Length contraction of falling things

harrylin

Yes, exactly. That is what is commonly done for atomic clocks in time keeping with the ECI frame: They are corrected for altitude, otherwise our time keeping on Earth would not work. But indeed "synchronise" is ambiguous, from the context it makes no sense to interpret it otherwise but I should have used the word "tune" from the start.
And my example isn't perfect, as the ECI frame isn't perfectly inertial - thus a very small difference will still be there. I now added a little footnote.
Regretfully yes. I now clarified that for later readers.

Austin0

i think you may be missing a significant point.

These are two very different descriptions of a single unambiguous physical reality.
The state of a photon during transit. The two descriptions are not themselves mutually exclusive. It could possibly be that both conditions pertain. I.e. The photon starts out with a higher frequency at a higher potential due to dilation and gains kinetic energy during transit to a lower potential.
It is the very fact that there is agreement on the observable results that renders them mutually exclusive.
AFAIK those results exactly correspond to the dilation factor corresponding to the difference in potential.
So if we accept the fundamental dilation concept as applied to clocks and consider it applies equally to electron emissions and receptions, then this both explains and necessitates those results completely, with no observable measurement left over to validate a posited frequency shift due to gravitational influence during transit.
If there was such a gradual change during transit then the results would inevitably be gamma + some additional factor. Such is the case with relativistic Doppler where the result is a composite of two effects.

So it is not simply a matter of semantics or interpretation , but a question of the actual physics and is unequivocal. Does a photon kinetically gain energy during transit or not?

The proposition that it does could only be true if our current conception of time dilation and it's effects on atomic periodicity is not correct.
Otherwise it would seem that it is not merely misleading but actually false. IMO

harrylin

Sorry for the ambiguity; I now added a footnote.

PeterDonis

No, it's not possible. Neither of these descriptions are "the state of the photon during transit". The state of the photon during transit is that its 4-momentum is parallel transported along its worldline, and its energy as seen by an observer is the contraction of its 4-momentum with the observer's 4-velocity, as I stated. That's the physics; everything else is interpretation.

You can *interpret* the above as the photon gaining energy (or frequency) as it "falls" in the gravitational field, or you can *interpret* it as the photon's energy "staying the same" but getting measured by an observer with a different "rate of time flow" than where it was emitted. Both of these are interpretations; neither one is the physics.

The "time dilation factor" corresponds to the "difference in potential" because that's how the two are defined. Both of them are defined in terms of the metric coefficient [itex]g_{tt}[/itex] in Schwarzschild coordinates. Or, if you want an invariant description, both of them are defined in terms of the metric as seen by observers who follow orbits of the time translation Killing vector field of the spacetime. Calling the difference in this with altitude "time dilation" is an *interpretation* of this particular feature of the metric; so is calling it a "difference in potential".

The time dilation concept is not fundamental. The EFE and its particular solutions are fundamental--or, equivalently, the spacetime geometry, the metric, is fundamental, as are the geometric objects in it, such as a photon's 4-momentum. "Time dilation" is a derived concept that applies in certain scenarios. So is "difference in potential".

harrylin

Obviously we disagree here. How can in my last example the explanation work that the frequency of the radio wave increases as it "falls" down towards the frequency analyser?

PeterDonis

To be strictly correct, I should have specified that I was talking about the "natural" frequency that would be measured by an observer, i.e., the frequency obtained by contracting the photon's 4-momentum with the observer's 4-velocity.

In your scenario you are artificially changing the "frequency" that your analyzer reports when it measures a photon. For your frequency analyzer, the rule I gave, that the measured frequency is given by contracting the photon's 4-momentum with the observer's 4-velocity, does not hold as I stated it; to compute the frequency your analyzer will report, you have to use a *different* 4-vector, basically the unit timelike 4-vector of the ECI frame. This is *not* the same 4-vector as the analyzer's 4-velocity; if it were, you wouldn't have to artificially synchronize the analyzer with the ECI.

jartsa

Not for many consecutive reflections, too complicated. But reflections at the bottom and at the top go something like this:



light - mirror collision time at the bottom:

k is the time dilation factor, or maybe inverse of that, k < 1
c is coordinate speed of light far away from gravity source
s is coordinate length of light pulse far away from gravity source
v is coordinate speed of mirror

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



And here is light-mirror collision time at the top of the box:


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


And frequency can be calculated as: number of wave crests in the wave pulse / collision time

Austin0

Originally Posted by Austin0

"state of the photon during transit is that its 4-momentum is parallel transported along its worldline"

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

"the photon's energy "staying the same" but getting measured by an observer with a different "rate of time flow" than where it was emitted"

So this description is a literal verbal translation of the definitions of 4-momentum and vector transport in this context. How then could you say this was not the physics of a photon in transit??


Originally Posted by Austin0

The point was not the self evident fact that dilation and potential correspond by definition.
But that the observed results correspond to the expected dilation factor.


Originally Posted by Austin0
.
this is just semantic antics. Would you say that time dilation was not a fundamental concept of SR? Time dilation is inherent in the metric just as it is in the Minkowski metric.
It does not just apply in certain scenarios but in all cases involving clocks at different altitudes , yes? And it is not just an interpretation but a word describing the relationships of clock rates . Whats that if not physics?

harrylin

The ECI frame can only work with what you call "artificial synchronization" and I did not refer to your analysis but to what you call an "interpretation and not the physics", as you put it:

"the photon gaining energy (or frequency) as it "falls" in the gravitational field".

Discussions here are about the physics and my example in post #43had a two-fold purpose: I also had in mind to illustrate in a very clear way that the physics of that interpretation is messed up.
So far, it looks to me a clear case of mixing up in one description two different perspectives according to the equivalence principle; in physics one cannot freely jump and mix reference systems*. But if, contrary to my thinking, that interpretation predicts without inconsistencies the observations of my example, then this will be less easy to do than I expected.

*For example it's nonsense if I say that you are gaining energy because I took off in an airplane. A physical interpretation can only be valid if it relates to a valid reference system for description of physics.

DaleSpam

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

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/Schwarz...zschild_metric
For elevations further in, one simply applies the appropriate correction factor - observed redshift is somewhat less then in general.
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 occuring 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

Harry - it really pays to read an entire thread before commenting critically on an initial part thereof - see you own further remarks below!!
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?
I have no idea what point you are trying to make here - please elaborate!
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!
Huh?

DaleSpam

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

Naty1

PeterDonis:
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

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.
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 ).

That sounds like a good analysis to me.

Q-reeus

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!
Tell me about it - are there local chapters of 'PhysicsForums Anonymous' offering addicts any hope of rescue?!
Thanks Harold - nice to get some positive feedback now and then.

PeterDonis

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.