Light behaves the same way, slowing down in the
medium and speeding back up at the interfaces with the vacuum. The
difference is the mechanism for slowing the waves down or speeding them up; v=λf
still holds. There is another way you can look at light: You may know that
light waves of some given frequency may be thought of as a swarm of
photons, each having an energy E=hf where h is Planck's constant.
Since f is the same in and out of the medium, the photons always
have the same energy whatever the wavelength of the corresponding waves.
QUESTION:
I'm a recent physics graduate, and I've realized that most of what I've done while studying physics is very theoretical, with the exception of some basic kinematics, circuits, and optics labs.
However, if at all possible, I would like to perform an experiment to show Relativity (or at least the underlying postulates) in the real world for myself. After all, it's different enough from our everyday experience to be difficult to accept on an intuitive level, even if you learn the math and such.
I figured I'd ask around and see if you had any ideas for such an experiment using either materials you can find at home or basic undergraduate lab equipment.
ANSWER:
Well, I couldn't think of a tabletop experiment
which would meet your criteria. So I did a quick Google search and found a
question basically the same as yours on
Physics Forums. I think you will find the ensuing discussion
interesting but probably nothing which would fit your wishes.
QUESTION:
Hello, I'm another 11th grader and I'm trying to understand how EM waves work.
The electric and magnetic waves are in the same phase. However, I was told that during induction the changing magnetic field induces the electrice one, and vice versa. From this I'd assume that B is shifted by compared to E, supposedly being its derivative, but that's not the case. I looked some things up on ChatGPT and the internet and found the following Maxwell equation:
After getting a rough understanding of gradients and partial differentiation (took a long time), I concluded that the rate of change in the electric field equals the negative rate of change of the magnetic field (divided by the change in time). Which still doesn't make sense! If E and B change alongside each other (beside some constant, or directional difference), then charged objects would gain a constant magnetic field and freshly magnetized pieces of iron would gain an electric field, which they could keep, until their magnetic field declines.
Besides the former, I'm struggling at understanding how energy is stored within photons, if E and B oscillate at the same time. Due to how inductors and capacitors can be charged, I am of the opinion that electric and magnetic fields contain energy, like in this equation about energy density: However, if E and B reach zero at the same time, that would mean the photon has no energy, which it clearly has, when E and B are at their maximum. This goes against the conservation of energy, which means I'm really missing something. Can you help me in this regard?
ANSWER:
[This question had equations and a figure which
I have edited out because I do not do LaTex symbol language. Also, because
the questioner is not really prepared to understand electromagnetism at
this level.]
I admire your quest to understand electromagnetism.
You are already in a good place in understanding qualitatively. Until you
have studied vector calculus and differential equations, however, be
satisfied that you have already achieved way beyond your years. I will
briefly give you some more information to help with your pursuit of this
difficult subjece. The first equation you show is
∇xE=∂B/∂t. This is not a gradient of E,
rather a curl of E. So you need to understand the
curl operator to understand the relative phases of E
and B.
The other equation you quote (corrected here) is
energy density of an electromagnetic wave,
U=½(ε_{0}E^{2}+B^{2}/μ_{0})
in empty space. What you are missing is that this
equation has been timeaveraged. And before it is time averaged any spot in
space would be oscillating between 0 and some maximum value; there is no
violation of energy conservation as you suggest. Regarding the photons,
this is not what the energy density is for them. If the frequency of the
wave is f
then the energy of one photon is hf where h is Planck's
constant. Hence, for photons U_{photons}=Nhf where N is the number of photons per unit
volume. In both cases U has units J/m^{3}.
QUESTION:
For the past several months, I (an aerospace engineer with an interest in quantum mechanics) have been trying understand Heisenberg's frame of mind when he first came up with his crucial insights in 1925. Everywhere I look I see essentially the same statement: "His leading idea was that only those quantities that are in principle observable should play a role in the theory, and that all attempts to form a picture of what goes on inside the atom should be avoided."
What none of the sources has conveyed to me is the following: granted that at the time Heisenberg developed his theory there was not a microscope powerful enough to see individual particles within the atom, but how did he conclude that seeing the positions and motions of the particles was "in principle" impossible? This was two years before he derived the Uncertainty Principle, which in any case is predicated on his original derivation of quantum mechanics. So without any of that "knowledge," he just arbitrarily decided that some quantities which had not yet been observed were in principle unobservable? I don't understand that. Can you help me?
ANSWER:
Your question is strange. What does
"Heisenberg's frame of mind" even mean? Anyhow, that isn't physics, that's
history of physics or maybe psychology of physics. In 1925 Heisenberg
constructed the first mathematically rigorous theory of quantum mechanics;
it was based on matrix algebra. Prior to that there was only the
BohrSommerfeld semiclassical description of what I refer to as quantum
physics. Within a year an alternate theory, based of differential
equations, the root being the Schrödinger equation which is a wave
equation. The latter was generally more accessible to physicists and is the
favored theory to introduce to students for that reason. Still,
Heisenberg's quantum mechanics is very elegant but more difficult to
calculate with. But the two were subsequently shown to be 100% identical in
rigor and both describe the exact same physics. What is different is the
mathematics used by each.
The uncertainly principle does not say some
quantities are unobservable. It says that some pairs variables are
'conjugate variables' and therefore you cannot measure both to arbitray
good accuracy, you cannot know both of them exactly. It is actually built
right into quantum mechanics, so no 'arbitrary' decisions needed to be
made.
I think you should give priorty to learning quantum
mechanics rather than Heisenberg's state of mind.
QUESTION:
Is it possible for a person to jump/drop onto an impact force measuring device and have the initial impact force be less than body weight?
Reference/context: Science world had a kinesiology exhibit, and one of the displays was about impact force measurement.
The person gets weighed and then climbs up to a platform approximately 1824 inches high. The person then jumps/drops down onto an impact force measurement device, and a display shows impact force by body weight.
After a couple attempts I achieved an impact force reading of 1.2 times my body weight, and I feel I could have done better. Could that number have been below 1.0?
ANSWER:
I need to know what the device actually
measures. "Impact force" is not a quantity in physics. A force is a force,
but an impact is the product of a force times a time. For example, if you
have a mass m, hit the device with a speed v and come to
a stop in a time t, your average impulse over that interval is
I=Ft=mv. where F is the average force you feel during the
collision time. So you could write F=mv/t. I suspect that
F is what this device measures. Assuming m is some fixed number,
how could you control how big or small F is? If you increase v
or decrease t, F will be bigger; if you decrease v
or increase t, F will be smaller. An everyday example is
consistent with this: If you fall and hit your head on a cement sidewalk,
you might get badly hurt because your head stops in a very short time; if
instead you fall on a thick foam mat, you would probably not get hurt
because it takes a pretty long time to stop.
So let's see if there is a way to get an F
which is less than your weight. Being a scientist, I must use SI units, but
I will try to convert to Imperial units when it seems helpful. Mass is
measured in kilograms (kg), length in meters (m), and time in seconds (s).
So suppose that your mass is 100 kg, the height is 0.5 m, and the time is
t. ; acceleration due to gravity is about g=10 m/s^{2}
so your weight is W=1000 N (a newton is mass times g and
is 1 N=0.224 lb). So, your weight would be about 224 lb. since v/t is your
acceleration, I finally come up with F/W=a/g
where a is your average acceleration. So you just need to have the
rate at which you are slowing down less than the acceleration due to
gravity. If you drop from 0.5 m your speed when you hit the device is about
v=3.16 m/s. So if you can make v/t=10, F/W=1.0,
so t=0.316 s. If you make the time greater than about a third of a
second you will get below 1.0; you can do that by watching what
parachutists do when they land—go into a crouch as slowly as you can.
Of course, I do not know how the measuring device works so my numbers may
be off, but anything you can do to increase the time from when you hit and
when you stop will lessen the average force you feel.
QUESTION:
When Hiroshima and Nagasaki where bombed with nuclear weapons in World War 2 there was total devastation. However within a relatively short period of time people were able to move back in and rebuild the cities. However when there is a nuclear power plant meltdown like Chernoble or Fukushima the area around the power plant remains contaminated with radioactivity for thousands of years. Why is that?
ANSWER:
I
have previously answered
this question.
QUESTION:
I'm having trouble with Hawking radiation. It seems to me that if you separate a pair produced at the "event" horizon, with one half of the pair falling into the black hole and the other half escaping, you have added mass\energy to the black hole rather than subtracted from it. The problem for me here is the assumption that the mass\energy to produce the pair in the first place is somehow subtracted from the black hole?! To me this seems to presuppose that the mechanism for the pair production is well understood and the reservoir for the mass\energy for this pair production can communicate across the "event" horizon.
ANSWER:
I have answered this
question before.
QUESTION:
Red shift/blue shift—light traveling toward us (Andromeda Galaxy as an example of blue shift) has a shorter wavelength than stars/galaxies that are moving away from us. These light waves are part of the visible part of the electromagnetic spectrum (right?), but what about the parts that are invisible to the eye—ultraviolet, infrared, radio waves, etc? Do these waves change frequency in the above examples? (I know about Doppler radar and how it works, so is the same principle at play here?)
ANSWER:
All frequencies of electromagnetic waves are Doppler shifted.
QUESTION:
Black holes because of their mass and gravitational
'pull' suck things in and there is no escape (at least as I understand it), and this includes light. But light has no mass, so what is it about light that makes it susceptible to the gravity of a black hole?
ANSWER:
Our best theory of gravity is general relativity. In general relativity a
photon, the quantum of the electromagnetic field, has energy. (Of course it
does because the electromagnetic field itself has energy.) The energy E
the photon has is E=hf where h is Planck's constant and
f is the frequency of the field. If you throw a ball straight up
in the air with some initial speed v, it initially has energy E=½mv^{2},
but the higher it goes the smaller v becomes until it is zero; so
when it runs out of energy it stops and, like any mass, will fall back to
the ground. The photon, however, still loses its energy but in another way:
Because of the red shift of light moving away from a heavy object, the
photon's frequency will get smaller as it rises. For most objects, a star
for example, the photons will end up with a pretty small shift in
frequency. For a black hole, however, the photon loses all its energy
before it can escape, that is, it simply disappears. If you are wondering
where the energy of the photon went, it went to increasing the mass of the
black hole by an amount m, mc^{2}=hf or the mass
of the black hole has increased by m=hf/c^{2}.
Incidentally, There are two kinds of red/blue
shifts. One is the one you are probably familiar with, generally called
"Doppler shift" and caused by motion of the source or observer. The other,
as discussed in this answer, is referred to as gravitational red/blue
shift.
QUESTION:
I've just generally been thinking about this for a while, and it's difficult to get an answer from Google. I think if I understand things correctly, it's actually vital for our universe that the speed of light is constant, and this constant effectively creates the electromagnetic spectrum we observe. Light can't go faster. It can only increase in frequency as more energy is added. If there was no constant speed of light, it might be the case that all energy in the universe would evaporate instantly? Maybe? I don't know. I'm just trying to conceptualize what's going on.
But what would be the physical mechanism by which the speed of light is constant? It must be a property of space? Do physicists have a solid understanding of what causes this?
ANSWER:
Go to the faq page. You will find "Why
is the speed of light constant to all observers?"
QUESTION:
I am an 11th Grade Student, and my teacher currently teaching us newton's laws of motion, in the above image there is a pulley arrangement, with all surfaces smooth! my teacher was discussing about this in the last few minutes of school and i couldnt understand it. He said that the big 10kg block will move towards the left! what? why would it move towards left? it would need a force in the left direction right? but here, i dont see any force on the 10kg block
which is on left direction... [followed by 364 unnecessary rambling
words!]
ANSWER:
Once again I am faced with a question which
violates site ground rules requiring "...single, concise, wellfocused
questions." Oh well, I have just edited it down and, rather than doing a
complete solution, I shall only address the student's primary concern: when
released, what happens to the 10 kg block? I always told my students to
"choose a body" when starting a Newton's second law problem. I choose the
pulley which, although not stated explicitly, must be frictionless and
massless. The next step is to find all the forces on that body.
Clearly the string is pulling on the pulley to the left by T_{1}
and down by T_{2}; because this is a
massless pulley we can assume T_{1}=T_{2}=T.
But, wait, is there anything else touching the pulley which might
exert a force on it. Yes! This setup would not work if the pulley were not
attached to the 10 kg block. The block must therefore exert some force on
the pulley to hold it on the corner of the block, F
in my diagram. If the block exerts a force on the pulley, the pulley exerts
an equal and opposite force on the block, and that force has components to
the left and vertically down. One more thing: The questioner indicates
that, when the system is released and the two 5 kg masses start
accelerating, the tension is the string is 5g. This is wrong.
QUESTION:
My question is about the way steam and heat works when food is cooking. For instance, if you cook food too long in a steamer, the food generally gets mushy if cooked too long. However if you heat up food in an airtight container in the microwave, the food does not get mushy if cooked for too long. In fact it will burn and get crunchy instead, despite having steam and lots of moisture in the container. Is there a physicsrelated reason that the two steaming methods vary so drastically in outcome if overcooked?
ANSWER:
In a steamer there is a reservoir of water
continually adding stream to the inside of the pot. The cooking happens
from the outside in. If you cook too long it will be overcooked. The same
happens simply boiling the food. When you microwave you heat up the whole
thing uniformly. And any steam that is in there came from the food itself,
not nearly as much as in a steamer. But if you do that too long you will
boil the water out of the food and it will certainly become "crunchy".
QUESTION:
I need to move an 80 pound speaker 28.5 inches
above the floor using a dolly on a 7 foot long plank. How much force will
I need to apply? (A very long question heavily edited by me!)
ANSWER:
Before you push there are two forces on the speaker, its 80 lb weight (red
force in the figure) which points straight down, and the force which the
plank exerts on it to keep it from breaking through the plank and which is
perpendicular to the plank (the green force in the figure). There is also a
frictional force which is parallel to the plank and pointing up the plane;
because you have managed to put the speaker on wheels, this force is
negligibly small for our purposes. When doing inclined plane problems it is
useful to resolve all forces into components parallel and perpendicular to
the plane; these are shown by the dashed forces. I won't go into detail
about resolving forces here, but these two components are completely
identical to the 80 lb force and show us that there is a 27 lb force trying
to push the speaker down the plane and a 75 lb force trying break the
plank. The plank is strong enough so the plank doesn't break; but if
nothing (this is where you come in) intervenes, the 27 lb force will
accelerate the speaker down the plane. If you now push up the plane with
just a little more force than 27 lb, it will go up the plane, as you wanted
it to do, instead.
QUESTION:
You have 4 hot hardboiled eggs you want to cool down quickly. They are in a pot. You throw in a handful of ice and then add cold water. Do you just cover the eggs with the water or fill the pot? Which is faster?
ANSWER:
I want to start out by noting this is an
academic question. The way cooks normally cool hot foods quickly to halt the
cooking is to plunge the hot food into an ice bath, water to which enough
ice has been added to bring the temperature of the water down to 0°C
(ice water).
This can get to be a very complicated system. I
will simplify it to include only what is going on with the water, ice, and
eggs. In other words, imagine that these three are enclosed in an
insulating box through which heat cannot flow. This is probably a good
approximation because cooling the food happens quickly enough that the
interaction with the bowl and air in the room is minimal. What happens is:

ICE: There are two
ways ice can absorb heat: (a) if the temperature is less than 0°C, it
can warm up until it gets to 0°C, and (b) at 0°C it will absorb heat
but not increase in temperature, instead melt. Most home freezers have a temperature of
0°F=18°C, so initially, the way the ice will absorb the energy is
by warming up; when the ice reaches 0°C it will absorb energy by melting.

WATER: The question seems to ask, mainly, how
much water should we add. I will look at several amounts of water. I
will start the water at 15°C, a typical temperature for tap water.

EGGS: The mass of an egg is about 50 grams and
we have four of them. They have just come out of boiling water at
100°C.
I struggled with answering this question. The main problem is that the
speed at which the temperatures change depends sensitively on things like
how heat diffuses from the centers of the eggs into the water and the size
of the pieces of the "handful of ice"; crushed ice will absorb energy much
faster than a big block of ice. So I cannot answer the question, which
presumes that only the amount of water determines the speed to come to
equilibrium—there are many more factors which play. I think it likely
that it would take longer times for large amounts of water to come to
equilibrium, so I could glibly suggest that the answer is the smaller
amount of water is faster; but the problem has too many variables to know
for sure. What I have decided to do is demonstrate the physics of how the
final temperature T is achieved and work that out for one example:
I will find the minimum amount of ice we must add to 1 liter of water so
that the equilibrium of the ice/water/eggs system will be T=0°C.
Heat is just the flow of energy. If an object has a mass M and a
temperature T_{1} and an amount of heat Q is
added its temperature increases to T_{2}. Then the change
of temperature is proportional to Q, Q=MC(T_{2}T_{1}),
where C is a proportionality constant called the specific heat of
the material absorbing energy. If the object is instead having energy taken
from it, T_{2}T_{1}<0 so Q<0.
If a solid is melting it is absorbing energy but not getting warmer; in
that case one needs a constant called latent heat of fusion, L which is the
energy absorbed per kilogram of material melted, Q=ML. So, the
following is the computation.

ice: C_{i}=2090 J/(kg·°C),

water: C_{w}=4182
J/(kg·°C),

eggs: C_{e}=3180 J/(kg·°C),
and the latent heat of
fusion for:

ice: L=3.33x10^{5} J/kg.
I will calculate the amount of ice which is
needed for the final temperature to be T=0°C
starting with 1 kg (a liter) of water.

M_{w}=1 kg of water at T_{w}=15°C,

M_{e}=0.2 kg of eggs at T_{e}=100°C,
and

an unknown mass M_{i} of ice at
T_{i}=18°C.
So, the equation (which is simply energy
conservation) to solve is:
Note that this is a linear equation for M_{i}
as a function of M_{w}.; this is shown in the figure. So,
1 kg will about cover the eggs and about a third of a kilogram of ice
will totally melt when T reaches 0°C. If you start with more ice, there
will simply be ice still unmelted in the pot when equilibrium is reached.
QUESTION:
I am currently studying for my physics exam. A student in the group chat asked us all a question and it got us stunned. Can you help?
Please be aware that this is NOT a homework question, we all just share different opinions on the answer.
Question:
A container with very thick/sturdy walls is on a frictionless surface. Inside the container, on the left side, there is a cannon with cannonballs. The cannon fires all its cannonballs against the right wall, where they fall directly to the ground upon contact (they do not bounce back). What is the effect on the container?

a) The container stays in place

b) The container moves to the right

c) The container moves to the left

d) Not enough information
I personally think the answer is A, but a lot of people are saying C, and others also say D. I'm lost.
ANSWER:
See the figure. I assume that the box and
cannonball are rigidly attached and have a combined mass of M; the
cannonball has mass m. Before the cannonball is fired, the whole
system is at rest and so its linear momentum is zero. Afterwards the cannon
ball is moving (relative to the ground which is where the momentum was
previously zero). But the total momentum must still be zero (conservation
of momentum) because there are no horizontal external forces on the system.
So, mvMV=0 or V=mv/M; when the cannonball hits
the wall its horizontal linear momentum becomes zero. There are still no
horizontal external forces so the motion of the box must also stop. Clearly
the answer is c).
There is another way you could look the problem.
Since after the the cannon is fired the cannonball, previously at rest, now
is moving. So the cannon/box must have exerted a force F
on the cannonball. But Newton's third law says that the cannonball exerted
an equal and opposite force F on the cannon/box
etc.
QUESTION:
I am a high school senior who writing software for the avionics of our school's model rocket for the TARC rocketry competition. I have taken AP mechanics (Calculus based) and had a difficult time figuring out the projectile path of a rocket.
After burnout the rocket will coast to a target apogee (around 800 ft) at subsonic speeds. The rocket has onboard an accelerometer (plus gyro), a really accurate clock, and an altimeter (with barometer). I was wondering if there was a way to predict the exact apogee of the rocket based soley on the accelerometer readings and the clock (while knowing all the constants: mass, drag coefficient, etc.)?
My team and I have tried several ways to do this but we get stuck in dependency loops. Drag is based on velocity while velocity is based subsequently on drag. We tried using energy, momentum, even modified kinematics (for changing acceleration), none of which worked.
What we need is a single function that takes in current vertical acceleration and the constants and predicts a delta height which we can added to the current height to predict apogee. Also, since air pressure is part of drag and that changes based on height according to a function, it would be nice if we could put that into the predictive function.
I really don't want to make this question too long but if I could add one more layer of complexity. We are using air brakes to control the altitude to achieve a target apogee. We planned to run the above mentioned predictive function in a loop and reverse engineering it to create another function to find out how much more drag area we would need to achieve the target altitude since the airbrakes are radial flaps that just increase our cross sectional area.
ANSWER:
First of all, you have not given me nearly
enough information about what you are trying to do. Is the time the rocket
is burning part of your calculation? It is very different (and harder) than
the trajectory after burnout for several reasons. Is the trajectory
onedimensional, i.e. straight up and back down? Or does the
motion of the rocket also have a horizontal component? Does the
accelerometer give you the direction of the acceleration vector? And, if
not, it is not very useful. Do you know the mass of the rocket before and
after burnout?
All the above is necessary for me to give you any
guidance. But the way you are approaching the problem, common among
newcomers to physics, is all wrong. One of the most important things to
consider when solving a problem in the real world (not the ideal models of
ontheblackboard problems in a physics classroom) is to determine what is
important and what can be approximated without serious error. Except in
textbook problems, you should never expect to be able to get the
appropriate equation in just one step, you need to make sure you understand
things step by step. You should be starting out by imagining you are doing
a textbook problem—ignore drag and ignore the change of mass during
the burn. Then you might put in drag. Here is a little tip: A fairly good
approximation for air drag D is D≈¼Av^{2}.
Here A is the area presented to onrushing air and v is the speed;
the vector D must be
opposite the vector v. The catch is that you must use SI units here and
D will come out in Newtons. The estimate for D does not look
dimensionally correct; that is because lots of stuff is included in that
little factor ¼, like that the fluid is air, the density is for atmospheric
pressure at sea level, the shape is a generic sphere, etc.
Including D in your calculations will tell you if it is actually
very important for your problem.
That's probably all I should do at this point. If
you answer my questions about what you are trying to accomplish, maybe I
can give you some other instructions.
By the way, if you know the acceleration as a
function of time you should be able to determine the position and velocity
as functions of time if you know the initial position and velocity. It
would require numerical integrations because a will not be constant.
REPLY:

I am only simulating the rocket after burnout. As such, the problem is more like finding the trajectory of a cannonball midflight.

Like a cannonball, the projectile will have velocities in the horizontal and vertical direction, however only the vertical component is needed as I am trying to predict apogee.

The problem is that at an arbitrary time during the ascent of the cannonball, we will be given vertical acceleration (drag and gravity with direction), current altitude, and time (if that is needed though it might not be), and we must determine the apogee time and altitude.

Constants like mass (after burnout), cross sectional area, drag coefficient, etc. are known.

Drag is quadratic (v^2)
Without drag, the apogee is simply found with kinematics (constant acceleration), but since drag is a changing force that imparts a changing acceleration unlike the constant 9.81 of gravity, no one on the rocket team could figure it out.
Also, I have found other equations online that include drag and give altitude as a function of time, but the equation we need should just give out a number for max altitude based on the given values stated above. That's why we tried using energy formulas to solve the problem since at the top there is no kinetic energy, only potential (again only vertical components).
ANSWER:
There is a good reason why you are having so much trouble: You are high school students who, at most, have had introductory differential and integral calculus. As I will show, this problem
will require differential equations. And not just simple linear
differential equations, but coupled secondorder nonlinear differentials.
Such equations can only be solved numerically, almost never analytically. I
have given a little thought and have thought of a way that just might work
for you. It won't be easy. One thing before we get started: You said "only
the vertical component [of the velocity] is needed." This is
not correct because to get the behavior of the vertical component you must
also know the horizontal velocity.
One thing which I am not clear on is what you are
given. It is imperative that you know the initial velocity, both magnitude and direction, to
solve this problem. When you say "we will be given
vertical acceleration" I think you must mean "we will be given
acceleration" because the acceleration is not vertical. When you say "drag
and gravity with direction" I assume that the drag force, both vertical and
horizontal components are given. The drag force is usually written
D=Cvv,
so
D_{x}=C√[v_{x}^{2}+v_{y}^{2}]v_{x
}D_{y}=C√[v_{x}^{2}+v_{y}^{2}]v_{y}_{
} From these equations you can deduce that
v=√(D/C) and
v_{y}/v_{x}=D_{y}/D_{x}=tanθ
where θ=tan^{1}(D_{y}/D_{x}) is the angle v makes with the horizontal.
From the initial drag you can now get the two components of v,
v_{x}=√(D/C)cosθ and
v_{y}=√(D/C)sinθ.
So this is how you get the initial values (t=0) values of the components of
the velocity. I will choose that the initial values of x and y
are both zero, i.e. I choose the origin of my coordinate system to
be at the initial position of the rocket.
So now we are ready to do some physics. What's key
in classical mechanics? Newton's second law, F=ma.
m(d^{2}r/dt^{2})=Cvvmg1_{y}
where 1_{y}
is the unit vector pointing vertically upward, m is the mass,
r is the vector pointing from the origin to the
rocket at time t, and g is the acceleration due to
gravity. I realize that vector calculus is something you have not yet
studied, but it's not so bad because we just write it in component form:
m(d^{2}x/dt^{2})=Cvv_{x}=C√[v_{x}^{2}+v_{y}^{2}]v_{x }
m(d^{2}y/dt^{2})=Cvv_{y}mg=C√[v_{x}^{2}+v_{y}^{2}]v_{y}mg.
These are secondorder (second derivative), nonlinear (square root),
coupled (v_{x} and v_{y} both appear in
both equations) equations. But we can make them much more tractable by
noting that (d^{2}x/dt^{2})=(dv_{x}/dt)
and (d^{2}y/dt^{2})=(dv_{y}/dt).
Finally we have
(dv_{x}/dt)=β√[v_{x}^{2}+v_{y}^{2}]v_{x
}and
(dv_{y}/dt)=β√[v_{x}^{2}+v_{y}^{2}]v_{y}g
where β=C/m. Now we can't solve these analytically but we can
do it numerically (that's where you come in!). This will just involve
arithmetic.
Δv_{x}={β√[v_{x}^{2}+v_{y}^{2}]v_{x}}Δt
and
Δv_{y}={β√[v_{x}^{2}+v_{y}^{2}]v_{y}g}Δt.
So here is your computer programming problem. You know β, g,
v_{x}, and v_{y} at t=0, they
all are just numbers. Choose some Δt, say Δt=1 s
and then you can calculate the changes in the velocity components. Now you
know the (approximate) values of the component velocities at t=1 s. Now you
calculate with those and get the values of the velocities at t=2 s.
Continue until you find the time when v_{y}≈0; going
past this t v_{y} will become negative. You will probably
need to try several values of Δt until it is small enough that the
final answer doesn't change (you are, after all, simulating what you would
get with infinitesimal Δt). After you find the total time to
the top, you still haven't answered where it is. What you need to do is to
integrate into your program calculation and tabulation of Δx=v_{x}Δt and Δy=v_{y}Δt
at each interval, and calculate running sums. The final sum of Δy
is your answer.
I know that this is a lot for a high schooler to
digest. But I have tried to simplify as much as I could. As I warned you at
the beginning, when you have air drag to contend with, you have a much more
complicated problem than with constant forces. If you are able to pull it
off I would love to see your results.
ADDED THOUGHT:
To check that I have not made an algebra error
somewhere I have written a PYTHON code to do the calculations I suggested
above. I randomly chose reasonable boundary conditions, β=0.001, v_{x}(t=0)=100 m/s,
v_{y}(t=0)=200 m/s, g=9.8, and Δt=0.25 s.
Shown are the components of the velocity, the components of the position,
and the path of the rocket.
QUESTION:
We have read for years that the universe is composed of certain percentages of visible matter, dark matter, and dark energy.
I can understand that there are ways of reasonably estimating the amounts of visible and dark matter. They are described for us
in published articles. How does anybody estimate the amount of dark energy?
And what is it anyway? Is dark energy expected to have all the characteristics types that regular energy has  kinetic, thermal, etc.?
All of the articles that I have seen simply present a percentage associated with dark energy. There don't seem even to be some
typical characteristics of dark energy that would allow us in the public to form a concept of what it is. Falling back on the Classical
Physics model, dark energy has to be the ability to do work on dark matter, if that helps!
Regardless of what dark energy actually, is one still wonders: how is it estimated?
ANSWER:
Site ground rules clearly state that I do not do astronomy/astrophysics/cosmology. I can tell you though that it is not clearly understood what either dark matter or dark energy actually are.
FOLLOWUP QUESTION:
But as a last thought, they tell us that dark matter can be evidenced by its observed effect on real matter. And then they quote actual percentage estimates for the
amounts of dark matter and energy.
And yet dark energy, I think, has not even been detected. There must be a process used to arrive at these numbers. That's what I am really after. Can you direct me to anyone or any organization that I can communicate with to find out just how it is done? My thought is that, if I can understand the process, I can more clearly
Understand what "they" think these things are. Maybe not.
ANSWER:
Dark matter is thought by most astrophysicists to be some kind of particle which interacts only via gravity. It is very hard to "observe" something if you cannot interact
with it using electromagnetism. There are numerous anomalies in the
universe which seem to not be described by our current "best" theory of
gravity, general relativity. You may make a model of hardtodetect
particles with mass but which only interact with the rest of the universe
via gravity, but if that model works it does not mean that you have
"detected" dark matter, you have only fitted the data, not necessarily one
which tells you what dark matter is; at this level it is
empiricism. It may be that dark matter is not stuff, but an indication that
we don't understand gravity as well as we think we do.
Dark energy is a different kind of thing. You say
that it "…has not even been detected…"
That is not correct, it has been "detected" just as surely as dark matter
has. The evidence for it is that very distant galaxies are observed to be
not just moving away from us but accelerating. In other words you can
assume that there must be some universal repulsive force pushing the
universe apart. In my opinion, dark energy is in just as detected as dark
matter is.
There is an interesting back story about dark
energy. Circa 1918, when Einstein was putting the finishing
touches on his theory of general relativity, it had not yet been discovered
that the universe is actually expanding; it was assumed to be static. That
presents a big problem since gravity is a longrange attractive force which
should have been causing the universe to be collapsing. So he added a term
to general relativity to hold everything in place and it became called the
cosmoligical constant λ. Much later in his life, when it was discovered that
the universe is not static, he said that λ was his "biggest blunder".
In 1998 when acceleration of distant galaxies was observed, new life was given to
λ! If you are interested in delving into this more deeply,
check out this
link.
QUESTION:
I was wondering if a feather and a bowling ball actually pull towards the earth at the same rate? Why would the gravitational fields produced by the feather and the bowling ball not increase the rate of their individual attraction towards the earth?
ANSWER:
First we need to specify that this notion
requires that there be no forces (like air drag in the real world) other
than gravity acting, the earth must be assumed to be enormously more
massive than the feather and bowling ball (which is true), and the
gravitational force which the earth exerts on the objects is uniform (which
is true only close to the surface of the earth). Also, you seem to be a bit
muddled as to what is going on. The reason that the two masses accelerate
is that each feels a force due to the gravitational field of the earth, it
has nothing to do with the gravitational fields of the objects themselves.
You seem to be thinking that the bowling ball feels a bigger force and that
is true; the force anything feels (which is called the weight of that
thing) is proportional to (depends on) its mass. But the effectiveness of
that force, what acceleration (the rate at which its speed will change
while falling) it falls with is inversely proportional to its mass. So, if
the mass of the ball is 1000 times bigger than the feather's mass, the
force on it will be 1000 times bigger. But, the more massive an object is,
the more force required to give it a certain acceleration is. It just so
happens (because of Newton's second law that the force necessary is
proportional to the mass. Therefore although the ball has 1000 times the
mass, it takes 1000 times as much force to accelerate it as it takes to
accelerate the feather. The acceleration is g=9.8 m/s^{2}.
If you want a more mathematical discussion, see an
earlier answer.
QUESTION:
One of the issues YouTube physicists discuss regarding interstellar travel is that the high speeds we would achieve would render the spaceship more vulnerable to constant bombardment by particles, gases, and dust, requiring a thicker hull in the direction of travel. Assuming interstellar travel is possible with current or nearfuture tech (not a given), is it true that more particles and gases would impact in the opposite direction of travel? That would seem to assume that most particles are more or less at rest in the frame of reference we are coming from with Earth. Wouldn't they all be going at random velocities up to near the speed of light relative to us?
Within the solar system, are particles mostly at rest relative to Earth's frame of reference? Within the gravitationallybound Local Group?
ANSWER:
Unfortunately, the space between planets, stars galaxies are not true vacuums as we often think of them. So, inevitably, atoms and molecules will be present in
varying densities, sometimes just moving around in relative vacuum, sometimes congregated in clouds. And you are right, they will be moving in random directions with a distribution of speeds, but the
speeds are all very small compared to the speed of light, c. But a ship traveling to distant stars would have to have speeds which are not so tiny compared to c. And
to someone on that ship, all the particles will be moving mainly at speeds
close to the speed which the ship has. When you say "mostly at rest" you do
not define what that means. What is true is that their speeds are very slow
compared to c; if that's what you mean by "mostly at rest", then yes most
particles which you encounter will end up coming toward you with a speed
very close to the speed you are traveling.
FOLLOWUP QUESTION:
Thank you for answering my question!! I have refined it to clarify the point you brought up.
We are measuring the velocity of the spacecraft relative to Earth's frame of reference. Therefore, we are measuring the velocities of the particles it might encounter from the same frame of reference  Earth's. You can't compare any particular velocity to the speed of light without a frame of reference. You must have an observer with its own frame of reference to measure the speed.
Why should we assume that particles outside the Solar System or Local Group travel at very low speeds compared to the speed of light from our frame of reference? Shouldn't we see the same number of highspeed particles bombard the ship from all directions, regardless of its velocity relative to Earth?
Does the idea that the spacecraft will encounter more particles in the direction of relativistic travel rely on the fact that all particles in the Local Group are gravitationally bound together?
ANSWER:
My bad! I should have specified my reference
frames. Let's call the earth our "rest" frame. First some numbers:

Typical atom/molecule densities most places in
our galaxy is 10100 particles/cm^{3}

Typical atom/molecule densities in intergalactic
space is 1 particle/m^{3}

Typical temperatures of most places in our
galaxy is 10100 K

Typical RMS speed of molecules is slow, e.g.,
CO_{2} at 10 K is about 300 m/s

So if we say that probably particles with speed
10 v_{RMS} are the fastest with any appreciable
number, those have speeds 3000 m/s=10^{5}c.
Since the biggest speeds relative to the earth are
so tiny compared to the speed of light, they look like they might as well
all have been at rest compared with the motion of the ship. Therefore, on
the ship with some velocity U,
every atom or molecule will be seen moving
with a velocity very close to U.
Suppose that you are driving in a forest at 50 mph.
Every tree from your frame of reference is moving backwards at 50 mph, not
just trees in front of you.
QUESTION:
Since the discovery of the Higgs Boson, which is said to be what makes matter essentially makes things be physical. Then wouldn't it make sense that there is indeed a graviton? In where the particle is in a wave function until check and collapsed into a particle? Not unlike the double slit. Couldn't this be the case? And it is my opinion that this could fit nicely into an current theory.
ANSWER:
I don't understand hows Higgs or double slits have anything to do with whether "…there is indeed a graviton…".
I will assume that the main issue is you want to know is whether "…there is
indeed a graviton…". As you may know (see the next question) general
relativity is generally depicted as a geometric model, deformation of
spacetime by mass. So gravity is not a force, merely geometry? If that is
the case, why seek a quantum theory of gravity where the quanta (gravitons)
would be "messengers of the force" when there is no force, only geometry?
But a few years ago I came across a short
essay which made the whole thing clear to me. A theory does not have to
have only one framework of characterizing it. For example, when quantum
mechanics was in its infancy, two competing mathematical frameworks were
used, both of which seemed to come to the same result when calculating:
Matrix mechanics was formulated by Heisenberg, Born, and Jordan;
differential equation formulation of quantum mechanics was introduced by
Schrodinger. The two were eventually shown to be identical. Similarly
general relativity has the properties of a field theory and can be
interpreted as having forces and being quantized. Hence, there should be a
theory of quantum gravity. If you want to read more see my
earlier answer
and be sure to look at the
essay linked there.
QUESTION:
Hi. I've been pondering gravity lately. Why is it that  although Einstein's general relativity best explains what we call
"gravity" one still sees statements such as "all masses attract each other—you and your pet, the earth and the moon . . ." etc., etc. What magical
"attraction" are they talking about? Also, could you recommend a
good illustration of what actually happens when an object near the earth
"falls"?
ANSWER:
The theory of general relativity (GR) is quite complicated mathematically.
To understand a complicated physical theory requires that we have a
qualitative explanation of some effects predicted to visualize. GR is a
theory which uses advanced geometry formalism, a fourdimensional spacetime
(time being the fourth dimension). It therefore becomes a theory we think
of as a geometrical theory and envision massive objects to deform the
spacetime around them resulting in attraction. But, there's a problem—since
we live in a 3dimensional world we really can't visual a 4dimensional
world. So what is usually done is to imagine a 2dimensional plane which,
if we put a bowling ball in it, gets deformed. If there is a marble nearby
it will roll toward the bowling ball; I often refer to this as the
trampoline model and it is a cartoon not to be taken too seriously but to
provide an example of an object deforming some space around it. You can
also give the marble a little shove and it will orbit the bowling ball as
shown in the figure. The marble also deforms the space around it but since
it has so little mass compared to the bowling ball, it is not apparent in
the figure.
QUESTION:
Recently I've got interested in one question. Lets assume we have 2 balls with same mass in space on some distance from each other (no other objects around). At the initial moment of time they are staying still (velocities = 0). Then, due to gravitational attraction, they start moving toward each other. At some moment they collide and collision is elastic—so they move away from each other, slow down and then again start moving toward each other (then collide ets.)
The question is—will they bounce forever (because of conservation of energy) or they eventually, like, stick to each other?
ANSWER:
The problem with your question is that you ask for an answer in an ideal situation, perfectly elastic collisions and perfectly empty space. Under those conditions, the two balls would bounce
forever. The fact is, though, that
no macroscopic balls can collide perfectly elastically and no vacuum is
perfect. So in the universe we live in, the balls would lose a little
energy every time they collided and quickly stop, "…stick[ing]
to each other…". And even if you had perfectly elastic collisions,
actual "empty space" (intergalactic space) contains about one molecule per
cubic meter; it might take eons, but eventually the balls will collide with
those molecules giving some of their energy to them and eventually sticking
together. And don't forget, there are lots of photons zooming around which
could add or subtract energy when they hit the balls.
QUESTION:
A while back I was reading an article about the electromagnetic spectrum, the visible light spectrum specifically. And a question raced through my mind. If I understand correctly the colors we perceive are based on reflective properties of objects in our environment. Certain reflective properties reflect different wavelengths of light and we perceive them as colors. My question is in the absence of all light does anything truly have a color? And if so what do you think that looks like? An infrared photo is the closest that I can imagine.
ANSWER:
If a tree falls in the forest and there is nobody nearby to hear it, does it make any sound?
Your question is similar—both are about semantics, not physics. If
you say that sound is your perception of it, the tree makes no sound if
there is nobody to hear it; on the other hand if you say sound is the
presence of compression waves traveling through the air, there is sound.
Color is trickier, though, because the color you perceive depends on how it
is illuminated. We usually think of white light as a distribution of all
the wavelengths of sunlight. So just saying something is red is a grossly
undetermined description because something red in one illumination may look
pink in another. So physics avoids this confusion and simply uses the word
color in a qualitative sense, just like you might describe something as
warm to denote temperature. So if you want to know what "color" something
is you specify the distribution of wavelengths emitted from an object when
it is illuminated by a different distribution of wavelengths.
QUESTION:
I was trying to bounce my Walmart ball at the ground and have it bounce off the ground, hit the ceiling, bounce off the ground and then hit the ceiling again. I noticed
that no matter how hard I threw, it would never hit the ceiling twice. I assume it is impossible for a physics reason and would like to know specifically what. The ceiling is about 8 ft high, ball is about 42 inches around.
ANSWER:
The "physics reason" is that your arm is
incapable of giving the ball enough energy. Each time the ball has a
collision with ceiling or floor it loses a fraction f of its
energy, so if its incoming energy is K, its outgoing energy is
K(1f); as it travels from the floor to the ceiling it loses
energy V=mgh where m is the mass of the ball, g
is the acceleration due to gravity, and h is the distance the
center of the ball travels; as it travels from the ceiling to the floor it
gains energy V. What I did is to calculate the energy stepbystep
to find the minimum initial energy K_{0}
needed for the final kinetic energy to be zero, just as it reaches the
ceiling for the second time. Then I solved the equation I got for K_{0}.
The result was
K_{0}=V(F^{2}F+1)/F^{3}
where F=1f, the fraction of energy retained in a
collision.
The energy is just the kinetic energy,
K_{0}=½mv^{2}=mgh(F^{2}F+1)/F^{3}
so v=√[2gh(F^{2}F+1)/F^{3}].
So let's look at your case, the radius of the ball
is about 6" so h is about 7'=2.1 m, g is 9.8 m/s^{2}, and I will
guess that F is about 0.5 which means half of the energy is lost
in each collision (I dropped a similar ball and it bounced back about
halfway from where I dropped it). Therefore, v=15.7 m/s=35.2 mph. I doubt that you
could give it that velocity.
QUESTION:
I'm doing my level best to finish my first speculative fiction novel, and this is a question to which I'm unable to have anywhere close to 100% confidence in my answer. I'm hoping your much more vast breadth of knowledge might help give me a bit more certainty with it; even if you have to do some educated guessing  which is more than I'd be able to do (though I've shared my guess below). So here goes:
I would like to have as close to an accurate sense of how water would move on other worlds; specifically, were all atmospheric conditions identical to Earth's, how would water (and most other liquids then) move on the Moon ("The Expanse" showed liquids moving much more slowly on the Moon) and on Mars?
My guess is, if they were right  I assume Astronomer/producer Naren Shankar has an pretty solid sense of it  and atmospherics are essentially the same, the biggest factor is the strength of the gravity; more than the speed of rotation (but this is a factor I'm uncertain about). Stronger gravity will make liquids move more quickly, I believe, but if they were right on that show, then twice as fast as on the Moon  or if rotation speed is a bigger factor than I'm thinking, that's something I'll have to educate myself about.
ANSWER:
The speed v in deep water is v_{deep}=√(gL/2π) where
L is the wavelength of the wave and g is gravitational acceleration; 'deep' means
h≥½L. So the speed is proportional to the
square root of g and to the square root of L. For example, a water wave on
the moon would have a speed of about v_{moon}=√(1.6/9.8)v_{earth}=0.4v_{earth}.
On Mars, v_{Mars}=√(3.7/9.8)v_{earth}=0.61v_{earth}.
For shallow water, h≤0.04L, the speed is approximately
v_{shallow}=√(gh); here the speed is
proportional to root h rather than root L. There is even
an approximate expression for the intermediate depths, v_{inter}=v_{deep}tanh(kh)
where k=2π/L. This is probably more than you
need to know, but if you want to see the whole reference which I used, go
here.
QUESTION:
I understand that time slows down for objects moving at relativistic speeds. However, I don't understand why the slowdown in time directly affects things. Why are biological reactions slowed down (so I age more slowly). Or why are chemical or nuclear reactions slowed down? Even though they call time the
'fourth dimension', it seems to have a much more direct effect of things than space does. If space where increased or decreased by 50% in extent, I wouldn’t become 50% bigger or smaller. If physical objects were directly affected by the growth or reduction of space dimensions, how would you ever be able to measure that change?
ANSWER:
You misunderstand time dilation. If you are
moving with some large speed relative to someone else, your clock runs at
some rate which is totally the same as you would at any speed. You
do not see that clock running at a different rate because you were moving
relative to someone else. If your ship had no way to look outside, you
would not even know you were moving. What is different is the rate which
you clock runs is different than the clock which is at rest relative to you
runs. You should read my description of the
twin paradox.
QUESTION:
So I am having problem in understanding if the acceleration must always be constant when we think of Free Fall. For example, imagine if I were about 70,000km above Earth's surface and the only object in my vicinity is Earth, then I’d only feel gravitational force of earth. Suppose I am falling towards Earth, then by definition what I experience is free fall. But when I get closer to Earth, say when I am 1,000km above the surface, the acceleration I'd feel would not be the same as what I felt when I was about 70,000km above the surface.
Having said that, can I conclude that the only reason we experience constant acceleration when we experience free fall on Earth is because the height for the motion is significantly less than the radius of Earth and therefore, we can ignore the minute changes in gravitational acceleration?
ANSWER:
The gravitational force F on a mass
m at some distance r from the center of a spherically
symmetric sphere of radius R and mass M is F=MmG/r^{2
}where G is Newton's universal gravitational constant.
Because of Newton's second law, a=F/m, so a=MG/r^{2}.
If you set M and R to be the mass and radius of the earth
you will find that a is about 9.8 m/s^{2}. That is the
only r which has this as the value of a; in other words,
the acceleration is never uniform. But, as is often the case in
science, a good approximation can yield perfectly adequate approximate
results provided that you understand the limitations. It is clear from your
question that you already understand that very well. I will give you a
quantitative way of estimating how high h in altitude you can go
before the approximate acceleration differs from the groundlevel
acceleration:

First, a(r=R)=g=MG/R^{2},
a(r=R+h)=g'=MG/(R+h)^{2}

Next, write (R+h)^{2}=R^{2}[1+(h/R)^{2}]
so that we can write g'=g/[1+(h/R)^{2}]
or g/g'=1+(h/R)^{2}

Suppose we want there to be only a 1% error at
some h; what is h?

(gg')/g=g(1g/g')=0.01g=g(h/R)^{2
}so h=√(0.01R^{2})=0.1R
QUESTION:
If the Laws of Thermodynamics and the Conservation of Matter state thate neither energy nor mass can be created nor destroyed, why do theoretical physicists fixate on the big bang as the creation of the universe?
To me it seems logical that both energy and mass (being interchangeable) are eternal, and that the big bang that caused the present universe may just be the latest of the likely cycle of eternal big bangs. What do you think, is it logical?
ANSWER:
There is no such thing as "conservation of matter" because matter can be created or destroyed. The total amount of energy in the universe cannot be changed although the amount of mass may change. However, I see no reason
why the "eternal" amount of energy in our universe should preclude "fixating" on the big bang.
QUESTION:
But energy cannot exist without matter and matter cannot exist without energy.
As both would be stagnant in the absence of the other.
The laws state that in an isolated system neither mass nor energy can be created or destroyed. Mass is a body of matter of indefinite shape or size.
Therefore mass is matter and cannot be created or destroyed.
How then, do you validate your proposition that it can be?
ANSWER:
This is utter nonsense. "The laws" state that in
an isolated system the total energy must remain constant. Mass is just a
form of energy, E=mc^{2} . You are not alone in believing
that conservation of mass is a law of physics; I often get questions based on this assumption. So. where did this "law"
come from? Modern chemistry was developed about 400 years ago. Experiments
showed that if you measured all the mass which contributed to a chemical
reaction and then measured all the mass after the reaction, the two, within
the accuracy of measurement, would be equal. The catch is that the accuracy
of the measurement is too poor to say that the two are precisely equal.
When special relativity was developed, it became clear that mass could be
converted into energy because E=mc^{2}. In an exotheric
reaction, energy is released in the form of heat and the origin of that
energy is a tiny decrease in the mass. Indeed, even with the most sensitive
instruments it is virtually impossible to measure this; chemistry is a very
inefficient source of energy. But if you look at nuclear reactions, the
mass change is easy to measure; e.g., the mass of a helium nucleus
(two protons and two newtrons) is about 3% lighter than the summed mass of
two protons and two newtrons. Stars, like the sun, are using nuclear
reactions to release energy from mass. If the sun were burning using
chemistry, it would have been totally reduced to ashes millions of years
ago.
QUESTION:
I was wondering how the oceans can absorb CO2 from the atmosphere if they have achieved equilibrium.
ANSWER:
Lot's of reasons. First, the CO_{2} doesn't just swim around but iteracts chemically so that
the water dynamically reduces levels allowing more CO_{2} to replace that decrease.
If the amount of CO_{2} in the atmosphere is increasing which
causes the equilibrium to be undone whether the CO_{2 }is
decreasing or not. So the real equilibrium is dynamic, not static. By the
way, the chemistry causes the ocean to become more acidic, not desireable.
QUESTION:
I have come across this paper written by a person called John Mandlbaur. He claims:
"My papers are rejected because they come to a conclusion which the editor doesn't believe. That is what rejected without review means.
Numerous physicists have described my maths as perfect and nobody can defeat it."
Now I do not believe this because although I am no physicist, I know physics well enough that papers are not rejected without good reason so I am sure his is rejected because the calculation and/or reasoning is fatally flawed but being no physicist myself I am not qualified to comment on the paper. If however you identified the relevant flaws in straightforward English, I could post your reply on Quora and hopefully prevent others from being taken in by these claims of Mr Mandlbaur's.
If this is not for you, but you can recommend someone else to explain why the paper is refuted in plain English please let me know.
ANSWER:
I
interacted with this fellow about 7 years ago.
We had a very lengthy backandforth discussion of his "ideas" which
eventually became so outrageous that I relegated him to the
"OfftheWall" Hall of Fame". I
don't want
to call names, but there has never been a questioner more worthy of
offthewall status than him. It is very sad that he has held on to these
delusions for such a long time. I think it was Bill Murray who said "It’s hard to win an argument with a smart person, but it’s damn near impossible to win an argument with a stupid person."
I am sure that if you go over these ramblings you will have plenty of
examples to convince yourself that Mr
Mandlbaur is a crackpot.
QUESTION:
I am writing a science fiction story, and I'd like the science to be accurate. So: if a space ship starts from Earth and steadily accelerates at 2 G's, how far will it be from Earth
in 3 months? I'm writing a love story intertwined with the difficulties of
interstellar commerce.
ANSWER:
Before you do anything you should read the four
previous answers
which you can find in the faq page. If you don't care to get into the
details you can just go to
this link which will let you know how I did the calculations for you. You first need to understand that acceleration is not a very useful quantity in physics if velocities are comparable to light speed;
this is discussed in the previous answers I refered to above. Since you are
interested in high speeds, we need to first determine whether classical
Newtonian physics can be used or whether we need to use special relativity.
We first need to specify who is measuring the acceleration since it is not
the same as seen by everyone with high relative speeds; I assume that you
are interested in where the ship is and how fast it is going after three
months. The figures show calculations I have done assuming g=10
m/s^{2}, for both relativistit (red) and Newtonian (black)
mechanics, and an acceleration observed from the rest frame of 2g.
and essentially velocity as a function of time. The first figure calculates
up to about t=23 months, the second to 3 months. Note that the
first calculation has the relativistic calculation approaches c at
large t whereas the Newtonian calculation increases linearly
forever, already at about twice light speed at about one year. But, as
shown in the second figure, the two calculations are approximately equal at
3 months. Therefore at 3 months you could say that the speed of the ship is
about half of light speed and the distance traveled was about d=½at^{2}
=10x(7.78x10^{6})^{2}=6x10^{14} m=0.06 light years.
It would be much more complicated to calculate the distance
relativistically.
QUESTION:
My question is about spinning black holes. Or rather spinning singularities. Like singularities are one dimensional objects right? A point of infinite density.
Anyway. don't you need like...another dimension to be capable of spinning? Like what axis does the point spin around? How can you rotate when there is no geometry? I Thought it was a point. How does a point rotate? It wouldn't make sense though. The thing would suddenly stop spinning because there was no more geometry to rotate.
Anyway, everytime I try to think about it I sorta reach this block in my head.
ANSWER:
Not one dimension, zero dimensions. Still your questions are pertinent. The first thing you should accept is that we do not know what the physics of a black hole is. The laws of physics are
not known inside the Schwarzschild radius. But reasonable calculations which can be done seem to lead to a true mathematical singularity.
I feel that most physicists are uncomfortable with the literal idea of
infinities. That said, there is no reason why a point particle cannot have
angular momentum. When black holes initially form from the gravitational
collapse of a dying star and the star originally had angular momentum, that
property will be conserved. From that point of view, one would expect just
about every black hole to be "spinning". You are thinking classically, but
most elementary particles have intrinsic angular momentum (spin) but if you try to interpret it as the
physical spinning of some mass distribution about some axis you will fail.
Electrons are thought to be point particles and they certainly have spin.
QUESTION:
I am trying to figure out how quickly (approz) a tangential velocity (I think) something would have to have, rotating around the earth at an altitude of about 400kms, to have a centripetal force of 1g ie 9.82 newtons. I think that's the right terminology?
Here's my thinking:

I think escape velocity at that distance is (very) roughly 27,360kph?

So if you're going at just or just under that speed you're in constant freefall round the earth, ie experiencing zero gravity, in a vertically constant orbit.
(First hurdle: I don't even know if that's correct!)

FYI, I'm imagining a hypothetical band looping entirely around the earth at ~400kms altitude, for example, creating its own centripetal force.

If you stood on the inside of that band, head towards the Earth, and this is where my thinking gets v v doubtful,
would you then have to spin whatever the whole artefact (I think it comes out at something like a tangential velocity of 29,366kph to create a 1g centripetal force?) on top of the originally calculated escape velocity in order to create 1g's reactionary force?

So, is the end tangential speed the band would need to rotate approx 56,726kph?
ANSWER:
I am sorry to tell you that most of the things you say are nonsense.
Let me just tell you how things are rather than picking your question
apart. First, qualitatively: A circular orbit of radius r (r
being the distance from the center of the earth) will have a particular
velocity associated with it; you seem to think that you can choose the
velocity you would like to have in that orbit. The gravitational force on
any orbiting object gets smaller the farther away from the center of the
earth you are; as you know, the gravitational force on an object of mass
m at r=R_{E} is
F(R_{E})=9.82m
where R_{E}=6.37x10^{6}
m is the radius of the earth. So the acceleration is g(R_{E})=9.82
m/s^{2}.
But, in general g(r)=M_{E}G/r^{2 }where
G=6.67x10^{11} N·m/kg^{2} is Newton's universal gravitational constant
and M_{E}=6x10^{24} kg. So now you can see that every orbit is in free fall
except the acceleration of that fall is dependent on the radius of the
orbit and is only approximately equal to what you think of as g at
nearearth orbits. We can also calculate the speed at any orbital radius
r
from Newton's second law
F=mg(r)=mv^{2}/r=mM_{E}G/r^{2}.
So, solving, v(r)=√(M_{E}G/r).
You are interested in the altitude h=4x10^{5}
m so r=R_{E}+h=6.41x10^{6}
m. Doing the arithmetic, g(r)=9.74 m/s^{2} and v(r)=7.9x10^{3}
m/s=28,440 km/hr.
ADDED THOUGHT:
Rereading your question it appears that you might want to be making "artificial gravity" like the big donutshaped space ships which you see sometimes in scifi movies like
2001 A Space Odyssey. But I don't know why you would want to do it
around the earth where there is already gravity present. If you are
interested in rotating space ships to create the appearance of gravity
because of the centrifugal force, I have answered questions along those
lines in the past. Find links on the
faq page.
QUESTION:
I suppose this is an off the wall question, but it is quite practical, and I'm not sure where to ask it. A web search turns up nothing of interest.
I've convinced myself that my coffee cup is interfering with the 2.4GHz signal between my mouse and its receiver, which is plugged into the back of my display. The spatial orientation of the three objects is not ideal, but my desk setup is a bit limited, basically thus:

2.4GHz wireless mouse on pullout keyboard tray, mouse on the right.

My ceramic coffee mug on the surface of the desk, just above and beyond the mouse, also on the right.

My display is just above and beyond that on the desk, with the mouse receiver plugged into a USB port in the middle of the back of the display.
Ordinarily, these three things tend unconsciously toward an unfortunate syzygy*. I suspect the coffee cup is interfering with the signal, causing severe mouse lag.
So far unexceptional. I don't see why the ceramic mug or its mostly water contents might not be reflecting the signal.
But there doesn't seem to be any interference when the mug is full. Only after drinking half or more do I notice the lag. If I move the mug way over to the left, that seems to eliminate the lag.
So my question is, does the water level in the mug change the overall resonant frequency of the mug such that it can interfere with the signal at some levels but not others? Is it like filling a jug to different levels to get different tones when you blow across the opening? And considering that microwave ovens work on a similar frequency by heating water, would water be an especially problematic interloper at 2.4GHz?
ANSWER:
Interference and blockage of electromagnetic
waves are tricky to understand and track down. Like sometimes when I am driving and come to a stop there will be a specific spot where a local
radio station will suddenly almost be totally blocked. All the things you have tried are reasonable: I salute you because physics is, at its root, an experimental science. If you could tolerate moving your cup to the
left side, that is likely your best solution. The 2.4 GHz seems like a
likely culprit here, particularly since there seems to be negligible
blockage once the water is removed. Cups of different construction would be
useful to look at, e.g., a metal cup with and without water. I
doubt that it is anything to do with interference like the blown over
bottle because sound waves are of the same orderofmagnitude wavelength as
the resonating object whereas the GHz waves are nowhere close to the size
of the cup. Good luck!
*If you, are puzzled, as I was, by the funny word
syzygy, it means when three or more objects line up as in eclipses for
example.
QUESTION:
Sir,My doubt is in POTENTIAL ENERGY .I am really struggling in understanding this topic .please help me out.Sir my question is that "is potential energy just a term to make sums look easier?" Sir, taking an example  if we throw a ball with v velocity perpendicularly upward then at the highest point the velocity becomes zero therefore the kinetic energy becomes zero but at this point we say that the energy has been transferred into potential energy but acutually where did the energy go ? And thus can we say that we defined potential energy so that our famous rule does not gets violated i.e. "energy can neither be created or destroyed it can only be transferred from one form to another " is it so ??? What does it mean that energy is stored???(potential energy)?????
ANSWER:
My, you really are struggling. It is clear that you are just beginning your physics education, so I will keep my answer addressed only at that level. In fact, I will only discuss
problem involving a uniform gravitational field. First, suppose we know nothing about potential energy. So, the example you give is taking a ball of mass m and throwing it up vertically with some velocity v.
Then it starts with energy E_{1}=½mv^{2} and rises to some height h where all its energy is gone, E_{2}=0. So energy was not conserved; why not? Because some force must have done work on the mass to take its energy.
Of course, that force is gravity which is mg. Because the force is constant and pointing down and the displacement is positive is positive, the work is negative, W=mgh. The work done is equal to the
change in energy W=E_{2}E_{1}=mgh=0½mv^{2}. You could use this to find out how high the ball will go,
h=v^{2}/(2g); you have probably done such a problem
before you knew anything about potential energy.
I next want to generalize this problem. The ball is
thrown vertically from a height y_{1 }with a
velocity v_{1} and at a later time it is at different
height y_{2 }and now has a
velocity v_{2}. So now the work done is mg(y_{2}y_{1})
and the change in energy is
½mv_{2}^{2}½mv_{1}^{2}.
So, rearranging, I can now write (½mv_{2}^{2}+mgy_{2})(½mv_{1}^{2}+mgy_{1})=0.
If there are no forces other than gravity which are doing work on the ball,
the quantity ½mv^{2}+mgy never changes! So
now I get the brilliant idea (just one way of looking at this problem) to
call K=½mv^{2 }the energy due to motion
(kinetic energy) and V=mgh the energy due to position (potential
energy). So you can call gravitational potential energy a clever trick
which allows you to never worry about calculating the work done by gravity.
Another great reason to use potential energy is that energy is a scalar
quantity whereas force is a vector quantity.
There are some things you have to be careful of. If
you are using V=mgy for gravitational potential energy, you must
choose a coordinate system where +y is up; e.g., if the
top of a cliff is at y=0 m, the bottom of the cliff cannot be at
y=100 m. You cannot use potential energy for just any old force,
it has to be a conservative force. A conservative force is one which does
zero total work over a closed path (ending where you started). E.g.,
gravity does negative work on the ball and positive work on the way down.
If you push a book across a table you must do work against friction; but
now if you push it back it will be the same amount of work so you cannot
have a potential for friction. Your "famous rule" is not very useful. I
think you should think in terms of the energy equation, E_{final}=E_{initial}+W_{ext},
the energy you end up with is the energy you started with plus what you
added. W_{ext} is the work done by all external
forces, forces for which you have not introduced a potential energy; forces
represented by potential energy functions are not external, you have
internalized them.
QUESTION:
I saw an older question on your site regarding how fast Flash could run a quarter mile.
Using similar assumptions, in Science Fantasy  verse Warhammer 40000, there are Space Marines who are biologically enhanced super warriors with 1922 implanted extra organs that make them able to overwhelmingly outperform a regular human. Space Marines are roughly 78.5t tall (depending on if it is earlier variant with 19 organs or later one with 22 organs then the height is on the taller end of the spectrum) with a mass of 750 pounds unarmoured.
Space Marines have been decribed running 87 km/h taking six meter strides.
Seeing that much shorter and less massive Flash could run quarter mile in ten seconds giving a velocity of 144 km/h, using the formula, how fast would you put the maximum velocity a Space Marine could run given their mass of 750 pounds?
Can you give one estimate using the rubber boot sole, and one for ceramite boot sole that their armour is made of?
And when a Space Marine wears armour, his mass is increased to 5001000kg, so how does wearing armour affect the maximum velocity when affected by laws of friction, gravity, etc...?
ANSWER:
Oh my, you missed the whole point of that
caluclation. The mass does not matter because the frictional force he can
exert without slipping is proportional to his mass. And his acceleration is
inversely proportional to his mass; just the same reason why the
acceleration of a falling mass (in vacuum) is independent of mass. And
strength doesn't matter either if he can exert the maximum frictional force
without slipping, μ_{static}mg.
QUESTION:
How an ellipsoid object, say an American football...ball is thrown through the air. It shows projectile motion, but the other interesting observation is that the front of the ball points in the direction of travel. I have thought that air resistance possibly applies unequal torques on different sides of the ball perhaps, but I read somewhere that it has to do with gyroscopic precession  that this axis is different to the one for angular frequency, and somehow that explains it. I'm not sure though, any simpler way to understand this?
ANSWER:
If we were to pass the football with spiral in a
vacuum, the spin axis would remain pointing in the same direction because
of conservation of angular momentum; this is the case for the lower
trajectory shown in the figure. In air, though, the angular momentum mainly
remains pointing in the same direction as the velocity of the ball. I always am surprised to find an everyday phenomenon the physics of which is unexplained or explained only recently. This is such a case and was explained only recently (2020)
in a paper in
American Journal of Physics. A more accessible description of the
phenomenon can be found at
this link. Your hunch that the air has something to do with it was
spoton.
QUESTION:
so, tell me if I'm wrong but if an object is energy it doesn't have mass yet it is still affected by gravity, but you need mas to be affected by gravity 1. why is energy still affected by gravity 2. is there any way whatsoever for something without mass that isn't energy be affected by gravity.
ANSWER:
No, you do not "need mass to be affected by gravity".
The only object we know which has no mass is the photon (a quantum of
light). And because it moves, it has kinetic energy which is
wellknown; you shouldn't say that a photon is energy though,
rather it has energy. When a photon finds itself in a strong
gravitational field it does not go in a straight line (as you see things)
which you might have expected, but rather it follows the curved lines of
the space where the gravity is; in other words, in a curved space a curved
line may be the shortest distance between two points. Your second question
refers to something which has neither mass nor energy; that isn't
really something, is it?
QUESTION:
I hope this question isnt stupid, also English isnt my first language, but here I go: People always say that you can't travel faster than light speed, but what is the point from which you count the speed (0 m/s)? Because the earth is already moving through space, so would we need to accelerate less when accelerating in the same direction and more if we accelerate in the direction it comes from?
Follow up, may be explained by the answer to the first question: does the light emitting from let's say a LED travelling at 60mph reach a target 100 miles away faster than a staionary LED (assuming we turn both on the second they are next to each other, so when the moving one passes it)?
ANSWER:
One of the triumphs of the theory of relativity is that its success validates
the principle of relativity: The laws of physics are the same in all frames
of reference. From this principle we can deduce that the speed of light in
a vacuum is independent of the motions of the observer or the source. So
that answers your 'followup': a bystander would see the two light flashes
arrive at the same time. Your main question asks which frame of reference may a moving object not exceed the
speed of light? Since all observers see the same speed of light, the answer
is any frame.
QUESTION:
Dad I have a (possibly) physics question: how come tempered glass just spontaneously combusts without any apparent cause? I remember my old roommate's shower fully falling apart during the pandemic when no one was even there
ANSWER:
Hi Hannah! I don't think you mean "spontaneously combust" because that would mean burning/flaming. Glass does spontaneously shatter or break in unexpected circumstances,
however. I have answered
variations of this question several times ( e.g.,
1,
2,
3 );
1 is probably the most complete.
QUESTION:
I have been thinking about a question, like how we can write the change in unit vectors over a finite time interval, now it
may seem like a homework problem but trust me it's not it's a genuine question that I have been thinking about. I know about the change in unit vectors in an infinitesimally smalltime interval 'dt' and that it's given by the magnitude of the really small angle that the unit vector has turned about, but what about a change in the unit vector over a finite time interval how can we write that.
I tried to find the answer in many sources, but I was unable to find any.
ANSWER:
It is not really clear to me what you are asking. If you are talking about Cartesian coordinates, they never change. But if you have a coordinate system like spherical polar coordinates, the directions
of the unit vectors are very dependent on where a point happens to be. I will use a two dimnensional polar coordinate system where the vectors have both radial and tangential components
to keep the answer from being too complicated. The figure shows a red dot
moving along some path (red dotted line) which shows the path taken by the
particle between two times t_{1} and t_{2}.
The unit vectors are shown in green. If you know those two locations you can see that the unit vectors all
rotate through an angle θ_{2}θ_{1} in the clockwise direction.
The path taken by the particle will be described by some vector function
f which has components f_{r} and f_{θ}
as shown in the figure. So the average rate of change of direction of unit
vectors is ω_{avg}=Δθ/Δt=(θ_{2}θ_{1})/(t_{2}t_{1}).
QUESTION:
A ship at the speed of light turns on headlights, understand that a person on that ship sees light go out at the speed of light, relativity. What does a person "see" from outside that area of relativity? Can they see that, is that currently possible, what if, if not current? Do they see a ship with no lights, light but no ship, a spectrum of both that equals the speed of light? I am, and with more questions, being as specific as possible: I find it hard to articulate a concept from the written word, please accept and understand my ignorance/lack of understanding!
ANSWER:
First of all, no ship can travel "…at the speed of light…" and I do not answer questions which stipulate at
or faster than the speed of light. So I will answer your question assunming
v=0.99c, 99% the speed of light. As you say, an observer
on the ship sees the light in front of him having a speed of c
regardless of how fast the ship is moving. An observer relative to whom the
ship is moving with speed 0.99c also would measure the speed of
the light to be c. This is one of the postulates of the theory of
special relativity: The speed of light in a vacuum is independent of the
motion of the source or the observer. However he would not
"see"
the same "color" of the light because of the doppler shift. Suppose that
the light as seen by the ship had a wavelength of 450 nm=4.5x10^{7}
m, blue; for 0.99c speed of the source the doppler shifted wavelength would
be about 4.5 nm, lowenergy xrays. Hence the stationary observer would see
nothing but if he could measure the speed of the xrays coming out the
front of the ship he would see them as having a speed of c.
QUESTION:
Lenses/Mirrors form real / virtual imgs. Let's consider real images.
My query is how can mere interaction of light waves create the exact same picture of the source from which they originated? Does this mean that light carries some information in itself and when it interacts / superimposes, it creates the beautiful picture called an "image". To elaborate, light itself is not coloured, but why can we see coloured images (when the light is reflected from the object) ?
ANSWER:
Imagine an object to the left of a spherical lens
as seen in the figure. There is an imaginary line drawn perpendicular to
the lens and through its center call ed the optic axis. There is some
object consisting of many colors. I have chosen to look at just three
points in this object, one red, one blue, and one green. Each rays from these
points illuminates exactly one point where the image is formed; although I
have only drawn three rays for each point, every ray passes through the the
same single point on the image. Apart from size and whether the image is
inverted or not, the image is an exact copy of the object. Your statement
that "…light itself is not coloured…" is not correctthe
color of light is determined by its wave length.
QUESTION:
Assuming no engineering challenges like parasitic losses in materials etc…. Is it theoretically possible to take:
Speaker A. Play a tone. 440hz. Digitally. Using EXACTLY 100 watts of power.
Speaker B. Play a tone. 440hz. Digitally. Using EXACTLY 100 watts of power.
Have speakers A and B facing each other head on. EXACTLY 180 degrees.
The 2 speakers are playing the same note. Same signal, timbre, frequency amplitude pitch etc….
Would these two “sounds” cancel out perfectly such that “in between” there would be zero “sound”. I understand that small changes in angles and frequencies will have enormous effects on there and how the sound will dissipate if this system is not perfectly balanced from the get go. Very loud. But if the experiment were set up perfectly, is it theoretically possible to zero out the wave? Full 100% destructive interference? Chaos? Feigenbaum?
ANSWER:
No, the result is a standing wave of the same frequency.
In the animation the original waves from the left (blue) and right (green)
combine to make the black waves which have the same frequency and
wavelength but zero velocity. This is due to the superposition principle
which states that if two or more waves are moving in the same medium, the
net displacement is equal to the sum of the individual displacements at
each time and location. Note that the black curve is sometimes totally
zero, when the two traveling waves are out of phase with other. If two
identical waves are traveling in the same direction they could completely
cancel if they were 180° out of phase. In the second (nonanimated) gif,
imagine the green and blue are moving in the same direction. The black
would be the resultant wave.
QUESTION:
I am trying to understand heat, that is, mostly the idea of hot
and hotter. My understanding is that heat is basically motion, and
heating something up (cooking, let's say) is about transferring motion
from one place to another. So that the changes that take place in
cooking are really caused by the molecules of stuff rubbing against
each other. So hotter really means MORE motion, because the actual
temperature at which things burn, like natural gas, can't be
increased just because you burn more of it.
Does that make some sort of sense to you? Clearly my mind wanders
in the kitchen, and this has been nagging at me for a long time.
ANSWER:
I believe that your question is answered by an
earlier answer.
You refer to motion; in my earlier answer the kinetic energy referred to is
K=½mv^{2} where m is the mass of
some atom or molecule in your flame or pot or food, and v is its
speed (motion).
QUESTION:
I have a question about fluid velocity/acceleration and its effects on tubular friction.
Is the relationship linear? Meaning , if the fluid velocity is being increased, is there a moment or a few moments where the friction increases more than what it would’ve been if the relationship was linear?
ANSWER:
From what I have been able to learn while researching your question is that "friction" as in two solids sliding on their surfaces is not applicable
to fluid inside a pipe. The fluid exactly on the surface is not sliding but
rather adheres to the surface and is at rest. The motion of remainder of
the fluid has some velocity just outside the surface increases to a maximum
at the center but certainly not linearly. Everything about this velocity
profile is due to the viscosity of the fluid, sort of the fluid version of
friction. More detail can be seen in
this link and links which it contains. The figure shows a figure from
that link and shows the shape of the velocity profile, v(r).
Eventually, at high enough speeds the fluid becomes turbulent and "all bets
are off".
QUESTION:
How should I get to calculating the force that needs to be applied to a piston to compress a container of gas? Suppose the temperature is constant, and by default, as the volume decreases the pressure increases. Let's suppose the volume gets halved.
ANSWER:
The force depends on the area A of the piston. And, if the gas cannot be well approximated as an ideal gas,
you would need detailed properties about the gas. For an ideal gas, PV/T
must remain constant and if T is constant, PV is
constant. Now, it seems easy, like if you halve the the volume you need to
double the pressure; that would give you the answer for the force to hold
the piston there once you got it there, F=P_{2}A=2P_{1}A.
Be sure that you understand that this is what you mean because if you
simply started at the beginning with this force and continued pushing with
that force, when you got to the halfvolume position you would go right by
it and it would start slowing down until finally it would turn around and
start coming back.
QUESTION:
For reference I'm 31.
Am I closer to the age 60 than I am 20?
I think I'm closer to the age 60 because time is moving forward and we cannot move backwards to my age at 20. So wouldn't I be closer to 60 because time can only move in a linear motion towards 60 and we are constantly distancing ourselves from the age of 20?
ANSWER:
You are 11 years older than 20 and 29 years younger than 60. You are closer to 20 by almost a factor of 3.
If you had asked which you would be closer to being, you would be closer to
being 60 than 20; but you didn't.
QUESTION:
Is a mirror VERY different to how people perceive me in reality?
I am bothered by the fact that I never realised before but I guess there's a reason…
How impactful is the reversal effect?
I always thought that's how I look to others.
ANSWER:
The image you see is not how others see you, but a "mirror image". If you look at the image of yourself in the mirror, the image of your right hand is the left hand of the mirror you. If you want
to see an image of yourself which is what others see, you need to use a corner mirror.
However, since most of us are pretty much leftright symmetrical, there is
seldom much difference.
QUESTION:
My eight year old has a question.
Get ready.
If two particles are in entangled, and one is on the earth and the other has passed the event horizon of a black hole.
Would the entanglement be broken or even if it's unmeasurable still be entangled.
Does the medium "the ether" transcend the event horizon?
Silas Massicotte (8) and Malachi Massicotte (6).
ANSWER:
Seriously? Your 8year old understands quantum mechanics and cosmology?! That would be pretty amazing. I usually don't do, as stated on the site, astronomy/astrophysics/cosmology,
but I did a little research and think I can give you an answer. When one of the two entangled particles is totally destroyed, the other one will not know at all because there is no communication between the two.
Since the wave function of the destroyed one simply disappeared, the wave function of the other will remain in the mixed state it was in originally, but it will no longer be entangled with anything.
If a measurement is made on the survivor, there will be no change in
whatever remains of the other particle. There is a pretty good explanation
at this
link. There is no such thing as the ether.
QUESTION:
how can electromagnetic waves travel in vacuum?
how can something without any physical body and dimensions travel in nothing?
it should need something to mediate its propagation
like sound waves require air molecules.
if its pure energy then what is pure energy?
ANSWER:
This is exactly the way physicists were thinking
at the end of the 19th century. Serious searches for the medium required
for light to be able to propogate were being made. The best known search
was the MichaelsonMorley experiment which failed to find it. It was called
the luminferous æther. The punch line is that it doesn't exist at
all. Light can travel through empty space. One way to understand this is
through electromagnetic theory, quite well understood at the begining of
the 20th century. James Clerk Maxwell unified all that was known about
electromagnetism into four equations, today known as
Maxwell's equations.
One of those equations says that timevarying magnetic fields cause
electric fields; another says that timevarying electric fields cause
magnetic fields. Using all four you can show that one possible solution is
a wave equation for both electric and magnetic fields. Since waves are
timevarying, these waves will "keep each other going" which has nothing to
do with any medium. If you go to the link above, you can follow other links
telling you more.
QUESTION:
I have a question about a calculation in the USPSA (United States Practical Shooting Association) rulebook. The power factor of ammunition is measured by the weight of the bullet in grains times the velocity in feet per second, divided by 1000. A 165 grain bullet at 1000 fps would be 165 power factor.
This is intended to keep recoil the same between competitors. Felt recoil is subject to a lot of variables that have more to do with the firearm than anything else. But the rule is to ensure that everyone has a minimum level of recoil to manage.
Does a 110 grain bullet at 1500 fps, or 165 power factor exert the same force backwards as a 250 grain bullet at 660 fps, or 165 power factor? All else being equal, is power factor an accurate measure of recoil? The muzzle energy is much higher for the lighter bullet.
ANSWER:
The quantity power factor is nothing more than linear momentum
(mass times velocity) in physics but in unusual units, milligrain feet per second:
mgr·ft/s); physicists usually express it in
kg·m/s. But the units you use don't affect the physics so I will use
yours. The thing about linear momentum is that if there are no external
forces on a system, it never changes its magnitude; this is called
conservation of linear momentum. So, if the rifle has a mass of M=3 lb=21,000 gr=21,000,000
mgr and the
two bullets both have equal power factors of 165 you can calculate the
speed V of the rifle after the gun is fired: Before the gun is
fired momentum is zero, so it must be zero afterwards: 0=16521,000·V
or V=0.00786 ft/s=0.094 in/s. Now, you asked for the force on the
shooter. There is not enough information to calculate this because that
depends on how quickly the rifle takes to stop—if you have a bony
shoulder it would hurt a lot less if you stopped it more slowly by having a
squishy pad between you and the rifle butt. Suppose it takes 1 ms to stop;
then the average force on you over that time would be [3 lb x 0.00786 ft/s
/ 0.001 s] = 24 lb. But you said "all else being equal" so I presume that
the recoil force would be the same for both scenarios because the masses of
the guns and the recoil velocities are the same.
QUESTION:
When text book gets to quanta, it would say E=hf. It would say f = n * fundamental frequency. Ok, I can understand there are 1Hz and 2Hz and so on. But couldn't one build a circuit that makes 1.5Hz? At the end, it is a voltage completing a cycle within some time.
And if f can be decimal, then it is continuous. So E is continuous.
I must be missing some concept here.
ANSWER:
Not everything is quantized, but
even if everything were in some way quantized, you could only detect with
your senses in some situations. Usually we think of quantum physics as what
happens in very small systems, systems on the general order of atomic
sizes. The thing which is bothering you, I think, is that something which
is macroscopic which shows no signs of being quantized means that an a
microscopic version of that system would not be quantized either. Take, for
instance, a mass m hanging on a spring. So if you pull it down to some
amplitude A, say 1 cm and let it go, it will oscillate up and down
with some frequency f determined by the stiffenss of the spring
k and its energy will be E=½kA^{2}
and its frequency will be f=2π√(k/m).
You could quadruple the energy by doubling the amplitude. But, of course,
you would be able choose any amplitude you liked? Like 1.2379 cm? Of
course, assuming you could actually measure that accurately. So quantum
mechanics is nonsense, right? No, I could contend that the macroscopic
oscillator is also quantized (theoretically at least). What is the spacing
of the energy levels of this quantum system? hf, of course. what
is hf if f=1 cycle/second? ΔE=6.6x10^{34}
J! Suppose that the spring in this oscillator had a spring constant k=1
N/m; then, since E=½kA^{2}, going to the
next allowed level would require a change of amplitude of ΔA=√[2x6.6x10^{34}/1]=3.6x10^{17}
m. That is about 1/10 the size of an atomic nucleus! I don't think you
would notice this constraint!
Your intuition tells you that quantum mechanics can't be
right because you have never noticed its effects. What is your intuition
based on? Your experience! Do you have any experience regarding such tiny
sized things?
QUESTION:
I am working on a piece of sciencefiction centred around the midcareer physicist, upon whose desk objects appear for significantly timed short periods, a significant pattern. At some point in the peace, the working assumption becomes That these objects are being transmitted from
"Parallel universe"/other reality in the Multiverse.
This may seem to be an odd question if such objects were assumed to be capable of being transmitted between parallel universes or other reality planes in a Multiverse, do you think they would be capable of being photographed and or scanned in some other forms such as xray?
ANSWER:
Of course, this is all
speculative, but I would say that if your eye can see them a camera could
see them because the eye is, essentially a camera. But you could also say
that maybe the eye is not really seeing them but rather only your brain was
seeing them, like a halucination. I remind you that I do not usually do
cosmology, the realm in which multiverse is included.
QUESTION:
I have a theory (or just an idea) concerning faster than light travel that I would like you to answer. My idea is that if FTL travel is possible, then the energy required to make a warp drive, wormhole, etc. work is exactly the same amount of energy as would be required based on Newtonian physics if there were no law of relativity.
For example, suppose you have a 1000 metric ton FTL spaceship that can travel four light years in four days. Please calculate how much energy it would take to accomplish this if only Newtonian physics were involved.
Then look up some FTL scheme. Here's one you can use https://newatlas.com/physics/ftlwarpdrivenonegativeenergy/ Or just pick another than you prefer. Then calculate how much energy it would take to move the 1000 metric ton spaceship four light years in four days. My idea is that perhaps the universe is designed in such a way that the energy required is the same. In other words even if FTL travel is possible, you don't save any energy by using it. In other words, there is no “free lunch.”
I donated $20 for your answer, since I suspect this will take a lot of work.
REPLY: (Other
readers note: I am only answering this question because the questioner has
sent a donation.)
It would been good if you had read the site ground rules before you had sent me a donation since I have no way to refund it. No faster than light questions is one of the ground rules. You can also be sure that anybody designing a "warp drive" is liable to be a crackpot and the amount of work I would have to do to evaluate any such design would be unreasonably large. I am willing to compute the energy using Newtonian mechanics
and also calulate the energy to do your fourday trip relativistically. I will assume negiligable time to get up to speed.
ANSWER:
One light year is
365 light days. Using Newtonian mechanics, the speed you would have to go
4 ly in 1 day is 365 times the speed of light, 365x3x10^{8}=1.1x10^{11}
m/s. The energy which a your spaceship would have at that speed is
E=½mv^{2}=½x10^{6}x(1.1x10^{11})^{2}=6.2x10^{27}
J=1.7x10^{21} kW·h.
Relativistically, the distance 1 light year (ly) would have to
be lengthcontracted to 1 light day (ld) would be 1=365√(1(v^{2}/c^{2}))
or v=0.999992494c or β^{2}=0.999984988.
Now, total energy is E=γmc^{2} where γ=1/√(1β^{2})=258.
But we don't want the total energy, we want the kinetic which is
K=Emc^{2}=(γ1)mc^{2}=257x1.x10^{6}x(3x10^{8})^{2}=2.31x10^{25}
J=6.42x10^{18} kW·h.
So you see, contrary to your idea, it takes less
energy to go relativistically than classically.
FOLLOWUP QUESTION:
I read of some warp drive scheme in which the author pointed out you'd have to convert the mass of Jupiter into pure energy via E=MC^2 in order to make the warp drive work.
Would you calculate this energy and compare it to the amount needed in my previous question regarding sending a 1000 metric ton ship four light years in four days? I would also like to know whether you think humans on the ship could survive the g forces of accelerating and decelerating on such a trip, once again assuming just Newtonian physics applied.
ANSWER:
M_{J}c^{2}=1.9x10^{27}x3x10^{16}=5.7x10^{43}
J, many orders of magnitude larger than previous calculations. Humans can
survive accelerations of about 10g and not for very long. You
would have to plan your trip carefully to avoid accelerations much greater
than g.
QUESTION:
I am a new and upcoming math and
physics teacher at a local high school. A more senior physics teacher
posed this question to me before he retired, and I am rather
embarrassed that I cannot solve it. Apparently, he said this system is
accelerating and that acceleration varies with the angle.
[COMMENT:
The questioner shared a copy of a textbook problem
with m=50 kg, M=100 kg, θ=13°, μ_{s}=0.41, and
μ_{k}=0.35. I chose to do it in general and we can
stick those numbers in later.]
ANSWER:
I would first choose the knot (massless) as the body; this is a good idea because we don't know whether the system is accelerating or not yet, but it doesn't matter because all the forces
on the knot must equal zero because it is massless. Because of the symmetry
of the system, the tensions in the strings attached to m and m must have
the same magnitudes, t_{1}=t_{2}=t,
so we have
ΣF_{y}=0=T+2tsinθ.
The xequation just tells us what we already know, that the xcomponents
cancel each other out. We now have one equation with two unknowns, T
and t, so we need to choose another body. I choose m on the left.
ΣF_{y}=0=Nmgtsinθ
ΣF_{x}=ma=tcosθf
f=μ_{k}N if m is moving or
f=tcosθ if tcosθ<μ_{s}N
Now we have five unknowns, T, t, N, a, and
f but only 4 equations. The other m is not going
to give
us any new information because of the symmetry, so we are left with only
M. This gives another equation,
ΣF_{y}=MA=MgT.
Note that I have chosen +y to be down so
that m and M are moving in their positive directions. The sixth equation will be the relationship between a
and A,
a=A·tanθ.
I also calculated that the net force on
m goes to zero
at 55° so the static coeffecient of friction would kick in and the
system would stop.
t=Mg/(2sinθ)
f=μ_{k}(mg+Mg/2)
tcosθf=0
and therefore,
θ=tan^{1}(500/350)=55°.
I leave it to the reader to do the algebra above.
When I did the calculation
using WolframAlpha it gave
me confidence that the equations of motion are all correct since A=0
at 55°.
So I now had 4 equations with 4 unknowns (I
incorporated the
aA and the fN relations to reduce the number of equations and unknowns to 4.) I put
in the value of θ=13° and the solution is shown below,
calculated in WolframAlpha. I should note that I checked that the
system is actually accelerating and not stuck by static friction at θ=13°.
The results of calculations over the whole range 055 are shown in the
graph below.
ADDED THOUGHTS:
Just a few comments. I have to admit that I have not rigorously shown that
a=Atanθ although I did enough qualitative thought to the relation that I feel pretty confident. Also, the behaviors of all
the variables are just what I expect them to be using that relation. It is
interesting that at θ=0, because when we usually think if a weight hanging on a horizontal string,
that the string would have to have infinite tension to not deviate from the
horizontal; that situation, though, requires the (massless) string to be tied between two immovable walls. The equations here have a kinetic frictional force acting on the "walls" so they are not immovable.
Since the string at 0° has no vertical component, T must be 0
N,
as the calculation shows; M must therefore be in free fall for the
instant that θ=0°. Since the frictional force is 0.35x490=171.5
N, the solution finds that t=171.5 N also; I guess technically
t could be anything and still have a=0 because of the
tangent factor, but small angle nonzero θ will have
whatever t is required. At θ=55° everything has
stopped and must be in equilibrium: both N and T are at
980 N and t=980x0.35/cos(55)=598 N. Because all these accurately
describe what we know they should actually be, I am pretty confident that
a=Atanθ is correct.
QUESTION:
My understanding is that atomic clocks circling the earth run
1) slower because they are moving fast
but
2) faster because they 'feel' less of earth's gravity
than clocks on the surface of the earth.
Which wins out, speed or gravity?
How much is the time dilation/contraction due to gravity and how much due to speed?
Could there be a planet of some mass that you could put a satellite in orbit around at a certain speed so that the two would balance out so that the orbiting clock would measure time at the SAME rate as one on the planet's surface?
And, how much math would I need to figure out the answer to that last question?
ANSWER:
I started trying to use the expressions for the two and doing the algebra
demanding that their magnitudes be equal. But then I stumbled on this
figure which addresses exactly what you were asking. The red curve is the
kinematic time dilation (slowdown) for circular orbits. The green curve is
the speedup due to gravitational time dilation. Also shown, in blue, is the
sum of the two, the function which you wanted to be zero somewhere. Lo and
behold! There is a spot, looks like it is about one third of an earth
radius above the surface. Note that where GPS satellites are, the
gravitational time dilation is much more important. If your GPS didn't have
general relativity corrections built into it, it would not work. The math
would not be very demanding, but the physics might be. You need to have an
analytic expression for the blue curve and set it equal to zero and solve
for r.
I suggest that you look at another Wikipedia article
on time dilation.
Near the end of the article the gravitational time dilation is discussed.
The location of the zero dilation is at 1.497 earth radii, about half an
earth radius altitude.
Everything below in this
answer is incorrect (except for the red curve). I was misled by an AI error, several incorrect
internet posts, undue confidence in myself to be able to do the GR
calculation! It is an excellent example of how incorrect an internet post
can be and how AI probably learns from its memorizing everything on the
web. I did more research because it seemed extremely odd that the
geosynchronous orbit and the zero dilation orbit would be the same and the
fact that my incorrect answer was about 10% smaller than the actual
geosynchronous orbit.
I am leaving it posted as an example of how wrong
the internet can be!
I really liked this question because it was one of
those questions which gave me the opportunity to learn something new. I
wanted to understand everything which went on in making the graph above.
Well, I worked on it for a number of hours and finally got where I
understood what should be the functions plotted. I got this graph from the
Wikepedia article on gravitational time dilation; I would recommend it
because it gives the form of the gravitational dilation (I leave it to the
reader to understand the notation here. t_{0} is at the
surface of the earth and t_{f }is at some distance r
from the center of the earth):
t_{0}=t_{f}√[1(2GM/(rc^{2}))]
What needs to be plotted is (t_{0}t_{f})/t_{f}
which must be zero on the surface. So I conclude that
(t_{0}t_{f})/t_{f}=√[1(2GM/(rc^{2}))]√[1(2GM/(Rc^{2}))]
The kinematical time dilation turned out to be
easier for a circular orbit of radius r. Without any details except to note
that GM/r=v^{2}
(t_{0}t_{f})/t_{f}=1/√[1(GM/(rc^{2}))]+1,
ensuring that an orbit at r=∞ have zero speed.
So, here's my calculation. The kinetic calculation
agrees very well with the Wikepedia graph but the gravitational one does
not. The result is that the zero net dilation is at about 6 earth radii,
but the Wikepedia calculation was at about 1.3 earth radii, a really big
discrepancy. I stared and stared and couldn't find an error. So I decided that
I would ask AI about this; here is my interaction:
Six earth radii is approximately where the
geosynchronous orbit, where all the communications satellites are. It would
appear that my calculations are right. I did check by googling and the
usual answer was the same as ChatGPT's.
I think AI got it backward, though: kinematic
dilation is slower, gravitational dilation is faster.
QUESTION:
Picture an attic that covers a house and its driveway. The area of the driveway and the house are the same, so they are each covered by 50% of the attic. In the center of both the house and the driveway are 2 identical circular holes into the attic above.
The temperature in the house is 68°F, attic is 86°F and driveway 104°F. Since warmer air flows upwards, will the driveway air get into the attic, or will the colder air in the house work as a barrier to keep the attic air from coming down?
ANSWER:
There are a few things we need to understand before we get to the answer:

The zeroth law of thermodynamics is that heat
always flows from a higher temperature to a lower temperature.

The zeroth law is true for an isolated system.
If there is some other agent which acts to add, subtract, or
redistribute the heat energy, it may not be true.

Particularly for gases, the notion of "warm air
rises" is a result of the fact that if you increase the temperature the
volume increases and so the gas becomes less dense; it rises because of
gravity so the warm "floats" on the cold just like wood floats on
water.

There are three ways that heat is transferred—radiation,
conduction, and convection.
First, lets consider these systems are totally
isolated—no gravity, perfect insulation all over the outsides, no
furnaces, ACs, etc. Looking at all three at once would muddy the
water, so I will imagine there is a perfect insulating wall in the middle
of the attic. Heat would flow from the hot to the cold by convection
through the holes and conduction through the ceilings. Eventually all the
air would be at the same temperature, somewhere between the two starting
temperatures, which would depend on the amount of air in each of the two
volumes. The same thing would happen for the garage/attic except heat would
flow up, not down. Next, turn on gravity; then there would be a continuous
increasing temperature from bottom to top. The ceiling would have the same
temperature top and bottom. Similarly for the garage/attic half. You might
want to imagine that the ceilings are perfectly insulating; the net result
would be the same but only convection through the holes would transfer the
heat.
Finally, your scenario, all three volumes connected
through the holes and no wall in the middle of the attic. With no gravity
everything would end up with the same temperature. With gravity added,
there would then be a continuous increase from floors to the roof of the
attic. Don't forget that this is the real world and any sources or sinks of
heat will change the details.
QUESTION:
I have searched the internet and not found an answer.
If there are 2 identical blocks of ice.
One is at 1 Degrees C
One is at 50 Degrees C
Do they melt at different rates? Will the colder piece take longer to melt?
I'm assuming the colder piece will take longer, but I don't actually know.
ANSWER:
I am assuming that each piece is in the same environment, say a room at constant temperature T somewhere above the freezing point or below the boiling point of water.
Heat flows from a higher temperature environment to a lower temperature environment. When the heat flows into the lower tempeature environment (the ices in your case) it is absorbed and can do
two things: it increases the temperature or it can cause a phase change (melting, freezing,
condensing, or evaporating) without a change of temperature.
(You might want to read a recent answer to a question similar to yours.) At
the beginning of the experiment each piece of ice will heat by increasing
its temperature. The 1°C piece will get to 0°C in some short time, t_{1}, then energy coming in is used to melt the ice until it is fully melted in a time t_{2}. As the ice is melting, the water from the melting
will start warming while the remaining ice will continue to melt without warming. Eventually all the water will be heated to the temperature of the room; call the time from when all the ice was melted
until all the water
was at T to be t_{3}. It will take the first piece t_{1}+t_{2}+t_{3}
to get to the end of the experiment. The only difference between the two
pieces is that it will take the colder piece fifty times longer to get to 0°C.
So it gets to the end of the experiment in a time 50t_{1}+t_{2}+t_{3}.
The times for the two pieces to completely melt will be the same, but it
will take the colder one longer before it starts melting.
QUESTION:
In the known universe, is there anything else in existence besides matter, waves, forces, movement and change?
In other words what are the very basic essential physical things and the effects, excluding any measures that we seem to give a reality to quite often?
Maybe waves could exclude matter or maybe waves are just matter in motion? Space itself seems not to be truly part of existence?
Time to me is a measure of forced motion and change and the same with energy.
What are the bare essentials of the known universe that I can detect through my senses, not just concepts alone?
Would you include fields?
What are the indisputables?
ANSWER:
This is a strange a very strange question. There is only one way I know to
encompus everything which exists. Mass is energy, light is energy, electric
and magnetic fields are energy, motion of a mass is energy, etc..
The answer is that the universe is ENERGY.
REVISED ANSWER:
It occurs to me that time also exists. Time is not energy, so the revised answer is ENERGY and TIME.
FOLLOWUP QUESTION:
What is energy and where has it come from?
Does the ability to do work come from energy or the forces or both?
It seems to me that of all the forms, shapes, colours, sizes, amounts of energy that can be stored or transferred but not created or destroyed.. energy is a quantitative measure of the movement that forces create on an object or particle.
Could energy be dispersing waves of just motion through fields since the big bang.
Anyway, if not a measure, what is it? Please not the ability to do work.
ANSWER:
You seem to have a talent for submitting questions which, contrary to
site ground rules, are not "…single, concise, wellfocused questions…"
You are expecting me to give you a complete tutorial on energy, something
which takes a couple of years of classes to fully understand. Essentially
we invent a quantity which, for any isolated system, is conserved,
i.e. remains constant no matter what happens inside it. By being
isolated we mean that there are no external forces doing work on the system.
[Sorry, there is no way to talk about energy without mentioning work.]
Conceptually, isolated means any system to which energy is neither
being added to nor taken from the system. Assuming that the entire universe
is all that there is (or not interacting with other systems outside the
universe), the total energy of the universe never changes.
QUESTION:
While water is freezing, energy is being released in the form of heat how does this jive with a temperature plateau during the freezing phase change?
ANSWER:
Suppose that we have a glass of water which has been put into a freezer which is at 10°C. Heat will flow from the relatively warmer glass of water to the cold freezer and will and will be
removed from the freezer into the room so as to keep the freezer at the set temperature. After a couple of hours all the water is frozen at a temperature of 10°C. Now there is a power
faliure and, since no freezer is perfectly insulated, heat will leak in from the room so the temperature will will start warming. When the temperature has
risen to 0°C heat will continue to leak in and the air will continue
heating and the ice will start melting. The air is using the heat to
increase its temperature but the ice will use the heat to melt the water,
not to increase its temperature. When all the ice has melted, it water will
start increasing in temperature.
Well, I see that my little story has not answered
your question, it is the reversal. So my story continues: The water and the
air are now at some temperature, say +10°C. Now the power comes back on
and the water and air start cooling, the freezer removing heat from both.
When the water gets to 0°C it starts freezing and the heat being
removed results in freezing, not cooling; but the air continues to cool.
Finally when all the water is frozen the ice now cools because heat is
being removed by the freezer. Eventually everything is at 10°C again.
QUESTION:
I have a question about time. I am a nursing student, taking anatomy courses and learning the atomic physiology of bodily processes. For example, how ions flow in and out of heart cells to make it contract. The whole process takes so much time and thought to track it all out, when in reality, it takes .8 seconds to actually complete a cycle.
My question is, how fast are these atoms moving in order to move at the speed to complete our bodily processes in the small fraction of a time we experience as humans? Or, is it that atoms are so small, they experience time different than we do? or is that not an accurate thought? thanks for the help!
ANSWER:
I do not know much about electric currents in the heart, but I do
understand electric currents in a conducting wire like copper. In a wire,
when you have a voltage between two ends of a wire, electrons will flow in
the wire with some average speed v. It is a common exercise in an
introductory physics class to have the students estimate the speed with
which those electrons move in the wire; a typical answer for household
wiring is about ½ inches per minute! So, if the switch for a lamp is 10
feet from the lamp, it would take an electron starting at the switch (10
ft)x[(12 in)/(1 ft)]/(0.5 minutes)=240 minutes=4 hours to get to the light.
This is, of course, nonsense. What actually happens is that when the switch
is turned on there is a voltage across the ends of the wire which
establishes an electric field* in the wire; the other end receives the
information that the field is there at the speed of light, almost instantly
at 10 feet. So all the electrons in the whole wire are moving, however
slowly, almost immediately.
Hearts are not conducting wires but they can conduct
electric currents because flesh is an electrolyte which plays the roll of
conductor and the charge carriers are ions (atoms or molecules either
missing an electron or carrying an extra electron). So when some action in
the heart requires a message to be sent from point A to point B, a voltage
between A and B is turned on at A resulting almost immediately in the pulse
arriving at B; an ion created at A does not get sent to B. I
believe that calcium ions are important charge carriers in the heart; I
take a drug which is called a Ca channel blocker but I couldn't tell you in
any detail how it works.
So the answer to your question is that the ions move
very slowly but the information which they carry arrives very, very
quickly.
*An electric field is the thing which causes there
to be a force on the charge carriers which drives the electric current.
QUESTION:
When light goes from, say, air to glass or glass to air at a non90° angle, it's refracted. Does that use energy? Are the air and the glass heated? And if so, is that what's happening when a microwave oven heats things the most at the interface between different things (like the ceramic plate and the ham).
ANSWER:
Read the Q&A right after yours. It should be clear that the photon emerging from the first medium has the same energy as when it entered that medium. Therefore no net energy has been lost at boundaries.
However, real materials also absorb some of the photons so a net loss of energy occurs when an ensemble of photons traverse the medium and it shows up as thermal energy.
QUESTION:
Concerns the speed of light in a medium. I understand the speed of light varies slightly with hot and cold earth atmospheres, primarily (apparently) with density, as it does in other mediums. Intergalactic space is a medium, however tenuous, and not a true vacuum. It would therefore seem that the speed of light would vary slightly even in it. The standard speed of light is calculated as in a true vacuum. Is this an answerable question?
ANSWER:
To understand refraction in classical E&M is difficult; essentially the
alternating electric fields induce vibrations in the atoms in the medium
and the result of complicated calculations is that the speed of the wave
changes. But it is easier to understand, at least qualitatively, in terms
of photons; the photons are absorbed by atoms and promptly reemitted.
This takes a tiny amount of time and is repeated enough times to cause a
change in the average speed of the photons. In intergalactic space the
density of hydrogen is about one atom per m^{3}. The likelihood of
there being an interaction which would measurably change the average speed
of a single photon are about as close to infinitesimal as we could
imagine.
QUESTION:
Air is moved downwards, how can momentum be conserved if the helicopter remains stationary? Should there not be an equal and opposite momentum upwards?
ANSWER:
Momentum isn't always conserved, only if there are no external forces
acting on system (or if all external forces add to zero). The rotating blades are pushing down on the air which
causes the air to accelerate downward. But, if the blades push down on the
air, the air must exert an equal but opposite force up on the helicopter
(Newton's third law). So, if we ask if there are any external forces
forces on the helicopter, the answer is yes— the earth exerts a downward force
W called the weight and the air exerts an upward force
L (often called the lift); we can therefore
conclude that momentum is not conserved for the helicopter. In the special
case L=W, and the
helicopter is either at rest or moving with constant speed down or up, the
momentum is conserved. For the air it really makes no sense to talk about
momentum unless you look at it microscopically; fluid dynamics is very
complicated and performed by computers. There is no doubt, though, that
the helicopter and air exert forces on each other.
PRELUDE:
This is a conversation I had with a questioner over the last week or so.
It is probably more than the average reader wants to plow through but is
interesting in terms of a physicist and mathematician struggling to
understand each other. The physics is at the end.
QUESTION:
My question is with regard to an apparent mathematical disagreement with inverse square law. Let's say you have point source in 3D space positioned at (x_0, y_0, z_0). It emits photons in uniform 3D directions.  I just want to exclaim that properly picking uniform 3D directions is not obvious! See Wolfram's Sphere Point Picking page . Anyway, you have an infinite length/height vertical y/z plane detector at x=1. One can derive the governing distribution of hits on this detector and it turns out to be the multivariate Cauchy distribution given by:
p(y,z) = 1/(2*pi) * (1x_0)/sqrt((1x_0)^2 + (yy_0)^2 + (zz_0)^2)^3
This density does not seem to follow the inverse square law. It has a 3/2 power in the denominator! It would need to be 2/2. I have derived densities for other 3D geometries (e.g. cylinder) that are not exactly inverse square law either. I am not a physicist but I am struggling with a physicist who claims it must be inverse square law. It is not. It is similar, but agrees less than the Cauchy density when compared to Monte Carlo simulation. What am I missing?
ANSWER:
Why would an infinitely large detector give you information on the r dependance of what ever the "source" is emitting? By definition, a point source of a vector force field creates a field pointing radially and having a magnitude which decreases like 1/r^{2}. I don't know what a multivariate Cauchy distribution is, or if I ever did I don't remember. You seem to be talking mathematics here and not physics.
REPLY:
The Cauchy distribution (maybe you know it as Lorentz?) is the distribution of where rays hit a vertical or horizontal line when emitted in uniform directions from a 2D point source (x_0, y_0). The bivariate one I wrote is the distribution of where rays hit a y/z plane (at x=1 in this case) when rays are emitted in uniform directions from a 3D point source (x_0, y_0, z_0).
ANSWER:
I have absolutely no idea what you are talking about! But whatever it is, it is not a means of observing or proving an inverse square law. If you want to find out what the field is, since whatever is coming from the source (I will call it "stuff") has to be isotropic, your detector should be either a sphere with the source at the center or a segment of that sphere. Since the area of a sphere increases like
r^{2} and the total amount of stuff hitting the surface of the sphere does not change (because there are no other sources or sinks), it follows that the density of the stuff must decrease like 1/r^{2}.
REPLY:
I think I have an answer for you! Though it has stumped me for more than a month now. If I look at the behavior of probabilities on a small area of the y/z detector plane and I allow the plane to move... say x=1,2,3,etc... Then the ratio of probabilities of detection in this small area will approximately follow the inverse square law. A PDF is not a probability itself. You have to integrate over an area of the detector plane to calculate probability of detection within that area. So if I calculate the probability of detection in, say, [0.1,0.1]^2 for the y/z detector plane positioned at x=2, then it's approximately 1/4th the probability of detection in [0.1,0.1]^2 for x=1. If I do the same for the y/z plane positioned at x=3, then it's approximately 1/9th the probability of detection relative to the x=1 scenario. And so forth and so forth.
I think this other physicist assumed you could just arbitrarily define a density function with inverse square law. This seems to only be true for histograms! The bins of a histogram can be thought of as the integral of some PDF over a small area.
Still, that Cauchy/Lorentz density is an inverse cube where the calculated probabilities behave like inverse square law. That's pretty weird and unintuitive to me!
OK. I'm a poor communicator at this stuff, I'm sorry. So the way this physicist thinks is that if a point source (x_0,y_0,z_0) emits a number of photons in 3D uniform directions, then the number of photons that should hit a specific location on a surface that is a distance r away from the source should be proportional to 1/r^2. So he asserts that you can just define a probability density of photons hitting a detector surface directly using the inverse square law. For example, if it's a y/z plane at x=1, then r=sqrt((1x_0)^2 + (yy_0)^2 + (zz_0)^2) and you get your inverse square law probability density function of p(y,z) = C*1/r^2 where C is a normalizing constant. That seems reasonable to me! And it's quite elegant, right? If you know how to calculate the distance to the detector surface, you can trivially describe the probability density of photons hitting any location of the detector surface. Well... but there's this Cauchy distribution that defines a distribution of photons hitting a planar surface when emitted from the point source. That one is proportional to 1/r^3... and it lines up almost perfectly with Monte Carlo simulation. So how can that be 1/r^3?
The physicist asserts inverse square law must be true. It's welltested and known to be true. I understand that. But the Cauchy distribution describes the simulation better than the inverse square lawbased density. From a math pointofview, it's so straightforward to derive the Cauchy distribution that it would be baffling if it wasn't the correct probability density. But the darn thing really has 1/r^3 in it!
ANSWER:
Since I can not understand your Cauchy approach or what it tells you about
about the point "source", I thought I would start at the beginning with a
point "source", in particular a point charge, and ask what can be learned
by asking about that field in the vicinity of an infinite plane.
I
believe I can square the confusion between my physics and your
mathematics; as you will see, your 1/r^{3} function
appears quite naturally. I will look at an infinite sheet occupying the xy plane and a
point charge Q, a distance d from the sheet. The
electric field has a magnitude E=kQ/r^{2} and
points radially away from Q. Although your calculations were in
Cartesian coordinates, cylindrical coordinates are much more natural. At the
surface of the sheet the electric field has no azimuthal component and so
the area element to consider is dA=2πρdρ. My aim here is
to get the total flux through the infinite sheet, but I will look at other
things as I go along. A few things we need are:
r=√(ρ^{2}+d^{2})
E=kQ/r^{2}=kQ/(ρ^{2}+d^{2})
cosθ=d/√(ρ^{2}+d^{2})
E_{z}=Ecosθ=kQd/(ρ^{2}+d^{2})^{3/2
} sinθ=ρ/√(ρ^{2}+d^{2})
E_{ρ}=Esinθ=kQρ/(ρ^{2}+d^{2})
dA=2πρdρ
The first thing to note is that E_{z} is
where your 1/r^{3} is; just because you happen to get a 1/r^{3}
behavior on the x,y plane does not contradict what the behavior of the
field is. E_{z}^{*}is what is important here if we are interested in something
spread over the entire sheet because for every E_{ρ} on the area
element, there is an oppositelypointing E_{ρ} 180° away which cancels it.
dA·Ecosθ is what physicists call electric flux and, taken over the area dA, the flux is dΦ=2πkQρdρ/(ρ^{2}+d^{2})^{3/2}.
Finally we can calculate the total flux through the sheet:
Φ=2π_{ 0}∫^{∞}E_{z }ρdρ=2πkQ
* The graph shows E_{z}
normalized to E_{z}(ρ=0, z=0) as a
function of ρ in units of d.
QUESTION:
I recently learned that at CERN they produced streams of protons by applying large magnetic fields to hydrogen gas.
This removes the electrons and these only the nucleus specifically protons.
This made me wonder as alpha particles are merely the nucleus of the helium gas, were you to apply a large magnetic field to helium gas would you merely be left with alpha particles?
ANSWER:
I don't where you "learned" this, but it is almost 100% wrong. In a proton
accelerator you need to do three things:

Ionize hydrogen gas to make a plasma. This is
done with a radio frequency electric field.

You then extract the protons from this plasma
using electric fields and inject those into an accelerator which speeds
them up.

Then you repeatedly speed them up, again using
electric fields and magnetic fields to steer and focus the protons
being accelerated.
Most accelerators, including for alpha particles,
work the same way. In fact, magnetic fields cannot ever exert the forces
necessary to accelerate charged particles. The reason is that the force
which a magnetic field B exerts on a charged
particle with a velocity v is perpendicular to
v.
QUESTION:
So I just read an article about virtual particles and Hawkins radiation, and how a black hole can take virtual particles, separate them, and take one of them while the other Hawkins radiation is ejected out and that's how a black hole can evaporate, which I still find hard to believe. If that is the case, and a virtual particle pops existence and one is taken away and the other is ejected what happens with that energy? So, if there's energy between the virtual particles and that energy can't be destroyed only transferred could that energy end up as dark energy?
I'll explain it again in the hopes of clarifying, near a black hole, virtual particle, can pop into existence, be separated, with one particle, being devoured by the black hole, and the other one being ejected out into space, if there is energy between the two particles at the moment of them being separated could that energy simply turn into dark energy or would it stay with the Hawkins radiation being pushed into outer space?
ANSWER:
If a particle/antiparticle pair is created near the Schwarzschild radius, and one of them is inside and one outside, the one inside cannot escape but the one outside can.
If this had happened either inside or outside the black hole, they would quickly have recombined to conserve energy consistent with the uncertainty principle.
But, now, because the particle outside has energy, energy would appear to be not conserved. But that
does happen, so that energy excess has to be provided by the black hole
which means that it must lose a bit of mass. The details doesn't matter of how this happen, it has to happen to conserve total energy.
QUESTION:
When a golf putt hits the edge of the hole, sometimes the ball will "lip out," whipping around the edge and heading off in a new direction. Some lip outs can look quite violent and travel surprisingly far from the hole.
My question: can hitting the edge actually increase the distance to the hole? Formally, if point H the center of the hole, is it possible for a lip out to finish further from H than an identical putt would if there were no hole punched in the ground?
Or does conservation of momentum make this impossible, and this is all just an optical illusion caused by the brain misinterpreting the abrupt change in direction?
ANSWER:
Two golf questions in a row! In an
earlier question I discussed
the motion of a putted ball in great detail. If you want to understand my
answer to your question, you must read this
Q&A first to see how I use
the frictional force which is what I used there. I will also assume that
the parameter
μ_{R}=0.093 and the mass of a golf ball is m=0.046
kg. So the frictional force of a rolling golf ball f=mgμ_{R}=0.045x9.8x0.093=0.042
N if the green is level which I will assume to be the case. The
acceleration of the ball along a straightline path will be a=f/m=0.91
m/s^{2}. The figure
shows the two paths you stipulate, one rolling a distance D_{1}; the second, rolling a distance D_{2}+D_{3}+D_{4}.
D_{3}, not labeled in the figure, is the path during
which the ball is "lipping out", moving on the rim around an arc of angle θ;
so D_{4}=Rθ where θ is in radians
and R=0.054 m is the radius of the hole. I will assume* that the
acceleration along the curved path is the same as along the straight
paths. Now, we can find D_{1} because, from the earlier answer, D_{1}=0.55v_{0}^{2}=where
v_{0} is the velocity which the ball has as it leaves the
putter. I will choose v_{0}=1.5 m/s which yields D_{1}=2.05
m. Since I have assumed all paths have the same acceleration along all
paths, each of the two putts must travel the same distance, so
D_{2}+D_{3}+D_{4}=D_{1}=2.05
m _{
}D_{4}=2.05D_{2}0.054θ
Since we are trying to find out whether or not it is
likely that the ball which lipped out is farther from the hole or not,
knowing D_{4 }will likely tell us. D_{2} is just the distance from the hole
center where the lie was; θ will probably not be bigger than 90°=1.6 radians
and, besides, it is very small in the scheme of things. If we choose D_{2}=1
m and θ=90°, D_{4}=2.0510.085=0.965
m. It makes more sense to look at a straight shot which just missed getting
lipped with the one that lipped by 1.6 radians. The second figure shows how
far each are from the center after stopping. The lipped one is closer by
about 8%.
*The question about what the rolling friction should
be in the arc was what worried me most when I started analyzing this
problem. But the normal force on the ball by the ground has to be bigger
than it is when going straight because of a centrifugal force acting
outward. But the conclusion would be the same that the lipped ball will end
closer to the center of the hole would would not be changed; if the normal
force is bigger, the friction is bigger, so it loses more speed and would
be even closer.
QUESTION:
I was trying to measure the bounce of a golf ball to determine whether its lifespan was up. In order to control the ball's bounce as to not get away I thought bouncing it in a tube with viewing hole would be a good environment. Unfortunately I noticed the balls bounce approx 50% less in the tube than outside of it.
I would like to know why the golf ball bounces higher outside than inside a 2 inch PVC tube from 5 feet off the ground? Is there anything I can do to alter the tube to get the same results (ie drill holes in it, cut out sections in the bottom to release air)?
ANSWER:
This is so different from how the freely falling ball falls, that I won't even
discuss air drag which is the main source of friction for the ball which
is not in the pipe. But something else is going on here. Imagine that the
diameter of the ball and the inner diameter of the pipe were exactly the
same and that there was no friction between the two; also imagine the
bottom is sealed. You release the ball and all the forces on it are its
weight down, the force of the atmospheric pressure on the top which is
also down, and the force of the pressure of the air (which starts as
atmospheric pressure) which is up. So the net force on the ball is
initially only its weight down; so it starts out just like the ball
without the pipe. But, as soon as its starts falling the pressure of the
air below it begins increasing. The farther it falls, the higher the
pressure gets, so eventually the net force from the pressures is up and
has a value exactly equal to the weight. But, when it gets to that point
the ball which has been accelerating downward begins to be more up than
down so the ball starts slowing down eventually coming to a momentary
stop. Now the ball has a net force up so it starts accelerating upward and
eventually goes back to where it started. It keeps oscillating up and down
just as if it were on a spring.
Now your situation is not like this but it is very
similar except it is leaky; the air will leak out as the ball falls, but
the pressure in the air below the ball will still increase. All the air
which the ball is pushing through has to squeeze through a much smaller
space than if the pipe weren't there. That is going to considerably impede
the fall of the ball. If you used a pipe much larger to keep from having to
chase the ball it would probably better match the freely falling ball.
QUESTION:
This is the 90 year old man in a nursing home and would like to know if it is possible to have a total vacuum in a closed space.
ANSWER:
In a very excellent vacuum system there are about 20,000,000 molecules/cm^{3}.
In intergalactic space there is typically 1 molecule per m^{3}. We
now understand that even if we had a volume with absolutely no molecules
in it, it would still not be truly empty because there are constantly
particleantiparticle pairs which pop into existence and pop back out,
called virtual particles; this is called vacuum polarization.
QUESTION:
I am in a staterun nursing home that has toilet facilities in a small enclosure with two doors measuring 5ft by 8ft and height 10ft. The temp rises to 6 degrees above the adjacent rooms. Is this due to the heating of the particles in the atoms creating energy in the form of heat?
ANSWER:
If the toilet room has its own heat register or radiator and this room is much smaller than the adjacent rooms,
the heat will increase the temperature more compared to larger rooms; and if the doors are closed for most
of the time or even less, it will naturally be hotter than a larger room on the same thermostat. This could be corrected by partially closing the register or partially reducing the water/steam flow to
the radiator.
QUESTION:
I just really confused myself with a strange hypothetical situation during the impulse/momentum unit at my school. If a kid is riding a sled on completely frictionless ice with negligible air resistance, and then separates himself from the side of the sled in a way such that both the sled and the kid are moving in the same direction, would the sled still speed up? I know the sled's mass would seem to decrease from the sled's perspective, so it should speed up to conserve momentum, but if the kid is still moving at the same velocity as before, wouldn't any increase in speed for either thing not conserve momentum for the system? I tried to explain this to my teacher, and we clearly didn't understand each other.
ANSWER:
This depends on how the kid separates from the sled. There are two things
you need to be aware in questions like this: Is linear momentum conserved
and is kinetic energy conserved. If there are no external forces
acting on a system the linear momentum will be conserved; most problems
you are likely to encounter have momentum conservation. If there are no
external forces doing work on a system the kinetic energy would be
conserved; if it is (which it usually isn't), it is called an elastic
event. I shall look at two scenarios:

What is the relation between the two final
velocities if we demand that linear momentum is conserved? If
the subscripts labeled 1 and 2 are for the boy and the sled, then
(m_{1}+m_{2})u=m_{1}v_{1}+m_{2}v_{2}
or v_{1}=u[1+(m_{1}/m_{2})](m_{1}/m_{2})v_{2
}where u is the speed before
separation and v_{i} are the velocities after._{
}For example, suppose that v_{2}=u/2, then
v_{1}=u[1+(½m_{1}/m_{2})].
As you can see, there are an infinite number of possible velocities
after the collisions depending just on how the kid got off the sled. Of
course, that is not surprising because we have only one equation for
two unknowns.

Now, suppose that we impose the condition that
energy must also be conserved. The derivation of the solutions for a
perfectly elastic event can be found in any introductory physics
textbook or go to the
Wikepedia article on elastic scattering in onedimension. Now we
have two equations for two unknowns which means of course there will
only be one solution. If both objects have the same speed
u before the collisions, then these
equations become
v_{1}=u
and v_{2}=u.
So, you see, the boy and the
sled have the same speed afterwards as they did before the boy got off
but only if he got off in such a way that no energy was added nor taken
away from the system.
It looks to me that you did not give me all the
information about the problem. If it had been something like "Suppose that
after the separation the sled had a speed of 3u/2, how fast is the
boy moving?" you could find out that the boy was moving more slowly, v_{1}=u[1(½m_{1}/m_{2})].
QUESTION:
I am a new adjunct instructor for a introductory physical science online class. My background is in chemistry. Anyway, I ran across a conceptual problem in OpenStax Physics 2e, that I am confused about. In the chapter reading, Coulomb's law is introduced followed by electric fields which are derived from Coulomb's law. Neither of which I find confusing. Here is the problem I am confused about: Compare and contrast the Coulomb field and the electric field.
ANSWER:
Electric field is a general term, Coulomb field is presumably referring
to a field which falls off like 1/r^{2}. The prototypical
Coulomb field originates in an isolated point charge q:
E=r_{o}kq/r^{2}
where r_{o} is a dimensionless vector of
magnitude 1 which points directly away from the point charge, called a
unit vector; the constant k is sometimes written as 1/(4πε_{o}).
This Coulomb field plays a crucial role in the determination of the field
of any charge distribution which is not a point charge. If you have a
chemistry background, you surely know calculus, so I will give you the
whole general picture so you can understand what goes on. If the course
you are teaching is "physical science" you surely will not teach it at
this level, probably the electric field due to 2 or more point charges. In
the figure a tiny (infinetesimal, really) piece of the blob of charge has
been focused on. The field at the point labeled P due to the tiny
(point) charge labeled q_{i} is a Coulomb field; now you
have to integrate over the whole volume for every point in 3D space.
Except for simple shapes like a sphere, this is an extremely difficult
calculation to do analytically. That is why your course will probably not
talk about anything much beyond point charges. You might also touch on the
field of a large uniformly charged plane which is pretty easy to
conceptionalize and introduces a uniform electric field.
QUESTION:
Einstein, in his theory of General Relativity, stated that Gravity is not a force. So, why the phisicists are still trying to develop a theory to explain Gravity as a force next to the other three fundamental forces of nature?
ANSWER:
I doubt very much that he did said that gravity is not a force.
Although thinking of spacetime warping is the most popular way to
visualize the results of his theory, it is not a unique way. General
relativity (GR) is also a field theory and interactions in field theory
are manifestations of forces. Einstein was aware of this alternate view,
in fact he embraced it—because GR is deterministic, he believed in
predestination. Also, the search for grand unification including gravity
is rooted in the hopedfor quantization of GR; field theories are
quantizable. See earlier
Q&As on this topic; be sure to see links in that Q&A.
QUESTION:
I am writing you to seek your expertise in the field of physics.
As you know, Newton's law of universal gravitation demonstrates that the gravitational force between two objects is proportional to the product of their masses and inversely proportional to the square of the distance between them.
The use of r^{2} instead of r arouses my curiosity.
Any further explanations would be highly appreciated.
ANSWER:
At the Newton was proposing his three laws there was a lot known about the
solar system. Most importantly, Tycho Brahe had made thousands of precise
measurements of planetary orbits, their shapes and periods. His
assistant Johannes Kepler continued after Brahe's death but he also used
those data to formulate the properties, the three Kepler's laws. Keep in
mind that Kepler's laws were purely empirical, nothing we would call a
theory. It was now known that the orbit of a planet was an elipse one
focus being at the sun, that the quantity T^{2}/a^{3} was the same
for all the known planets (T is the period and a is the
semimajor axis of the elipse); also, the line from the center of the sun
to the orbit swept out equal areas
in equal times. Newton's laws
introduced the notion of force and its relation to motion. He realized
that the planets moving around the sun implies that some force was acting
between the sun and the planets. And surely he realized that the force
would not be constant nor would it get bigger as the objects got farther
apart. Maybe he did try a force which went like 1/r, but if he did it
would not explain the data—Kepler's three laws were the data. But it
would not work. Nor would 1/√r, 1/r^{3}, etc., but a 1/r^{2} force did work.
There is another way you can look at forces. A force
usually may be thought of as a field where we think of lines of force
directions in the direction a mass would be attracted; the closer the
lines are together, the stronger the field; and total number of lines is
proportional to the mass. The red lines in the figure show the earth's
gravitational field but only in one plane. The blue lines show spheres
again only in two dimensions. You should imagine the lines coming in from
all directions and many imaginary spheres. The strength of the field would
be the number N of lines per unit area A. But every sphere is punctured by
exactly the same number of lines as all the others. The area of a sphere is
4πr^{2} so the number of lines per unit area, strength
of field, is N/(4πr^{2}). Eureka! There's your
1/r^{2}.
QUESTION:
I am a student in Belgium. Near my school, there is a wind turbine that has caught my interest. I have been wondering if it's possible to determine the wind speed by measuring the time it takes for the turbine blades to complete one full rotation, given the radius of the wind turbine. I am assuming a constant wind speed for this calculation.
I believe that the wind speed (V_{w}) can be expressed as V_{w} = (3.6 * 2 * π * R_{turbine}) / T (in kilometers per hour), where
R_{turbine} is the radius of the wind turbine, and T is the time it takes for the blades to complete one rotation. However, this formula does not seem to yield realistic results. Despite conducting some research on this matter, I haven't been able to find a satisfactory answer.
ANSWER:
You would have been right that the rotational speed of the blades depends
on the speed of the wind if you were looking at a simple pinwheel. There
are no other significant torques on this machine than that caused by the
wind on the blades so, to a pretty good approximation, the only thing
affecting the rotational speed is the wind. The wind turbine, however, is
a very complicated machine. Here are some things which are interesting:

The blades must move slowly, usually around
1020 rpm, because of structural limits and noise polution.

So there are mechanisms which keep the
rotational speed within those limits.

Because this is too slow for the generator to
work there is a gear box to increase rotation to the generator by
a factor of about 100.

The blades do not work like paddles or sails,
rather like the wings of an airplane or rotor of a helicopter.

The speed of the tips of the blades is more
important than the rotational speed of the blades themselves for
maximizing efficiency.

Don't forget that, unlike the pinwheel, energy
is being constantly taken away from the blades and converted into
electrical energy which will be sent out to the world; so there must
some energy balance where the energy taken out is equal to the energy
being supplied by the wind minus energy lost to friction, inefficiency,
etc.
There is a very good
webpage going into the details of wind turbines; be sure to look at the
video too. There is another
video which is
also pretty detailed.
QUESTION:
if pi measures a circle and and it never repeats it self, but its used to measure a circle doesn't that mean the numbers would eventually go around the circle and repeat.
ANSWER:
This isn't physics. π does not "go around the circle",
whatever that means; it is a measure of the ratio of two lengths, the radius R and circumference C of a circle, π=C/(2R).
QUESTION:
For matter that is falling into a black hole, even before it gets to the event horizon, it would be impacted by both gravitational and velocity time dilation in accordance with standard and generally relativity. This time dilation can be very extreme at these extremes. So would this mean that through the perspective of the matter falling in, it's still hasn't reached the center or even the event horizon? Because from our perspective outside the black hole it is going to take them billions or even trillions of years to get there. So while for them it's a quick decent, for us it's a very long time. And since we are here now could that possibly mean that all that matter in black holes has yet to reach the center because they are nearly frozen in time dilation? If I'm wrong what aspect is not working the way I think?
ANSWER:
Your question is not really a "single, concise, wellfocused" question as
required by site ground rules. I suggest this
link to get a
tutorial about falling into a black hole.
QUESTION:
Is infinite energy possible in a frictionless surface
ANSWER:
There is not an infinite amount of energy in the whole universe. The only
way it could get infinite energy is if it goes with a velocity equal to
the speed of light and that is impossible exactly because you cannot get
an infinite amount to give to it.
QUESTION:
I was learning about a STEAM activity for my children. This activity involved cutting up pool noodles and putting toothpicks in them to stick the noodle pieces together. I was wondering, why doesn't the toothpick fall out? What force(s) are acting on it to keep the two together? It seems strange to me, because the pool noodle is porous. I want to know so I can teach my children the physics behind the activity.
ANSWER:
I have raised 4 children, and I never heard of this—must be pretty new. It
is friction between the toothpick and the noodle piece that keeps it in.
The frictional force can be felt because if it takes a force to push it
into the noodle (or pull it out). An interesting thing is that the
frictional force between two surfaces depends on how hard the surfaces are
pushed together—when you push the toothpick in it pushes the noodle over
to make room for itself. Since the noodle stuff is elastic, (if you
squeeze it, it springs back), the two surfaces are pushed together pretty
hard which gives you a lot of friction. If you
were to drill a hole exactly the same size as the toothpick is thick,
there would be much less friction. If you think about it, that makes sense
because you know it would much easier to push the toothpick into that hole
than to push it into the noodle without a hole.
QUESTION:
If sound like light travels on earth, does the sound like light just continue on continuously.
ANSWER:
Sound does not travel anything like light does. The only thing they have
in common is both are waves. The most important difference is that sound
needs a medium through which to travel, air in the case you are thinking
of I presume. Light does not need a medium, it can travel through a
perfect vacuum. So light keeps right on going when the atmosphere
disappears but sound does not because there is no air to travel in.
QUESTION:
As a solid object (ex a box) travels in a circular path, the side of the box closest to the inside of the curve moves a shorter distance than the side of the box towards to the outside of the curve. Yet, the sides of the box remain in the same relative position to each other............
How can one side of the box travel a greater distance than the other and the box still remain intact?
ANSWER:
That's
just the way it is in Euclidian geometry—every part of a rigid body which is rotating about some axis remains locked to the where it is in the rigid body. In fact, that is what a rigid body is defined
to be. Think of a CD rotating about its central axis. Any point a distance r from the axis has a speed v=rω where ω is the angular velocity in radians per second. So the farther out
any point is the faster its speed is. But if the spinning object is not a rigid body it will not remain intact. If you start with a spinning pancake made of soft putty it will spread out. If the box you were
thinking about was made of rubber it would not retain the same shape if you started to spin it.
FOLLOWUP QUESTION:
Thank you so much for replying to my question, but....
I'm sorry, even tho your answer makes perfect sense, it did not answer my question .
How can one side of the box travel a greater distance than the other and the box still remain intact?
Imagine the box is going around a circular race track (like a race car). The track has a diameter of 1000' at the center of the track  the circumference of the center of the track is 3141.59' . Say the box is 4' long and 2' wide. In one trip around the track, the center of the box travels 3141.59'. The outside of the box (at a diam of 1001') travels 3144.73'. The inside of the box (at a diam of 999') travels 3138.44'.
So, the outside of the box has travelled 6.29' farther than the inside of the box....
How can one side of the box travel a greater distance than the other side and the box still remain intact?.
I have wondered about this for many years of my life and have never been able to find someone to explain it to me.
(straightline physics is so much easier than curvedline physics....)
Thank you so much for your attention to this question.
ANSWER:
The
answer is actually simple, if one part of the box is farther from the
center it moves with a larger speed than another part of the box closer to
the center. What I think the problem is that you do not know the basics of
rotational motion. Central to rotational kinematics is what the
linear velocity v of any object rotating about some axis with
angular
velocity ω and at some point a distance d from the
axis: v=ωd. The angular velocity must be measured in
radians/second (s^{1}); since there are 2π radians in a circle,
an angular velocity of 1 rotation per second has ω=2πd.
I have drawn a diagram of a stick rotating around one end with angular
velocity ω showing one end having a speed v but the middle having a speed of
v/2 and the end it is rotating around is at rest; v=Lω.
Your instinct that if it wasn't strong enough to have those differing speeds, remain a rigid body, it will break. A good example of
that is when a very tall chimney falls, it breaks before hits the ground.
QUESTION:
I know that electrical current is in units of coulombs/s. Each coulomb contains a bazillion electrons. For a electrical transmission line with an AC signal, I think of the mean value of electrons alternating back in forth at the AC rate, since the current is alternating. What happens to these electrons when they feed an antenna and the power is radiated? Or do I have all my assumptions completely wrong?
ANSWER:
Let's just look at one electron. It oscillates as if it were attached to a
tiny spring. When an electron is oscillating it radiates electromagnetic
waves which carry energy away from the electron so it will get smaller and
smaller amplitude of vibration until it stops. But in a radiating (as
opposed to receiving) antenna all the electrons are being pushed around by
the transmitter, so they don't stop but move and radiate in the way that
the transmitter pushes them around.
QUESTION:
I suspect that Elon is not producing trucks because of the way electric vehicles brake.. Am I correct in saying that there is a transfer or energy to the road surface on EVs, not on normal vehicles that rely on pads.a big rig, heavy.. those wheels, the weight, the method of slowing, I believe that e trucks will not suit our roads. Do you get me?
ANSWER:
Yes,
you are wrong that there is "…a transfer of energy
to the road surface…" when EVs brake. Let's discuss what braking
does. When a vehicle is moving it has energy by virtue of its motion
called kinetic energy. In order to stop the vehicle from moving, you must
get rid of that kinetic energy. The traditional way to do that is to use
friction of brake pads (or shoes) rubbing on a metal disc (or drum). Where
does the kinetic energy of the vehicle go? Into heat energy because the
brakes get very hot. Wouldn't it be nice if you could capture some of that
energy and store it to use to change back to kinetic energy later rather
than get the energy from the burning fuel. There is a way other than
friction that could be used; that method is called regenerative braking
(regen) and involves using electromagnetism to create an electric current
which could be stored by sending the current to a battery (not to the road
surface!) Before recent times there were no batteries to store all this
energy except the usual 12 V battery which gets recharged but not by a
braking system but by the alternator. But with hybrids or EVs, there are
big batteries which are hungry to grab any energy they can to lengthen the
time before they next need recharging. But no vehicle relies solely on
regen because regen might not slow the vehicle down fast enough or it
might fail altogether; all hybrids and EVs have both conventional brakes
and regen brakes and usually you can adjust what fraction of the
braking is done by each. Furthermore, if the vehicle is not braking fast
enough with just regen, conventional brakes would jump in. Therefore, as
you can see, braking is not an issue at all in whether the vehicle is
feasible. But adding regen to any vehicle is to your advantage if you have
the ability to store the electric energy.
QUESTION:
I read that quantum spin is intrinsic angular momentum. I do not understand what intrinsic angular momentum is. Could you please tell me? In ignorance, I guess that it is some number which quantum physicists derive from the positions, movements and interactions of electrons, other particles and photons, and which predicts their positions, movements and interactions, but does not refer to classical rotation because the number is so high that under the equation E = MC2 the electron, other particle or photon would have infinite mass, which is wrong.
ANSWER:
I believe that it is always good to have a clear picture first of what a
quantity is in classical physics before trying to understand corresponding
quantities in quantum physics. We often break angular momentum (AM) into
two pieces, orbital angular momentum (OAM) and intrinsic angular momentum
(IAM). IAM corresponds to the AM an object has by virtue of spinning about
an axis in the object itself. OAM corresponds to AM which an object has by
virtue of its moving relative to some other point. An example is the earth
which has OAM by virtue of its motion around the sun and IAM by virtue by
virtue of its rotation about its axis. A rough, but often successful model
of the atoms can be understood by thinking of electrons being in orbits
around the nucleus (OAM) but also you must assume that the electrons have
IOM to understand what is going on in detail. Electrons have IOM which
never changes. In general, however, in quantum mechanics the angular
momentum of a system cannot be just anything you like (for example, if you
spin a ball there is no limit on how fast it spins, 1, 23, 32.5, 0,
etc. RPM); instead they can only have certain discrete values which is
called quantization of AM. Elementary particles have one fixed AM but the
total angular momentum of a quantum system in some particular state which
has quantum numbers either integers n=0, 1, 2, 3… or halfodd half integers,
n=1/2, 3/2, 5/2… The first set are called bosons and the second set
are called fermions. The angular momentum J of a system with
quantum number n is J is given by J=ℏ√(n(n+1)) where ℏ is the rationalized Planck's constant.
IOM is often referred to as spin. But if you try to make a semiclassical
model of an electron as a little spinning ball you get impossible results,
like the surface of the ball traveling faster than the speed of light. So
we can think of it like a spinning ball but have to keep in mind that it
is not a classical object and just has a property that behaves sort of
like classical objects do.
QUESTION:
I am trying to understand something about gravity. To it seems that the gravitational pull equations I was taught is flawed. It is stated that it is directly proportional to the mass of the object and inverse to the distance from the object. The distance makes sense. The mass does not because gravity itself is needed to make the mass. Without it, mass would not exist. It is like trying to describe the color red using crimson.
ANSWER:
You have it backwards, gravity is the result of the presence of mass (or
any energy density for that matter), not vice versa. Also,
gravitational force is inversely proportional to the square of the
distance.
QUESTION:
Light is able to be bent by both gravity and it easy bent when traveling through clear objects like a prism. Could it be that the secrets to gravity and antigravity might be found in light?
ANSWER:
These two bendings are for two entirely different reasons. So the answers
are no and no.
QUESTION:
I'm not sure if this counts as astrophysics, but if an object was falling towards something, like the earth (no air resistance), and then the earth disappeared, would it keep the inertia from the fall? Because if considering gravity as just spacetime curvature, it disappearing seems like it would leave the object motionless, as it always was motionless in freefall, because standing on the earth is the same as accelerating up?
ANSWER:
I
seem to always be saying...if you ask about velocity you have to specify
velocity relative to what. If you are watching this scenario in a frame at
rest relative to the earth, the object will move with the velocity it had
at the instant earth disappeared.
QUESTION:
My question is does it matter if we have two Dynamos which produce
the same voltage but different frequency does the frequency matter and how?
ANSWER:
It
depends on what you are powering with the dynamos. But most devices are
designed to operate at a particular frequency.
QUESTION:
Is it theoretically possible for a black hole to contain other smaller black holes?
ANSWER:
If you mean by contain inside the Schwarzschild radius, the answer is yes, although it wouldn't last for very long. If you mean inside the black hole itself, what does it mean for a point to "contain" anything?
Keep in mind that I, as stated on the site, do not normally do astronomy/astrophysics/cosmology.
QUESTION:
How fast would a missile be flying if it was fired from an SR71 blackbird flying mach 3.2?
ANSWER:
As always with questions about velocity, you must specify velocity with
respect to what. If the blackbird has a velocity
v with respect to the ground and the speed of this particular missile
has a velocity u with respect to the ground
if fired from the ground, and the
velocities are in the same direction, then the missile has the speed
u+v with respect to the ground if fired
from the blackbird.
QUESTION:
So let's say my buddy and I are driving along the highway in our Pontiac Trans Ams at 99% the speed of light maybe a couple hundred yards apart. I'd like to chat with him on the CB radio. Are we traveling faster than the radio waves can be received?
ANSWER:
It sounds like you and your buddy are traveling in the same direction with the same speeds. In that case, according to the principle of relativity, there is no difference from your both
sitting at rest separated by "a couple of hundred yards apart."