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.

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 non-90° 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 re-emitted.
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 multi-variate Cauchy distribution given by:
p(y,z) = 1/(2*pi) * (1-x_0)/sqrt((1-x_0)^2 + (y-y_0)^2 + (z-z_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 multi-variate 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((1-x_0)^2 + (y-y_0)^2 + (z-z_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 well-tested and known to be true. I understand that. But the Cauchy distribution describes the simulation better than the inverse square law-based density. From a math point-of-view, 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} =E cosθ=kQd /(ρ ^{2} +d ^{2} )^{3/2
} sinθ=ρ /√(ρ ^{2} +d ^{2} )
E_{ρ} =E sinθ=kQρ /(ρ ^{2} +d ^{2} )
dA =2πρ dρ

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 oppositely-pointing E_{ρ} 180° away which cancels it.
dA·E cosθ is what physicists call electric flux and, taken over the area dA , the flux is dΦ =2πkQρ dρ/ (ρ ^{2} +d ^{2} )^{3/2} .
Finally we can calculate the total flux through the sheet:

Φ =2π _{ 0} ∫^{∞} E_{z } ρ dρ= 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 straight-line 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} =2.05-D _{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} =2.05-1-0.085=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
particle-antiparticle pairs which pop into existence and pop back out,
called virtual particles; this is called vacuum polarization.

QUESTION:
I am in a state-run 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 one-dimension. 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 3-D 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 space-time 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 hoped-for 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
10-20 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, well-focused" 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.
(straight-line physics is so much easier than curved-line 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 half-odd 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 anti-gravity 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 space-time 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 SR-71 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."