Length contraction of falling things

[Dale]

Does this not mean that its 4-momentum vector (energy) remains constant (stays the same)along that worldline?

First, the four momentum is not the energy. Energy is the timelike component of the four momentum. So even if you interpret parallel transport as keeping the four momentum constant you cannot conclude that the same applies for energy.

Second, it is hard to interpret parallel transport as keeping a four vector constant. If you parallel transport a vector on a closed loop it does not generally wind up in the same orientation. It is an interpretation that can be useful over small regions of spacetime where you can neglect curvature, but it can cause problems if taken too far.

 

[Q-reeus]

The spin rate declines according to which observer? The local or the observer at the initial elevation?

By using the term 'coordinate determined' it is automatically implied the observer is at an initial elevation out at 'infinity'. See e.g. the definition of coordinate time given in this Wiki article: http://en.wikipedia.org/wiki/Schwarzschild_metric#The_Schwarzschild_metric
For elevations further in, one simply applies the appropriate correction factor - observed redshift is somewhat less then in general.

Could you elaborate on the assumption that "rest mass/energy declines with lower potential " ?

Sure. Annihilate a positron-electron pair out in gravity-free space and all we have left is a pair of gamma rays with a given emitted frequency. That is a fully conservative process of conversion of rest mass to EM energy. Now lower (not free-fall drop) the same positron-electron pair down into the potential well of some gravitating mass. Then annihilate as before. The gamma rays received 'at infinity' have been redshifted (a la Pound-Rebka experiment) owing to the annihilation process occurring in that gravitationally effected environment. Which can be equated with that in the process of lowering the electron-positron pair down the gravitational potential well, energy is being extracted to the lowering agency (say a winch). Everything is thus consistent with overall conservation of energy/momentum - but there has been a redistribution going on.

 

[Q-reeus]

Apparently you assume that no light ray reflects off the bottom or the top during motion. That's not likely IMHO. And thus its effect should be analysed. Is then work done or not? And equal but opposite?

Harry - it really pays to read an entire thread before commenting critically on an initial part thereof - see you own further remarks below!

Q-reeus: "[..] the trapped EM field energy [..]"
I would say trapped radiation energy - not field energy!

You wish to make some fine distinction here? As I assume you are a retired comms engineer or similar, you aught to be fully familiar with that a cavity field (leaving aside TW resonators) is just equipartitioned cyclicly alternating and phase orthogonal electric and magnetic fields. Disagree?

Q-reeus: "Say for a cubical cavity resonator, resolve cavity field into three mutually orthogonal sets of counterpropagating traveling waves. Two sets propagating transverse to radial direction, one along that axis. Sets need not be equal amplitude. It should be obvious the two transverse sets have no appreciable coupling to the only conceivable factor that hauling-up process could bring to bear as locally measured [..]"

As discussed in parallel, analyses with "local" measurements easily complicates analysis. It can be useful to develop a theory but when you already have the theory, needless frame jumping is often counter productive.

I have no idea what point you are trying to make here - please elaborate!

Q-reeus:" [..] from a Machian pov that seems suspect. [..]"
I find Machian POV's themselves suspect.

You are in turn welcome to that pov - but how would you then explain the process of assembling a matter shell around a spinning flywheel, without it implying a transfer of angular momentum to that assembled shell? HINT - there is a gravitationally analogous 'magnetostatic' field generated by that flywheel, and considering the shell assembly as having to radially draw in mass from infinity, one might suspect a 'magnetic Lorentz force' analogue is operative on that inwardly moving shell matter. Just my humble outlook of course - not out for a full-on intellectual stoush with your emminence! In the past we have a history of mostly agreeing on things, but hell here's a chance to really differentiate! Go for it Harry!

And I think that we have largely covered the topic of this thread.

Huh?

 

[Dale]

collision speed = light speed+mirror speed=k*c + v
collision length = length of light pulse = k*s
time= collision lenght/collision speed = (k*c+v)/k*s

Where do these equations come from? I don't recognize any of them as standard SR or GR equations.

 

[Naty1]

PeterDonis:

...In another thread I used this type of model to describe how one could extract work from a box filled with photons by slowly lowering it instead of letting it freely fall, and how extracting the work would lower the "photon temperature" inside the box, whereas a freely falling box of photons would have the same "photon temperature" inside as it fell (relative to the box, in both cases). But in either case, the "average" motion of the photons inside the box is no longer null, but timelike--it's just the motion of the center of mass of the box+photons.

Very interesting, but I don't get it. why does the photon temperature change dependent of the rate of fall?

The simple minded approach I thought would work: forget the details... you can't hide from the change in gravitational potential [PE] regardlesss of how the photons are are lowered...freefall, slowly, etc and if the PE changes than so must the KE of the photons as total energy remains constant...

Can you explain the last sentence...what's the context...are you saying the freely moving photons have no reference frame, but bouncing ones do...have an average...??

thanks. always interesting!

 

[harrylin]

[..] I have no idea what point you are trying to make here - please elaborate!

When you mentioned "local measurement" that suggested to me that you use different reference frames, with the risk of mixing them up or not transforming correctly between them. It's often better to avoid that.

[..] how would you then explain the process of assembling a matter shell around a spinning flywheel, without it implying a transfer of angular momentum to that assembled shell? [..]

I'm sorry, I meant that we've already finished discussing the topic (which is Length contraction of falling things!) and I already spent too much time on topics that I chose to discuss (I'm not retired, and I have a life). So I'm now trying to reduce (but physicsforums is like an addiction, worse than TV ).

[..] Annihilate a positron-electron pair out in gravity-free space and all we have left is a pair of gamma rays with a given emitted frequency. That is a fully conservative process of conversion of rest mass to EM energy. Now lower (not free-fall drop) the same positron-electron pair down into the potential well of some gravitating mass. Then annihilate as before. The gamma rays received 'at infinity' have been redshifted (a la Pound-Rebka experiment) owing to the annihilation process occurring in that gravitationally effected environment. Which can be equated with that in the process of lowering the electron-positron pair down the gravitational potential well, energy is being extracted to the lowering agency (say a winch). Everything is thus consistent with overall conservation of energy/momentum - but there has been a redistribution going on.

That sounds like a good analysis to me.

 

[Q-reeus]

When you mentioned "local measurement" that suggested to me that you use different reference frames, with the risk of mixing them up or not transforming correctly between them. It's often better to avoid that.

A misunderstanding then - I simply meant if one rides the box back up elevator style, there should be no detection of frequency shift enroute. Sans point raised later re flywheel!

I'm sorry, I meant that we've already finished discussing the topic (which is Length contraction of falling things!) and I already spent too much time on topics that I chose to discuss (I'm not retired, and I have a life). So I'm now trying to reduce (but physicsforums is like an addiction, worse than TV ).

Tell me about it - are there local chapters of 'PhysicsForums Anonymous' offering addicts any hope of rescue?!

That sounds like a good analysis to me.

Thanks Harold - nice to get some positive feedback now and then.

 

[PeterDonis]

Very interesting, but I don't get it. why does the photon temperature change dependent of the rate of fall?

It's not the "rate of fall" that's critical: it's the difference between the box being in free fall and the box feeling acceleration. Also there is an underlying assumption that the time of flight of photons across the box is short enough that tidal effects are negligible.

Consider a photon that bounces off the upper wall of the box, travels to the lower wall, and bounces off it. Look at this process in the instantaneous rest frame of the box, which, by our underlying assumption, can cover both "bounce" events in a single local inertial coordinate patch.

If the box is in free fall, then it can be considered to be at rest in the local inertial frame for both bounce events, so the momentum exchange of the two events cancels. So there is no energy exchange between the photons and the box.

If the box is accelerated upward, then it is moving upward for one of the bounce events. That means that there is more momentum exchanged when the photon hits the lower wall of the box than when it hits the upper wall of the box. So on net there is work being done. In the case where the box is being slowly lowered, the work will be done by the photons on the box (and that work is extracted by means of the rope that is holding the box), so the photons are losing energy relative to the box; hence the "photon temperature" relative to the box goes down (where "photon temperature" is just the average energy per photon).

I should note that I also assumed in this that the energy of the box itself was negligible. For a real box, of course, that would most likely not be the case, and one would have to consider the motion of the atoms in the box itself, not just the photons in it.

 

[Naty1]

PeterDonis:

If the box is accelerated upward, then it is moving upward for one of the bounce events. That means that there is more momentum exchanged when the photon hits the lower wall of the box than when it hits the upper wall of the box. So on net there is work being done.

That seems clear.

Originally Posted by Q-reeus

[..] Annihilate a positron-electron pair out in gravity-free space and all we have left is a pair of gamma rays with a given emitted frequency. That is a fully conservative process of conversion of rest mass to EM energy. Now lower (not free-fall drop) the same positron-electron pair down into the potential well of some gravitating mass. Then annihilate as before. The gamma rays received 'at infinity' have been redshifted (a la Pound-Rebka experiment) owing to the annihilation process occurring in that gravitationally effected environment. Which can be equated with that in the process of lowering the electron-positron pair down the gravitational potential well, energy is being extracted to the lowering agency (say a winch). Everything is thus consistent with overall conservation of energy/momentum - but there has been a redistribution going on.

"Now lower (not free-fall drop) the same positron-electron pair down into the potential well of some gravitating mass..."

This means lower the pair and then place them at rest at some point down in the potential well? Is this different than letting the pair free fall, stopping their fall, then letting them annihilate?? I don't see how the gamma ray energy is any different at the top and bottom of the well: local observers each see the same energy locally, top and bottom, right?; but such an observer at the top will see redshift from the bottom gammas, at the bottom, blue shift from the top gammas.

 

[harrylin]

[..] If the box is accelerated upward, then it is moving upward for one of the bounce events. That means that there is more momentum exchanged when the photon hits the lower wall of the box than when it hits the upper wall of the box. So on net there is work being done. In the case where the box is being slowly lowered, the work will be done by the photons on the box [..]

Thanks that looks like an accurate (although still qualitative) description to me - and I would like add that the momentum exchange is accompanied by a corresponding Doppler frequency shift. Those are consistent with each other as well as with the analysis of Q-reeus. Such a box will likely never exist, but it can serve as a model for what happens when an object is lowered.

 

[Q-reeus]

Q-reeus: "Now lower (not free-fall drop) the same positron-electron pair down into the potential well of some gravitating mass..."

This means lower the pair and then place them at rest at some point down in the potential well? Is this different than letting the pair free fall, stopping their fall, then letting them annihilate?? I don't see how the gamma ray energy is any different at the top and bottom of the well: local observers each see the same energy locally, top and bottom, right?; but such an observer at the top will see redshift from the bottom gammas, at the bottom, blue shift from the top gammas.

No real difference in outcome Naty1 - that process is 'path independent'. Just making sure no-one unreasonably seized on any lack of specificity to infer it might have referred to annihilation during a free-fall situation (which would have resulted in a further redshift factor reduction in received frequency 'out there'). I find it pays to have a lawyer's attitude to wording on occasion!
[Oops - just to further clarify that last bit about further redshift in free-fall. Was meant to refer to that ray received by a distant observer inline with radial in-fall. There is also blueshift of the other ray and overall there would actually be no net redshift at all. I'l make a lawyer yet!]

 

[harrylin]

[..] "Now lower (not free-fall drop) the same positron-electron pair down into the potential well of some gravitating mass..."

This means lower the pair and then place them at rest at some point down in the potential well? Is this different than letting the pair free fall, stopping their fall, then letting them annihilate?? [..]

I did not linger on that part... but it might have been better to not include that! I guessed that Q-reeus wanted to avoid any issues with thermal energy.
Oh OK, the reason was different, as I see now. But the conclusion remains the same, sometimes less is better.

 

[Naty1]

No real difference in outcome Naty1 - that process is 'path independent'.

good..forgot to ask about paths...

yeah:

I find it pays to have a lawyer's attitude to wording on occasion!

you mean always!
One tough thing about these forums is that our posts don't get objective editing before posts like a textbook does...even when the poster really knows the facts. I am writing myself an explanation of some of this and thought I really had it down...until I read my own notes the next day and realized.."oops that's not quite right' and 'fixed' it..only to discover upon checking on a third day 'oops,again, did not think about this/that issue...It makes one appreciate a clearly written textbook...though not the pirces!

 

[Austin0]

Originally Posted by Austin0

Does this not mean that its 4-momentum vector (energy) remains constant (stays the same)along that worldline?

First, the four momentum is not the energy. Energy is the timelike component of the four momentum. So even if you interpret parallel transport as keeping the four momentum constant you cannot conclude that the same applies for energy.

How are you distinguishing between momentum and energy in this context. Both are just a function of frequency and the same constant (reduced Planck).
So if the momentum remains the same then the frequency must remain the same. If the frequency remains the same how could the energy vary??

Second, it is hard to interpret parallel transport as keeping a four vector constant. If you parallel transport a vector on a closed loop it does not generally wind up in the same orientation. It is an interpretation that can be useful over small regions of spacetime where you can neglect curvature, but it can cause problems if taken too far.

Austin0

the definitions of 4-momentum and vector transport in this context.

This context being; along a geodesic. Free falling photon.

I notice you did not comment on the actual point of the post. Or the logic leading to the expressed conclusion.

 

[PeterDonis]

Second, it is hard to interpret parallel transport as keeping a four vector constant. If you parallel transport a vector on a closed loop it does not generally wind up in the same orientation. It is an interpretation that can be useful over small regions of spacetime where you can neglect curvature, but it can cause problems if taken too far.

Hmm, this is a good point which I hadn't considered when I was talking about parallel transport before. You're right that the effects of curvature mean that even parallel transported vectors don't "stay the same", in the sense that there is a change around a closed curve. However, parallel transport is as close as we can come to a vector "not changing" along a curve in any invariant sense; we don't have any better definition to use. Also, in the cases we've been discussing, parallel transport does at least preserve properties "at infinity", such as constant energy at infinity along a geodesic.

 

[jartsa]

Where do these equations come from? I don't recognize any of them as standard SR or GR equations.

With those equations I describe what a static observer sees, when observing a light pulse that bounces up and down in a box that is being lifted.

And what the observer sees is:
1: Light moving down contracts and slows down.
2: Light moving up expands and speeds up.
3: The frequency of light bouncing off a mirror changes by an amount that is proportional to change of collision time caused by the motion of the mirror.

collision time = collision start time - collision end timeWhether falling light blue shifts or not, doesn't matter here, because the observer does not see any blue shift actually happening anyway.

 

[Dale]

With those equations I describe what a static observer sees, when observing a light pulse that bounces up and down in a box that is being lifted.

And what the observer sees is:
1: Light moving down contracts and slows down.
2: Light moving up expands and speeds up.
3: The frequency of light bouncing off a mirror changes by an amount that is proportional to change of collision time caused by the motion of the mirror.

collision time = collision start time - collision end time

Whether falling light blue shifts or not, doesn't matter here, because the observer does not see any blue shift actually happening anyway.

That is nice that you are sticking to your story, but none of that answers the question: where did those equations come from? I have never seen any of them. Did you read them in a manuscript that you can reference? Did you derive them from the Schwarzschild metric, or the parallel transport equation? Did you just pull them out of thin air because they sounded cool? To me, it looks like the last.

 

[Dale]

How are you distinguishing between momentum and energy in this context. Both are just a function of frequency and the same constant (reduced Planck).
So if the momentum remains the same then the frequency must remain the same. If the frequency remains the same how could the energy vary??

The energy is the timelike component of the four-momentum and the momentum is the spacelike component of the four-momentum. The four-momentum of a photon is a null vector, so the spacelike components must vary together with the timelike components, as is indeed implied by the fact that they are both proportional to the frequency. All three (energy, momentum, frequency) vary together.

I notice you did not comment on the actual point of the post. Or the logic leading to the expressed conclusion.

True. I rarely reply to an entire post when I think that replying to one key misconception is sufficient (or I run out of time). I don't do it to be rude, it is just a mental "triage". If there is a short specific point that you would like me to address and I neglected, then please mention it.

 

[jartsa]

That is nice that you are sticking to your story, but none of that answers the question: where did those equations come from? I have never seen any of them. Did you read them in a manuscript that you can reference? Did you derive them from the Schwarzschild metric, or the parallel transport equation? Did you just pull them out of thin air because they sounded cool? To me, it looks like the last.

I don't know what Schwartschild metrics is.

This equeation from physics book I did find very useful: time = distance / speed

There is also the inverse of time dilation factor k. That kind of takes care of the General Relativity part.EDIT: k is time dilation factor, not the inverse of it

 

[pervect]

I don't know what Schwartschild metrics is.

This equeation from physics book I did find very useful: time = distance / speed

There is also the inverse of time dilation factor k. That kind of takes care of the General Relativity part.

time = distance / speed works very well in an inertial frame. In General Relativity calculating distance becomes rather involved, and the relation time = distance / speed won't work with the usual definition of speed, which is usually defined by local measurements.

 

[harrylin]

[..] One tough thing about these forums is that our posts don't get objective editing before posts like a textbook does...even when the poster really knows the facts. I am writing myself an explanation of some of this and thought I really had it down...until I read my own notes the next day and realized.."oops that's not quite right' and 'fixed' it..[..] It makes one appreciate a clearly written textbook..[..]!

I thought exactly the same when I clarified my example in post #43. Happily the time to edit has been increased, so that it's sometimes still possible after a night's sleep to "fix" things for the benefit of later readers

 

[harrylin]

With those equations I describe what a static observer sees, when observing a light pulse that bounces up and down in a box that is being lifted.
[..]
3: The frequency of light bouncing off a mirror changes by an amount that is proportional to change of collision time caused by the motion of the mirror.

collision time = collision start time - collision end time [..].

Like Dalespam, I don't think so. Common theory of moving mirrors gets by fine neglecting "collision time" which is thought not to affect frequency. As I remarked in post #70, you can easily infer the frequency change from Doppler shift (qualitatively that is very easy).

 

[harrylin]

[..] The simplest assumption is that the walls of the box are perfectly reflecting mirrors; [..]with the photon's momentum perpendicular to the box wall reversing direction (and a corresponding change in the momentum of the box). In another thread I used this type of model to describe how one could extract work from a box filled with photons by slowly lowering it [..]

Could you find back that post by you? I ask, because now, when I went swimming, I realized that I too hastily agreed with your following explanation:

[..] Consider a photon that bounces off the upper wall of the box, travels to the lower wall, and bounces off it. Look at this process in the instantaneous rest frame of the box, which, by our underlying assumption, can cover both "bounce" events in a single local inertial coordinate patch.

If the box is in free fall, then it can be considered to be at rest in the local inertial frame for both bounce events, so the momentum exchange of the two events cancels. So there is no energy exchange between the photons and the box.

If the box is accelerated upward, then it is moving upward for one of the bounce events. That means that there is more momentum exchanged when the photon hits the lower wall of the box than when it hits the upper wall of the box.[..]

If I now correctly understand you, here you are effectively using the equivalence principle, so that you transformed (and I'm making it more concrete) to the frame of a space shuttle with a box inside. If the box is free falling together with the shuttle then the local observation should be as in an inertial frame of SR. OK so far.

Next its gets tricky: you say that when accelerating the box upward, more momentum and thus no energy is exchanged when the photon hits the lower wall of the box than when it hits the upper wall of the box. But you did not really show this, and I'm not sure if you mean relative to the Earth (as I originally thought) or relative to the falling frame.
Relative to the falling frame, a box in rest on Earth is also accelerating, and I think that no energy should be exchanged.
And contrary to my earlier thinking, that analysis doesn't yet answer the question of moving a box to a different height whereby the box accelerates, has a speed and decelerates.

 

[jartsa]

Like Dalespam, I don't think so. Common theory of moving mirrors gets by fine neglecting "collision time" which is thought not to affect frequency. As I remarked in post #70, you can easily infer the frequency change from Doppler shift (qualitatively that is very easy).

By time I meant duration.

Pick A collides with 6 guitar strings during 0.1 seconds time,
Pick B collides with 6 guitar strings during 0.2 seconds time

different sound results from these two collisions

 

[harrylin]

By time I meant duration.

Pick A collides with 6 guitar strings during 0.1 seconds time,
Pick B collides with 6 guitar strings during 0.2 seconds time

different sound results from these two collisions

Well, what can I say?

 

[jartsa]

Well, what can I say?

When you are sawing a log, the frequency of the vibrations changes if the speed of the saw changes. If the log does its own motion that also changes the frequency, if the saw contracts or expands that changes the frequency too.

But we can calculate the frequency in a simple way:

number of saw teeth / duration of one sawing stroke

 

[Dale]

I don't know what Schwartschild metrics is.

Well, once you learn some GR then please come back and try your derivation again.

 

[Cosmoknot]

Well, once you learn some GR then please come back and try your derivation again.

How much does a metric really matter when we are discussing light? Light only has 2 spatial dimensions to deal with since distance for it has completely contracted. Keeping it symmetrical then requires a second time dimension at C, making it 2+2 spacetime dimensions. Relativity means symmetry, right?

 

[Dale]

How much does a metric really matter when we are discussing light?

Since the metric is what determines which paths are null geodesics, it matters a lot. The rest of your post is really nonsense.