^{2} )
as seen by an outside observer. The speed must still be
c . So, if the component along the direction of the ship
is c_{x} =v , you must have c_{y} ^{2} +c_{x} ^{2} =c ^{2} =c_{y} ^{2} +v ^{2}
or c_{y} =c √[1-(v /c )^{2} ].
Regarding what you would see in the ship, the principle of
relativity states that the laws of physics are the same in
all inertial frames of reference; if your ship is at rest,
we could agree that anything aimed straight acoss will go
straight across. In other words, there is no such thing as
absolute rest —any inertial frame with constant
velocity relative to another inertial frame would have the
same laws of physics.

QUESTION:
When we say that a clock in moving frame runs slower than a clock in stationary frame. What does it mean?

ANSWER:
If you are moving relative to me, I observe your clock to be
running more slowly. The easiest way to understand this, I
believe, is the "light clock" which you can read about in an
earlier answer .
(Key to understanding this is the notion that the speed of
light one measures is independent of how you are moving.) It
would also help you to understand if you read about the "twin
paradox ".

QUESTION:
Hello, I'm not a student or anything I was watching a TED Talk video and it inspired a curiosity question. He was talking about antimatter accelerators and transverse seeing the universe at various speeds of light, and he said that if the ship could approach 99.99 percent the speed of light then we could reach the edge of the viewable Universe in 30 years relative to the spacecraft. However 13.9 billion years would have passed here on Earth.

My question is—how would a radio signal behave if this ship pointed a transmitter at the earth when it started this journey and broadcast a constant radio signal as it grew closer and closer to light speed traveling further away?

ANSWER:
Well, you got your numbers wrong somehow. 99.99% of c is not
really that big for such a long distance. If you go at that
speed, the distance (as seen by the ship) would be 13.9x10^{9} √(1-0.9999^{2} )=2x10^{8}
ly. Since the speed of the ship is 0.9999 ly/yr, the time
necessary would be 2x10^{8} /0.9999≈2x10^{8}
yr, not 30. I have seen this kind of estimate before and the
situation being described was probably that the ship has a
constant acceleration (as seen by the ship) equal to the
acceleration of gravity (about 10 m/s^{2} )
essentially forever; so the speed (as seen from earth) gets
to be much larger than 0.9999c . You can see the
calculation of the speed of accelerating objects as seen
from an inertial frame in an
earlier answer . If you want a really detailed discussion
of getting to the edge of the universe in a few decades, see
this post by
Dave Goldberg ; his number is more like 99.9999999999999999998% of
c !

But your question does
not depend on the details of how far you went and how long
it took you to get there. The radio signals being received
on earth will be Doppler shifted to lower frequency, the
shift getting larger as the speed gets larger. This is the
same idea as the Doppler shift of a siren going away from
you getting changed to a lower pitch. The expression for the
shift in frequency is f _{observed} =f _{source} √[(1-(v /c ))/(1+(v /c ))]=√[(1-0.9999/(1+0.9999)]=0.0007f _{source} ,
a much lower frequency.

QUESTION:
With respect to special relativity, is it correct (or not) to say that time dilation and length contraction are "real" because Nature has them in place to protect the upper speed limit "c"? Is there a better way to say this?

ANSWER:
First of all, what does it mean for something to be "real"?
I would argue that, for example, length contraction is real
because experimental measurements confirm its reality.
Second, I find anthropomorphizing "Nature" to be fatuous;
"Nature" does not "place" or "protect". Finally, you have it
exactly backwards: time dilation and length contraction are
the result of the fact that the speed of light is a
universal constant, not vice versa . I would also
point out that these would also be true if the speed of
light were different from what it is; as illustrated by
George Gamow's book
Mr.
Thompkins in Wonderland , they would be much more
evident if the speed of light were very small.

QUESTION:
I am writing a novel and am trying to understand how my interstellar spacecraft will behave. It is going to Iota Persei. 34.38 ly away. I was thinking it would travel at .6c (somewhat arbitrary but...). If I accelerated the space craft at a consistant 9.8m/s, it would take ~212.44 days at a distance of ~.1744 ly. Presumably, it would take about the same time and distance to slow the craft as it approached Iota Persei, roughly speaking.
While I've no idea of my craft's mass, it would be large, like 40,000 people large, and be propelled by an anti-matter drive. Would the energy output of the antimatter drive need to increase due to the increased mass as speed increased? Is it feasible to suggest that an antimatter drive would be able to produce enough energy to accomplish the above scenario?

ANSWER:
I have answered variations on this question several times in the past.
The most useful for you to read and understand compares
acceleration of an object as measured from both a rest frame
and from the frame of the object; I recommend you read both
this and all the links in
that answer . [The derivations of
the equations below can be found in the earlier answers.]
From the perspective of your space ship the velocity v
as a function of time t is v =(a _{0} t )/√[1+(a _{0} t /c )^{2} ] where
a _{0} is the acceleration as measured by the ship (g in your case). In order to achieve this acceleration, you must exert a force
F=ma _{0} where m is the mass of the ship.
The velocity is plotted in red in the graph; reading from
the graph, the time when v /c =0.6 is when a _{0} t /c ≈0.8=9.8t /3x10^{8} . Solving, t =2.45x10^{7} s=0.78
yr=284 days. The position as a function of t is x =(mc ^{2} /F )(√[1+(Ft /(mc ))^{2} ]-1)=(c ^{2} /g )(√[1+(gt /c )^{2} ]-1);
for gt /c =0.8, x =2.58x10^{15}
m=0.273 ly. At this point you turn off your engines and
coast at v /c =0.6. The distance you need to
go to get halfway to the star is 16.9 ly and it will take
you 16.9/0.6=28.2 years. The second half of the journey will
take the same time, so the time for the whole trip is T =2(28.2+0.78)=58
years.

Your answers (212
days, 0.17 ly) are different from mine (284 days, 0.273 ly)
because you have assumed that the acceleration, viewed from
the earth, is the same as the acceleration (9.8 m/s^{2} ) seen from the ship. The
black curve in the graph shows the ratio of the acceleration
seen from the earth compared to the acceleration seen from
the ship; reading from the graph, at 0.6c the ratio
is about 0.5, 4.9 m/s^{2} . Be sure to read the
discussions of acceleration in the
earlier answers if you want to understand this issue
with acceleration in special relativity.

Since there is no such
thing as an antimatter drive engine, I can make only a rough
assessment of how feasible this is. The force you need to
apply during the 0.78 year accelerations is F=mg. What is m?
The space station has a mass of about 400,000 kg and you
have 40,000 passengers. Let's say each passenger needs one
space station of mass to support his/her needs, so the
required force would be 1.6x10^{9} x9.8=1.57x10^{10}
N. The energy you have to supply to the ship to v /c =0.6
I figure to be about 36x10^{24} J and the amount of
mass converted with 100% efficiency into energy would be
about 4x10^{8} kg and you would need that much again
to stop it. So your ship of mass 1.6x10^{9} kg would
have to carry 8x10^{8} kg of fuel, one third of the total
weight! Obviously, I have not included this mass change in
my estimates of time. And, of course, there is the issue of
how you are going to store 4x10^{8} kg of
antimatter. The whole scenario certainly does not look
feasible to me!

Finally, note that all
the distances and times are as measured by the earth bound
observer. In other words, the passengers will not be 58
years older when they arrive at their destination. Because
of length contraction, they will have a shorter distance to
travel. Ignoring the accelerating periods, I can estimate
the time of the trip by just assuming they travel 0.6c the
whole trip. In that case, T' ≈T √(1-.6^{2} )=58x0.8=46.4
yr.

QUESTION:
We could induce artificial gravity through centripetal acceleration. For example a ring-like structure in a spaceship could rotate about 1.34 rpm if the radius to the centre is 500 meters. This will give 1 g at the edges of the ring.
However we can also induce artificial gravity in the spaceship through its propulsive power and hence a constant acceleration at 1 g of the spaceship is required.
But what if the spaceship is accelerated at more than 1 g in order to achieve 75% the speed of light, even though the spaceship consists of the 500m radius ring-like structure that rotates at 1.34 rpm. What type of artificial gravity will the occupants feel? Will the excessive acceleration cancel out the centripetal effects in the ring? Is there an equation or formula that combine both linear acceleration of an object while the objects is also rotating?

ANSWER:
You seem to think that you need an acceleration greater than
g to get to 0.75c , but any acceleration will do. But,
since you seem interested in a greater than g acceleration, I
will choose a _{0} =1.5g. So now a person of
mass m in the ship sees two fictitious forces, mg
which is radially out and 1.5mg which is
toward the rear of the ship. A person in the ship would experience a net
force of about 1.8mg pointing out and back. This would not be a
comfortable situation. If you are just going to accelerate constantly, I
would recommend not messing with the rotation at all, set your
acceleration to g , and make all the "floors" in the ship
perpendicular to the acceleration. If you are interested in how long it
would take to reach 0.75c with a _{0} =g ,
you can read off the graph (which I took from an
earlier answer ) that gt /c =1.8,
so t =5.5x10^{7}
s=1.7 yr.

QUESTION:
Where can I get a graph (or other information) about the increasing "relativistic resistance" to the acceleration of a particle (an electron, hopefully) as its velocity is increased to near-relativistic speeds? If such a graph is not available, then how can I calcuate this increasing force that resists a particle being accelerated when it is already traveling at some relatively high percentage of the speed of light, like maybe 70%?

ANSWER:
It would definitely behoove you to read an
earlier
answer first which has lots of details and discussion of
acceleration. There are two ways to approach this problem:

Suppose
the observer in the inertial frame is doing the pushing and wants to
know how hard to push the particle to achieve a particular
acceleration.
Resistance to acceleration is usually called inertial mass and the
inertial mass m of a particle with rest mass m _{0}
and speed v is m=m _{0} /√[1-(v /c )^{2} ].
The first figure above plots
m /m _{0} as a function of v /c .
So, to achieve an instanteous acceleration of a , a force of
F=ma =m _{0} a /√[1-(v /c )^{2} ];
so for your example of v /c =0.7, you can read off
the graph that you would need to exert a force 1.4 times larger than
you would if the particle were moving slowly. This is probably what
you want if you are interested in electrons accelerating since you
would be accelerating them.

Suppose
the observer was on the particle and the particle was a rocket ship.
You adjust your engines so that the force F which they
exert on the ship causes a constant acceleration of a _{0} =F /m
where m is the rest mass of the ship. What
acceleration a does another observer in an inertial frame
(on earth maybe) see when the rocket has a speed v ?
Starting with the velocity derived in the
earlier
answer , v =(a _{0} t )/√[1+(a _{0} t /c )^{2} ],
we can calculate the acceleration by differentiating with respect to
t , [dv /dt ]/a _{0} =a /a _{0} =(1+(a _{0} t/c )^{2} )^{-3/2} .
Now, reading off the second graph, when v /c =0.7,
a =0.36a _{0} ; the stationary observer will
only see 36% of the acceleration seen on the ship.

ADDED
NOTE:
It is important to note that the acceleration observed depends
on the inertial frame you are in. In way#2 above, two different inertial
frames, say with v /c =0.7 and 0.9, will observe the
ship having different accelerations. This is why acceleration, and
therefore also force, in special relativity do not play an important
role as they do in Newtonian physics where all inertial observers see
the same acceleration. This is discussed in the
earlier
answer .

QUESTION:
My dad told me about your website, very interesting reading. My question deals with molecules. When a molecule emits a photon, the mass of the molecule decreases to account for the energy in the photon. So, the mass of the molecule as a whole decreases, but this mass does not come from the "parts" of the molecule. In other words, the mass of the constituent electrons does not decrease, the mass of the protons does not decrease, so the energy must come from the electric field between the electrons and protons.
But the electric field has energy, not mass. Now mass is a form of energy, but I don't think you can say that the field has mass? But yet, it is said that the mass of the molecule decreases. The electric field contributes to the mass of the molecule, but yet it is incorrect to say that the field has mass?

ANSWER:
Actually, this is not as complicated as you are trying to make it. It
all boils down to the fact that mass is a form of energy and must be factored into
any energy conservation that occurs in an isolated system. You say that
the masses of the protons and electrons do not change, but that is not
right. Look at the simplest case, a hydrogen atom. If you measure the
mass of this atom it will be less than if you measure the masses of a
free electron and a free proton. Here is how you can see that: if you
pull the electron away from the proton, that is you ionize the atom, do
you have to do any work? Of course you do because the electron and
proton are bound together. So, you have added energy to the system (p+e)
and that energy shows up as mass. In a system as complicated as a
molecule you cannot say which particle or particles changed their
masses, but you can say for sure that the total mass of the molecule
changed by exactly the energy of the emitted photon divided by c ^{2} .

QUESTION:
This is my 2nd question related to the issue of time dilation - this one being related to the issue of motion [the other being based on gravity]. Since time dilation occurs for all moving objects, and considering further that the Earth has been revolving around the sun at 30 km per second for the last 4 billion years -- and further that our solar system is moving at roughly 45K mph through space, can't it be said that, compared to other objects in the universe, time dilation has occurred to a significant degree for our planet over those 4 billions years? And that really every object in the universe likewise has its own unique time dilation associated with it? Can't it also be said that every consolidated arrangement of matter in the universe is moving along at different "rates of time?" Wouldn't, over the course of several billion years, these "pockets" of different time spans become more and more "incompatible" with each other?

ANSWER:
The two speeds you quote are about the same (45,000 mph ≈2x10^{4}
m/s and 30x10^{3} km/s=3x10^{4} m/s). So let's just
choose the larger one and see how much time dilation there is. Relative
to the sun, an elapsed time T =4x10^{9} y would
correspond to T'=ΥT wwhere Υ= 1/√(1-(v /c )^{2} )=1/√(1-(3x10^{4} /3x10^{8} )^{2} )≈(1+0.5x10^{-8} ).
Therefore T'≈ (T +20), a difference of 20 years. This
may sound like a pretty long time to you, but relative to 4 billion it
is less than 10^{-6} %. That is not "significant" to my mind.
You are right that every object has its own clock which, relative to
other clocks, is not necessarily the same; every object also has its own
meter stick, not necessarily the same as other meter sticks in the
universe. The important thing is that you always must talk about
velocity with respect to what.

QUESTION:
I have a question about light please. I was thinking about one of Einstein's thought experiments. Let me set some the situation first - there is a stationary observer at a train station. Einstein is on a train travelling at 99.99999%c and is about to pass the train station. As the front of the train reaches the train station, the stationary person emits a light beam in the same direction as the train while at the same time, the train turns on its headlight. Based on what I understand, both beams of light should be travelling at C but I am interested in understanding what each person sees. I can understand that the stationary person will see both light beams leave him at C, along with the train at 99.99999%c. But what about Einstein on the train? I understand that you cannot add the velocity of the train to its headlight beam. But I can imagine that he would see the beams moving ahead and away from him at .00001%c since he is already travelling at 99.99999%c. But that seems to violate the idea C being the same for everyone in every reference frame. I can also imagine that due to that rule, he would see both light beams moving ahead and away from him at C. But if that's true, how can he possibly be going 99.99999%c if the beams are moving away from him at C and how could the stationary person see both the light moving at C and Einstein moving at 99.99999%c? I need your insight to understand this please.

ANSWER:
This is a very long question, but the answer is very short. All
observers, regardless of their motions, see the speed of any
electromagnetic radiation to be c . Your statement that the
train rider would see the beams having is speed 0.00001c is
simply wrong. That is not to say that the beams are unchanged —they
would be Doppler-shifted to different wavelengths, but still move with
speed c .
See the faq page to help yourself to accept that this is true.

FOLLOWUP QUESTION:
Thanks! I do accept what you are saying is true. Acceptance was never the issue, understanding was.
Since the doppler is shifted then what Einstein would see with his eyes would be the light around him becoming redder and redder the closer to C he gets?

ANSWER:
I hope you did go to the faq page and now have an "understanding"; if
not, you should look at these links
there . You might also read an
earlier answer about relativistic velocity addition. You are right,
if light with frequency f is moving in the same direction as
you are, you will see a lower frequency light f' ,
red-shifted. The Doppler equation is f'=f √[(1-β )/(1+β )] where β=v /c.
Suppose we take your case where β= 0.9999999=10^{-7}
so √[(1-β )/(1+β )]=2.24x10^{-4} and choose λ =550
nm (yellow light). Then f=c /λ =3x10^{8} /5.5x10^{-7} =5.45x10^{14}
Hz and f' =2.24x10^{-4} x5.45x10^{14} =1.23x10^{11}
Hz=123 GHz. This frequency is in the short-wave radio range, nothing he could "…see
with his eyes…"! The Doppler effect has red-shifted the wave
way beyond visible red!

QUESTION:
Dear Physicist, my question relates to special relativity and time dilation. This question may sound dumb but it continues to bother me. I have a space ship which I am sending to Alpha Centauri, the distance is 4.3 light years. I am able to push the ship to .7 the speed of light.
According to my understanding of time dilation, events on the ship will slow down due to the increase in the mass put into the system because, I am approaching the speed of light (a fixed constant) is it correct events back on Earth, which is moving at a much lower percent of the speed of light, will pass at a faster rate? If so, then doesn't that mean increasing the velocity of the ship to get there faster is really self defeating because instead of reaching Alpha Centauri in 5.59 years, time dilation will cause the actual time measured in Earth time to be much longer?

ANSWER:
I am afraid you have some serious misconceptions. Time dilation says
that if you are moving with respect to me, I observe your clock to
run slowly . Do not mistake that
for saying that your clock appears
to run slowly —it might
appear to run faster or slower
than mine, but is running slowly. Furthermore, it has nothing to do with
mass increasing with speed. But how do you observe things? Your clock
appears to run just the same as if you were at rest; but wait, you are
at rest relative to yourself and you see me moving and therefore find
that my clock runs slowly! Now, the time that I see you getting to the
star is 4.3/0.7=6.14 years. But, you see it as a considerably shorter
time. How can that be if your clock is running normally and you clearly
see your speed as 0.7c ? The answer is length contraction—you
observe the distance to the star to be less than 4.3 light years by the
gamma factor, √(1-0.7^{2} )=0.71, or 3.1 light years so the
time will be 3.1/0.7=4.43 years. If you could get going a speed 0.9c ,
you could get there in less than half a year by your clock but still
6.14 by mine. I have given you a couple of links to the
faq page , but you might want to peruse that page
for more relativity answers which might interest you, particularly the
twin paradox.

QUESTION:
If any particle drop falling from height h, then velocity increases but in case of photon with light veloctiy c falling drop then what changes occurs, it will increase velocity or not?
If velocity of photon increase then how its possible because no one have
velocity is greater then speed of light?

ANSWER:
A photon gains energy when it "falls" in a gravitational field, just
like a rock does. But, it does not speed up because the speed of light
is a universal constant. The energy of a photon is proportional to its
frequency, so as its energy increases its frequency increases. As light
travels (with constant speed) in the direction of a gravitational field
(down) it shifts toward the blue end of the spectrum and this is called
a blue shift. As light travels (with constant speed) opposite the
direction of a gravitational field (up) it shifts toward the red end of
the spectrum and this is called a red shift.

QUESTION:
Two frames xOy and x'O'y' are in uniform motion along their x axes. We will
consider for simplicity the first frame to be "fixed" while the second one
moves to the right with a velocity v. From the origin of the moving frame
x'O'y' two rays of light are emitted simultaneously, one along the axis O'x'
and the other one at an angle of 60 degrees with the O'x' axis. Two mirrors
are placed at the same distance l on the two tracks and the light gets
reflected. Obviously, the two reflected rays, as observed in the frame x'Oy'
return to 0' at the same time. I made some calculations for v = c/2 and l =
1,000,000 m. Surprisingly, the two rays, as observed from frame xOy, do not
appear to meet return to the origin at the same time. I can share my
calculations with you. It is possible that I made a mistake. However if my
calculations are correct, this would be a very strange thing indeed.

ANSWER:
First of all, it would not be " …a
very strange thing indeed… " to find what is
simultaneous in one frame is not in another. One of the keystones of
special relativity is that simultaneaty of two events depends on the
frame of reference. I have not included your solution to the question
because I prefer to work it out myself rather than troubleshoot your
work. The figure shows the situation as seen by each observer. The
primed system moves in the +x direction while the red-drawn
distances to the mirrors are at rest in the unprimed system. The rest
lengths of the two distances are L and the angle one makes with
the x -axis is θ . The round-trip time along the
x -axis is t _{1} =2L /c and
along the other length is the same, t _{2} =2L /c .
In the moving system all lengths along the the x' direction are
reduced by a factor √(1-β ^{2} ) where β=v /c ;
therefore L'=L √(1-β ^{2} ) so t _{1} '=2L' /c =2L √(1-β ^{2} )/c=t _{1} √(1-β ^{2} ).
For L" only its x' component is contracted, L" cosθ' =L √(1-β ^{2} )cosθ
while the y" component remains the same, L" sinθ' =L sinθ ;
from these you can easily show that L"=L √(1-β ^{2} cos^{2} θ )
and tanθ'= (tanθ )/√(1-β ^{2} ).
Finally, t _{2} ' =2L" /c =[2L √(1-β ^{2} cos^{2} θ )]/c =2t _{2} √(1-β ^{2} cos^{2} θ )≠t _{1} ' .

ADDED
COMMENT: I see that I misread your question. I see that
you have put the lengths at rest in the moving frame whereas I have them
in the stationary frame. But in special relativity it makes no
difference which is moving and which is at rest. So, you have found that
the times in the frame not at rest relative to the lengths are not the
same, just as I have; I reiterate that this is expected, not surprising.

FOLLOWUP
QUESTION: Thanks a lot, your competent and prompt
answer is much appreciated!

I would like to reiterate that for two events that occur in the same place to be simultaneous in one frame but not in the other frame is really very strange, the non-existence of absolute simultaneity being a central tenet of Einstein's relativity not withstanding. For suppose that two light pulses when they arrive simultaneously to O' (according to a frame x'O'y' observer) they trigger an event, perhaps an explosion for which the simultaneous arrival of the two pulses is a necessary condition. So the explosion is the result of the simultaneous arrival of the two pulses for the x'O'y' observer but not for an observer in the frame xOy (who observes the explosion, too), since the latter sees the two pulses arriving at different times and thus cannot correlate the explosion to the two light pulses. The x'O'y' - observed concurrent emission of the two light pulses that go out from the very same point O' may be in an analogous way correlated to some event in the x'O'y' frame, but not in the xOy frame.

One can think with some justification that this scenario means that the x'O'y' system is somehow more "special" than the xOy system, pretty much in contradiction, if not with Einstein relativity postulate proper, at least with its spirit. In other words, in my opinion, the relativity theory is not devoid of inner contradictions, or at least it leads to conclusions which seem to invalidate the most fundamental notions of cause and effect or of things being what they are irrespective of where we observe them. And to me it is debatable whether a physical theory, however legitimately counter-intuitive it may be understood to be, can lead to this type of conclusions and be nevertheless considered a valid description of the physical world.

ANSWER:
Again, I would like to reiterate that it is neither strange nor
unexpected! First of all, the two events in the moving frame are at
neither the same time nor at the same place; only in the stationary
frame are the two events simultaneous and at the same place. By
"stationary frame" I mean here the frame where the apparatus is. Now,
your followup expands the apparatus to include also a "trigger" and a
bomb. The trigger will be some sort of electronic device to detect the
simultaneaty of the two light pulses' arrival at the origin and will
detonate the bomb in that case. An observer in the moving frame will not
see any contradiction, he will not say
that the bomb will not detonate because the device which detonates it is
in the stationary frame and he will agree that the pulses in that frame
are indeed simultaneous. Is the stationary system in some way "special"?
Of course it is —that is where the apparatus resides and all
frames will agree that the bomb will be triggered in that frame.

QUESTION:
An iron ball made of only iron ,no hollows no any other material,just iron. So first we measure its weight at room temperature and then we heat it up to higher temperatures near to melt but not not melting. Then we measure its weight. Does weights are different or same. If different please indicate which is higher.

ANSWER:
For purposes of calculation, let the ball have 1 kg of mass at room
temperature, 20°C. Because mass is a form of energy, the iron ball
has energy E=mc ^{2} =9x10^{9} J. The melting
point of iron is 1540°C and its specific heat is 449 J/kg·°C,
so the enegy to heat it from 20 to 1540°C is 1x1520x449=6.8x10^{5}
J. So, the ball would have increased its mass by 6.8x10^{5} /9x10^{9} =7.6x10^{-5}
kg. You would be hard-pressed to actually measure so small a change in
mass.

QUESTION:
I have been trying to understand this for years. In Einstein's theory of time verses speed I believe he used the situation of a man (lets call him man A) standing in a train yard. A second man is on a train (lets call him man B) and the train is moving. They both drop an object at the same time. If I standing in the yard with man A to me the object man A drops would appear to me in normal time. Man B's object would appear to take longer causing the difference in time. I understand this. My main question is if a man C was on a different train going the same direction as man B's train the time difference between me and man A to man B and man C. I understand that to man B and Man c they would be the same. What if man C's train was going in the other direction? To me and man A the would seem the same, but man B to man C I believe the difference in the dropped object would be appear to be doubled to them between them. Since I believe it would appear doubled between man C and D. I think there should be a double time difference. From man A's view the age difference would be the same, but between man B and man C there should be double the time difference. How can there be doubled the time difference when they are traveling at the same speed?

ANSWER:
Your question boils down to what is called velocity addition .
In classical physics, v _{CB} =v _{CA} +v _{AB} ;
the notation is "v _{IJ} is the velocity of I relative
to to J". I have written this so that it corresponds to your question —A
is you, B and C are the trains; it is more convenient for you if we
rewrite the equation as
v _{CB} =v _{CA} -v _{BA}
which we can do because v _{AB} =-v _{BA} .
If the trains have speeds v in the same direction, v _{CB} =v-v =0;
in opposite directions, v _{CB} =v- (-v )= 2v— each
sees the other moving with speed 2v . But, this form of velocity
addition is wrong for very high speeds (see an
earlier answer ).
The relativistically correct velocity addition equation is
v _{CB} =(v _{CA} -v _{BA} )/[(1-(v _{CA} v _{BA} /c ^{2} )]
which reduces to the classical equation for the speeds much less than
c . So, for the trains moving in opposite directions, v _{CB} =2v /(1+v ^{2} /c ^{2} );
for example, if v =0.5c , v _{CB} =c /(1+0.25)=0.8c .

Now, you seem
to think that if you double the speed, you double the time difference.
This is not correct— time dilation goes like the gamma
factor, 1/√(1-v ^{2} /c ^{2} ).
So, for your situation with v =0.5c , t _{A} =1.33t _{B} ,
t _{A} =1.33t _{C} , t _{C} =1.67t _{B} ,
and t _{B} =1.67t _{C} . These are
confusing, I admit. They are meant to denote, for example, that t _{B} =1.67t _{C
} means that B sees his clock tick out 1.67 s when C's clock ticks
out 1 s; B observes C's clock to be slow.

QUESTION:
While reading about the twin paradox, I've been told at the end of the traveling twin's journey, he begins decelerating in order to land back on Earth, and he, the traveling twin, observes his brother's clock on Earth to SPEED UP. This makes sense to me except for one problem: This suggests that the light pulse in the Earth clock would be percived by the traveling twin to be moving faster than C. Of course, the traveling twin is no longer in an inertial frame. I thought perhaps that since he feels himself moving now, he would also measure himself and the light having a CLOSING SPEED greater than C, even though he would see the light moving across the ground on Earth to equal C? If so, at what rate would a clock in frame S behind the traveling twin run at, faster or slower? Is it also possible that acceleration, from the point of view of the traveling twin, causes length contraction perpendicular to the the ship's vector, shortening the distance the pulse has to travel?

ANSWER:
There is no need to discuss acceleration to understand the
twin paradox . Acceleration
just makes everything harder to understand. Basically, just assume
necessary accelerations (departing, turning around, landing) occur in a
vanishingly short time. But, you are not really interested in the twin
paradox, you are interested in how things appear in an accelerated
frame. I am sorry, but nothing in your question after " …clock
on Earth to SPEED UP…" makes any sense. First of all, the fact
that any observer will measure the speed of light to be c is a
law of physics and the general principle of relativity states the laws
of physics are the same in all (not just inertial) frames.
Second, to measure a speed in your frame of reference you must use your
clock, not somebody else's. And third, how another clock looks is really
irrelevant because how it appears to run and how it
is actually running are not the same.

I would like
to address how the clocks look from the perspective of the
Doppler effect . The relativistically correct Doppler effect is
usually expressed in terms of the frequency of the light; for our
purposes, it is more convenient to express it in terms of the periods,
T _{observer} =T _{source} √[(1+β )/(1-β )]
where β=v /c and β is positive for
the source and observer moving apart; it makes no difference which is
the observer—each twin will see the other's clock running at the
same relative rate. Let's illustrate with a specific example, β =-0.8,
the traveling twin coming in at 80% the speed of light; T _{observer} =T _{source} √[(1-0.8)/(1+0.8)]=T _{source} /3
so the observer will see the source clock running fast by a factor of
three. But special relativity tells us that moving clocks run slow, not
fast; 1 second on the moving clock will be 1/√(1-0.8^{2} )=1.67
seconds on the observer's clock. How the moving clock looks is thus
demonstrably not a measure of how fast it is actually running. Now, if
the incoming twin puts on the brakes such that he slows to β =-0.6,
T _{observer} =T _{source} √[(1-0.6)/(1+0.6)]=T _{source} /2
so the observer will see the source clock running fast by a factor of
two, apparently slowing down. Now, 1 second on the moving clock
will be 1/√(1-0.6^{2} )=1.25 seconds on the observer's
clock, speeding up compared to when β =-0.8.

QUESTION:
If gravity is not a force but just a curvature of spacetime then how does a massive object (like earth) affect a much less massive object (like a tennis ball) when they are not in motion relative to each other?
For example, if I were able to travel out to space far enough away from earth and then stop so that I am not in motion with respect to earth and let go of a tennis ball it will stay stationary.
But if I traveled to say 100,000 ft above sea level and were able to hover there so I am not in motion to earth and then let go of a tennis ball it will immediately begin to move towards earth.
In other words, how is it possible for the curvature of spacetime to affect bodies that are not in motion to other?

ANSWER:
You have some misconceptions here. First of all, no matter how far away
you get, there will always be a small force toward the earth. It may be
so small that you would have to wait a millenium to see it move a
millimeter, but it is still there. The curvature of spacetime justs gets
smaller as you get farther away. It is certainly true that if you drop
something from an altitude of 100,000 feet it will accelerate toward the
earth. But, what does that have to do with motion? Even if you gave it
some motion, say throwing it horizonatlly, it would still accelerate
toward the earth just the same as dropping it, but now it would also
have a velocity component parallel to the earth as well. Maybe you have
never seen the "trampoline
model " which illustrates (in a simplistic, not literal way) how
warping the space (the surface of a trampoline) by a massive object
(bowling ball) will cause a light object (marble) to be attracted to the
massive object.

QUESTION:
So if black holes suck in everything in including light that must mean everything is getting pulled in as fast or greater then the speed of light. So if light is weightless and it is sucked in. What happens to any mass as it is sucked in. Would the mass of the object then cease to have mass? Because im pretty sure anything traveling at the speed of light has to be mass-less correct? And how does gravity effect something with no mass? I dunno if it is a good question or not but i couldnt find a whole lot on the subject.

ANSWER:
Nothing ever goes faster than the speed of light and only light
can go at the speed of light. When an object merges with the black hole,
its energy, E=mc ^{2} , is not lost and the black hole
becomes more massive by the amount of the mass energy. When light is
swallowed by the black hole, the mass increases even though the light
does not have any mass because light does have energy and the energy
shows up as increased mass of the black hole, m=E /c ^{2} .

QUESTION:
When speaking of particle accelerators,the accelerators keep adding energy to the particles, even though they cannot speed up any further. But where does the energy go?

ANSWER:
Well, they never really get to a speed where they cannot go any
faster because they never reach the speed of light. I have always
thought "accelerator" was a misnomer for very high-energy machines
because the acceleration (rate of change in speed) is almost zero; I
would prefer "energizer". The easiest way to think about it is that the
particles acquire mass and that is "…where…the energy
go[es]". The energy of the particles is E=mc ^{2} =m _{0} c ^{2} / √[1-(v /c )^{2} ]
where m _{0} is the mass at rest. For example, if v /c
increases from 0.99 to 0.999, that is only a 1% change in speed; but the
energy increases from 7.1m _{0} c ^{2}
to 22.4m _{0} c ^{2} , more than triple.

QUESTION:
Does mass of an object increase its mass exponentially as it approaches infinitely close to the speed of light?

ANSWER:
If df /dq =Aq for some function f (q ),
where A is a positive constant, f is said to be
exponentially increasing with q . For your question, f
is m and q is v . The expression for m (v )
is m=m _{0} / √[1-v ^{2} /c ^{2} ]
where m _{0} is the rest mass. Calculating the
derivative, dm /dv =mv /[c ^{2} -v ^{2} ].
Thus, although m increases without bound as v
increases, it does not increase exponentially.

QUESTION:
A starship pilot wants to set her spaceship to light speed but the crew and passengers can only endure a force up to 1.2 times their weight. Assuming the pilot can maintain a constant rate of acceleration, what is the minimum time she can safely achieve light speed?

ANSWER:
This question completely ignores special relativity. It is impossible to go
as fast as the speed of light. Furthermore, acceleration is not really a
useful quantity in special relativity and you must use special relativity
when speeds become comparable to the speed of light. I have
earlier worked out the
velocity of something which would correspond to occupants of your spaceship
experiencing a force equal to their own weight due to the acceleration which
I will adapt to your case later. (See the graph above.) First, though, I
will work out the (incorrect) Newtonian calculation which is presumably what
you want. The appropriate equation would be v=at where v =3x10^{8}
m/s, a =1.2g =11.8 m/s^{2} , and t is the time to
reach v ; the solution is 2.5x10^{7} s=0.79 years. For the
correct calculation, you cannot reach the speed of light; from the graph
above (black curve), though, you can see that you would reach more than 99%
of c when
gt /c =3.
To make this your problem, we simply replace g by 1.2g and
solve for t . I find that t =7.7x10^{7} s=2.4 years.

QUESTION:
I am a high school student interested in relativity and I recently read an
article
about relativity. The article stated that the "The combined speed of any object's motion through space and its motion through time is always precisely equal to the speed of light." For an object moving at 90% the speed of light, it should only be moving at 10% the speed of light through time. I assume that means that time should pass only at 10% the actual speed of time for the object (correct me if I am wrong). However based on the Lorenz factor, time passes at 0.43 the actual speed of time for the object. Why is this so?
Furthermore when time dilation occurs it is only seen by someone else in a stationary frame of reference. In the moving object's frame of reference time does not slow down at all. Does this mean that the combined speed of the moving object's motion through space and time can be more than the speed of light in the moving object's frame of reference? Can the moving object go beyond the speed limit in its frame of reference?

ANSWER:
Look closely at the example given: "To get a fuller sense of what Einstein found, imagine that Bart has a skateboard with a maximum speed of 65 miles per hour. If he heads due north at top speed—reading, whistling, yawning, and occasionally glancing at the road—and then merges onto a highway pointing in a northeasterly direction, his speed in the northward direction will be less than 65 miles per hour. The reason is clear. Initially, all his speed was devoted to northward motion, but when he shifted direction some of that speed was diverted into eastward motion, leaving a little less for heading north. "
The speed in the northward direction will now be v _{N} =65cos45^{0} =46
mph; his velocity in the eastward direction will be v _{E} =65sin45^{0} =46
mph. His total velocity will be √(v _{N} ^{2} +v _{E} ^{2} )=65
mph, not (v _{N} +v _{E} )=92
mph. In relativity N and E now are x and t for the
stationary observer and x' and t' for the position and
time of the moving observer as seen by the stationary observer. The catch,
though, (which I presume is why the author of your article avoided these
details) is that time and space are slightly different from space (N) and
space (E) in that to get the length of the vector you calculate the square
root of the difference of the squares instead of the sum. So the speed
through spacetime would be √(c^{2} -v^{2} ).
So, the stationary observer will see a "spacetime speed" of v _{spacetime} =√(c^{2-} v^{2} )
and the "space time distance traveled" by the moving observer as seen by the
stationary observer would be d _{spacetime} =v _{spacetime} t'
where t'=t / √(1-(v ^{2} /c ^{2} )).
If you do your algebra you will see that d _{spacetime} =ct
indicating that the stationary observer sees the moving observer having
a speed of c through spacetime.

QUESTION:
When a photon loses energy ( Redshifts) climbing up thru a gravitational field, does the photons decreased energy go into the mass of the gravitational field itself? According to GR, the gravitational field itself contributes to the mass of the system. For earth, it is very small. When I elevate a massive object in a gravitational field, I can say the "potential energy" is in a very small increase of the mass of the object. But photons have no mass! And are never at rest! The only thing I can think of is that the photon "transfers" energy to the gravitational field itself, which appears as a small increase in mass of the field.

ANSWER:
Let's think of a star emitting a photon of frequency f . Initially
an energy of hf is removed from the star. But, by the time that
the photon is very far away the star, it has lost some amount of energy Δhf ,
so the net loss to the star and its field is
hf -Δhf . At the instant the photon is created, the
mass of the star is reduced by hf/c ^{2} . But when the
photon, which is losing its energy to the field, is far away, the energy
of the field will have increased by Δhf /c ^{2} ;
I would not use the terminology "mass of the field" since the field has
energy density, not mass. But now, it seems to me (not a cosmology/general
relativity expert) that there is a bit of a paradox: we always associate a
given mass with a particular gravitational field, so the field should have
the energy content associated with a mass M'=M -hf/c ^{2} .
But, in fact, the field would have energy content associated with a mass
M'=M -hf/c ^{2} +Δhf /c ^{2} .
I am guessing that the field and the mass somehow "equilibrate" so that the
final mass of the star is consistent with the energy of the final field. (I
could easily be wrong! Perhaps it is only meaningful to look at the total
energy of the star and its field. Whatever the case, I would not talk about
mass of the field.)

QUESTION:
Sir,I am a mechanical engineer by profession but very interested in reading physics fundamentals. Recently I went through the fundamentals of electro magnetism and I got this doubt. Consider two charges each of charge +q rigidly fixed in a train moving with a constant velocity, V. Let the train speed be negligible compared to speed of light so that we can treat the problem in non-relativistic terms. The fixity condition ensures that it overcomes electrostatic forces and remain motionless. A traveler in train sees both of them at rest and there wont be any magnetic forces developed between the charges.
Now consider an observer in platform. For him, both charges are moving with a constant velocity, V equal to the train velocity. Each charge will develop magnetic field according to this observer as per Biot-Savart law and there will be mutual attraction as each charge is moving under the magnetic field of other.Thus, an observer in train sees no magnetic forces whereas an observer in platform sees mutual magnetic attraction. How do you explain this?

ANSWER:
There is no rule which says that the either the electric or magnetic field
must be the same in all frames of reference, even slowly moving frames like
you specify. The real root of your problem is that electromagnetism is
intrinisically relativistic, even at slow speeds; the electric and magnetic
fields of classical electromagnetism are really both components of the
electromagnetic field which is a tensor and when you change inertial frames,
you cause a transformation of that tensor into another where both the
electric and magnetic field pieces of it are different. In your second case
you would also find that the electric fields were slightly different from
their original values but the differences would be very tiny; the magnetic
fields, though, are nonzero but small, but small is very big compared to the
original magnetic field of zero.

If you
are interested, I will give here the electric and magnetic fields for one of
the charges moving with velocity v in the +x direction.
E' =i E_{x} + γ (j E_{y} +k E_{z} )
and B' =- (v xE' )/c ^{2
} where
γ= 1/√[1-(v /c)^{2} ] and i , j ,
k are unit vectors; the vector E is in the frame
where q is at rest and E' and B' are when
q is moving.

QUESTION:
Is it true that a person flying in an airplane is actually living a shorter amount of time, than a person standing on the ground?

ANSWER:
You must specify "…living
a shorter amount of time… "
with respect to whom . The
person in the airplane sees time progressing at a normal rate. However, the
person on the ground will see the clock of the person on the airplane run
slowly, so he will perceive the traveling person to be " living
a shorter amount of time", i.e. she will have aged less when she
returns to earth. You should read FAQ Q&As on the
twin paradox ,
how clocks run , and the
light clock .

QUESTION:
I think if an object is turning, it has more gravity than an object which is not turning

ANSWER:
You are right. However, it really has nothing to do with the turning, per
se . When something is turning it has rotational kinetic energy and
therefore a spinning planet has more energy than an otherwise identical
planet not spinning. In general relativity you usually hear about gravity
being caused by mass warping spacetime. However, mass is just the most
obvious source of gravity and what it is which really warps spacetime
is energy density and mass has a lot of energy (E=mc ^{2} ). The energy due to the turning
is infinitesmal compared to the mass energy of the object so you would never
be able to distinguish the difference due to the turning by looking at the
gravity.

QUESTION:
Suppose light is travelling in a straight line parallel to Y-axis takes time t to reach from y1 to y2 in a frame of reference S. Let there be another frame S' which is travelling parallel to X-axis with velocity v relative to frame S. Then a seen from S' the light ( which was travelling parallel to Y-axis in frame S ) now takes a longer path which is greater than y2-y1 (as now it will be travelling obliquely in frame S' ). Now due to time dilation time taken in frame S' is lower than t as seen by frame S due to time dilation. Then speed of light in frame S' as seen from S will be greater than c as length is higher and time is lower. then how to explain the constancy of speed of light in this case ?

ANSWER:
You are making this too hard! The fact that c is a universal constant
is where you should start, not end. When you move with your S' frame with
speed v in the positive x direction, the light beam acquires
an x -component of c _{x} =-v . If the speed of the
light is to remain constant, this must mean that the y-component must
therefore decrease such that c ^{2} =c _{x} ^{2} +c _{y} ^{2} .
So,
c _{y} =c √[1-(v ^{2} /c ^{2} )]
which means that the angle θ which the light makes with the y -axis
is θ= tan^{-1} [β /√(1-β ^{2} )] where
β=v /c. See an earlier question for
an example of this phenomenon.

QUESTION:
if black hole has enormous
gravitational pull, can it attract light?? if yes, does this prove photon
has a dyanamic mass as gravity attracts the mass only??

ANSWER:
Any gravitational field will "attract" a photon, it does not have to be a
black hole. For example, a photon passing by a large mass (like the sun or a
galaxy) will be deflected. What is special about a black hole is that if a
photon gets closer than a critical distance called the Schwartzchild radius,
the light cannot escape. However, you are wrong when you say that gravity
only attracts mass. A photon has no mass. See the
faq page.

QUESTION:
OK, I am having trouble with this speed of light thing. If a train left earth at near the speed of light and turned on it's light, the light would leave the train at the speed of light relative to the train, correct? But doesn't that mean that those photons are traveling away from Earth at waaaay more than the speed of light? If so, could we launch a ship (hypothetically) and get to %90 the speed of light and then launch another smaller ship from that ship and travel %90 of light speed relative to that ship? A ship that would be distancing itself from Earth at waaaay more than the speed of light?
This is where our slow speeds in space bother me. Since space is a vacuum (nearly) and rockets only push on themselves from the inside and speed is relative, wouldn't a ship that is coasting 40,000 mph away from the Earth, suddenly fire a rocket and accelerate no matter what? Without wind resistance and with the slight help of solar winds, I would think we could travel a probe much quicker. Might have to build it in space but 8 saturn Vs connected together ought to make a difference. No?

ANSWER:
The problem you are having is that you are thinking in terms or classical
physics or intuitively; intuition is based on experience and you have no
experience with trains going nearly the speed of light. Whenever you
extrapolate what you "know" to be true of trains and light to situations
outside your experience, you need to be cautious. You assume that if
the train has a speed v relative to the earth and the light has a
speed c relative to the train, then surely the speed of that light
relative to the earth will be c+v . This is called Galilean velocity
addition and is only approximately true when all speeds are very small
compared to c . In the theory of relativity the velocity addition
expression is u '=(v +u )/[1+(vu /c ^{2} )]
where v is the speed of one thing relative to the earth, u is
the speed of another thing relative to the first thing, and u ' is the
speed of the other thing relative to the earth. In your first case, v
is the speed of your train and u is c , the speed of light, so
c '=(v+c )/(1+v /c )=c .
In your second case v '=(0.9c +0.9c )/(1+0.81)=0.99c .
I urge you to look at the FAQ links, why c is the
universal speed limit , and relativistic
velocity addition .

QUESTION:
I understand that because of time dilation that we
could never see something actually entering a black hole. That is to say,
from our earth view, time would appear to stop at or near the event horizon
and observed objects would appear to move slower and slower until they
seemed to be standing still (to be clear, I certainly understand that the
object does not actually halt its motion, and will continue to accelerate
and fall into the hole). I understand (layman's level) the basic concept of
time dilation, and that it is how the speed of light remains constant for
all observers, but the idea of something appearing to halt its motion is
pretty mind-blowing...I picture (when it becomes technically possible to
actually see one) an event horizon cluttered with incoming matter but that
matter never descending (according to our observation) into the black hole.
That seems far fetched to me, so I am sure I'm missing something.

ANSWER:

First of all, once an object passes the event horizon, you cannot see it at
all. I would say that we really do not know how a clock would behave
inside the event horizon because, as shown below, the field is two
intense for the usual gravitational time dilation expression to be
valid. You might want to look at an earlier answer on
gravitational time dilation . The rate at which the clock runs slower
in a gravitational field (compared to your clock in a very weak
gravitational field) is
√[1-(2 MG /( Rc ^{2} ))]
where R is the distance from the center of a spherically
symmetric mass M . Note that if R =2MG /c ^{2}
this is zero and the clock stops. That magic R is called the
Schwartzchild radius and so if you are watching this clock, you see it
stop at this R . From the point of view of someone traveling with
the clock, though, time clicks on as usual and you enter inside R .
Your fate would be that as you got closer to the black hole (imagine you
are falling head first), the gravitational force would be much larger on
your head than your feet; this kind of gradient of force is called a
tidal force and it would rip you apart. A nice little animation may be
see
here .

QUESTION:
Theoretically speaking, If you had a super powered light bulb Initially located 1 light year away from an observer, but the observer moved at a constant 99% the speed of light away from the light source, how long would the observer observe the light if the light was turned on for a brief moment of like 4 minutes? would the observer observe only 4 minutes of an illuminated bulb or because
of the speed of the observer moving away from the light source would it take
longer for all of the light to reach be observed thus making it seem like
the light would be illuminated for much longer?

ANSWER:

This is an example of the relativistic Doppler shift. If the period of the
light waves at the source is T and at the observer is T' ,
the relationship between them is T'=T √[1+β )(1-β )]
where β=v /c. For your question, T'= 14.1T. It
follows, therefore, then that the time it takes for the wave to pass you
is 4x14.1=56.4 min. The light has been extremely "red-shifted".

QUESTION:
Suposing a Man Went To Another Planet That Has a Gravity
More Than Earth Hundreds Of Times Wich Means That Time Is Too Slower There Than Earth.
If We Say That 1 Hour In That Planet Is 10 Years On Earth And He Stayed 2 Hours There Will He Comeback To Earth Looking 20 Years Younger Than Other People Who Were at The Same Age When He Left ?

ANSWER:

No, not hundreds of times, billions of times more. I calculate that if the
planet were the size of earth, its surface gravity would be such that
the man's weight would be approximately ten billion times greater than
on earth. That man will not be coming back from this planet since his
own weight will crush him. If you had some indestructable clock which
you could put there and pull back when that clock had run for 2 hours,
it would return to you in 20 years as measured by your clocks. Although
I have not done the calculation, it is likely that this would, in fact,
be a black hole and that it would not be possible to return your clock
if it were inside the Schwartzchild radius.

QUESTION:
When an object is picking up speed while free-falling in a gravitational field, an accelerometer doesn't register any acceleration because there is none. The objects velocity is changing but it is not accelerating. How do you describe this state in which the velocity of an object is changing but it is not considered to be accelerating?
When an object is in free-fall in a gravitational field are there considered to be any forces acting on it? Of course when an object is not in free fall it experiences the force of gravity as it is pressed against the earth. When it is falling however, it feels no force so is it correct to assume that it has no force acting on it while in free-fall .

ANSWER:

You have just encountered what is called the equivalence principle. Since
you have stipulated that the object is in free fall, the answer to your
second question is that, yes, there is a force on the object, its own
weight. The equivalence principle states that if you are inside a box in
empty space with acceleration a , there is no experiment you can
perform which will tell you that you are not in a box at rest in a
gravitational field with gravitational acceleration a . A
corollary is that if you are in free fall in a gravitational field,
there is no experiment you can perform which will tell you that you are
not at rest or moving with constant velocity in empty space. That is why
the accelerometer reads zero in your first question, not "because there
is none"; the object is actually accelerating but cannot detect it.

There is a story about Albert
Einstein, probably apocryphal, but amusing and instructive. As you
probably know, his first job was as a clerk in the Swiss Patent Office
and he could easily complete all his responsibilities quickly and then
spend the rest of the day thinking about physics. One day, while
watching a workman paint a building across the square, the workman fell
from his ladder. Einstein thought to himself, "there is no experiment
which he can do which could distinguish whether he was in free fall in a
gravitational field or was in an inertial frame in empty space".

QUESTION:
I watched a documentary recently about extending human life, and my space loving head went to next stage with the idea and started pondering how that might prove useful for exploring the universe outside our solar system.
So to get to the point, the documentary spoke of a theoretical way to extend human life to 3000 years long. So naturally this would mean that traveling to other stars wouldn't be such a horrible ''life eater'' activity and make traveling to other stars more durable.
So in my head I took the next step. Travel to other galaxies.
The highest velocity on a spacecraft propulsion system I found was anti matter propulsion at 80%C.
So let's say our future spacecraft traveling at 80%C to Large Magellanic Cloud is traveling there with its crew. So plugging in relativistic effects, would traveling at 80%C mean that a person who was a baby when the craft left Earth, would still be alive (+and hopefully in a condition to increase the population in the destination) after the trip?

ANSWER:

3000 years sounds pretty crazy to me. And, don't you think it would be
pretty boring? Anyhow, we can estimate the time (on the ship) required
pretty easily. The gamma factor would be γ =1/√(1-0.8^{2} )=1/0.6
and the distance is about d =1.6x10^{5} ly. So, the
length-contracted distance would be d '=d /γ= 96,000
ly so the time to get there would be t=d '/v =120,000 years.
3000 years ain't gonna get it!

QUESTION:
If we heat a metal for a certain period with a
certain heat it will radiate EM waves (Light). If we continued to heat it
for a long time will the mass of the body gets affected?

ANSWER:

The higher the temperature, the more energy the metal contains. So,
technically, a higher temperature means a higher mass. The effect,
though, is so small that you would be hard-pressed to measure it.

QUESTION:
If the Higgs Boson is a hundred times heavier than a
proton, how are they produced by colliding two protons? Where does the extra
mass come from?

ANSWER:

Good question. Relativity tells us that the mass of an object which is m _{0}
when at rest and is traveling with a speed v becomes m=m _{0} /√[1-(v /c )^{2} ]
where c is the speed of light. In the LHC the speed of the
protons is
0.999999991 c ,
so their mass is m=m _{0} /√[1-(0.999999991 )^{2} ]=7456m _{0} ,
plenty of mass available!

QUESTION:
When light enters into a black hole does the velocity of light
become greater than the speed of light in vacuum due to extreme gravity?

ANSWER:

As light falls into a black hole, it gains energy like a mass does, but it
does not do that by increasing its velocity. The energy of a photon is
hf where h is Planck's constant and f is the
frequency of the light. So, the energy of the light increases by the
frequency increasing (or equivalently, by the wavelength decreasing).

QUESTION:
Clearly more massive objects create a greater gravitation
field, but what about objects that are equally massive yet have more energy?
For example, what if a certain object had an outrageous amount of heat, or
rotational energy? Would these objects produce an equally strong
gravitational field as their equally massive (yet cold and stationary)
counterparts?

ANSWER:

It is my understanding that general relativity predicts than any region in
space which has energy density creates a gravitational field, it need
not be static, cold mass. Since you must add energy to heat something
up, it would have more mass but very little more mass.

QUESTION:
The faster you move, the more massive you get, due to E=mc2, right? But do I ACTUALLY get heavier? Does the chemical energy in my cells also have mass? So if I run, therefore spending some chemical energy, and gaining some kinetic energy, do I not weigh the same when standing still and when running? (Assuming of course perfect consumption of energy to move me).

ANSWER:

First of all, it is not due to E=mc ^{2} , it is due to the
redefinition of linear momentum in special relativity. There are lots of
links on the faq which address the question of relativistic mass (1 ,
2 ,
3 ).
Basically, think of mass as inertial mass, the resistance to
acceleration. It is well known that objects cannot exceed the speed of
light, no matter how hard you push them. Therefore, inertia must be
increase as you go faster. In general relativity, Einstein was able to
show that gravitational mass is identical. So you have more mass (and
therefore weight) when you are running but by an immeasurably small
amount (unless you can run at speeds near the speed of light). All your
rambling about chemical energy, kinetic energy, etc . just
confuses the issue�you are either moving or you're not, don't worry
about how you got there.

QUESTION:
In case of electron the total energy will be equal to half m v square but it will also be equal to m c square according to Einstein's equation but substituting these we get v of electron more than c(speed of light) about root 2*c what do this mean

ANSWER:
What it means is that you do not understand what �mv ^{2}
and mc ^{2} mean. K ≈�mv ^{2} is the
kinetic energy of in classical mechanics of a particle of mass m and
speed v ; it is only (approximately) true for v<<c . E =mc ^{2}
is the energy which a mass m has, by virtue of its mass, when it it
at rest. The correct kinetic energy of a mass m with speed v
is K=mc^{2} [1/√(1-v ^{2} /c ^{2} )-1].
It is fairly easy to show that, if v ^{2} /c ^{2} <<1,
K ≈�mv ^{2} . Letting β=v /c ,

QUESTION:
Does special relativity theory seriously claim that there is no objective physical world independent of how variously it might be observed, as from frames with relativistic velocities? In other words, do physical objects and the distances between them physically contract? If not, why not clarify that observer-dependent measurements only show *apparent contraction,* not physical "shrinkage?"

ANSWER:
What does it mean to "physically contract"? If you mean is it
the same as if we put it in a vise and squeeze it, no, it is not the same.
In the vise case, the object gets shorter in its own frame whereas something
going past you at some high speed will still be its original length in its
frame but shorter in ours. In physics it is important to have objective, not
subjective, definitions of what we mean by something like length. Here is
what length means in physics: you measure the positions of the ends of the
object at the same time and the difference is the length. I think you
would agree that this is a perfectly reasonable definition of length. It is
not that things appear shorter , it is that things are shorter .
You might be interested in an
earlier answer .

FOLLOWUP QUESTION:
If a frame of reference approaches Earth at 86% of light speed, special
relativity claims it will measure its diameter to be about 4000 miles,
half that established by earth science. Please answer my question about
*apparent* contraction vs. *physical* contraction. Earth's diameter
obviously does not contract to 4000 miles or other variations according to
different speeds and directions.

ANSWER:
It is clear to me that you have made up your mind. I cannot imagine that you could have read my answer and have written what you have.
Normally, I would not have bothered to answer, but let me try one more thing.
The diameter of the earth is about 12.7x10^{3} m and the 0.86 c
is about 2.6x10^{8} m/s. Special relativity says that the diameter,
as measured by the moving frame is 12.7x10^{3} √(1-0.86^{2} )=6.5x10^{3}
m. Now, the time which the moving frame observes that it takes for the earth
to pass is 2.5x10^{-5} s. If the diameter of the earth was 12.7x10^{3}
m, the time would have been 4.9x10^{-5} . That is how long an
observer on the earth observes the moving frame to pass the diameter of the
earth. Both length and time depend on your reference frame. Experiments
equivalent to this have been done and everything I have written in this
paragraph has been experimentally verified. (If you had read my original
answer, you would have seen that I did answer your question about apparent
vs. physical contraction.)

QUESTION:
Can momentum of light be increased?

ANSWER:
Yes. For example, a photon falling into a black hole gains
energy as it falls. Its energy E is given by E=hf and its momentum
p is p=E /c . Therefore, the photon gains momentum by
increasing its frequency f .

QUESTION:
I want to know is there any expression for relativistic force as we have for momentum?

ANSWER:
In classical mechanics, force F only only has meaning in
terms of the acceleration a it causes some mass m to have, F=ma .
This can be more generally stated as the force is the time rate of change of
linear momentum, F =dp /dt . Using the relativistic
expression for p , this defines force in special relativity. Usually,
though, it is much more useful, both in classical mechanics and relativistic
mechanics, to solve problems using the potential energy V (r )
associated with a force, V (r )=-∫F∙ dr .

QUESTION:
If you were able to experience a time dilation field and you were to place
a living organism half way in to the field, what would be the hypothetical
result?

QUERY:
What is a time dilation field?

REPLY:
It's a field where inside it, time is either going by super slow or super fast relative to outside the field. It is used in scifi quite a bit so I'm not sure if it's just science fiction or if it is theoretical

ANSWER:
There are two ways you could achieve this. First, have a very
fast-moving spaceship. Inside the time would run much slower than outside.
Second, you could have a region of extremely intense gravitational field
adjacent to a region of very weak gravitational field; I cannot imagine how
you could achieve this. In the first case, how are you going to have your
organism half moving and half not? In the second, the strong gravitational
gradient across the boundary between weak and strong gravitational fields
would rip the organism apart before time could make any difference.

QUESTION:
I have seen some questions on your site about the different weaponry in Science Fiction. You have mentioned in many of them that it is ok to use the KE=�MV ^{2} on them because they are slow in terms of relativity. So my question is at around what velocity should the kinetic energy of an object be calculated with the relativistic kinetic energy calculation? Like I have heard that relativistic effects become noticeable at around 15% lightspeed, so around there? And at what velocity would you say that an object becomes noticeably relativistic as a rule of thumb?

ANSWER:
This depends on how noticeable is noticeable. Usually the gamma
factor is used as a guide, γ =1/√[1-(v /c )^{2} ].
At v/c=0.15, γ =1.01; a nonrelativistic calculation would only have
about a 1% error. The graph to the right shows that relativistic effects do
not really start kicking in until around v /c =0.5.

QUESTION:
I remember from college the common explanation of why massive bodies cannot exceed the speed of light: as objects travel faster, they acquires more mass, and as they acquires more mass, they require more energy to accelerate further. At the limit of the speed of light, it would take infinite energy to accelerate the object.
However, gravitational acceleration (in a vacuum) does not depend on mass--the energy required to accelerate an object still scales with its mass, but the gravitational force scales equally.

ANSWER:
Your supposition that gravitational acceleration is independent of mass is incorrect. This is only true at low velocities. To find out how a mass would move in a uniform gravitational field, see an earlier answer.

QUESTION:
I have recently gained an interest in physics, and have been looking into it over the past year. One of the areas I found interesting was special and general relativity, as I had always known the terms, but had no idea what they involved. As I have put more thought into time dilation, I have come up with a scenario that seems to be a bit of a paradox, and am hoping you can clear some things up for me. The scenario I have thought up is that of a clock on a spaceship that is travelling at a significant percentage of the speed of light. This spaceship is also set up with a camera, recording the clock, that is broadcasting live to another ship that is motionless. I imagine the moving ship to be going in circles around the motionless ship, but I'm not sure what effect, if any, this would have. So my question is, would the broadcast on the motionless ship run at normal speed, or would it run slower? Would people in the motionless ship see the broadcasted clock running slower on their screens, or would it run at the same speed because it's a live feed? To me, it seems that the broadcast would run at the same speed, because the image transferred from the camera on the moving ship to the screen on the motionless ship would travel at the speed of light, causing no significant delay. Also The camera is on the moving ship, so it is motionless relative to the clock, so it seems it would record and broadcast the clock at normal speed. But if time is running slower on the moving ship, how would it be possible for the motionless ship to see the moving clock running at the same speed? Let's say, according to the ship in motion, this task lasted 2.5 years, but it ran for 5 years on the motionless ship. So if the live feed was running the whole time, what happens? The camera was broadcasting the whole time, but that is only 2.5 years for the moving ship and 5 for the motionless one. So the image would have to slow down, wouldn't it? I assume I'm missing something here, so I hope you are able to clear this up for me.

ANSWER:
If the clock moves toward or away, your question involves the question of how time
is and how time appears and involves only special relativity. See an
earlier answer . If the moving clock is moving in a circular orbit of
radius R , it is more complicated and involves both general and
special relativity. In that case there will be time dilation due to its
speed, v , but also due to its acceleration a=v ^{2} /R
. To calculate the time dilation due to the acceleration you would use
the
equivalence principle and say that it would be the same as the
gravitational time dilation in a gravitational
field with acceleration g =a=v ^{2} /R . For the
orbit, there would be no difference between how time is and appears to be
because the distance from observer to clock is constant. Questions like
yours must be considered if GPS systems are to be accurate.

FOLLOWUP QUESTION:
Thanks for answering my question, you helped me to better understand what would be happening to the ship in orbit, so now I see that there would be more at play in that scenario. However, you have not answered my main question.
My main concern was what would show on the stationary ship's tv screen. Would it show in slow motion? If so, why does that happen? With my basic understanding it seems like the broadcast would run at full speed, but how could that be possible over a significant amount of time? Is it that the effects from the moving ship's speed and acceleration causes the transmission to slow down?
Thanks for answering my question, you helped me to better understand what would be happening to the ship in orbit, so now I see that there would be more at play in that scenario. However, you have not answered my main question.
My main concern was what would show on the stationary ship's tv screen. Would it show in slow motion? If so, why does that happen? With my basic understanding it seems like the broadcast would run at full speed, but how could that be possible over a significant amount of time? Is it that the effects from the moving ship's speed and acceleration causes the transmission to slow down?

ANSWER:
Well, you did not read my answer very carefully. If you had read the
q&a about the father and son, it
would have told you how sound would be sped up or slowed down and the picture of the clock would behave the same way. I guess I could have added for the circular orbit part that both the special and general time dilations are to slow down, so the
video of the clock would be be slower than the local clock. However, to have a significant slowing would require an enormously large orbit to have a large enough speed for special time dilation
to be significant and a small enough acceleration that it did not crush you;
the reason that the transmitted clock video is slower is simply because the
clock being recorded is running slower in the frame of the "stationary"
ship. Of course there is the time delay due to the time of transmission in
all cases, so if you synchronized all clocks at the start you would see a
delay when comparing clocks, so it would be important to look at the rates
of the clocks, not their absolute readings.

QUESTION:
Are there any theories as to why the speed of light is 186,000 mps? That is, why not more, why not less, what dictates this value?

ANSWER:
Maxwell's equations are four equations which contain everything
(except photons) that can be known about electromagnetism. They
predict waves which propagate with a speed
of 1/√(μ _{0} ε _{0} ) which just happens to be
exactly the measured speed of light. The two constants, μ _{0}
and ε _{0} , are the permeability and permittivity of free
space and essentially quantify the strengths of electric and magnetic
forces. You might also want to read earlier related answers linked to on my
faq page.

QUESTION:
If one stood on a typical neutron star and turned on a flashlight pointed horizontally, would the light bend down to the surface much like water coming out of a garden hose here on earth?

ANSWER:
The density of a typical neutron star is about 5x10^{17}
kg/m^{3} and the radius is about 12 km, so the mass is about M =5x10^{17} x4xπ (12x10^{3} )^{3} /3=3.6x10^{30}
kg. To find the acceleration due to gravity, g=GM /R ^{2} =6.67x10^{-11} x3.6x10^{30} /(12x10^{3} )^{2} =1.7x10^{12}
m/s^{2} . So your weight on the surface would be about a trillion
times what it is on earth. You would be crushed totally flat. We can
estimate how far light would "fall" in such a gravitational field. The time
it would take to go 100 m would be about t ≈100/(3x10^{8} )=(1/3)x10^{-6}
s. The distance it would fall would be y =�gt ^{2} ≈10 m.
On earth the speed of something which dropped 10 m in 100 m would be about
71 m/s≈160 mph, quite a bit faster than a garden hose. But qualitatively,
yes the light would fall quite a lot.

QUESTION:
[Gravitational time dilation] From a mechanical point of view I understand how a clock can tick slower if its closer to the gravitational mass. But how does that affect our biology so that we actually age slower?

ANSWER:
Actually, you do not understand how a clock can tick slower
because it is not something gravity does to the clock, it is something
gravity does to time. All clocks run slower because time itself runs
slower as the gravitational field gets stronger; this includes biological
clocks. You would never notice it for biological clocks in the earth's
gravitational field because the effect is very tiny and biological clocks
are not accurate enough to detect the slowing. You could detect it if you
were close to a black hole where gravity was much stronger.

QUESTION:
If my friend were to travel toward me near the speed of light, then length contraction would mean he would appear smaller, correct? And from his POV I would appear smaller than him. So if I were to hold up my arm to the side of me and my friend were to run into it, what would this look like? From my POV he has shrunk and would therefore go under the height of my arm untouched. But from his POV, my arm would be near his knee height and he would trip up?

ANSWER:
You have stumbled on an example which argues that length
contraction happens only along the direction of relative motion. So you and
your friend are each thinner to each other but the same heights. (Regarding
your expression "appear smaller", lengths do not necessarily appear
shorter but they are shorter. See an
earlier answer .)

QUESTION:
If an object had zero mass (technically it could not exist) but for the sake of the question let's assume it can, and could withstand any physical affliction and accelerated at 1 km/h would it accelerate at an infinite speed or travel a certain speed only?

ANSWER:
An "object" with zero mass can exist; it is called a photon.
However, you must be thinking of this object classically using Newton's
second law, a=F/m and so any force will cause an infinite
acceleration. In fact the theory of special relativity requires that any
object with zero mass can move only at the speed of light in vacuum. So a
massless object at rest cannot exist.

QUESTION:
An accelerating rocket ship, with an observer at the top (with a clock) and an observer at the bottom (with a clock) will observe the other's clock running at a different rate from than their own. The observer at the top of the rocket will see the "bottom" observer's clock running slow, of course this difference is going to be incredibly small, but we have really good clocks.
The observers also have accelerometers. If the top observer views the bottoms observers accelerometer to be the same as his, acceleration same on top as bottom, but the bottom observers clock running slower than his, how would the observers reconcile the fact that there is no relative velocity between the two? The same acceleration, over different amounts of time would result in a relative velocity between the two, but how can there be a relative velocity between the two of them? This can't be a "reading of the clock" error, as in the bottom observer's reference frame, his clock is running fine, it is the top observer's clock that is running fast. Would you have to average the times, and apply this time to the center of mass of the rocket?

ANSWER:
First, let's be sure we understand why the clocks run at
different rates. If we apply the equivalence principle, there is no
experiment that can be done on this accelerating rocket ship which can
distinguish between acceleration a and being at rest in a
gravitational field with local gravitational acceleration a .
Therefore, it is as if the two were at rest in a uniform gravitational
field. Then it is a consequence of general relativity that the clock
"higher" in the gravitational field will run faster. Now, you suggest having
accelerometers at each end and that they will record the same; that is true
because it is no different from being at rest in a uniform field and
measuring the field at both ends. Now, if you are at rest in a field, there
is no problem because there is no acceleration and therefore no velocity
change. So the results are required (by the equivalence principle) to be the
same if you are in a zero field with an instantaneous acceleration. I think
the problem is that since acceleration is not really a useful quantity in
relativity, you should not conclude that equal accelerations in unequal
times necessarily implies a relative velocity between two observers in the
same frame. In fact, it clearly cannot because both are at rest in the same
frame.

FOLLOWUP QUESTION:
One of the stipulations of the Eqivalence Principle, is that that you can not distinguish between an acceleration of a frame at "g" and being in a UNIFORM gravitational field of the same "g". The word Uniform here implies that the acceleration field at point "a" is the same value as the acceleration field at point "b" some distance away, in other words, no gravity gradient.
But planets do have gravity gradients ( although the earths is a weak gradient). If I was in a gravity gradient, that was really steep, I would feel this gradient as my feet being pulled at with a different acceleration then my head.
How does a steep gravity gradient change the equivalence principle?

ANSWER:
If the field is not uniform, it is just a little harder to
compute the relative rates that clocks run in the field. What determines the
rates at two points in space is the potential difference between the two
points. It does not change the equivalence principle. There is still no
experiment you can do to determine whether you are in a uniform field or an
accelerated frame. Incidentally, the effect of field gradient is called the
tidal
force . You may have read somewhere that when you fall into a black hole,
you get "all stretched out" as you fall; this is due to the extreme tidal
force from one end of your body to the other.

QUESTION:
Since everything is relative in special relativity, it is equally valid to consider the ''Earth'' to be
moving toward ''stationary'' particles in the upper atmosphere. In that case, time slows down for Earthbound observers. The particles then decay at their usual half-life pace in their stationary reference frame while only a fraction of these half-time passes for the speeding observers on Earth. Then, just as the speeding astronaut in the Twin Paradox returns to find a much older twin, the speeding Earthbound observers would encounter an extremely old population of cosmic ray particles, which means that they should have long since decayed, and should not have been ''detected''.

ANSWER:
The results are perfectly symmetric regardless of whether you
put the earth in a rest frame or the particle in a rest frame. In the earth
frame, the lifetime is lengthened by a factor γ =1/√(1-(v /c )^{2} ),
T' =γT (T is the lifetime for a particle at rest) but in
the particle frame, a distance toward the earth is shortened by the factor
1/γ , L'=L /γ (L is the distance for the earth at
rest). To check that this makes sense, the earth sees the particle
move with speed L /T' =γL' /γT=L' /T and the
particle sees the earth move with a speed L' /T =(L /γ )/(T' /γ )=L /T' ,
both the same.

QUESTION:
Say I am a stationary observer and observe a rocket launched from earth to Proxima Centauri which is about 4.35 light years away from us the rocket has a constant acceleration of 1g or 9.8m/s^{2} for half the trip and -1g for the other half. Newtonian kinematics states x_{f} =x_{i} +vt+at^{2} /2 yet over the course of time my velocity will become relativistic and the equation no longer applies.
What is the correction needed for me, a stationary observer, to solve how long it will take the rocket to reach its
destination?

ANSWER:
It turns out that I have already solved
this problem ;
I have found a solution for the distance as a function of time for a
constant force F rather than the time as a function of distance, but
that can easily be inverted. I showed that x =(mc ^{2} /F )(√[1+(Ft /(mc ))^{2} ]-1);
in your case, F=mg , so x =(c ^{2} /g )(√[1+(gt /(c ))^{2} ]-1).
Solving for t , t =√[(x /c )^{2} +(2x /g )].
For the first half of the trip, x =2.175 ly and I calculated that g =9.8
m/s^{2} =0.11 ly/yr^{2} , so t =6.58 yr. Given the
symmetry of the situation, the total time for the trip, as observed by you,
is 13.2 yr. Using the result from the
earlier answer , v =(gt )/√[1+(gt /c )^{2} ],
I find that the speed at the midpoint of the trip is 0.89c . The
average speed for the whole trip was 4.35 ly/13.2 yr=0.33c .

FOLLOWUP QUESTION:
What is the correction to determine how much time the rocket will have
perceived to have passed?

ANSWER:
The mathematics gets a little difficult here but it is a common
problem which has been worked out. I will just give you the final result.
For more detail, see the blog by John Baez on the
relativistic rocket . For a rocket with acceleration g halfway to
a distance D in light years and then with deceleration g the rest of the way, the
elapsed time T in years on the rocket is T =1.94∙arccosh[(D /1.94)+1)].
So, for D =4.35 ly, T =3.58 yr.

QUESTION:
Suppose i am inside a uniformly moving spaceship and I send a light beam perpendicular to the direction towards the opposite wall. since the spaceship is moving and according to relativity light doesn't feel any sort of "kick", shouldn't the light beam be off the target and thereby letting me know that
I am moving?

ANSWER:
If you aim at a spot on the wall you will hit it. You will observe the light beam going straight across the ship.
The reason is the principle of relativity which states that there is no
experiment that you can perform which can distinguish whether you are at
rest or moving with constant velocity. You could also fire a rifle at the
opposite wall and hit the target. An observer outside your spaceship,
however, will see the light with a component of its speed in the direction
you are moving but will still see the speed of the light as being c
and the light hit the target. Your experiment would be different, however,
if you were accelerating forward; the light would miss the target for the
same reason that if you were to aim a rifle directly at a target on earth it
would fall some on the way there. You might be interested in the
light clock . Your space ship could be thought of as a light clock which
ticks once when the light hits the target.

QUESTION:
I've been wondering about the hazards of traveling at high speeds in
space for a while and the challenges it would bring for us in terms of
spaceship design. My main interest is interstellar travel and how meteors
might affect the safety of such an exercise. So if we had a spaceship that
would accelerate to let's say 7 percent c and we sent it along with some
colonists to Alpha Centauri and the ship while traveling 7 percent c hit a
meteor the size of a man's fist, how catastrophic would an impact of this
nature be to the spaceship?Let's say the meteor was 10cm across and was
comprised of iron.

ANSWER:
There is no way I can even begin to do a calculation here. It
would depend on the design of the space ship. Even if I had more details,
this would be more of an engineering problem than physics. You can use your
imagination, though. The shuttle Columbia was going 545 mph=11 m/s≈0.000004%
of c and a piece of foam insulation (much softer than iron) had
catastrophic results. You might be interested in another
recent answer .

ADDED ANSWER:
I estimate your meteor would have a kinetic energy of about 2x10^{17}
J, about the same as 2,000 Nagasaki atomic bombs. Curtains for this space
ship!

QUESTION:
What is the Cosmological Constant and why it couldn't be described perfectly by Einstein?
Was only because of it that Einstein thought the universe was static?

ANSWER:
When Einstein proposed the theory of general relativity, around
1918, it was generally believed that the universe was static. General
relativity is the theory of gravity, and if gravity is the main interaction
among stars and galaxies, this is not possible; with only gravity, the
universe would have to be either expanding and slowing down or compressing
and speeding up. Therefore Einstein had to introduce something to balance
the universal gravitational force; this something was called the
cosmological constant. It was later discovered that the universe is
expanding and he later denounced the cosmological constant as "�my greatest
blunder�" Interestingly, several years ago it was discovered that the
expansion of the universe is actually speeding up implying some kind of
repulsive force, akin to the cosmological constant, often referred to as
dark energy; this has reignited interest in Einstein' "greatest (or maybe
not) blunder". I do not know what you mean by "�couldn't be described
perfectly�"

QUESTION:
A space craft travelling at very near light speed sets a course for a distant object.
So travelling at near light speed, could a navigation computer make corrections quick enough, and alter the trajectory for objects in its path. Could these calculations be made quick enough to miss the object?
Assuming computer calculations happen at the speed of light the craft could be on a collision course with an object before (a:) it saw it and (b:) before it could recalculate an evasive action missing the object.
With the added disadvantage that human thought isn't any where near quick enough to react to a blocking object.
So how could 'we' travel to distant objects without fear of ploughing into the first path crossing object?

ANSWER:
You raise an interesting and important question here. But it is
much worse than you think. For one thing, because of length contraction, an
object which an outside observer would see as being, for example, 1 light
year away from you would only be 1x√(1-.99^{2} )=0.14 light years
away as you observed it if going 99% the speed of light; that gives you a
lot less time to maneuver. But there is a much more important barrier to
being able to navigate at very high speeds. The upper picture to the left
shows the space craft as seen by an outside observer; light from a distant
star comes with velocity c making an angle of θ
relative to your velocity v . But, if you now are on the space
craft, you observe the star at a different location specified by θ'
because of the velocity transformation to the new light ray c' .
Note that the magnitudes are the same, c'=c , but the directions are
different, θ'≠θ. It can be shown that θ'= tan^{-1} [sinθ √(1-(v /c )^{2} )/(cosθ+ (v /c ))].
I have plotted this for several values of v to the right. The effect is
quite dramatic. For example, at 99.9% the speed of light everything in your
forward hemisphere and much from behind you will appear inside a 5^{0}
cone in front of you! Just having optics which could give good enough
resolution to see anything but an extremely bright light in front of you
would be a challenge in itself. But wait, it even gets worse! There will be
significant Doppler shifts and much of the light at very high speeds will be
shifted out of the visible spectrum. A nice little animated gif below (which
I found on
Wikepedia ) demonstrates this but only up to 89% of c .

QUESTION:
Does mass affect kinetic energy?

ANSWER:
Of course. Classically the kinetic energy K =�mv ^{2 } where
m is mass and v is speed. Relativistically, K=E-m _{0} c ^{2}
where c is the speed of light, m _{0} is the mass of
the object when not moving, the total energy E =√(p ^{2} c ^{2} +m _{0} ^{2} c ^{4} ),
and the linear momentum p =m _{0} v /√(1-v ^{2} /c ^{2} ).
So, you see, mass appears all over the place. The momentum of a massless
particle, the photon, is p=E /c ; its energy is all kinetic, and
K=E=pc
since a photon has momentum even though it has no mass.

QUESTION:
Suppose two friends break their arm at the same time and a doctor says it will take both six weeks to heal. On the same day one friend is due to board a rocket that travels 99.999% speed of light. They have an uncanny ability to count in sync, in their heads. They agree to close their eyes and count up to six weeks, second by second.
The traveler only has to hit a reverse button once he counts exactly 3 weeks, and he'll return to the exact spot he left his friend.
After they have both counted six weeks in their heads, they open their eyes...
Is the traveler back with his friend??
The friend who did not travel finds his arm has healed. Has the travelers arm also healed??
The obvious answer is yes to both. The traveler has traveled at the same speed one way as far as the other (counting 3 weeks each way).
The traveler has counted the same amount as the non-traveler (3,628,800 seconds) so therefor his arm should be healed.
Special relativity would suggest his arm has not healed.. it would suggest a lot less than 6 weeks has passed for him! It suggests that not only time slows down, but the travelers mind & body slow down too...does he count more slowly in his head? Does his breathing slow down also? Wouldn't he die from breathing so slowly??!!

ANSWER:
You have this a bit muddled. Let us first find out how far d'
the traveling friend travels in 3 weeks (as measured by him). d'=vt' =0.99999c ∙3=2.99997c
or d' =2.99997 lw each way (1 lw is a light week, the distance light
travels in 1 week). The earth-bound friend, however, sees a much longer
distance that his friend has traveled because the friend sees the distance
length-contracted by a factor √(1-.99999^{2} ); so the distance which
the earth-bound friend sees is d =2.99997/√(1-.99999^{2} )=670.8
lw each way; the time someone traveling 0.99999c takes to go this distance
out and back is 2x670.8/0.99999≈1342 weeks. The earth-bound friend has been
healed for about 1336 weeks, more than 25� years, when his friend returns
with a just-healed arm. You still cannot really accept that moving clocks
really do run slowly, can you?!

QUESTION:
If I shine a light it emits a light particle outwards. Assuming it is travelling in vaccum all the way out and hits no light absorbing medium.
The farther it travels it starts losing energy and shifts to the red. Now how long in Earth years will it take for that photon particle to lose "ALL" its energy. How far will it travel in distance (light years) and what happens to that particle at the end of it all.. i.e it just cannot "disappear" or become void or null.
Or perhaps it will all be converted to some other form of energy but then what would it lose its energy to given it was hypothetically travelling in a vaccum/void all the way out?

ANSWER:
Let's first consider the particle to be a ball. If you throw it
up it will begin immediately losing energy and eventually stop. You can now
think of the earth as having acquired the energy the ball had, now in the
form of potential energy. Being a ball, it then falls back and extracts all
its energy back. However, if you throw the ball hard enough it will keep
going forever and the farther away it gets, the less energy it loses,
essentially moving with constant speed when it has gotten pretty far away.
The same is true of a photon, as you note, except for two important
differences: it cannot stop and it has no mass, just pure energy. The red
shift you refer to is the gravitational red shift and as the photon becomes
redder it is, as you stated, losing energy (also to the mass it is leaving)
but not speed. For the earth (and almost every object in the universe) the
effect of gravity is insufficient to ever cause the photon to lose all its
energy. Only black holes can rob a photon of all its energy; if a photon is
inside the event horizon of a black hole, as it tries to escape it gets
"redder and redder" until it has finally lost it all and it disappears. The
energy it had will now be found in a slightly increased mass of the black
hole.

QUESTION:
If I am at a point in space where very, very, very little gravitational pull exists, what happens to the time clock on my spaceship as opposed to clocks on earth?

ANSWER:
A clock in a gravitational field runs slower.

FOLLOWUP QUESTION:
What I am asking, if you were at a point in space, where the gravitational forces were almost 0, would your clock, in your space craft be running extremely fast as observed from earth?
I was thinking about a spot in the universe were the gravitational effect
would be almost opposite of a black hole. Like the trampoline example you used on the web site. I set 5 or 6 bowling balls on the trampoline so each makes a dent. I then use a marble and find a spot to set it where it is not affected by the bowling balls. Now I know my ship is moving always towards some mass but there has to be a spot where gravitational effects all almost 0. Maybe that spot in space would be inaccessible to my ship, kind of an anti event horizon where I would not have enough power to get too. a spot where Light can not even enter.
I Know that my question is flawed, but I can't help to think about a black hole as a gravity pit, and that makes me think of a spot in space where there is a gravitational mountain (Anti-Black Hole, with an anti-Event horizon). In this spot time runs so fast as not to exist....

ANSWER:
First of all, all your thrashing around with black holes etc .
trying to find a spot with small gravitational field is totally unnecessary;
all that is required is that you need to be far away from any mass and
nearly the whole universe satisfies that criterion. We generically call such
places intergalactic space . So, choose some place, maybe halfway
between here and the Andromeda galaxy, our nearest galactic neighbor. For
all intents and purposes the field there is zero and a clock would run at
some rate. Now, take that clock and put it on the surface of a sphere of
radius R and mass M . The
gravitational time dilation formula you need to compute how much slower
the clock would tick at its new location is √[1-(2MG /(Rc ^{2} ))] where G =6.67x10^{-11}
N∙m^{2} /kg^{2} is the universal constant of gravitation and
c =3x10^{8} m/s is the speed of light. Putting in the mass and
radius for the earth, I find that the space clock runs about 7x10^{-8
} % faster than the earth clock, certainly not "extremely fast"! The
bottom line here is that, except very close to a black hole, gravitational
time dilation is a very small effect. An interesting fact, though, is that
corrections for this effect must made in GPS devices because they require
extremely accurate time measurements to get accurate distance measurements.

QUESTION:
From a spaceship that moves away from you very quickly (say 0.33c), a "photon" is emitted when the spacecraft clock marks the time t0 and a length measurement performed on the ship tells your distance is 1 light year.
How long does the photon to reach you? with times measured by you and by the ship?

ANSWER:
The distance from you to the ship, as measured by you, at the
time the photon is emitted is 1/√(1-.33^{2} )=1.06 ly and so the time it takes to reach
you is 1.06 yr. An observer on the ship says it takes 1 yr.

QUESTION:
Two people are on a high speed, constant velocity train on opposite ends. One person standing on the platform. The two people aim a laser beam at the center of the train at the same time in their reference frame. In the center of the train, there is a special gun that will shoot and kill whoever's laser beam reaches it first; but if the two beams reach it at the same time it will be disarmed. From the person on the platform's standpoint, one of them dies due to the relativity of simultaneity. From the two people on the train's standpoint, they both live. Later, the two people get off their high speed train and meet with the person on the platform. Is this not a paradox? The platform person clearly saw one of them die.

ANSWER:
The only thing that matters is what the "special gun" sees; it
is in the same inertial frame as the shooters, so it does not fire because the two pulses arrive simultaneously in
that frame. The guy on the side sees the pulses not arriving simultaneously but the gun not firing.

QUESTION:
I am inside a train in which all the sides/windows are sealed i.e., I cannot see anything from inside. If the train moves in a friction-less rail, i.e., without jerking, how can I prove that there is motion taking place or there is displacement?

ANSWER:
There is no such thing as absolute velocity, so there is no way
you can determine if you are moving with a constant velocity. The principle
of relativity says that the laws of physics are the same in all inertial
frames of reference so there is no experiment you could devise telling you
what your speed is. If your train were on earth, then you would not be in an
inertial frame because the earth is curved and rotating and very careful
measurements could detect this, but I believe you meant that the train is
moving in a perfectly straight line and no earth motion effects

QUESTION:
(This question refers to an
earlier
answer .)
With respect to your
ANSWER above, I have read elsewhere that one of the observations from particle accelerator experiments is that the mass of a particle increases the more it is accelerated AND that there is a concomitant release of energy (I recall reading �release� vs expenditure or other terminology, but the ol� memory could be faulty, too). I read in your
ANSWER above about inertia being what increases versus size/"amount of stuff" as a given mass accelerates. Per my outside reading I am attempting to conceive the source of the energy released, not in mathematical terms, but with creative logic, if there is such a thing. Maybe not. Math might be the only way to explain the phenomenon. My attempt with English in question form: Is the measurable release of energy due to the fact that so much more energy is required to continue accelerating the particle? Is the source of the "released energy" I read about actually the instrument itself, the particle accelerator? For some reason, prior to reading your
ANSWER , I conceived that as mass/inertia increases [as acceleration takes place] there was some inherent energy in the particle itself that with the added force of acceleration, inevitably had to be expressed / released. Creative huh? What IS the source of that energy "released" as the particle accelerates and its inertia increases? Or am I so way off you�re having a cleansing, releasing, hearty laugh? And, can this concept be explained in language that an extremely uneducated but by no means unintelligent person can understand? Are you that good? I�m giving you lots of credit, much deserved for your dedication to askthephysicist. If answering requires math, my reading said answer is not likely to be helpful so save yourself the time and move on with your day.

ANSWER:
I have never encountered reference to such an "energy release"
by relativistic particles as they were accelerating. While the particle is
accelerating, the force accelerating it does work which increases the energy of the particle. The only math I will use here is
E=mc ^{2} which I hope you can work with�this simply says that any particle
has an inherent energy due to is its mass m which is the mass (inertia) times the speed of light
c
squared. If a particle is at rest, it has energy because it has mass. If a
particle is moving, the classical view is that it also has what we call kinetic energy,
energy by virtue of its motion. There are two
ways you can look at it. One is that we observe that the mass m
increases as the particle speeds up, eventually approaching infinite mass as
the speed approaches the speed of light; in this view, kinetic energy is
simply folded into the quantity mc ^{2} . The other way is to
say that the particle has an inherent mass energy m _{0} c ^{2}
where m _{0} means the value that m has if the particle is at
rest; in this view we interpret the energy of the particle to be rest mass
energy plus the kinetic energy K it has by virtue of its motion, m _{0} c ^{2} +K .
These are just two ways of looking at the same thing; therefore, if the
particle retains its identity, if it now interacts with its environment it
can release the energy it previously gained by accelerating and if it ends
up at rest it will have released K Joules of energy. Any energy this
particle "releases" can only come from energy which it previously acquired.

ADDED
THOUGHT:
An accelerated particle
does radiate electromagnetic energy (light, x-rays, etc .) called
bremsstrahlung ; this is how an antenna works. So, when a particle is
being accelerated, it does "release" a bit of energy in this way. The rate
of this radiation is usually quite small compared to the rate at which
energy is being added and is often ignored when considering accelerators.
That which you "read elsewhere" might have been referring to
bremsstrahlung.

QUESTION:
What is mass?

ANSWER:
There two kinds of mass. Inertial mass is the property an
object has which resists acceleration when a force is applied; the harder it
is to accelerate something, the more inertial mass it has. Gravitational
mass is the property an object has which allows it to feel and create
gravitational forces; for example, the more gravitational mass an object has
the greater the force it will feel due to the earth's gravity�the more it
will weigh. It turns out that the two masses are actually identical; this
fact is one of the cornerstones of the theory of general relativity.
QUESTION:
W hat is the formula to calculate the total energy of an object that is moving at relativistic speeds? (Ignoring any potential energy)

ANSWER:
E =√(m ^{2} c ^{4} +p ^{2} c ^{2} )
where p=mv /√(1-v ^{2} /c ^{2} ) is the
linear momentum, m is the rest mass, v is the speed, and c
is the speed of light. This can also be written as E=mc ^{2} +K
where K is
the kinetic energy (which is not �mv ^{2} ). Comparing these two
expressions and doing some algebra, you could find an expression for K .
Also, it may be written E=mc ^{2} /√(1-v ^{2} /c ^{2} ).
QUESTION:
I read a fact that a compressed spring weighs more than a relaxed spring...why is it so??

ANSWER:
It is true but unmeasurable. Suppose the spring constant
is k =2,000 N/m and you compress it by x =1 cm=0.01 m. The work
you did is E =�kx ^{2} =0.1 J. This energy will end up as added
mass m=E /c ^{2} =10^{-18} kg.
QUESTION:
I was reading a novel the other day which takes a surprisingly realistic approach to theoretical space-combat (namely ships attacking each other with missiles from many millions of kilometers away), albeit with missiles moving at significant fractions of the speed of light (around 30-40%).
This made me wonder, say two space ships are fighting and one fires a missile traveling we'll say 50% the speed of light. I'm having real trouble understanding the effect time-dilation would play (and the author completely avoided the subject). I tried reading some of your past answers on time-dilation, but I just ended up confusing myself.
My thought is that the missile which is traveling at the significant fraction of c (while the ship is not) would have a huge advantage. With time passing slower the missile would have the net effect of having more time available to itself; to which it could use to account for the ship's attempts to evade, to itself evade any counter-measures the ship might employ, all in addition to being difficult to detect and track due to its speed.
Would the theoretical missile in this situation have those benefits, and the target ship be suffering a serious disadvantage; or does time-dilation not work that way?

ANSWER:
Time dilation really does not come into play here if I
understand things correctly. If you are thinking that the missle is robotic
and able to adjust to evasive movement of the target, there is no advantage
because the missle sees the distance to the target shrink by the same factor
that its clock slows down. As seen from the stationary fleets, there is just
a projectile with half the speed of light. Since the speed of light is
300,000 km/s, the attacked fleet will observe the missle coming with a speed
of 150,000 km/s and many million kilometers, let's say 150 million for
illustration, means that it will take 1000 s to get to the target as
measured by the fleets. The target fleet will not know it has been attacked
until 500 s before the missle arrives, but that still gives time for evasive
action. The main problem is that because of the high speed, evasive action
becomes increasingly difficult as the missle gets close but this has nothing
to do with time dilation. The advantage to the missle is due to its high
speed and the limited reaction time of the target.
QUESTION:
I'm in a space ship moving at the speed of light. I put my space suit on, go out on the deck of the ship and tee up a golf ball. Assuming I have the skill necessary to hit the ball, will I be able to hit the ball forward? It seems to me that in the vacuum of space, the energy required to hit the ball forward would not change with my vehicle's speed, but it's also my understanding that the speed of light is an absolute maximum speed. One of these is apparently wrong. I think you are going to tell me I can't hit the ball forward. Can you explain why?

ANSWER:
As I have said a thousand times, you cannot move at the
speed of light. See the faq page if you
have a problem with this. However, we can address your question by saying
your ship has a speed 99.999% the speed of light u =0.99999c .
Suppose that you can launch a golf ball at a speed of 10% the speed of
light, v =0.1c . (Of course, it is impossible to make it go that
fast because you are not strong enough, but it makes this example easier to
quantify.) If you were able to do this, there would be absolutely no reason
why you could not do it on the moving ship: you would simply see the ball
move away from you with speed of 0.1c just as if you were here on
earth. Now, classical physics (and your intuition) would have the speed seen
by someone watching this to be V =(0.99999+0.1)c =1.09999c ,
almost 10% faster than the speed of light. However, classical physics does
not work here; again, see the faq
page. Instead, V =(0.99999+0.1)c /(1+0.99999x0.1)=0.99999182c .
QUESTION:
My question deals with the "potential energy"
of a gravitational field, in relativistic terms. As an object falls in a
gravitational field, a tiny bit of its mass gets converted to kinetic
energy, a small reduction in mass. However, according to Special Relativity,
as an object accelerates, gets closer to the speed of light its mass (how it
interacts with spacetime) increases. So, does an object accelerating
(increasing kinetic energy) in a gravitational field lose mass, as a result
of falling in a gravitational field, or gain mass as a result of approaching
the speed of light? OR is there some breakeven point were the reduction of
mass from falling stops, and then mass starts increasing again as a result
of approaching the speed of light. If the mass starts increasing because of
approaching the speed of light, wouldn't the gravitational potential
approach infinite?

ANSWER:
You are the same questioner from
earlier similar questions . The reason you do
not want to simply say that there is "a
small reduction in mass" if it is falling is that it is not at rest when you
look at it later. As I explained in the earlier answers, you need to
calculate things relativistically to find out how the mass changes. That is
essentially what the
last link I gave you in my earlier answers does, calculates
β= v /c
as a function of time for a constant force. To know how the mass varies (if
you interpret relativistic momentum that way), just calculate m _{0} / √(1- β ^{2} )
where m _{0} is the rest mass. It occurs to me, though, that
it might be of interest to redo that calculation for a force which is not
constant but which has a value mg =m _{0} g /√(1- β ^{2} )
where g is the acceleration due to gravity in whatever strength field
you wish to examine. I will not give all the details here, just the results.
We start by integrating the relativistically correct form of Newton's second
law,
m _{0} g /√(1- β ^{2} )=dp /dt =(d/dt )[m _{0} v /√(1- β ^{2} )]
integrated gives
gt /c =�ln[(1+β )/(1-β )]
solved for β gives
β= (1-exp(-2gt /c ))/(1+exp(-2gt /c ));
put this into the mass and get
m /m _{0} =1/√(1-β ^{2} ).
These are plotted to the right above.
Note that the
result for a constant force m _{0} g
is plotted in red for comparison. Potential energy is not a useful concept here. Note that the time to get to
near c , t ≈3c /g , is about 280 years for
g =9.8 m/s^{2} ! At that time the moving mass is about 10 times
greater. A more lengthy discussion of a mass in a uniform gravitational
field (including general relativity) can be seen
here .

QUESTION:
Around 1905 when Lorentz came up with his tranformation equation, it was assumed that c was constant, but was it also assumed that c is the speed limit of the universe?
Why didn't he assume that v could be greater than c in some cases? Why didn't he put a corrective constant next to c in the equation to allow for the case that time might stop BEFORE or After v reaches c?
The reason I'm asking this question is because the Lorentz transformation has time stop exactly when v=c. How did he know that should be the case?

ANSWER:
You have your history a little skewed, I think. Lorentz
was working with electromagnetic theory, Maxwell's equations in particular.
The transformation which bears his name was arrived at empirically with no
reference to the speed of light. He noticed that if you took Maxwell's
equations and transformed them into a moving frame of reference that they
took on a different form which was unacceptable because they had internal
inconsistencies which could not be borne out by experiments. He found that
if he used a somewhat different transformation, now called the Lorentz
transformation, the transformed equations were self consistent; he had
stumbled on a truth without understanding what he had. And he certainly did
not consider the speed of light constant because he was one of the foremost
proponents of the luminiferous �ther, the special medium which
supposedly supports light waves. To my mind it is unfortunate that the
transformation is named for him since it was an accidental empirical
discovery, not based on any fundamental physics. Some books try to refer to
the Lorentz-Einstein transformation, but that hasn't really caught on. To
understand why c is a constant and why it has the value it does, go
to my FAQ page; there you can also see earlier questions about why c
is the speed limit. Your question about a "corrective constant" makes no
sense since universal constants by definition do not need correcting. Since
no clock can ever reach c , time never "stops". I doubt that Lorentz
ever thought about time stopping in any context.
QUESTION:
I have points A, B, and C which are moving in 3 dimensional space. If I use point A as a frame of reference, I know the direction and velocity of points B and C. If my points are moving arbitrarily close to the speed of light, what equations would I use to calculate the relative velocity of C from B's frame of reference, and the difference in clock speeds between those two points?
I understand how I would do this if the speeds were nowhere near the speed of light, but things get a tad complicated after accounting for relativity.

ANSWER:
I have previously dealt with relativistic
velocity addition if the velocities of B and C are colinear: in your
notation, V _{CB} =(V _{CA} +V _{AB} )/(1+(V _{CA} V _{AB} /c ^{2} ));
read V _{XY} as velocity of X relative to Y. This may be
generalized
to noncolinear motion as V _{CB} =(V _{CA} +V ^{‖} _{AB} +γ _{CA} ^{-1} V ┴ _{AB} )/(1+(V _{CA} ∙V _{AB} /c ^{2} ))
where V ^{‖} _{AB}
and V ┴ _{AB
} are the components of V _{AB}
parallel and perpendicular, respectively, to V _{CA }
and γ _{AB} ^{-1} =√[1-V _{CA} ^{2} /c ^{2} ].
Note that if the speeds are small compared to c , V _{CA} ∙V _{AB} /c ^{2} ≈0
and γ _{CA} ^{-1} ≈1
so V _{CB} ≈(V _{CA} +V ^{‖} _{AB} +V ┴ _{AB} )=(V _{CA} +V _{AB} ).^{ } The reason that the transformations are different for parallel and
perpendicular motions is that there is length contraction for the parallel
motion but not for the perpendicular motion.

QUESTION:
How can the speed of light be the fastest thing in the universe if it is a relative measurement? I mean that if you had two light particles going in opposite directions, wouldn't one go twice the speed of light relative to the other?

ANSWER:
See my FAQ page. It is
a relative measurement, but your idea of the relative velocities is
incorrect at large speeds. You want to use
v' =( u + v ) ,
but the correct expression is
v' =( u + v )/[1+( uv / c ^{2} )]
where
c
is the speed of light; notice that if
u
and
v
are both small compared to
c ,
your expectation is very nearly correct. For your particular example,
u=v=c
and so
v' =( c + c )/[1+( c ∙ c / c ^{2} )]= c.
You can also find other answers about light speed, the maximum velocity
possible, etc . on the FAQ page.

QUESTION:
Suppose you have radiation detectors fixed on the ground on Earth. Will they detect radiation coming from a charged particle in free fall near them?
The first answer that comes to mind is: Yes, they will detect radiation because the particle is accelerated, and electrodynamics predicts that accelerated charges must radiate in this situation.
According to the Equivalence Principle, this situation is equivalent to detectors fixed on an accelerated rocket with acceleration g moving in the outer space and far away from the influence of other bodies. If the answer to the previous question is yes, then the detectors on the rocket should also detect radiation coming from a charge in free fall as observed by the reference frame of the rocket. But a charge in free fall in this reference frame is at rest in the inertial reference frame fixed with respect to the distant stars, and a charge at rest in an inertial frame should not radiate.
Is it possible that detectors fixed on the rocket detect radiation but detectors at rest in the inertial frame do not? Is radiation something not absolute, but relative to the reference frame?

ANSWER:
This is a fascinating question and points to an experiment
which would seemingly violate the equivalence principle. The answer to your
first question is an unequivocal yes, an electric charge accelerating in
free fall in a gravitational field radiates electromagnetic waves, an
electric charge not accelerating does not radiate. But, suppose that you are
falling along with the charge; relative to you the charge is not
accelerating and therefore not radiating. Or, equivalently, suppose that you
are in a spaceship in empty space with your rockets turned on. If you
release an electric charge inside, it will "accelerate" toward the rear of
the ship and therefore radiate because the equivalence principle states that
there is no experiment you can perform which can distinguish between the
accelerating frame and a static gravitational field. However, the charge
will move with constant speed relative to an inertial observer nearby and
therefore not radiate. In both cases we have an electric charge both
radiating and not radiating, a seeming paradox. Although I had not heard of this
paradox before, apparently it has been a topic of many articles. The most recent of
these, by
Almeida and Saa ,
has evidently laid the paradox to rest. They demonstrate in this article
that observers for whom the charge is not accelerating "�will not detect any radiation because the radiation ﬁeld is conﬁned to a spacetime region
beyond a horizon that they cannot access�" and "�the electromagnetic ﬁeld generated by a uniformly
accelerated charge is observed by a comoving observer as a purely electrostatic ﬁeld."
Like all "paradoxes" in relativity, there is not really a paradox; rather a
radiation field in one frame may be a static field in another. Basically,
you nailed it when you said "radiation
[is] something not absolute, but relative to the reference frame."

QUESTION:
How fast a person would have to move to be completely invisible to our eyes? Like Goku from Dragon Ball.

ANSWER:
Well, I never heard of Goku. I guess I am not up on Japanese
anime! The only way for him to be invisible would be for the visible light
coming from him being Doppler shifted out of the visible range which is
wavelengths about 390-700 nm. I will first do one-dimensional calculations,
he is coming directly at you or going directly away. The appropriate
equation is λ '/λ =√[(1-β )/(1+β )] where λ ' is the observed wavelength, λ
is the laser wavelength, and β is the ratio of the velocity
v of Goku relative to the speed of light c , β=v /c ;
β> 0 is for Goku moving toward you. When he is coming
toward you, β >0 and so he must be going fast enough that 700 nm
light is shifted to 390 nm light (blue shift). I find then that β _{toward} =0.53.
Similarly, when he is moving away from you, β <0 and so he must
be going fast enough that 390 nm light is shifted to 700 nm light (red
shift). I find then that β _{away} =-0.53. So, he needs to
be moving with a speed of about half the speed of light.

If he is not moving directly in a line to or from
you, the situation is a little more complicated. The more general
expression for the Doppler shift is λ '/λ =(1+β cosθ )/√(1-β ^{2} )
where θ is the angle between his path and your line of sight.
(Note that when θ= 0^{0} , this reduces to the equation
above.) When θ= 90^{0 } he is directly in front of you and
so λ '/λ =1/√(1-β ^{2} ),
always a red shift because 1/√(1-β ^{2} ) is
always greater than 1 so λ '>λ . Solving, β _{transverse} =0.83.

One more thing: he will be emitting other
frequencies beside visible. For example, he will be emitting infrared
radiation which will be doppler-shifted into the visible when he is moving
toward you. I have ignored this possibility and assumed that he only emits
visible light. After all, he is just a made up character anyway, right?!

QUESTION:
If you see a laser
beam as blue light of 420 nm from a spacecraft. How would you be able
to tell how fast the spacecraft is going and in what direction? I believe the wavelength of the laser pointer is 620nm.

ANSWER:
I will assume that the spacecraft is either coming straight
toward you or straight away from you; otherwise you need more information.
The one-dimensional Doppler-shift equation is λ '/λ =√[(1-β )/(1+β )]
where λ ' is the observed wavelength (420 nm), λ
is the laser wavelength (620 nm), and β is the ratio of the velocity
v of the spacecraft relative to the speed of light c , β=v /c ;
β> 0 is for the spacecraft moving toward the observer. Solving
this equation for β , β= -0.371. The negative sign means that
the spacecraft is moving away from you. This is called a redshift.

QUESTION:
This question should be fairly quick. Are there any know situations were momentum isn't conserved? I would say no, as momentum is always conserved if you make your system big enough. The only time momentum appears to be not conserved is when you put restrictions on the size of your system, and don't account for the momentum transferred outside your system. When you include the system " outside" your system, momentum is in fact conserved.

ANSWER:
First, what you learn in first-year physics is that linear
momentum is conserved for an isolated system; an isolated system is
one for which there are no net external forces acting. This is actually a
result of Newton's third law which essentially states that the sum of all
internal forces in a system must equal zero. This works really well until
you get to electricity and magnetism where it is easy to find examples of
moving charges exerting electric and magnetic forces on each other which are
not equal and opposite (you can see an example in an
earlier answer ). In that case you would say that such an isolated
system obeys neither Newton's third law nor momentum conservation. However,
looking deeper, we find that an electromagnetic field has energy, linear
momentum, and angular momentum content and, in the end, momentum
conservation, because of the momentum contained in the field, is still
conserved for an isolated system. Thus, Newton's third law is saved, but not
always in the simplistic "equal and opposite forces" language. Finally, if
linear momentum is p =mv , linear momentum is not
conserved in special relativity. But, physicists so revere momentum
conservation that in special relativity momentum is redefined so that it
will be conserved but still reduce to p =mv , for low
speeds: p ≡γmv =mv /√[1-(v /c )^{2} ];
this then unifies energy E and linear momentum p into a single
entity which is conserved, the energy momentum 4-vector: E ^{2} -p ^{2} c ^{2} =m ^{2} c ^{4} .
Not as quick as you expected!

QUESTION:
How fast must one move away from a typical light bulb so that it Doppler shifts into microwave radiation and cooks him/her?

ANSWER:
The expression for relativistic Doppler shift is δ≡f _{observed} /f _{source} =√[(1+β )/(1-β )]
where β is the ratio of
the relative speed of the source and observer to the speed of light. Taking
f _{source} ≈600 THz=6x10^{14} Hz and
f _{observed} ≈2.5 GHz=2.5x10^{9} Hz, I find
δ ≈4.2x10^{-6} . Now, the
Doppler equation may be rewritten as
β =-[(1- δ ^{2} )/(1+δ ^{2} )]≈-(1-δ ^{2} )^{2} ≈-(1-2δ ^{2} )=-0.999999999966,
pretty fast! I used binomial expansion, (1+x )^{n} ≈1+nx +�
if x <<1, a couple of times to approximate.

QUESTION:
My question is in regards to relativity and quantum mechanics. I�ve heard physicists say on t.v programs and in books that relativity and quantum mechanics are incompatible and describe two different phenomena. Do these two theories contradict one another? Is the universe contradictory or is there anything in the universe that can definitively be defined or described as a contradiction?

ANSWER:
You have to be careful that when somebody refers to "relativity", you know what they are referring to. The theory of special relativity,
which predicts things like time dilation, length contraction, E=mc ^{2} ,
etc . is perfectly compatible with quantum mechanics; there is
something called relativistic quantum mechanics. What is probably being
referred to by your sources is the theory of general relativity which is the
best current theory of gravity. No one has been successful in developing a
theory of quantum gravity. I would not so much describe them as
"incompatible" as I would call them "nonunified".

QUESTION:
When a subatomic particle is listed as having a voltage of, say 1.8 Gev does that mean approx or exactly 1.8 and are all such particles exact.

ANSWER:
The electron volt (eV), is not a measure of voltage but
rather a measure of energy; it is the energy acquired by a particle which
has one electron charge (1.6x10^{-19} Coulombs (C)) and has been
accelerated across a potential difference of 1 Volt (V). Since 1 C∙V=1 Joule
(J), 1 eV=1.6x10^{-19} J. Also, 1 GeV=10^{9} eV=1.6x10^{-10}
J. The energy of a particle can mean several different things, depending on
context. Most likely, since you used the word "listed", E =1.8 GeV
refers to the rest mass energy of the particle, E=mc ^{2} ; so,
you see, specifying this energy is equivalent to specifying the mass of the
particle at rest. Sometimes the mass is specified as, in your example, m =1.8
GeV/c^{2} . For example, the mass of the proton is 0.9383 GeV/c^{2} ;
every proton has exactly the same mass as every other proton because, as I
will explain below, a proton is a stable particle, it lives forever. If a
particle is listed with a mass of 1.8
GeV/c^{2} , however, not every one of those particles will have
exactly the same mass because I know for a fact that there is no particle
with that mass which is stable�it will decay with some lifetime. The reason
the lifetime makes a difference is the Heisenberg uncertainty principle
which states that the uncertainty of the energy of a particle can only
be zero if you measure that energy over an infinite time, or ΔE Δt >ћ
where ћ≈ 6.6x10^{-25} GeV∙s is the rationalized Planck
constant, ΔE is the uncertainty of energy (mass), and Δt
is, essentially, its lifetime. So, the shorter the lifetime of a
particle, the greater will be the spread of possible masses for that
particle. Suppose that your particle has a lifetime of 1 ps=10^{-12}
s; then ΔE ~6.6x10^{-25} /10^{-12} ≈6.6x10^{-13}
GeV. So there would not be much of a spread of measured masses but it would
not be zero.

Other things which
could be meant by specifying the energy of a particle are its total energy
E _{total} =mc ^{2} /√[1-(v /c )^{2} ],
where v is its velocity, or its kinetic energy, K =E _{total} -mc ^{2} .

QUESTION:
Let's say we have planet A and B. Planet A suddenly fly away from planet B at 1/2 the speed of light. Planet B also suddenly fly away from Planet A at a little bit more than 1/2 the speed of light. If we observe from Planet A, we can see Planet B because light travelled from Planet B to Planet A. However, after all the flyings, Planet A and Planet B will be 'separating' from each other at a little bit faster than the speed of light. Does it mean that after a while, observers from Planet A can no longer see Planet B? How does this workkkkk???????? :(

ANSWER:
For clarity, let us have a third planet D relative to which
your experiment happens. So the speed of A relative to D is v _{A} =0.5c
and the speed of B relative to D is v _{B} =0.6c let's
say. Your assumption is that the speed of A relative to B is 1.1c . (c
is the speed of light); that is wrong. In an
earlier answer I discussed this kind of problem. The speed v with
which A and B each see the other going away is given by
v =( v _{A} + v _{B} )/[1+( v _{A} v _{B} / c ^{2} )]=1.1 c /[1+( 0.5x0.6c ^{2} / c ^{2} )]=0.85 c .
Not intuitive, but true!

QUESTION:
How can a stationary radio receiver set to a particular frequency receive transmissions from a moving transmitter, say, from an aircraft?
Why doesn't the movement of the aircraft cause its radio transmission to be off frequency relative to the static receiver?

ANSWER:
As is also the case for sound waves, the Doppler effect is
not noticeable if the speed of the source is very small compared to the speed
of the waves. Someone walking toward you and talking does not sound like
Mickey Mouse; the receiver here is your ear and it is not sensitive enough
to detect the shift. The expression for the shifted frequency f of
electromagnetic waves relative to the source frequency f _{0}
is given by f=f _{0} √[(1+β )/(1-β )] where β=v /c
and v is the speed of the source and c= 3x10^{8}
m/s is the speed of light; β >0 is when velocity is toward you.
If β is much smaller than
1, it may be shown that the fractional change in frequency is |f-f _{0} |/f _{0} ≈β .
For example, suppose the source is an airplane with speed 300 m/s (about 700
mph) and the frequency is 90 MHz. Then |f-f _{0} |/f _{0} ≈10^{-6}
or the Doppler shift is |f-f _{0} |≈90 Hz. A tuner does not
have nearly the sensitivity to detect a 10^{-4} % change in
frequency.

QUESTION:
How can a fast moving physical ship with a slow ticking clock get to Andromeda 89 thousand times faster than light, as you claim?... 28 years on the ship's slow clock but 2.5 million Earth orbits around the Sun, just for a "reality check."

ANSWER:
Where do I claim that? I do not know what you are talking about.

REPLY:
So sorry! I mistook your site for Dave Goldberg's site,
Ask *a* Physicist , where I read his claim awhile back on his
"We Can Make It to the Edge of the Observable Universe in a Few Decades" article/blog. I take it that you do not agree with him.

ANSWER:
OK, I found Goldberg's site and everything he says is
perfectly correct. In fact, I found a
similar answer on my site
which we can use for comparison. The speed which Goldberg uses is 99.9999999999999999998% of
c . In my calculation I used a very fast speed
but much slower than his, 99.999% of c . When I calculated the
elapsed time on the ship I got about 11,000 years, he got only 28 years. My
ship was going the same speed the whole time but his started from rest and was constantly
accelerating, so he would have gotten a smaller time yet if his ship had
been going with a constant speed the whole time; if the Goldberg space ship
had been going 0.999999999999999999998 c
the whole way, it would get to Andromeda in about an hour, its time! Your
problem, I think, is how either his or mine could be right since it takes
2.5 million years for light to travel that distance, right? But, it depends
on who is measuring the time. An observer on earth will see the light take
2.5 million years to get there and the space ships arrive just a
little (compared to 2.5 million years) later. But an observer on the space ships will see a greatly
shortened distance to Andromeda, so he will observe both himself and light
to take a much shorter time to get there, the light always just beating
the space ship. You are right, nothing can travel faster than light, but
everybody sees the same light speed in their own frame; there is no
contradiction in saying one observer sees light taking 2.5 million years to
travel between two points in space while another observer sees light as
taking 10 years to travel between the same two points.

QUESTION:
In the TV-series Stargate SG-1 a Goa'Uld Ha'Tak is capable of going 50% lightspeed with sublight engines. Wouldn't these kinds of speeds already be relativistic and what happens excactly to the crew in the ship when the relativistic effects start? And what is the minimum relativistic speed in which relativistic effects become noticeable? A friend of mine said that at 8% lightspeed relativity kicks in making space travel above 8% lightspeed really challenging.

ANSWER:
Where you need to start doing calculations relativistically
depends on how accurate you want to be. The factor which determines the
magnitude of relativistic effects is usually γ= √(1-β ^{2} )
where β is the ratio of the speed to the speed of light. For example
if β= 0.1, 10% the speed of light, then γ= 0.995 which would
mean only about 0.5% error using nonrelativistic calculations. For β= 0.5,
50% the speed of light, γ= 0.867,
still only about 13% error. But, there are no "relativistic effects" for
somebody on the ship; everything appears normal on the ship. They would,
however, see distances to objects toward which they are traveling shrink. I
do not know what it means that "space travel�[is] really challenging". Of
course, it is challenging to get to relativistic speeds in the first place,
but I do not see any hazards "kicking in" at
β= 0.08.

QUESTION:
Is it possible, and if so how to calculate the magnitude of a 4-dimensional vector?

ANSWER:
I assume that you are asking about 4-vectors in special
relativity. This kind of question can depend on the mathematics of the
multidimensional coordinate space you are working with, but in special
relativity if a vector has a space-like part (A _{x} , A _{y} ,
A _{z} ) and a time-like part A _{t} ,
then the magnitude of the vector is A =√[A _{t} ^{2} -(A _{x} ^{2} +A _{y} ^{2} +A _{z} ^{2} )].
For example, the space-time vector has a magnitude of R =√[c ^{2} t ^{2} -(x ^{2} +y ^{2} +z ^{2} )]=√[c ^{2} t ^{2} -r ^{2} ]
and the energy-momentum vector has a magnitude of √[E ^{2} -(p _{x} ^{2} +p _{y} ^{2} +p _{z} ^{2} )c ^{2} ]=
√[E ^{2} -p ^{2} c ^{2} ]=m _{0} c ^{2} .
If you want a bunch more detail, see the Wikipedia article on
4-vectors .

QUESTION:
Is there a Doppler shift between the centre and outside of a wheel.
Say for example a standard merry go round you find at a sideshow, if I stand in the middle and look at something on the outer does my image Doppler shift.
The reason it is confusing to me is because your angular rotation is the same for both people but not your linear speed.

ANSWER:
If you were asking about sound rather than light, there would
be no Doppler shift because for normal waves there is no effect if there is
no component toward or away between the source and the observer. For light,
however, which is what you ask about, there is a Doppler shift. If your
question had asked for a source and and observer moving on straight parallel
paths with different speeds (sort of what you asked), then you could
relatively compactly write the Doppler shift equation relative to the
observer. In the figure to the right, the observer will see the source
moving with a speed v and light comes at an angle θ as shown.
In this case, f'=f /[γ (1+(v cosθ /c ))] where
γ =1/√[1-(v ^{2} /c ^{2} )] and f and
f' are the frequencies at the source and observer, respectively. If
θ= 90^{0} ,
f'=f /γ ; this is called the transverse relativistic Doppler
effect. But, in your scenario, each will see the other to be at rest. But
both are actually accelerating relative to an inertial frame, so the
situation is not so simple. You can see a discussion of this scenario on
Wikepedia . Suppose that the source is at a distance r _{1}
from the center, the observer is at a distance r _{2} , and the
angular velocity is ω. Then assuming that the
distance traveled by the observer in the time it takes the light to travel
between the two is very small, I find that f'≈f √[(1-(r _{1} ω/c )^{2} )/(1-(r _{2} ω/c )^{2} )].
I just realized that you actually asked an easier question, namely that r _{2} =0
and r _{1} =R where R is the radius of the
merry-go-round, so, f'=f √(1-(Rω/c )^{2} ). This is
exactly the same as the transverse Doppler effect above, f'=f /γ.
[Sorry that the answer is much longer than needed. Whenever I misread a
question and do a lot of interesting and more general work, I am loathe to
delete it! The concise answer for you is that f'=f /γ. ]

QUESTION:
I regularly contribute to another website that I will not mention here, but we had a discussion on a particular subject matter and one topic which came up was kinetic energy of an object moving close to the speed of light.
Specifically an assertion was made that an object traveling 85% of c would have kinetic energy equal to a matter / antimatter reaction of the same mass as the traveling object.
0.5mv^2 did not produce compatible numbers unless one assumed quite low efficiency of the M/A reaction. It was then stated to me that formula is not valid to the situation, as it fails to account for the increased mass of an object traveling at such speeds.
I'm no physics professor, but that seemed a little fishy to me. I attempted to divine an answer to this with the use of some Google-Fu, however it seems my Google-Fu is not strong enough, as I found multiple conflicting answers (and even multiple conflicting formulae to determine the kinetic energy in such a situation.)
So since my Google-Fu has failed I figured I would ask you. Do you need a different formula to calculate the kinetic energy of an object traveling at such speeds, and if so what is that formula?

ANSWER:
You have come to the right place ! From your
discussion, that unnamed website is the wrong place to get physics right.
The expression �mv ^{2} for kinetic energy is true
(approximately) only for speeds much less than the speed of light. There is
no mystery about what kinetic energy is in special relativity, but I will
not go through the derivation. The total energy E of a particle with
rest mass m is E = E=mc ^{2} /√(1-v ^{2} /c ^{2} ).
If the particle is at rest, v =0 and E=mc ^{2} . If the
particle is not at rest, it still has its mc ^{2} energy and
everything else must be kinetic energy K : K=E-mc ^{2} .
That is all there is to it.

[It is not at all
clear what is meant by " �a
matter/antimatter reaction of the same mass�" in your question; I will just
briefly discuss particle/antiparticle pair creation and required kinetic
energy.]
For your case (v =0.85c ):
E =mc ^{2} /√(1-0.85^{2} )=1.90mc ^{2} ,
so K =0.90mc ^{2} .
There is therefore not enough kinetic energy to create another particle of
the same mass (which I am assuming the remark in your question about
"antimatter" refers to). The way to create an antiparticle, for example an
antiproton, is to collide a fast proton with a proton at rest; the energy
required to create one particle of the same mass takes quite a bit more than
mc ^{2} of kinetic energy because momentum must also be
conserved, so the particles cannot be at rest after a collision. Also, you
cannot simply create a mass out of kinetic energy, you have to create a
particle-antiparticle pair so in the proton experiment you end up with three
protons and an antiproton. It turns out that the incident proton kinetic
energy energy needs to be 6mc ^{2} . The advantage of a
collider is that the momentum is zero so you only need to have each proton
with kinetic energy mc ^{2} to create a proton-antiproton
pair.

Here is an added detail since I am sure that the
expression for K above looks unfamiliar to you. If v <<c ,
1/√(1-v ^{2} /c ^{2} )≈1+�v ^{2} /c ^{2} +�;
this is just a binomial expansion and, if you are not familiar with
that, try it yourself: e.g . if v /c =0.05, 1/√(1-v ^{2} /c ^{2} )=1.001252
and 1+�v ^{2} /c ^{2} =1.001250. So now we can
write K ≈mc ^{2} (1+�v ^{2} /c ^{2} )-mc ^{2} =�mv ^{2} .
Tada!

FOLLOWUP QUESTION:
Just to clarify a bit the discussion which led me to asking you arose out of one concerning hypothetical weaponry. One individual asserted that a projectile traveling 0.85c would have kinetic energy equal to a matter-antimatter bomb with a similar warhead yield. [Thus a 500kg projectile would have energy equal to the energy released by the reaction of 250kg of particles annihilating with 250kg of antiparticles] I figured being a real physicist you would not want to waste too much time drawn into a discussion on the purely hypothetical topic. Thus my question came to you, since my original calculation for kinetic energy using 0.5mv^2 with m=500kg provided significantly less energy than E=mc^2 with m being 500kg.
The website in question is TVtropes.com which analyzes elements common in fictional works literary, TV, Film, and others.

ANSWER:
Well, my answer gives you almost mc ^{2} of
kinetic energy. The speed to get exactly mc ^{2 } is 0.866c .
But, you have no comprehension of how much energy is involved here. I think
you would be interested in an earlier answer along
these lines based on the movie Eraser .

QUESTION:
how much kinetic energy does a beam of hydrogen atoms moving at 5% the speed of light have.

ANSWER:
Well, you do not give me any information on the intensity or
length of the beam, so I cannot give you a number. I can show you how to
calculate the kinetic energy K of one hydrogen atom of mass M and you
can figure out how many atoms are in your beam. K=Mc ^{2} (1/[√(1-0.05^{2} )]-1)=0.00125235Mc ^{2} =1.88228x10^{-10} J where c is the speed of light. Since the speed is quite small
compared to the speed of light, you could get a good approximation using the
classical value of K , K =�Mv ^{2} =1.87875x10^{-10}
J.

QUESTION:
Just read Ender's Game, and it got me thinking of relativistic travel. What would be the optimum speed (relative to Earth) that you could send a probe (say, 1 light-year), and return to Earth? Go too slow, and it takes a long time. Faster is better, but if you go too fast, then relativity starts to work against you.

ANSWER:
I am puzzled by your statement that "relativity
starts to work against you." And, what are the criteria which determine the
"optimal speed"? If you are on earth and want to get the probe back to earth
as quickly as possible, the faster the better. If you are on the probe and
want to get there and back as quickly as possible, the faster the better.
How does relativity work against you?

FOLLOWUP QUESTION:
So, say there is an object 1 light-year away from the Earth. I want to send a robotic probe there, spend 1 day gathering a sample, and then return it to the Earth. If I send it at 50% c, (if I did my math correct) I calculate it would take (from my point of view on Earth) ~ 4.6 years (plus one day) to get the sample back (with only 4 years + 1 day elapsing relative to the probe). But, if I set the probe speed to .999 c, it would take me ~ 44.8 years ( + 1 day ) to get my sample, which is almost 10 times longer than .5c. This is what I mean by relativity working against you.

ANSWER:
Well, I do not know what you did, but your answers are
certainly wrong. As I understand it, you stay on earth. So, using your clock
you see a probe going 0.5c to take two years to go one light year and
2 years to get back: 4 yr + 1 day. If the speed is 0.999c , the time
out (and back) is 1/0.999=1.001: 2.002 yr + 1 day. From your point of view,
relativity is irrelevant if all you are interested in is measured by your
clocks and your meter sticks.

You might ask how much
time elapses for a clock attached to the probe. (See my discussion of the
twin paradox .) Each probe will
see the distance you see (1 ly) Lorentz contracted, the slower by √(1-0.5^{2} )=0.866
and the speedier by √(1-0.999^{2} )=0.045 and so the elapsed times
will be 1.73 yr + 1 day and 0.09 yr + 1 day.

QUESTION:
It's well established that a rocket propelled spaceship accelerating forward will gain mass as it goes faster and faster. Appreciably so as it approaches light speed. My question is, how does this mass gain manifest itself? That is to say, does it occur by the atomic particles themselves gaining mass (protons, electrons etc.) or do new particles simply pop into existance due to the energies involved? Also, by what mechanism does this occur? How does energy being expelled from the rear result in the addition of mass to the ship and it's contents?

ANSWER:
When you think of mass, you think of "amount of stuff". When
I think of mass, I think of inertia. Mass measures how hard something is to
accelerate, and that is inertia. In classical mechanics, inertia does not
depend on on the velocity; a 1 N force will provide exactly the same
acceleration on a mass regardless of how fast it is going. In special
relativity, as an object with mass goes faster, it becomes increasingly more
difficult to accelerate it. In my view, special relativity requires only the
concept of rest mass, its inertia when at rest; beyond that we introduce new
definitions of energy and momentum which then take care of this inertia
issue.

QUESTION:
If I person could build wings for themselves like Icarus and fly at 99.999 percent the speed of light, all the way to the Andromeda galaxy, two and a half million light years distant, how much time would pass from their perspective? Would they experience said trip as taking two and a half million years, or would time pass more quickly for them? Would they even be alive upon reaching said galaxy?
Do we even possess the mathematics to work a problem like that out?

ANSWER:
What a peculiar way to ask the question most people ask
employing space ships! Of course, there is no air in space so no wings would
be needed to fly and, of course, breathing would be a problem. The painting
of Daedalus and Icarus shown here is the same that I remember from my Latin
book in high school. The important factor in relativity, called the Lorentz
factor γ , is γ = 1/√[1-(v /c )^{2} ]
which, for your case, is γ = 1/√[1-(0.99999)^{2} ]=223.6. For
Icarus, time would seem to pass as normal, but he would see the distance he
has to travel shrink by the Lorentz factor: D' =D /γ =2.5x10^{6} /223.6=11,180.7
light years. So, the time that would elapse on his clock would be
11,180.7/0.99999=11,180.8 years. I do not think he would live that long.
Note that only arithmetic was used here, so, yes we certainly "possess
the mathematics to work a problem like that out"!

QUESTION:
I am trying to estmate the energy of a subatomic particle like a proton,by making the v=c or extremely close to c,but i am having a problem that the momentum is becoming Zero when calculating momentum at relativistc speed ,I my completely making a mistake,or am i missng somethin,

ANSWER:
E^{2} =(pc )^{2} +m ^{2} c ^{4}
where m is the rest mass. Also, p may be written as p =mv /√[1-(v /c )^{2} ].
As v gets closer and closer to c , p gets bigger and
bigger until eventually the m ^{2} c ^{4 } term becomes negligibly small
and E≈pc . So, I guess I do not see where your problem is.

QUESTION:
I know Einstein said E=MC^{2} and basically all mater can be equated to some quantity of energy; then why do we go to the gas station to fill our cars? Why can't we use garbage, which is mass and has energy, to power our cars? How can we convert matter to energy? I know we can burn gasoline to use perhaps 1/4 the heat content in the form of expanding gas to apply pressure to the piston in the engine. Has anyone invented a converter that changes matter to energy yet? We eat food and basically run on sugar which fuels a chemical based process. Any other matter converters?

ANSWER:
Most of the energy mankind uses comes from chemistry. Burn
coal or gasoline, for example. When you eat and metabolize food, chemistry
is going on. The energy which is extracted comes from―guess what―mass! For
example, when you burn coal the main thing which is happening is that carbon
is combining with oxygen to form carbon dioxide. One carbon dioxide molecule
has a smaller mass than one carbon atom plus one oxygen molecule. So,
chemistry is the best known example of your "matter converter". The problem
is that an extremely tiny fraction of the mass is converted to energy,
something like 0.00000001%, so chemistry is a very inefficient source of
energy. Now, to get more efficiency we have work not with atoms but with
nuclei of atoms. If a heavy atomic nucleus can be induced to split (fission ),
the mass of the fragments is smaller than the mass of the initial nucleus by
an amount much bigger than with chemistry, something like 0.1% which is a
huge improvement over chemistry; this is how nuclear reactors work. Also
from nuclear physics, you can take very light nuclei and make them combine (fusion )
and get like 1% of the mass converted into energy; this is how stars work
and so, you see, solar energy comes from "matter converters" too and so does
wind energy since the sun is the energy which causes winds to blow. If you
want to get 100% efficient you have to go to particle-antiparticle
interactions in particle physics. When an electron and its antiparticle the
positron meet, their mass completely disappears and all the energy comes out
as photons. Did you ever see the Back to the Future movies? Doc came
back from the future where they had invented a small appliance called "Mr.
Fusion" to do what you want, to convert garbage into the huge amount of
energy needed to power the time machine.

QUESTION:
My Question is, if you were in a spaceship in space with no windows, what experiment could you perform in order to prove whether you were moving or not moving?

ANSWER:
There would be no such experiment, because "moving" only has meaning only relative to
something else. If you were not burning your engines (that is there were
no forces acting on your ship), Newton's laws of motion would be correct
laws of physics in your ship. However if you were burning your engines
but you did not know that for sure, there would be no experiment you
could perform to determine whether you were accelerating or whether you
were in a gravitational field.

QUESTION:
Hey, so muons falling down towards the Earths surface survive decaying before they reach the surface because of length contraction as a result of Special Relativity. But I read SR doesn't apply when the object is accelerating. Aren't the muons accelerating due to Earths gravity? What have I understood wrong?

ANSWER:
First of all, special relativity can be applied to
accelerating objects. Acceleration does not play a central role like it
does in Galilean relativity because it is not invariant; what this means
is that if two observers in two different inertial frames measure the
acceleration of some object, they get the same answer in Galilean
relativity but different answers in special relativity. Nevertheless, at
any instant the object (muon in your case) has some velocity and special
relativity may be applied at that instant. But, while that is important
to understand, it is essentially irrelevant in the case of muons
approaching the earth. A muon lives on the order of 10^{-6} s
and has a relativistic speed, say 2x10^{8} m/s; with an
acceleration of about 10 m/s^{2} , its velocity changes by about
10^{-5} m/s over its lifetime. Hardly accelerating, is it?

QUESTION:
I'm hoping you can settle a bet between my father and myself. We are
both movie buffs, and work together. While working we discussed the
Ah-nuld movie "Eraser". I'm going to make the assumption you have never
seen the movie which revolves heavily around the manufacture of
man-portable super-railguns that fire aluminum rounds at 'close to the
speed of light'. Being firearms enthusists as well, we discussed the
flaws in their idea (beyond the fact no technology in the near future
would allow such a weapon as the one depicted to be made, let alone one
man-portable.) Where we came to a disagreement was why for other reasons
it would be useless. While we agree that the mooks Ah-nuld tears through
as he does in every movie would be out of luck. In practical situations
the weapon would be useless but we disagree why. My father believes that
a 5.56mm / 0.00401753242 kg bullet (our assumptions) traveling 80% of c
would immediately flatten out and quickly lose all of its kinetic energy
due to its low mass and aluminin's relatively soft nature upon striking
anything solid such as a door, or brick. I say a round with the same
dimensions would explode on contact with something solid (again like a
brick) as the work-heating would vaporize a piece of aluminimum with
that mass. If it could even get there without being melted by
atmospheric friction. As I've submitted questions to you before we
decided we would --drum roll--- ASK THE PHYSICIST!!! Are either of us
correct? The loser has to buy the next Pizza we order.

ANSWER:
Neither of you is addressing the real issue here�a 4 gram
bullet traveling at 80% the speed of light has an almost incomprensible
amount of energy. So, saying that this weapon would "be useless" is way off
the mark. Saying that it is impossible to make such a weapon is a different
issue which I will address after talking about what such a speedy bullet
would do. You are probably not interested in the details, so I will just
give you the kinetic energy such a bullet would have�about 5x10^{14}
Joules. To put that in perspective, if you took that energy and delivered it
to the power grid over a period of a day, this would be the equivalent of a
6 gigawatt (6x10^{9} watts=6 billion Joules per second) power
station. The largest power station in the US has a power output of about 4
gigawatts. Or, the energy of the Nagasaki atomic bomb was about 10^{14}
Joules, 1/5 of the energy of your 4 gram bullet. So to argue whether the
bullet's flattening or exploding is the reason it would not do much damage
sort of misses the point, don't you think?! If the bullet takes 10 seconds
to deliver its energy to whatever can take it, you are still talking about
the energy of 5 WWII-era atomic bombs being delivered. I would not want to
be within 50 miles of that. Finally, it means that the "gun" has to deliver
all that energy in an unimaginably short amount of time (I presume that
since, if it is "man-portable", the gun must be no more than a few feet
long). I figure that a force on the bullet of more than 2 million pounds
would be required during the time it was in the gun; do you think an
aluminum bullet could withstand such a force? And, Newton's third law says
that, if the gun exerts that force on the bullet, the bullet would exert
that force on the gun. Imagine the recoil! This is such a ridiculous
scenario, I don't think we even need to beat a dead horse and talk about
what air friction would cause to happen as it flew to the target.

QUESTION:
The
clock that works with light pulses is commonplace when we're learning
about Einstein's Special Relativity. Unfortunately, I can't translate
the reasoning to a real clock, with a real pendulum. How different would
be the classical explanation, using a real clock?

ANSWER:
Why is the light clock not a "real clock"? It keeps perfectly good time
here in my reference frame, right? I just have to calibrate it so it says
tick each second and it is a good clock. Now, here is the trick: take some
other clocks (which you consider more "real") and put them next to the light
clock. They all run just fine, right? And some of them could be biological
clocks like a petry dish of growing bacteria or you growing older. Now,
observe all these clocks from a moving frame. It would certainly be strange
if another observer observed the clocks running at different rates wouldn't
it?.

QUESTION:
Why exactly does the speed of
light (c) come in to play in Special Relativity? It seems to be rather
arbitrary in the sense that the Lorentz Transformation would (from what
I can gather in my admittedly novice research) yield the same results
using any speed that is accepted as being constant, whether it be the
speed of light, sound, or the 4:17 to Charleston. Is light specifically
used for any reason other than it's bearing on optics?.

ANSWER:
Good question. I know the postulate that c is a universal
constant seems very ad hoc . In fact, that postulate is not needed.
Here is the reasoning: the cornerstone of special relativity is that the
laws of physics must be the same in all inertial frames of reference.
Maxwell's equations, the laws of electromagnetism, predict the speed of
light to be exactly what it is and to depend only on two fundamental
constants of nature (constants which quantify the strengths of electric and
magnetic forces). Since Maxwell's equations are laws of physics, they must
be the same in all frames of reference and hence the speed of light must be
the same in all frames of reference.

QUESTION:
Can the total mass of a closed system fluctuate at any given
time, while the total energy remains of course constant, considering the
existence of mutual conversions of matter and energy ?

ANSWER:
Yes. Since we understand mass as just a form of energy, it may change
into a different kind of energy thereby changing the mass of a system.
Examples are:

There is an
electron and a positron (an antielectron). They encounter each other and
annihilate resulting in two photons. The original mass was twice the
electron mass and the final mass was zero.
A uranium nucleus
undergoes fission. Afterwards there is less mass but more kinetic energy of
the fission products.
QUESTION:
I would like to understand why time slows down at high speeds. Any information i find on this topic seems to complicated to get my head around, even my physics tutor said its too complicated and i should just assume 'it just does'. Could you explain it to me in a way i may be able to understand?

ANSWER:
It is not hard to understand thanks to a clever idea by Feynman�the
light clock. Imagine a stick with a flashbulb and a light detector at one
end and a mirror at the other end. A flash of light goes up to the mirror,
bounces back and hits the detector. The detector is connected to a little
electrical circuit and a speaker which says "tick" whenever it detects a
flash of light and also tells the flashbulb to send out another flash of
light. So, the clock goes "tick, tick, tick, ..." Now, this is a perfectly
good clock, right? What you now do is make this clock go past you very fast
(velocity being perpendicular to the stick). Now the light has to travel
farther than when it was at rest, right? But, since the special theory of
relativity asserts that the speed of light is the same as seen by all
observers, the clicks are less frequent as seen by you because the flashes
have farther to travel. So, the moving clock runs slow. You can play with a
nice
simulation of a light clock.

QUESTION:
what is the real meaning of momentum & how we can relate it with photons which has no mass at rest and not defined mass while moving with speed of light?

ANSWER:
In classical physics, momentum is mass times velocity, p=mv.
One of the most important features of momentum is that the total
momentum of an isolated system never changes; this is called conservation of
momentum. However, in the theory of special relativity, if you choose
momentum to have the same definition, you find that momentum conservation is
lost. Conservation is such a powerful concept that we choose to redefine
momentum. The details of all this are given in an
earlier answer . The
result is that if a particle of mass m has an energy E , then
its momentum p is p=√{(E /c )^{2} -m ^{2} c ^{2} }
where c is the speed of light. So, you see, it is not necessary for a
particle to have mass to have momentum�it need only have energy.

QUESTION:
Hello I used to have a Feynman book that had this scenario and I
forgot how he explained it I have since lost the book and was wondering
if you could explain it. I have a spaceship moving at 180,000 kms inside
of that spaceship I have another spaceship moving at 180,000kms. To the
observer on the ground the second spaceship is moving at 360,000kms.
That exceeds the speed of light please explain what would happen.

MY FIRST
RESPONSE:
Your recollection is wrong. I am sure Feynman never said the speed of
the second space ship exceeds the speed of light because it doesn't.

FOLLOWUP:
ya, he may have never said this but can you explain what would happen in
that scenario?

ANSWER:
The equation which describes what is called "velocity addition" in
relativity is v' =(u +v )/[1+(uv /c ^{2} )]
where u is the speed of the first ship, v the speed of the
second ship, c the speed of light (300,000 km/s), and v' is
the speed of the second ship seen by the outside observer. Note that if u
and v are both very small compared to the speed of light, then
the quantity (uv /c ^{2} ) is very close to zero so that
v' =(u +v ), which is what you expect to be correct, is
approximately true. However, in the example you cited the speeds are not
small compared to c (they are 60% of c ). If you do the
arithmetic you will find that v' =265,000 km/s.

QUESTION:
What is the formula for displacement as a function of time for a particle undergoing constant acceleration? It's d=1/2 a*t2 at a Newtonian scale, but relativity kicks in as it gets closer to the speed of light. By "constant acceleration" I mean constant from an observer at "rest" relative to the accelerating particle (as would be the case of a particle falling towards a gravitational well) as opposed to "constant in the accelerating particle's frame of reference" (which would be the case for a rocket burning fuel).
Here's the function for the case I DON'T need:
http://math.ucr.edu/home/baez/physics/Relativity/SR/acceleration.html .

ANSWER:
Why is acceleration an important quantity in Newtonian physics? Because
it is what is called an invariant quantity; any two observers in different
inertial frames will measure the same acceleration of a third object. This
simple fact is the basis for Newton's second law because, since the force is
what causes the acceleration and therefore if there is some force on an
object all inertial observers must measure the same acceleration.
Acceleration has no such hallowed position in special relativity and, as you
may know, Newton's laws of motion must be revised by
redefining momentum .
Acceleration is essentially a useless quantity in relativity. In fact, it is
not possible to have uniform acceleration in the sense that you define it
(that is why it is defined differently in Baez's discussion). Think about
it: Suppose that the "uniform acceleration" were 1 m/s^{2 } and your
speed right now was 0.5 m/s below the speed of light; if acceleration stayed
constant, your speed in one second from now would be 0.5 m/s above the speed
of light, a forbidden situation.

FOLLOWUP QUESTION:
If you could drop a rock down an infinitely deep well with a constant gravitational "pull," what formula would describe its velocity ("v") in terms of time falling ("t")? I know it starts as v = at (where "a" is the gravitational pull) and approaches v = c (where c is the speed of light), but what does it do in between?

ANSWER:
Let us stay away from gravity since the definition of a uniform
gravitational field is problematical. But, I think what you are interested
in is what is the velocity if the applied force on a mass m is
constant. I know that an object with zero net force has dp /dt =0
where p =mv /√[1-(v /c )^{2} ]
(again, see the link to
relativistic momentum ). Now, I am going to define a constant
force F to be one for which the rate of change of momentum is
constant, that is dp /dt =F where F is the constant. Now, we
know that v≈ Ft /m
for small time and v ≈c
for large time. It is very easy to integrate dp /dt =F to get momentum as a function of time,
p=Ft . Putting in what p is in terms of v and solving for
v, I find v /c =(Ft /(mc ))/√[1+(Ft /(mc ))^{2} ].
This function has the correct properties at small and large t and is
shown in the graph at the right. This is the correct v (t ) for
a constant rate of change of momentum F. (For purists, I am talking
about 3-momentum.)

NOTE ADDED
LATER:
Someone expressed interest in the position as a function of time for
this problem. This is straightforward to do by integrating dx=vdt.
Doing this I find x =(mc ^{2} /F )(√[1+(Ft /(mc ))^{2} ]-1).
Note that this has the expected properties that for small time, x ≈�(F /m )t ^{2} ,
and for large time, x≈ct . (I assumed x =0 at t= 0.)

QUESTION:
Much is made of the fact that the speed of light is independent of its source speed (a speeding rocket, etc.). However, isn't the same true for other waves such as sound? No matter how fast a train is moving, its sound waves still travel at sound speed. I realize there is a
Doppler shift, but this does not change the wave speed. So why is the big deal made about light speed being constant?

ANSWER:
What much is made of is that the speed of light is a universal constant
regardless of any relative velocity between the source and the
observer. No matter how those two move, the wave speed is the same. Sound is
not the same since it moves relative to a medium (the air, normally). If the
source moves toward you, as you note, you will still measure the same speed
of the waves. However, if you move toward the source, you will measure the
waves moving with a velocity of your speed plus the speed of the sound in
still air. The key is that there is no medium with respect to which light
waves move; they may move through completely empty space, unlike sound which
requires a medium.

QUESTION:
According to Google the mass of an electron is 9.10938188 � 10 (power of) 31 kilograms. Does that mean that a charged capacitor has more mass than the same capacitor without a charge?

ANSWER:
First the simple answer. The same number of electrons added to one plate
of the capacitor are taken away from the other plate, so there is no net
charge and the net mass is unchanged. Now, the trickier answer. Since it
takes energy to charge a capacitor, the whole system has more energy than it
started with. Therefore, according to E=mc ^{2} , the whole mass of
the capacitor must have increased. However, since the energy stored is very
small, this increase of mass would be impossible to observe.

QUESTION:
This is something which came
to my mind. A book is on a table at rest, so the weight
of the book or downward force must be equal to reaction force or upward
force. The weight of the book=force given by the table on the book. Now if
we put another book on this book the equation goes: weight of first
book+weight of second book=force given by the table on the books. since
from above equations: force given by the table on the book=force given
by the table on the book or weight of the first book=weight of first
book+weight of second book then weight of the second book is zero(0) how
come its possible. plz plz make me clear.

ANSWER:
There are lots of clues in your question that you misunderstand Newton's
first and third laws. You refer to the upward force of the table on the book
as the "reaction force" which implies that you think that Newton's third law
(N3) is the reason it must be equal to the weight. The "reaction" force,
that referred to in Newton's third law, is never on the same body as the
"action" force; for a more thorough discussion of Newton's third law see an
earlier answer . The weight of the
first book is the force which the earth exerts on the book; the reaction
force is the force which the book exerts on the earth. The reason the force
which the table exerts on the book is equal and opposite to the book's
weight is that the book, being in equilibrium, must have the sum of all
forces on it equal to zero; this is Newton's first law (N1). I will try to
go through each of your two questions individually:

The book has two forces on it, its own weight and a force from the
table. Those are the only two forces. Because of N1 , the force
from the table must be up and equal in magnitude to the weight.
I will call the top book T and it has a weight W _{T} .
I will call the bottom book B and it has a weight W _{B} .
The table exerts a force on the bottom book and I will call that F .
Book B exerts a force up on book T and I will call that F _{TB} .
Book T exerts a force down on book B and I will call that F _{BT} .
Focus your attention on book T. There are only two forces on it, its own
weight and the force from book B. It looks exactly like the one book
problem (it is the one book problem!) and so the force F _{TB
} must be equal in magnitude to W _{T} and pointing
upward because of N1 . Now, focus your attention on book B. There
are three forces on book B, its weight W _{B} , the force
from the table F , and the force from book T F _{BT} .
These three must add up to zero because of N1 . But we do not know
F _{BT} . But wait, aha! We know that F _{BT}
is the force which T exerts on B and F _{TB} is the force
which B exerts on T, so N3 tells us that these must be equal and
opposite. Therefore the magnitude of F _{BT} must be equal
in magnitude to W _{T} and point down. Putting it all
together and using N1 , the magnitude of F must be W _{T} +W _{B}
and F , of course points up.
A final word of warning: never say something like "the weight of the top
book is a force down on the bottom book" because the weight of anything is a
force on that thing, not something else. Just because it works out here that
the magnitude of the force the top book exerts on the bottom book is equal
in magnitude to the weight of the top book does not mean the "forbidden
statement" above is true. If you had a book on the floor of an upward
accelerating elevator, the magnitude of the force of the book on the floor
would not equal the magnitude of the weight of the book.

QUESTION:
Can a massless particle have or carry energy?

ANSWER:
The energy of any particle is E =√[m ^{2} c ^{4} +p ^{2} c ^{2} ]
where p is the linear momentum. If m =0 then E=pc.
Massless particles have momentum. The only massless particle we know is the
photon which has an energy E=hf where h is Planck's constant
and f is the frequency. So the momentum of a photon is hf /c .

QUESTION:
Einstein's famous E=Mc2 doesn't seem to hold for a photon which is massless but has energy. It doesn't even hold for the creation of energy since the photon is not created by annilating mass, but rather by an electron shifting orbit. Further, the photon doesn't convert into mass when absorbed. What am I missing?

ANSWER:
I often get this question. It originates with taking a famous equation
and not understanding when it is applicable. E=mc ^{2} is the
energy of a particle of mass m at rest; a photon is never at rest and
therefore this equation is not applicable to it.
The energy of any particle is E =√[m ^{2} c ^{4} +p ^{2} c ^{2} ]
where p is the linear momentum. Note that if p =0, the particle
is at rest and indeed
E=mc ^{2} . If m =0 then E=pc.
Massless particles have momentum. The only massless particle we know is the
photon which has an energy E=hf where h is Planck's constant
and f is the frequency. So the momentum of a photon is hf /c.
Regarding the rest of your question, when an electron drops to a lower orbit
the mass of the atom decreases by exactly the right amount to supply the
energy to the photon. When a photon is absorbed by an atom, conversely, the
atom becomes excited and therefore more massive.

FOLLOWUP QUESTION:
In a previous answer you stated the mass of an electron decreases as it shifts to a lower orbit and releases a quantum (hf) of energy. How does this mass change occur? Does the electron divide itself and become smaller in order to shed the mass required for energy release? What's the mechanism at work here?

ANSWER:
I did not say that the electron mass decreases. I said that the mass of
the atom decreases. A bound system always has a mass less than the sum of the
masses of its parts. Consider a system of two bound masses, each of mass m
when alone. Their bound mass is less than 2m . You can say this because it
takes work to pull them apart. Where does this work go? Into mass. In
atoms, it is almost impossible to measure the mass difference because it is so
small. But in nuclei,
where the binding is much stronger, it is measurable. The mass of a ^{4} He
nucleus is measurably smaller than the masses of two protons and two neurtons.

QUESTION:
A vertical narrow tube has a photon emitter at one end, so that photons travel through the tube and are emitted along the verical axis of the tube (all in a frame moving left to right near the speed of light). Do the photons emitted from the end of the tube leave at an angle to the vertical, as seen by a non relativistic frame? If so, do the photons hit or move toward the wall of the tube on their way inside the tube?

ANSWER:
First, view from the frame of the tube. The photons travel straight
up the center of the tube and never hit it. Therefore, all observers have to
agree that the photons do not hit the tube. Certainly the photons, as seen
by the stationary observer, do not move vertically. Rather, they have a
component of their velocity which is v along the direction of the
motion of the tube but their total velocity is still c . The angle
they make with the vertical is therefore
θ= sin^{-1} (v /c ). What you are describing is
very similar to the light
clock.

FOLLOWUP QUESTION:
you said that both the moving observer and the stationary observer must agree on whether the photon hits the wall of the tube or not. You said the moving observer see the photon traveling straight up the tube.
but you also said the stationary observer sees the photon traveling at an angle to the tube wall. you didn't say whether that angled path resulted in the photon hitting the wall and how this discrepancy is resolved.

ANSWER:
There is no mystery here �the
tube has the same horizontal velocity as the photon does, v . So,
every time t that the photon moves a distance vt , the tube
also moves a distance vt and so the photon stays in the middle of the
tube

QUESTION:
During a discussion about basic physics I was having with my daughter, she asked an interesting question. She saw a story about helicopters that mentioned the helicopter blade tips could exceed the speed of sound creating the characteristic �whop, whop� sound, or small sonic booms. She wondered if it would be possible for the blade tips to ever exceed the speed of light. I was intrigued by this so I worked for a few hours to come up with an equation to answer her question. I am not a math expert, but using simple algebra I came up with the formula: r=(60V) / (2 π R) where r = radius of the blades in meters, V = velocity in meters per second, and R = blade RPM. From this I calculated that the blade tips would exceed the speed of light if the blades had a radius of 286.28 kilometers and were spun at 10,000 RPM. This of course assumes the materials used could withstand the forces involved. My question: is this formula accurate and given the radius and RPM mentioned, what is the formula to calculate the forces along the blades starting from the center moving to the tips (centrifugal, centripetal?)?

ANSWER:
Nothing can exceed the speed of light. The qualitative way to
explain why for this case is that a mass increases as the speed increases.
As you go farther out on the blade, the mass increases more and more until
you are, at the crucial length, requiring an infinite force to keep the mass
moving in a circle. The speed of something a distance r from the end
would be v=r ω =9.55Rr
where R is rpm as per your notation and ω is the angular
velocity in radians/s; it looks like your equation got 60/(2π )
where you should have had
( 2π ) /60.
So, I get the length would be just 3.14 km if it could work. Suppose the
blade had a linear mass density λ kg/m and a length L . So, the
mass of the whole blade would be Lλ. If you calculate the force
classically , ignoring relativity, you would get F (x )=(L-x )λxω ^{2}
where x is the distance out to the point where you are calculating
the force necessary to keep the mass beyond from flying off. But,
relativistically, λ depends on x , λ =λ _{0} /√[1-(ω ^{2} x ^{2} /c ^{2} )]
where λ _{0} is what the mass density is if the blade is at
rest. So, you see, if ωx=v=c , the mass of the whole blade becomes
infinite meaning the force needed anywhere is infinite.

QUESTION:
Have there been any theories presented by rational and peer-accepted scientists, trying to pair the combining force of Gravity with the separating force of Dark Energy into a single equation or system of equations?

ANSWER:
This already exists, has existed ever since the origin of general
relativity. At the time Einstein developed general relativity to explain
gravity, it was believed that the universe was static. He therefore included
a term in the general relativistic equations called the cosmological
constant. Later, when Hubble found that the universe was expanding, he
removed the constant calling it his "greatest blunder". If now it is
reinserted, dark energy is simulated. However, this is far from a complete
theory or explanation, seems to be pretty empirical.

QUESTION:
The formula I learned in school for kinetic energy is E = mv^2 / 2. But the famous atomic energy formula is E = mc^2... why doesn't the 1/2 apply?

ANSWER:
Because they are totally different things, one is kinetic energy at
low velocities and the other is the total energy of a particle at any
velocity. In Einstein's equation m=m _{0} /√(1-v ^{2} /c ^{2} )
for a particle with velocity v ; here m _{0} is the
rest mass, the mass something has when it is at rest. I have shown it
before and will not do it again, but it can be shown that if the speed
is much smaller than c, the energy is approximately given by E ≈m _{0} c ^{2} +�mv ^{2} ,
the energy something has because of its mass plus the energy it has
because of its motion (kinetic). So relativity reduces to Newtonian
physics at everyday speeds.

QUESTION:
This may be a stupid question but i was just curious about this. I know with modern technologies and advanced medicine people live longer. My question is even in the slightest amount do you think that the way people get around today; seeing as our modes of transportation are a lot faster than they were say 60 years ago, are people living longer because of it? Or does the twin paradox just relate to actual travel in space and not on earth and if so what factors keep that from being true on earth?

ANSWER:
No, it is not a stupid question. You just need to get perspective on
the magnitude of time dilation. Any time something moves there is time
dilation. The amount by which moving clocks slow down is given by
γ =√[1-(v /c )^{2} ] where v is the
speed of the clock and c is the speed of light. Consider the
speed of a near-earth satellite, about 18,000 mph≈8000 m/s. Then γ =√[1-7x10^{-10} ]≈1-3.5x10^{-10}
which means that a clock on the space station runs about 0.000000035%
slower than your clock.

QUESTION:
Doesn't the fact that a black hole can bend light prove that something can travel faster than the speed of light? As light is pulled toward the black hole it would accelerate, since it is already traveling at the speed of light the moment it started moving toward the black hole it would be going faster than the speed of light would it not? Just curious.

ANSWER:
No, the light does not speed up as it falls into the black
hole. What happens is that, as you would expect, it gains kinetic energy as
it falls but light's energy is all kinetic. But, your idea of kinetic energy
is probably
�mv ^{2} , but this obviously cannot be true for light since
it has no mass. The energy of a photon is hf where h is
Planck's constant and f is the corresponding frequency of the
electromagnetic wave. So, what happens when a photon gains energy is that
the frequency increases; this is known as a gravitational blue shift (the
color of the light moves to shorter wavelengths) and happens when a photon
approaches any massive body, not just a black hole.

QUESTION:
My daughter has a stumper for me. It comes from a first-year survey university course on time. What if B and A were to take off in opposite directions to planets both 8 light years away, and travelling at 80% of the speed of light. They can see each other�s ship. What would A�s clock read when she reached her destination? At the same time (for A), what would B�s clock show? Now assume that they both turn around immediately to return to Earth, and at the same speed. At what location and date would A see B arrive at her destination?

ANSWER:
Interestingly, this is identical to a question I answered
several years ago. It is posted in an
earlier answser which you should read so that you understand what the
picture below is showing. At the midpoint each observer's clock reads six
years to himself and he sees the other clock reading less than one year (I
will let your daughter work out exactly what since this is her assignment).
When they return to earth, each sees both their own and the other's clock to
read twelve years. It is important to note that how the other's clock is
seen to be and what it is are different and this is not due to
relativity, it is because of the time it takes for the light you see to
reach you. The similarity of the question to my answer makes me think the
instuctor must have used AskThePhysicist.com as the source of the question.

QUESTION:
i know that nothing can travel at or faster than
the speed of light. but, just simply why? what equations or whatever
says no...

ANSWER:
Because the mass of an
object, that is its inertia, increases as the velocity increases.
Therefore it gets harder and harder to accelerate it as it goes faster
and faster. The expression for the mass of an object m as a
function of its velocity v is m =m _{0} /√(1-(v ^{2} /c ^{2} ))
where c is the speed of light and m _{0} is the
mass when it is at rest. Note that as v approaches c , m
approaches ∞ so it is impossible to push beyond c .
Another way to look at it is from the perspective of energy. The energy
of a particle is E=mc ^{2} =m _{0} c ^{2} /√(1-(v ^{2} /c ^{2} )),
so the energy required to accelerate the mass to the speed of light is
infinite and there is not an inifinite amount of energy in the universe.

QUESTION:
we were discussing time travel in class and learned that it is theoretically possible. We were thinking about this and wondered if it is possible to actually perform an experiment to prove this. Have there been any experiments that demonstrate this phenomena. Will traveling at great speeds actually cause a temporal shift? Also, does going back in time and "changing the past" affect present events or is it impossible to affect the present. We are aware of the "paradox of time travel" (going back and killing Mozart...would his music cease to exist?....or....going back and preventing your parents from uniting, would you cease to exist?)

ANSWER:
Here is what physics has to say about time travel: it is entirely possible
to travel forward in time but impossible to travel back in time. In the
theory of relativity, moving clocks run slow which is what allows
forward time travel; see my earlier discussion of the
twin paradox . Moving clocks
running slow (called time dilation) has been
verified experimentally in many ways. Since, as far as we know from
physics, backward time travel is not possible, we do not need to worry
about the kinds of paradoxes you ask about.
QUESTION:
I was wondering if you could explain how length contraction works. I've already done some background research and I understand the mathematical reasons my text book gives me, I was just wondering if you could give some kind of analogy that would enable me to picture the effects of length contraction, and better yet allow me to explain it to my friends in a way they can understand.

ANSWER:
I don't know a simple one-step way to
intuitively understand length contraction. But, I know a good two-step
way starting with intuitively understanding time dilation.
Time dilation is pretty easy to understand
in one simple example, the light clock. See my
earlier answer about
the light clock and be sure you understand that and have a good
intuitive feeling for why moving clocks run slow. From that point,
length contraction can be understood as a natural consequence of time
dilation. Here is how it goes. Imagine a bomb which has a fuse of 1 s.
That is a clock. If that bomb is moving by us with a speed of 99% the
speed of light, the elapsed time before detonation will be 1/√[1-(0.99)^{2} ]=7.09
s. So, the distance it will travel is 0.99x3x10^{8} x7.09=2.1x10^{9}
m. But, the bomb will measure in his own frame that he should last 1 s
and go 0.99x3x10^{8} x1=2.97x10^{8 } m, only about 1/7 the
distance we measure. If you think about the distance we measure as a
long stick of length 2.1x10^{9} m, then the bomb sees this stick
moving by him with speed
99% the speed of
light, so to reconcile his results with ours he must measure that length
to be
2.1x10^{9} x√[1-(0.99)^{2} ]=2.97x10^{8
} m.
QUESTION:
I'm reading "Relativity" by Albert Einstein and in it he says that simulteneity is relative and uses the example of two lightning strikes at points "A" and "B" and an observer at the mid point between the two "M". Einstein states that an observer on a train moving from "A" to "B" observing the lightning strikes just as the observer reaches "M" will see that strike "B" occurs before "A" (as the train it moving toward the light from "B" and away from the light coming from "A").
Does this mean that an event only becomes real when the light from it reaches the observer? Or could one legitimately says "These two events happened simultaneously, but I saw the flash from "B" before I saw the flash from "A"?

ANSWER:
The relativity of simultaneaty is not "how
it looks", it is how it is. In relativity you have to be careful how you
define a time interval so that you correct for such things as the time
it takes light to reach you. The converse is also true, a distant star
is seen right now but as it was long ago. I always find the following
example more convincing than the lightning at the ends of the train
example. In the center of one of the cars there is a flashbulb which
flashes. Light reaches the front and rear walls of the train at the same
times. However, an observer on the side of the track watching the car go
by sees the rear wall come forward to meet the oncoming light and the
front wall running away from the light coming at it. Hence the event
corresponding to the flash at the back wall occurs earlier for the
observer at rest. (It is crucial to realize that both observers see all
light moving with the same speed.)
QUESTION:
We have two clocks starting out in an the same inertial framework, and
then accelerate one of them (Clock A) to some constant speed V, but the
two clocks are also tethered by a rope. At some point, the accelerated
Clock A will tug on the rope and begin accelerating the clock that is
still at rest in the original framework (Clock B). Clock A will begin to
decelerate and clock B will begin to accelerate until they are both
traveling at the same speed V2. The question is, will they both show the
same time (from the point of view of someone still at the initial
framework)?

ANSWER:
I have been puzzzling over how to best explain this one! I believe that
the problem, as stated, is equivalent to clock A moving away at constant
speed for a while and then stopping. In both cases, clock A starts in the
same inertial frame as B, then jumps out of that frame and stays there for a
while, and then jumps back into B's frame. I recommend first reading my
earlier explanation of the twin
paradox to see what my paradigm is for the explanation illustrated by
the graph at the right. The graph represents clock A traveling at 80% the
speed of light to a point 8 light years from the earth; then clock A stops
moving. The graph shows clock A's ticks (once a year) with black crosses,
and clock B's ticks (once a year) with red crosses. Clock B measures 10
years for the trip [(8 light years)/(0.8 light years/year)]. But clock A
measures only 6 years because of length contraction (the distance is only
4.8 light years for clock A). Thereafter, the clocks run at the same rate
but clock A is 4 years behind clock B.
QUESTION:
Say there is a father and son. The father is the first astronaut to attempt to "time travel" into the future by flying his rocket around a black hole at extremely high velocities, so that for a given period of time (say an hour) it is a lot longer on earth (say 10 hours). My true question is, if the father could somehow communicate to his son in real time, what would this conversation sound like? Would the son hear his dad talking extremely fast? or would the distance between them make up the distance in "time travel". If it was the distance that is the factor, what if the dad was on a super train traveling at similar speeds on Earth?

ANSWER:
Below I have copied a figure from my
earlier answer where I talk about
the twin paradox, the situation you allude to in your question. Here there
are specific numbers (given in the caption) to make it concrete. As
explained in the earlier answer, the traveling twin (the father in your
question) takes 6 years to go each way while the stationary twin (son) has
ten years elapse. Each (father and son in your case) sends out one light
pulse each year and by looking at the spacing of those pulses you can deduce
how a conversation would sound. Here is a summary of how each sounds to the
other:
On the trip out, the father hears the son slowed down by a factor of
3 (2 yearly signals from home in 6 years).
On the trip home, the father hears the son speeded up by a factor of
3 (18 yearly signals from home in 6 years).
For the first 18 years, the son hears the father slowed down by a
factor of 3 (6 yearly signals from dad in 18 years).
For the last 2 years, the son hears the father speeded up by a
factor of 3 (6 yearly signals from dad in 2 years).
Of course, it cannot really be a conversation because of the long transit
times of the signals; rather each is just speaking, reciting poetry or
something. Higher speeds would lead to more extreme numbers but similar conclusions.
Overall, note that the father has aged 12 years while the son has aged 20
years. I always like to emphasize that how time appears to elapse on
a moving clock is not the same as the time which actually does
elapse; for example, during the last two years for the son, the father's
clock looks like it is running faster than the son's whereas it
actually is running slower.

QUESTION:
Due to the length contraction, you notice that a passing train appears to be shorter than when it is stationary. What do the people in the train observe about you?
If you are on a train that is going really fast, do the people on the ground look shorter, longer, or the same?

ANSWER:
Length contraction causes the lengths parallel to the direction of
motion to be shortened. So, a fat man standing at the station would
become a skinny man (side to side, but not front to back) but no shorter
as measured by someone on the train. Similarly, as measured by someone
on the platform, a fat lady standing on the train would become a skinny
worman but no shorter. You will note that I did not say that these folks
"look skinnier" because physicists normally do not care how something
looks , they care about how something is . This is a very
important distinction and one which even authors of physics books often
fail to make. How something looks may be very different from how
something is. Hence, your question is incorrectly stated ("� appears
to be shorter� ")
although I believe I know what you meant.

I want to, in this answer,
provide a very detailed discussion of how moving objects look (I will
restrict this to one-dimensional objects like sticks moving along the
direction of their lengths, directly toward or away from the observer). I will be able to refer to this answer when
similar questions are asked in the future. When a physicist talks about
the length of something, here is what he/she means: measure the
positions of the two ends of the object at the same time ; the
difference of those positions is the length. When you look at a stick,
you are not observing the stick ends where they were at one time but you
are seeing the farther end as it was sometime earlier than when you see
the closer end. Of course, this does not matter if the stick is at rest,
but if it is moving it does matter. For everyday moving sticks, there is
no perceptable change in the
apparent
length of sticks because speeds are much less than the speed of light.
But what if the speed is really big, let's say 80% the speed of light?
Then, as I will shortly show, the effect is really big. But before we go
into how long the stick looks or appears , we better be sure we
understand how long the moving stick is . The result from special
relativity, using the definition of length I gave above, is that the
moving stick is shorter by a factor of
√(1-(v ^{2} /c ^{2} )), so if v =0.8 c
(i.e . 80% the speed of light), the length of the moving stick is
only 60% its length when at rest. (This effect is called length
contraction.) So now, the first picture shows the situation if the stick
is coming toward you (you are on the right). Light (red arrow) leaves
the far end of the stick and does not catch up with the near end of the
stick until the stick has gone a long way (four stick lengths) and now
light from the far (red arrow) and near (green arrow) ends move forward
to your eye. So the stick looks to be 5 times longer than it actually is
and 3 times longer than if it were at rest! Now, if the stick is moving
away from you, the situation is very different. The moving stick is
still 60% its at-rest length, but now the near end moves away to "meet"
the light from the far end; the result is that the stick, as shown in
the second figure, looks much shorter than it is. It now appears to be only 1/3 the
rest length or 5/9 the actual (moving) length. Note that in neither case
does the stick appear to be its actual length. (The scales of the two
figures are different; note the different rest lengths. I had to do this
so the "much-shorter" and "much-longer" figures would be about the same
size.) So, maybe you can now understand why I often make a big deal
about relativity being about how things are, not how things appear .
The same kind of arguments may be made about
time dilation : moving clocks run
slower, they may or may not appear to run slower.

QUESTION:
I understand that nothing with mass can travel the speed of light because an infinite amount of energy would be required to accelerate the mass and there is not an infinite amount of energy in the universe. As a statement, it makes perfect sense. But I don't understand the math:
". . .
The expression for the mass of an object m as a
function of its velocity v is m =m _{0} /√(1-(v ^{2} /c ^{2} ))
where c is the speed of light and m _{0} is the
mass when it is at rest. Note that as v approaches c , m
approaches ∞ so it is impossible to push beyond c .
Another way to look at it is from the perspective of energy. The energy
of a particle is E=mc ^{2} =m _{0} c ^{2} /√(1-(v ^{2} /c ^{2} )),
so the energy required to accelerate the mass to the speed of light is
infinite and there is not an inifinite amount of energy in the universe. "
Can you explain without math why an infinite amount of energy would be needed?

ANSWER:
No, you really cannot do it totally without math, but maybe I can make
it more explicit. Let's calculate the quantity 1/√(1-b ^{2} )
for various values of b. If b is the ratio of speed to light speed, then
this factor is what determines how big the energy of the particle is.
For example, if the speed is 50% the speed of light, b =0.5, b ^{2} =0.25,
and 1/√(1-b ^{2} )= 1/√(1-.25)=1.15.
Go faster, say 80% the speed of light, then b =0.8, b ^{2} =0.64,
and 1/√(1-b ^{2} )= 1/√(1-.64)=1.67.
Go faster, say 99% the speed of light, then b =0.99, b ^{2} =0.98,
and 1/√(1-b ^{2} )= 1/√(1-.98)=7.07.
Go faster, say 99.999% the speed of light, then b =0.99999, b ^{2} =0.99998,
and 1/√(1-b ^{2} )= 1/√(1-.99998)=223.6.
Go faster, say 99.99999% the speed of light, then b =0.9999999, b ^{2} =0.9999998,
and 1/√(1-b ^{2} )= 1/√(1-.99998)=2236.
Can you see where this is going? When (never) we go 100% the speed of
light, then b =1, b ^{2} =1, and
1/√(1-b ^{2} )= 1/0=infinity.

QUESTION:
If a plasma rocket were able to maintain an acceleration that mimicked 1
G to humans on board, how soon would the rocket reach the closest star?

ANSWER:
This is a somewhat complicated question because the rocket can never
exceed the speed of light so, although the force on it is constant, the
acceleration (to an outside observer) is not. I have actually worked
this problem out in an earlier answer which you can have a look at. In
that answer I give the position as a function of time to be x =(mc ^{2} /F )(√[1+(Ft /(mc ))^{2} ]-1).
In your case, F=mg . Solving for t , I find t= √[(x /c )^{2} +(2x /g )].
I chose to calculate t for 4 light years, approximately the
distance to the closest star, and found t ≈5 years. This is the
time elapsed on earth for the trip. For a passenger on the rocket the
time will be much shorter because the velocity for much of the trip is
apparently near the speed of light (it takes only about 5 years to go 4
light years).

QUESTION:
I have seen at MIT site that "A ping-pong ball moving near the speed of light still looks spherical." I thought that the ball would contract in the direction of movement. Why would it still look spherical?

ANSWER:
As my loyal readers know, one of my favoite points to harp on is that
relativity is about how thing are , not how they look . But
it is, nonetheless, interesting to ask how things look. I tried to
illustrate that things look different than they are in a simple example
of a stick moving at high speed directly at or away from you in an
earlier answer .
The key is that when you look at the object, you simultaneously see
light from different parts of the object which were emitted at different
times; hence, if it is moving at a speed comparable to the speed of
light you are not seeing the object as it is at some time but how
different parts of it were at different times. I found a
wonderful movie on the web which illustrates this for a moving
object which would be a sphere when at rest (like your ping pong ball).

QUESTION:
I have been told that as an object accelerates to the speed of light that it becomes more 'massive'. I have also checked some other answers on the site and they use the term 'increases in mass'
My question is how can something increase in mass? it can't right, there are only some many atoms in the object? Perhaps they are confussing weight with mass?
Also Force = M*A so M=F/A therefore if anything as A increase M decreases?

ANSWER:
Mass is inertia, the resistance to acceleration when you push on it. An
increase in inertia is exactly what we see when an object has a very
large speed. If you are thinking of mass as the count of how many atoms
there are, this assumes that the mass of an individual atom is a
constant, doesn't it? As the speed approaches the speed of light, the
mass approaches infinity which means it would take an infinite force to
accelerate it any faster. Regarding your F=ma argument, you are
confusing acceleration with velocity�you
can have a large acceleration without having a large velocity. Anyway,
Newton's second law, in the form
F=ma , turns out to be wrong
at high speeds. If you want to read an answer which gives the big
picture leading to why inertia increases with speed, my best answer is
probably
this one .

QUESTION:
Two trains 200 km apart are moving at a relative speed of 50 km/h. A fly takes off from one train, flying straight to the other one at the speed of 75 km/h, touches and flies back to the first train. How far will the fly have gone when the trains collide?

ANSWER:
I suppose this might be somebody's homework, but I have waited long enough
that the homework was probably due long ago! It takes the trains 4 hours
to collide. Something going 75 km/hr will travel 300 km in that time.

QUESTION:
I understand that matter gains in mass as you accelerate it towards the speed of light, and that as you approach the sped of light, mass tends toward infinity, meaning you would need an infinite force to "push" it beyond the speed of light, and that's one of the reasons why nothing can exceed the speed of light. However, I also understand that with increased mass comes increased gravity, so... When they accelerate those sub-atomic particles to velocities approaching the speed of light at CERN, why don't they suddenly start exerting lots of gravity and sucking the entire CERN campus toward them?

ANSWER: First, a
little perspective. If I were to bring a baseball into the CERN campus, I
think you would agree that it would not "suck in" everything
gravitationally. So, an accelerated proton better be a whole lot more
massive than a baseball! Suppose the speed of the protons were 99.99999999%
(that's ten 9s) of the speed of light. Then you can calculate the mass of a
proton going this fast as m =1.7x10^{-27} /√(1-0.9999999999^{2} )=1.2x10^{-22}
kg. The proton has increased mass by a factor of about 70,000 and that is
still far from a baseball mass.

QUESTION:
How fast do you have to travel before you experience significant relativistic effects. For the purposes of this question, let's say 1 hour of travel at "X" velocity equals 1 hour plus 10 seconds.

ANSWER:
The quantity which tells you how important relativistic effects are is
called gamma and is
√[1-(v /c )^{2} ] where v is the speed and
c is the speed of light. I am glad you gave me a concrete example (10
seconds for an hour) since "significant" depends on how accurate you need to
be. For example, although satellites used for GPS navigation are nowhere
near the speed of light, not correcting for relativity leads to
significant errors in the computation of where you are. Your situation is
about 0.3% time difference, so √[1-(v /c )^{2} ]≈0.997≈1-�(v /c )^{2} +�
where I have used a binomial expansion of the square root because I assume
that v /c is small compared to 1. Solving, v /c ≈0.055,
so the speed would about 1/20 the speed of light.

QUESTION:
Doesn't a photon have to have a mass equal to its energy divided by the speed of light squared?

ANSWER:
The trouble with having a well-known equation like E=mc ^{2}
is that it is often used when not appropriate. If you write E=mc ^{2} ,
then this is the energy of a mass which is at rest; or else it means that
the mass has a different meaning from what you usually think of when it is
moving with speed v , namely m =m _{0} /√(1-(v ^{2} /c ^{2} ))
where m _{0} is the mass of the particle at rest, what you
usually think of as inertial mass. As I have said in many
earlier answer s, I
prefer to not think of mass as increasing with velocity, m to me just means
rest mass. The correct equation for energy is E= √ (p ^{2} c ^{2} +m ^{2} c ^{4} )
where p is the linear momentum. So, if a particle is at rest, momentum is
zero and
E=mc ^{2} ; if the mass is zero (as is the case for a
photon), E=pc. So a photon has momentum even though it has no mass.
One thing to be careful of, as explained in my
earlier answer s, is
that momentum is no longer mv but rather mv /√(1-(v ^{2} /c ^{2} )) .

QUESTION:
What causes gravity? How can gravity be explained?
General Relativity, as I understand it, says that gravity is not a
force or interaction. Rather that spacetime is "curved" by the presence
of mass, and that this curve "tells" other matter ( a test mass?) how
to behave. Have I got that right? But the question remains does it not?
Accepting what GR says is one thing, but in reality the real question
is why or how does mass cause spacetime curvature? Am I thinking
correctly here? I teach astronomy at a local school, and some of those
kids come up with some tough (for me) questions.

ANSWER:
General relativity starts
with a simple premise, the equivalence principle: there is no
experiment you can perform which can distinguish whether you are in a
gravitational field or in an accelerating frame of reference. For
example, if you were in an elevator which was accelerating and a beam
of light entered through the side it would follow a curved trajectory
to the opposite wall; this is exactly what would happen if you were
sitting still in a gravitational field. This principle, coupled with
the principle of special relativity (the laws of physics are the same
in any inertial frame of reference) leads to the general principle of
relativity, the laws of physics are the same in any frame of reference.
One implication of this theory is that mass deforms spacetime which is,
as you state, how gravity works; mass deforming spacetime is simply a
consequence of the postulates of the theory. Is it the last word?
Probably not because gravity has not been reconciled with quantum
theory and the quest for a theory of quantum gravity is one of the holy
grails of physics. I would not say that gravity is not a force just
because we understand the mechanism for that force. Asking "why or how"
mass causes the curvature is essentially equivalent to asking what is
mass, why do objects possess it? The current well-publicized quest for
the Higgs boson is important because this is the particle which
physicists think is responsible for endowing the elementary particles
of nature with mass.

QUESTION:
I have a question regarding special relativity. I
was pondering over a thought experiment that I came up with using a
modified version of a light clock. This modified light clock, with every
tick, sends a signal to a space ship which is moving towards it at a
uniform speed. This is where I get a little confused. It would seem that
since the spaceship is moving towards the light clock that the frequency
of the signals being received would increase. Intuition would tell me
that the rate of the ticks has therefore increased. And yet I find it
hard to reconcile with the fact that, since the light clock is in motion
relative to the spaceship, that the ticks ought to be slowing down. I
realize that my argument is flawed somewhere as special relativity has
already been proven through countless experiments. I ,however, need help
understanding where exactly that flaw is. Do you have any thoughts on
this?

ANSWER:
You have stumbled upon, as many of my loyal readers will tell you, one
of my favorite points of emphasis in special relativity. How things are
is not necessarily how things appear to be . To answer your question,
read my explanation of the twin
paradox where I emphasize that the rates of clocks as you see them is
not the same as the rate at which clocks are ticking. See also
another answer . Time
is defined as the time interval between two events at the same point in
space and the clock coming toward you is not at the same places when it
sends you sequential ticks, so the ticking you observe is not the time
interval between ticks on that moving ship. If the space ship is going away
from you the observed frequency of ticks would be less than the real
tick rate. The same kind of phenomena are observed for length contraction.
Length is defined as the distance between two points in space measured at
the same time. But if a stick is moving toward you, it appears longer
than its rest length even though it is actually shorter and the stick moving
away from you looks shorter than its actual contracted length. Again,
see an earlier answer and links
there.

QUESTION:
I am not getting the idea of length contraction.Does length of an object actually shrink when in motion or is it just apparent to the observer which is at rest.Is moving rods shrinking a measurement problem?

ANSWER:
First you have to define how you are going to measure the length of
something. In physics, length is the difference between the positions of two
points in space which are measured simultaneously. So, if a stick is moving
by you, you have to be careful to measure the positions of the two ends of
the stick at the same time. If you agree that this is a reasonable
definition of what length is, then the answer to your question is yes, the
object actually is shorter if it is moving. What is apparent to the observer
is an entirely different thing. How long a stick looks is not the
same thing as how long the stick is . For much more detail, see an
earlier answer .

QUESTION:
When we heat something, the particles in it move faster and faster till they reach the speed of light. But what would happen if we keep heating? The particles can move faster than the light, or there's a maximum value for the temperature (like a minimum: 0 Kelvin), or what?

ANSWER:
No, anything with mass can never reach the speed of light, but
neither can it go any faster. I have
answer ed this
question before and the answer is that there is no upper limit. I have
actually been corrected since there is an upper limit imposed by the total
amount of energy in the universe, but that begs the question. If you have
something, there is no limit to the amount of energy you can put into it,
assuming you can get that energy somewhere. There is a minimum temperature
possible, as you note, but that cannot actually ever be reached because of
the laws of quantum mechanics. So the speed of particles in an object are
constrained by 0<v<c corresponding to temperatures
of 0<T < ∞ and the extremes are
unreachable.

QUESTION:
Why are only mass, length, and time changed as an object approaches the speed of light? Why not other "fundamental" properties of matter e.g., charge? Does this imply that charge is not fundamental or is somehow very different from mass, length, and time?

ANSWER:
The elegant way to put it is that some quantities are invariant
under a Lorentz transformation and some are not. This means that some
quantities are the same whoever measures them, some are not. Electric charge
is invariant and so is the total energy of a particle. It would be my
inclination to say that invariant quantities are more "fundamental" than
those that are not invariant.

QUESTION:
What is the medium through
which electromagnetic waves propagate? It's easy to see physical waves
go through water. And I can understand sound waves traveling through
air. But what is the stuff that carries electromagnetic waves?

ANSWER:
This is one of the classic questions in the history of physics and its
answer resulted in one of the great revolutions in physics, the theory
of special relativity. The simple fact is that, since it is electric
and magnetic fields which are "waving" and these fields may exist in
perfectly empty space, that electromagnetic waves may propogate through
perfectly empty space. They are the one wave which requires no medium
through which to travel.

QUESTION:
My 11 year old has asked me,
"Does gravity bend light?" I did highschool physics about 20 years ago,
so am very rusty and not up to date with current thought. I have looked
at this discussion topic on the archives of many forums, but have ended
up very confused by the differing opinions/explanations. I would really
like an answer which is easy to explain to a child, but yet not so
simplistic that it is inaccurate. Am I asking the impossible? She has
read about the nature of light and also about gravity, and can't
understand how light can be affected by gravity, when it has no mass.
Is it because photons are energy and so can be used instead of mass? Or
is it that the gravitational pull around massive stars affects the
"space" around it and light just follows the stretched paths? If this
is true, how can some authors say that the light is still moving in a
straight line even when it is following a curved path?

ANSWER:
Here
is one explanation, probably the easiest for your daughter to
understand: Light being affected by gravity is a result of the
principle of equivalence in general relativity. This states that there
is no experiment which you can perform to distinguish between your
being in a gravitational field or in an accelerated frame of reference.
Thus, for example, imagine that you are in an elevator which
accelerates upward; if light enters through a hole in the side of the
elevator it will clearly appear to fall like a projectile because of
the acceleration of the elevator. So, the same thing will appear to
happen in a gravitational field the acceleration due to which is
exactly the same as the acceleration of the elevator. Hence, light
will "fall" in the earth's gravitational field with an acceleration of
9.8 m/s^{2} . You might be interested in the answer to an earlier question .

Here is
another: If we look at the world as having a Euclidean "flat" geometry
and watch a ray of light pass a very massive object, we see the light
bend. But, the way that general relativity describes the world
says that, if we are in the vicinity of a massive object the space
itself is not Euclidean but is curved; in this space the light follows
a "straight line" in that non-Euclidean geometry.

QUESTION:
Taking the light as a
guiding agent (its invariant speed) in several thought experiments,the
lorentz transformation of coordinates can possibly be completely
derived,and time dilation,
lenght contraction,non synchronisation of simultaneous events,and such
things can be understood therefrom.But mass increase seems to be
detached from this continuity of explanations.
The explanation of "Mass" gets carried away by referring to Sir J.J
thompsons electron experiments.
Hence the question arises-
Is it possible to deduce the effect of mass increase
("Mass increases by gamma factor as velocity increases"{where
gamma=((1-(v/c)^{2} )^{0.5} )})
by purely referring to lorentz transformations?without referring to
explanations and definitions from electromagnetism.
The relativistic addition of velocities seems to provide some clue.
Also an explanation as to how the time part of four momentum can be
treated as energy is needed.

ANSWER:
All the quantities which you can derive from
the Lorentz transformation are what we call kinematic quantities. Mass,
force, and more importantly linear momentum and energy are what we call
dynamic properties. So, just like in an introductory physics course
where, after we learn how to describe motion (kinematics), we next want
to understand how motion can be changed; in classical physics this
leads us to Newton's laws. What happens in relativistic physics is that
we quickly find that Newton's second law, in the form F =ma
is no longer a true law of physics; that is, if two observers both
measure the acceleration of a mass m they will get different
answers for a , so that would mean that force is no
longer a useful concept in that context. So what we do is look for the
relativistic equivalent of Newton's laws. To do this, write Newton's
second law as F =dp /dt where p =mu
is the linear momentum and u is the velocity of the
particle, so for an isolated system we expect to find dp /dt= 0,
that is the momentum is conserved. If mu is the
definition of momentum, where m is what classical physicists
call the inertial mass, we find that momentum conservation is no longer
a true thing for isolated systems. So, what we do is to redefine what
we mean by momentum such that momentum is conserved and the new
definition becomes the old definition for small speeds. If we define
momentum as p = g mu we find that momentum is
conserved in an isolated system and p @ mu
for small u . So, you see, the gamma factor comes from
redefinition of momentum, not redefinition of mass. Almost all
introductory physics texts say that it is mass which increases, and
this is certainly a possible interperatation of the new definition of
momentum. I prefer to say that m is the inertial mass of an
object at rest and that p = g mu . Your
question about energy being the fourth component of four momentum is
too involved for this site.

QUESTION:
if two people are moving
past each other at a constant speed in an infinite empty space, to
either one the universe would be static and the other person would be
in motion (relativisticly).
if two people were at a constant unmoving distance from each other, but
one of them was rotating at a constant speed, to either one the
universe would be static and the other person would be in motion,
correct?
so if they were far enough apart, wouldn't the rotating person see
their friend travelling faster than the speed of light (tengentially)?

ANSWER:
The two situations are not equivalent.
Assuming that in your first scenario each was an inertial frame (one in
which Newton's first law is true), then both are inertial frames and
each can rightly claim that he is at rest. In the second scenario, if
one is an inertial frame then the other (revolving around it) is not,
so the rotating frame cannot claim to be at rest.

QUESTION:
I was recently watching a
programme about the Hubble Space Telescope and the pictures it had
captured. It mentioned that the pictures it captured were of galaxies
forming millions of years ago-If this is so- Could it then be possible
for man, should he a find a way to manipulate space travel, be able to
position himself at a certain area of space, at a certain distance and
witness the birth of the milkyway itself? or indeed the birth of
Earth?..........perhaps even the birth of man?

ANSWER:
Suppose that something happened here on
earth 500 years ago and you were located at a distance of 500 light
years from the earth with a very good telescope. Then you would be able
to witness that event right now. But, here is the rub. You cannot get
there from here before the light from the event reached there because
it is physically impossible to travel faster than the speed of light;
in other words, you could get there in just under 500 years and you
would be able to witness what happened here tomorrow.

QUESTION:
I’m a PhD student in
philosophy interested (but not working in) the philosophical
foundations of relativity. I have a couple of questions dealing with
the speed of light and special relativity. The main question is number
3, the other two are auxiliary. Your response would be greatly
appreciated. I will also greatly appreciate it if you could let me know
whether there is some bibliography available dealing with this kind of
issues.

Why is the relativistic
length contraction – as it appears in the Lorentz transformation – a
function of the speed of light?
Conceptually, it is possible
that the speed of light had a different value than it actually has. How
would that have affected special relativity (in particular length
contraction for fast-moving bodies)? I recommend your looking into the
Mr. Tompkins in Wonderland books by George Gamow.
Suppose we will discover a
“form of light” (call it light*, symbolized c*) whose speed is double
the speed of light, and it is also invariant. This would enable us to
send signals at this speed and to use these signals to synchronize
clocks so that we may provide an empirically meaningful definition of
simultaneity. Do we need to revise special relativity and base the
Lorentz transformation for length on this new value (c*), instead of c?
Which of the following possible responses should be the case and why:
a) we can do that; b) is it impossible to do that; c) we must do that.
ANSWER:

The answer is that the fact
that light has the same speed in all frames of reference leads to this
result. It is simple algebra and, if you are seriously interested, you
should take the time to learn the basics of special relativity.
Special relativity would have
the same form it does. If the speed of light were 100 miles/hr, effects
like time dilation and length contraction would be everyday phenomena
which we would not find puzzling at all. I recommend your looking into
the Mr.
Tompkins in Wonderland books by George Gamow.
This seems extremely unlikely
since light is just a manifestation of electromagnetism and its speed
is predicted by Maxwell's equations. For a new kind of invariant speed
to exist, there would have to be another fundamental force in nature
the theory of which predicts radiation of that speed. Since there is no
good quantum theory of gravity, that is a remote possibility; gravity
waves, however, have never been directly observed although their
existence has been inferred from energy loss of binary systems. It is
usually assumed that gravity waves move with the speed of light, but
that is unverified. Regarding your simultaneity question, there is
never any problem defining when two events are simultaneous. The
problem is that events simultaneous in one frame of reference are not
simultaneous in other frames; having some other universal speed is not
going to change that.
QUESTION:
Why is the speed of light given by 1/sqrt(permittivity
*permeabillity)? What is the great mistery behind such a simple
relation?
How these two parameters combine to give the speed of light?
Why does the vacuum (nothing...) has physical properties such as
permittivity and permeability?

ANSWER:
This is the great triumph of Maxwell's work
in the 19^{th} century. There are laws of electromagnetism
which can be summarized in four equations, now known collectively as
Maxwell's equations. The quantity e _{0}
(permittivity of free space) is just a proportionality constant which
tells you how strong the electric force is and, of course, it appears
in the equations. Similarly, the quantity m _{0}
(permeability of free space) is just a proportionality constant which
tells you how strong the magnetic force is and, of course, it appears
in the equations. (In this context, there is nothing wrong with empty
space having permittivity and permeability because one certainly does
not need matter between charges or currents for them to exert forces on
each other.) When Maxwell messed around with the equations he
discovered that they could be rewritten as wave equations and that the
speed of these waves had to be 1/[e _{0} m _{0} ]^{1/2} . That this happened
to be the speed of light was the point in the history of physics that
we understood what was doing the waving in light waves--electric and
magnetic fields.

QUESTION:
i was once concidering the
effect of propelling an object backward off a moving vehicle with the
exact velocity that you are travelling. From my point of view the
object is moving away from me, but to an observer standing at the side
of the road at the instant i throw the object it merely drops to the
ground. This made me wonder if it is possible for light to behaive in
the similar manner. In this hypothetical situation an object must be
travelling the speed of light. Would the light from the object
travelling in the opposite direction of travel merely stand still in
space becoming "dead light" or would it accelaterate to it's regular
pace. In this situation we would also have to assume that the object is
generating it's own light since light itself would not be able to
reflect off it due to it's speed.

ANSWER:
It seems like I answer this question (in
various forms) at least weekly! There are two problems with your
hypothesis, both violations of the theory of special relativity. First,
no material object may have a speed equal to or greather than the speed
of light; the reason, simply, is that to accelerate it to the speed of
light you would have to supply an infinite amount of energy, obviously
not available. Second, one of the basic postulates of the theory of
theory of special relativity is that the speed of light in a vacuum is
the same for all observers. Thus, even if your object were moving with
99.99% the speed of light (which is possible), an observer at the
roadside would see the light from its back end moving with the speed of
light, not 0.01% the speed of light. The theory of special relativity
is completely verified experimentally and no serious scientist doubts
its correctness.

QUESTION:
I have read
that the speed of light cannot be broken. I have also read that all
motion is relative. Suppose point A is a stationary reference point.
Point's B and C are start near point A and travel away from
each other at 51% the speed of light relative to point A. Does this not
mean that relative to each other, points B and C are traveling apart
from each other at 102%? Do you not add their speeds together as if two
automobiles are each traveling away from each other at 60 mph are
traveling at 120 mph relative to each other?

ANSWER:
No, your intuitive way of adding velocities does not work in
the theory of special relativity. Because of the fundamental
postulate of special relativity, that the speed of light is the same in
all frames of reference, the correct formula for velocity addition is v= (v _{1} +v _{2} )/(1+v _{1} v _{2} /c ^{2} )
where c is the speed of light. So, for your example, v _{1} =v _{2} =0.51c
so v =0.81c , 81% the speed of light. Note that the
formula reduces, to an excellent approximation, to your v= (v _{1} +v _{2} )
if the speeds involved are very small compared to the speed of light
because the term v _{1} v _{2} /c ^{2
} is extremely small compared to 1.

QUESTION:
Here's a question about the Length Contraction Aspect of the
Special Theory of Relativity - say there's a Railway Car sitting
Motionless on a Track, with a Lamp on the East End of the Car - the
Lamp is Switched On and the Time is Measured for the Light to Reach the
West End of the Car - then the Car begins Rolling Eastwards, and again
the Light is Switched On and the Time Measured again - this time it
Takes Less Time for the Light to Reach the West End of the Car, so the
Special Theory would say that, since C is Constant, the Length of the
Car has Changed - the Car's Motion caused the Car to become Shorter?

Is all
that right? - if so, then what about if the Car begins Rolling West,
and again the Light on the East End of the Car is Switched On - this
time it would take Longer for the Light to reach the West End, so by
the same Reasoning, it seems like the Special Theory would say that now
the Car's Length has Increased!

Is that a
Flaw in the Special Theory? - maybe the Wording should be something
like "Sometimes an Object in Motion becomes Shorter, but Sometimes it
becomes Longer"

Do I
deserve a Nobel Prize? - or am I Overlooking Something?

ANSWER
I will answer your last questions first--no, you do not
deserve a Nobel prize! And it is not so much that you have
overlooked something, but that you have computed the length of the car
incorrectly. In order to measure the length of something, you
need to measure the positions of each end at the same time, not
different times. So first you must use special relativity to get
length contraction so that you know what the length of the moving car
is and then you can analyze your experiment.

You
cannot deduce the length of something by simply measuring the time it
takes light to travel between its ends (which is what you have
discovered here!). Let's do your problem but do it
correctly. Suppose we call the length of the car, in its own rest
frame L _{0} . Then, if it has a speed v relative
to some observer, its length as measured by that observer is L=L _{0} [1-v ^{2} /c^{2} ]^{1/2} .
For your first experiment (car at rest) the time is t =L _{0} /c .
For your second experiment, (car moving east) let us call the time t _{1} .
Looking at the first picture (dashed lines show where the car has moved
to when the light, red, strikes the other end), it is clear that L=L _{0} [1-v ^{2} /c^{2} ]^{1/2} =vt _{1} +ct _{1} ,
so t _{1} =t [1-v /c]^{1/2} (you will
have to do a little algebra to get here). t _{1}
is, as you state, indeed less than t but not because L
is shorter than L _{0} , although it is. In your
third experiment, refer to the second picture and call the time t _{2} .
Now we have that L=L _{0} [1-v ^{2} /c^{2} ]^{1/2} =ct _{2} -vt _{2} ,
so t _{2} =t [1+v /c]^{1/2} . So
now, t _{2} >t but not because the car is any
different length from experiment #2 (it has exactly the same length);
rather because the light had to travel farther than the length of the
car.

QUESTION:
I don't knwo if anyone if farmiliar with "A Brief History In
Time," but it has leaft me confused. Can you please explain the 4
dimension rule. How Time and Distance are related. Also how a clock on
the ground of earth would be slower than one on the top of a mountain.
I understand the slower frequency rule due to gravity, but i don't
understand how that affects how the clocks tell time. Their mechanisms
which keep time shouldn't be related to waves.

ANSWER:
You need to learn the theory of special relativity. It
requires only algebra to understand its basic concepts.
Essentially, if we demand that the speed of light in vacuum is the same
for all observers (by now a well-documented experimental fact), the
inescapable conclusion is that space and time are not separate entities
but are "entangled". The "entanglement" is much like that of the
three spatial dimensions. An example is the rotation of a
coordinate system shown at the right. Note that the components of
the vector in the black coordinate system (x,y ) are not the same
as in the rotated (red) coordinate system (x',y' ).
In fact, (x',y' ) depend on what (x,y )
are:

x'=x cosq +y cos q
y'=-x sin q +y sin q .

The
rotation "mixes up" the spatial dimensions. It turns out that in
special relativity, the transformation between space and time if one
system is moving with speed v with respect to the other looks
very much like a rotation in "four-dimensional space" because x
(the direction of the relative velocity) and t get mixed up:

x'= g (x-vt )
t'= g (t-vx/c ^{2} )^{
} g =(1-(v/c )^{2} )^{-1/2}

The fact
that this transformation mixes up these two dimensions leads us to
recognize that space and time are no longer separate unrelated
concepts. Instead, we should think of "space-time" which includes
both time and the three dimensions of space, thereby suggestive of a
four-dimensional world.

Regarding
your second question, I found an interesting
example on the web which might be instructive for you.

QUESTION:
This question is purely theorical. If you somehow got a
ship moving at 4/5 the speed of light in one particular direction. Then
the same ship is accelerated to the same speed in a direction
perpendicular to the original direction. So given that, wouldn't the
resultant speed be greater than the speed of light? And if so wouldn't
that go against the theory that faster than light travel is impossible?

ANSWER:
To answer your question you must know the Lorentz
transformation for velocities. I shall call the direction of the
velocity of the ship relative to the ground, before the acceleration in
the perpendicular direction, the x -axis; so the velocity of the
ship in its own rest frame is, of course, u'_{x} = 0 and
the velocity of the ship in the ground frame is u_{x} =0.8c
(where c is the speed of light). Now the ship
is given a velocity in the primed frame (the one moving in the +x
direction with speed 0.8c ) of u'_{y} = 0.8c .
(It may be easier to visualize the ship firing a projectile in the y'
direction with speed 0.8c and asking what the speed of the
projectile seen from the ground is; it is equivalent to your
question.) The inverse Lorentz transformation for components of
the velocity perpendicular to the direction of relative motion (i.e.
the transformation which tells you u_{y} if you know u'_{y} )
is

u_{y} =u'_{y } /[ g (1+(u'_{x} v /c ^{2} ))]

where g =1/[1-(v /c )^{2} ]^{1/2}
and v is the speed of the moving frame (in the x
direction), 0.8c in your question. It is now a simple
matter to compute u_{y} =0.48c . Therefore
the speed of the ship in the ground frame is u =[u_{x} ^{2} +u_{y} ^{2} ]^{1/2} =0.933c
which is, as it must be, less than the speed of light.

Let me
speculate where you may have gone astray in your reasoning here.
You were probably assuming that u_{y} =u'_{y }
because the Lorentz transformation for lengths in the y direction
is y=y'. As shown above, however, this is not the case
because of the fact that the two observers use clocks which run at
different rates to measure the velocities in their own frames.

QUESTION:
In 1998, researchers at Stanford University's Linear
Accelerator Center successfully converted energy into matter. This feat
was accomplished by using lasers and incredibly strong electromagnetic
fields to change ordinary light into matter. The results of this
experiment may allow for the development of variety of technological
gadgets. One such development could be matter/energy transporters or
food replicators that are commonly seen in some of our favorite science
fiction programs.

My
question is, if we were to build a transporter that worked by
converting people into energy, would they be able to theorically
survive the procedure once their energy has been reconverted back into
matter, since both are interconvertible?

ANSWER:
Changing electromagnetic energy into mass has been seen for
decades. It is called pair production and happens when a very
high energy photon spontaneously creates an electron-positron pair.
Conversely, any time a particle meets its antiparticle the mass usually
totally disappears and electromagnetic energy appears. Regarding your
question, it has no short answer. But, there is enormously more
to the transporter idea than simply destroying and creating mass--you
have to put it back together the way it was. I should also note
that it has never been possible to take some macroscopic thing like a
person and totally annihilate all the mass. Anyway, if you want
to read a good discussion of the physics of why it is extremely
unlikely that a transporter will ever work, I would recommend a book
called The Physics of Star Trek by Lawrence M. Krauss.

QUESTION:
1. Can you tell me how and why did the special theory of
relativity lead to the atomic bomb and revealed the secret of stars
?
2. How did general relativity give us space warps , the big bang and
black holes ?

ANSWER:
The answer to your first question is that one of the
consequences of the theory of special relativity is that there is a
connection, an equivalence, between mass and energy. That is,
mass may be converted to energy and vice versa. As a
result, if you cause a very heavy atomic nucleus to split (nuclear
fission) energy is released because the mass after the fission is less
than that before; this is the principle behind a nuclear reactor or a
conventional atomic bomb. But if you take two very light nuclei
and fuse them together (nuclear fusion) you also get energy released
because the mass after the fusion is less than that before; this is the
principle behind how stars generate energy and also behind the hydrogen
bomb.

The
answer to your second question is that general relativity is the theory
of how gravity works. In essence, in order to satisfy the
postulates of the theory, the structure of space and time themselves
are altered by the presence of gravitational mass; this alteration is
often described as a warping of the space around massive objects which
then explains why massive objects attract each other (they simply fall
along "straight-line" paths in this warped space). I don't think
that I would say that general relativity "gave us" the big bang or
black holes. They just happened. Of course, to properly
describe a black hole, you need to use general relativity. But
the reason a black hole forms is that when a star is all "burnt out"
the force of gravity (which is always trying to compress any large
object) has no force opposing it which is strong enough to keep the
object from collapsing.

QUESTION:
what is "energy" really? heat, motion, capacity to do
work.....all of these?! i am very confused about the concept of
energy....any info where it is looked into philosophically?

ANSWER:
Yes, all of these and more! When I teach the concepts
of energy to students in introductory physics, I always emphasize that,
in many respects, energy is just a clever trick to either make problem
solving easier or to allow us to solve harder problems. But it is
not a new fundamental thing; it is just casting Newton's laws (which
are the fundamental things in classical mechanics) into a new
form. In more advanced courses, energy arises naturally from
mathematical manipulation of the differential equation which is
Newton's second law. The "elemental" form of energy, the one most
easily grasped, is kinetic energy, the energy something has by virtue
of its motion. The way that kinetic energy can be changed is to
do work on the system. Kinetic energy is useful in the
kinetic theory of gasses and solids--temperature is nothing more than a
measure of the average kinetic energy per molecule. The next
thing we introduce in mechanics is the potential energy which
attributes energy to something by virtue of its position; I always tell
my introductory students that potential energy is basically a
bookkeeping device--it automatically keeps track of work done by some
force which is always present. What you gain with energy
compared to forces is that energy is a scalar which leads to easier
problem solving; the price you pay is loss of information about the
time development of the system. An important thing we learn from
studying energy in classical mechanics is the beauty and power of
conservation principles--some things, like the total energy of an
isolated system, are constant with respect to time.

Physics
at more advanced levels attributes more importance to energy than a
cute trick or a bookkeeping device. In the theory of special
relativity, the energy associated with the mass of an object emerges as
an unexpected important property of material objects (E=mc ^{2} ).
In electromagnetism we are often interested in the energy contained in
electromagnetic fields. In quantum mechanics, energy is one of
those variables in nature which you can not know to arbitrary precision
(that is, a Heisenberg uncertainty principle applies to energy).
Even empty space contains energy (called vacuum fluctuations) whereby
the vacuum is viewed as particle/antiparticle pairs popping into and
out of existence (since the uncertainty principle allows conservation
of energy to be violated if for a short enough time).

I do not
know much about philosophy of science (although I do a lot of
philosophizing on my own!), so I cannot point you to a philosophical
reference on energy. Surely there must be good ones, though.

My answer to an earlier question might also be of
interest to you.

QUESTION:
What is the speed of gravity? If you don't have a exact
speed calculated, is it faster than the speed of light?

ANSWER:
No one has ever measured the speed at which the gravitational
force propagates. In Newtonian physics we calculate orbits by
assuming that the forces are instantaneously transmitted, but nobody
believes this--it just works pretty well because realistic propagation
times are bound to be very short over the distances of interest.
A better theory of gravity is general relativity. Here the theory
predicts that gravity is due to distortions of spacetime caused by the
presence of mass and these distortions are predicted to propagate with
the speed of light. In general, all the physics we know forbids
any information propagating faster than light speed. Finally, we
believe that a theory of quantum gravity will someday be found in which
the quanta which transmit the force will be massless particles called
gravitons (similar to photons which transmit the electromagnetic
force). Massless gravitons, like all massless particles, would
have a speed equal to the speed of light so the force which they
transmit would move with that speed. There is a very enlightening
essay at http://math.ucr.edu/home/baez/physics/Relativity/GR/grav_speed.html

QUESTION:
I have heard through books that I read and other sources that
objects cannot excede the speed of light. Why is this? If you are
travelling at any speed and you turn on a flashlight, how fast is the
light moving? Is it moving at C + the speed you are moving? Or is it
just moving at C? Please send I response, I would be thankful.

ANSWER:
The cornerstone postulate of the theory of special relativity
is that the speed of light is a universal constant. Therefore, no
matter who measures the speed of light coming out of the flashlight you
allude to, the measured speed will be the same, 3x10^{8}
m/s. This is hard to swallow because it conflicts so seriously
with our "Newtonian-Galilean" intuition, but it has been shown with
great accuracy to be true. With this postulate under our belts,
we can reformulate mechanics. One of the consequences of this
reformulation is that the amount of energy it takes to accelerate a
mass m to a speed v is given by

E=mc ^{2} / (1-v ^{2} /c ^{2} )^{1/2}
- mc ^{2 }

where c
is the speed of light. Notice that if we want to accelerate the
object up to v=c, an infinite amount of energy needs to be
supplied. Sorry, but nobody has an infinite amount of
energy! You need to find a popularized book on special relativity
(there are many); it does not take very sophisticated mathematical
ability to understand the theory.

QUESTION:
Contrary to the usual description of the twin paradox, where
one of them stays on earth and the other is travelling at relativistic
speeds, consider the situation where both of them ( say A & B ) are
travelling at 'identical' speeds but away from the same point ( some
arbitrary point in space, away from the influence of any gravitational
field so as to neglect the effects of General theory of relativity ) in
diametrically opposite directions and then returning to the same point.
Then, according to a third 'neutral' observer at rest with respect to
the 'point' from where A & B started off, both of them would have
aged equally. But in A's Frame Of Reference (FOR in short ) B has been
constantly travelling ( except at the moment where they stop and start
retreating ) and hence B has aged according to A; the same applies even
to A from B's FOR. Here, though we ( the 'neutral observers' ) know
that they are of the same age how is one going to resolve the paradox
keeping in mind the views of A & B ?

ANSWER:
My favorite way to understand the twin paradox is to suppose
that each twin, using his own clock, sends a light pulse to his brother
once a year. Each brother receives all the pulses from the other
but the moving brother sends fewer since, because of length contraction
of the distance to the object he travels to, he has less far to travel
in his frame than his brother observes. The diagram below shows
how this works for a brother who travels to a star which is 8 light
years away and back with a speed of 80% the speed of light. Since
the traveling brother sees, because of length contraction, a distance
to the star of only 6 light years, he sends out six light pulses on the
way out and six on the way back, and all are received by the earthbound
brother, so both agree that he has aged 12 years. The earthbound
brother, of course, sends out 20 pulses and the traveling brother
receives them all so both agree that he has aged 20 years. It is
interesting to note that this graph illustrates that the rate at which
clocks run is not the same thing as the rate at which they appear
to run. The earthbound brother sees his twin age only 6
years in his first 18 years and then age 6 more years in only 2
years. The traveling brother sees his twin age only 2 years in
his first 6 years and then 18 more years in his remaining six years of
travel.

So now we
come to your question. How would we analyze the situation if one
twin went on the original trip and the other twin went to a star 8
light years in the opposite direction and back? The graph below
illustrates this scenario. It is clear that each twin sends and
receives 12 pulses so each sees both himself and his twin age 12 years
over the duration of the trips. Again, note the interesting
disparities between actual aging and the appearance of aging.
Neither twin receives any pulses until the return trip.

QUESTION:
I want to know (among other things i consider wrong) why you
believe light is massless? Do you believe these to be true: light bends
around stars, light can't escape from black holes, light is able to
push light-craft such as futuristic light-sailships and spherical craft
that WAS able to be propelled 200 feet into the air by a high-powered
laser, or that we can see light at all? Well MY point is that anything
that acts on anything else must have mass, otherwise there is 0
momentum to transfer onto the next item. If light was to be attracted
by gravity, wouldn't it have to have some sort of mass that the gravity
acts on since the amount of attraction between two objects by gravity
is related to the mass of BOTH objects?

ANSWER:
None of the things you cite require light to have mass. The
problem is that you are thinking classically. In fact, even in
classical electromagnetic theory, light has momentum and energy but no
mass. In modern physics, the electromagnetic field is quantized, i.e.
the light behaves like photons in certain circumstances which are
massless particles with velocity c =3x10^{8} m/s, energy
E=hf (f is frequency, h is Planck's
constant) and momentum p=E/c. Light being affected by gravity
is a result of the principle of equivalence in general relativity. This
states that there is no experiment which you can perform to distinguish
between your being in a gravitational field or in an accelerated frame
of reference. Thus, for example, imagine that you are in an elevator
which accelerates upward; if light enters through a hole in the side of
the elevator it will clearly appear to fall like a projectile because
of the acceleration of the elevator. So, the same thing will appear to
happen in a gravitational field the acceleration due to which is
exactly the same as the acceleration of the elevator. Being able
to see something has nothing to do with mass; the detection only
requires that we have a detector sensitive to the electric or magnetic
fields associated with the light.