Calculation of gravitational redshift in an accelerating elevator

[externo]

TL;DR

Problem of comprehension in the method of calculation.

When we derive the formula of the redshift by the equivalence principle we imagine a light ray which goes from the bottom to the top of the elevator and which would take a duration t = h/c to make the journey, with h = height of the elevator. I don't understand why t = h/c, because while the ray goes from one partition to the other the elevator goes on, and the light therefore has a greater distance to travel than h.

 

[Dale]

I agree with you. Can you link to the source you are using? This seems like a mistake or an approximation. If it is an approximation then they should justify it.

 

[externo]

I have found several sources but in french.
There is more or less the same calculation here :
https://fr.wikipedia.org/wiki/Décalage_d'Einstein
or here
https://forums.futura-sciences.com/archives/26694-redshift-redshift-gravitationnel.html

so during this time t = h/c, the elevator increase his velocity by gh/c and therefore there is a doppler shift making df/f = gh/c²

 

[PeterDonis]

I don't understand why t = h/c, because while the ray goes from one partition to the other the elevator goes on, and the light therefore has a greater distance to travel than h.

While this is true in principle, that does not mean the correction involved is significant. If the time between two successive light pulses (or crests of a light wave) is small enough compared to $h / c$, the correction is negligible. My understanding is that this assumption is implicitly made in arguments of the kind you refer to, although, as @Dale notes, assumptions like this should be made explicit and not left for the reader to surmise.

 

[externo]

https://www.mathpages.com/home/kmath722/kmath722.htm

"But later, in Section 7.2 (Some Consequences of the Principle of Equivalence), Feynman gives another derivation of the frequency shift, and claims that it does not rely on the relation E = mc2. He says

"The principle of equivalence also tells us that clock rates are affected by gravity. Light which is emitted from the top of the accelerating box will look violet-shifted as we look at it from the bottom… The time that light takes to travel down is to a first approximation c/h, where h is the height of the box. In this time, the bottom of the box has acquired a small additional velocity, v = gh/c. The net effect is that the receiver is moving relative to the emitter, so that the frequency is shifted [by the Doppler factor] 1 + v/c = 1 + gh/c2. Note that this conclusion does not depend on E = mc2 and on the existence of energy levels, which we had to postulate in the argument we have given previously. The conclusion is based on the expected behavior of classical objects; the calculation results from geometry and kinematics, and makes a direct physical prediction from the postulate of equivalence."

As explained previously, this argument is incorrect, because..."

 

[Dale]

You will need to look in detail at your french sources, but the mathpages quote is correct. There they explicitly say "The time ... is to a first approximation ..." So they are telling you that this quantity is an approximation, specifically a first order approximation. That is ok.

For a receiver at the front has a worldline $(c t,h+1/2 \ g t^2)$ and a pulse of light from the back has a worldline $(c t,c t)$. Setting those equal gives $$t=\frac{c-\sqrt{c^2-2gh}}{g}=\frac{h}{c}+\frac{gh^2}{2c^3} + \frac{g^2 h^3}{2 c^5} + \frac{5 g^3 h^4}{8c^7}+ O(h^5)$$

 

[PeterDonis]

the mathpages quote is correct

Do you mean the part that is quoted from Feynman (which is what you refer to in your post), or the part after the quote from Feynman, in which the author of the mathpages article argues that Feynman's argument is incorrect?

 

[Dale]

I mean that it correctly identifies that the quantity the OP is complaining about is an approximation. It is up to the OP to determine if the french sources similarly identify it as an approximation, which is OK.

 

[PeterDonis]

I mean that it correctly identifies that the quantity the OP is complaining about is an approximation.

Yes, Feynman correctly identified that. But the counter-argument the mathpages author is making is that, in this approximation, there is no redshift, because the travel time of each light pulse (or crest of the light wave) from source to receiver is the same, and the argument that there is a redshift depends on their being different. So according to the mathpages author, specifying the approximation, in itself, is not sufficient to make the argument for the redshift correct.

 

[Dale]

I am not worried about the whole argument, only the part that the OP is complaining about.

To my understanding the OP is not learning from the mathpages source, but from the french sources. I am only showing them what to look for in their french sources.

 

[PeterDonis]

while the ray goes from one partition to the other the elevator goes on, and the light therefore has a greater distance to travel than h.

While this is true in principle, that does not mean the correction involved is significant.

the counter-argument the mathpages author is making is that, in this approximation, there is no redshift, because the travel time of each light pulse (or crest of the light wave) from source to receiver is the same, and the argument that there is a redshift depends on their being different

To reconcile these seemingly different statements that I have made: I think there are actually two different possible arguments that could be made for deducing gravitational redshift from the equivalence principle.

(1) One could argue starting from Newtonian mechanics. That seems to be what the second Feynman quote given in the mathpages article is doing, and the argument that the mathpages author is refuting. The key to the mathpages author's counterargument is basically that the scenario is stationary, and in a stationary scenario the travel times of successive light pulses (or wave crests) from source to receiver is independent of time. That, plus the fact that in Newtonian mechanics, the only source of Doppler shift is a change in the travel time of successive light pulses (or wave crests), implies that Newtonian mechanics predicts no gravitational redshift in a stationary scenario.

(2) One could argue starting from Special Relativity. Doing this changes the argument in two ways:

(2a) In Newtonian mechanics, a scenario in which both the bottom and the top of the elevator have the same proper acceleration is stationary. But in SR, it is not: in order for the scenario to be stationary in SR, the top of the elevator must have a slightly smaller proper acceleration than the bottom.

(2b) In SR, we have a known exact model for the stationary scenario of the elevator: the Rindler congruence of worldlines. And it is easily shown that in that model, there is redshift of light signals going from the bottom to the top of the elevator, even though the scenario is stationary (and even though the top has a slightly smaller proper acceleration than the bottom). So the EP argument for gravitational redshift in this context is straightforward: a small enough patch of any curved spacetime looks like flat spacetime, and a small enough patch of the stationary worldlines of the bottom and top of a room sitting at rest on the surface of a gravitating body like the Earth will look like the Rindler worldlines of the bottom and top of the elevator, and hence gravitational redshift will be present. This does not require any heuristic or approximate reasoning of the kind often given in previous arguments in the literature (including Einstein's own arguments); it is a simple, exact result.

Argument (2b) above is, IIRC, at least implicit in some discussions in the literature, but I think it is fair to say that it is not emphasized as much as it should be.

 

[PeterDonis]

I am not worried about the whole argument, only the part that the OP is complaining about.

Fair enough. I am concerned with making sure the OP realizes that even fixing the part the OP is complaining about does not, IMO, make the argument correct. I have given my viewpoint in post #11.

 

[Sagittarius A-Star]

(1) One could argue starting from Newtonian mechanics. That seems to be what the second Feynman quote given in the mathpages article is doing, and the argument that the mathpages author is refuting.

The mathpages author wrongly accuses Feynman of describing Newtonian mechanics in combination with the emission theory "provided they are emitted at a constant speed relative to the source". In this case, described in an inertial frame, every consecutive light pulse would be emitted slower than the previous one and have the same travel time. But in the cited text, Feynman didn't write that.

 

[PeterDonis]

The mathpages author wrongly accuses Feynman of describing Newtonian mechanics

Yes, good point, the Feynman Lectures on Gravitation, which is what the mathpages author is quoting from, were (at least for the most part--see below) based on relativity, not Newtonian mechanics.

However, Feynman does discuss Newtonian mechanics in Chapter 7 as well, and he does not make use of the exact SR solution I mentioned for an accelerating elevator (the Rindler congruence), so it's quite possible that the mathpages author did not read Feynman's section 7.2 as making an argument based on SR.

 

[Dale]

For a receiver at the front has a worldline $(c t,h+1/2 \ g t^2)$ and a pulse of light from the back has a worldline $(c t,c t)$. Setting those equal gives $$t=\frac{c-\sqrt{c^2-2gh}}{g}=\frac{h}{c}+\frac{gh^2}{2c^3} + \frac{g^2 h^3}{2 c^5} + \frac{5 g^3 h^4}{8c^7}+ O(h^5)$$

If they are looking for a fully relativistic derivation then the worldline of the receiver is $$\left( c t, h + \frac{c^2}{g}\left( \sqrt{1+\frac{g^2 t^2}{c^2}} -1 \right) \right)$$ Then solving for $t$ as above gives $$t=\frac{h(2c^2-gh)}{2c(c^2-gh)}=\frac{h}{c}+\frac{gh^2}{2c^3}+\frac{g^2 h^3}{2 c^5}+\frac{g^3 h^4}{2c^7}+O(h^5)$$
They actually wind up being the same up to 3rd order.

 

[PeterDonis]

If they are looking for a fully relativistic derivation then the worldline of the receiver is $$\left( c t, h + \frac{c^2}{g}\left( \sqrt{1+\frac{g^2 t^2}{c^2}} -1 \right) \right)$$

This looks like a Bell congruence worldline, not a Rindler congruence worldline. But the assumption is the scenario are that it is stationary, which requires a Rindler congruence worldline.

 

[Dale]

This looks like a Bell congruence worldline, not a Rindler congruence worldline. But the assumption is the scenario are that it is stationary, which requires a Rindler congruence worldline.

It is just a worldline of a single constant-proper-acceleration receiver so there is no difference between Bell and Rindler. There is no congruence here. But you are right that a Rindler congruence could not use a simple $+h$ to distinguish the receiver from the transmitter

 

[PeterDonis]

It is just a worldline of a single constant-proper-acceleration receiver so there is no difference between Bell and Rindler. There is no congruence here.

Hm, yes, technically this is true for what you are computing because you are only including a single point on the emitter's worldline, which you can simply assume to be the origin $t = 0$, $x = 0$.

However, in order to make the argument being discussed in this thread, it is not sufficient to include only a single point on the emitter's worldline. You have to include at least two: the emission points of two successive light pulses (or wave crests). That brings in the issues I discussed in post #11.

 

[Dale]

However, in order to make the argument being discussed in this thread, it is not sufficient to include only a single point on the emitter's worldline. You have to include at least two: the emission points of two successive light pulses (or wave crests). That brings in the issues I discussed in post #11.

It is sufficient to make the point that $h/c$ is a valid first order approximation whether you use no acceleration, Newtonian acceleration, or relativistic acceleration.

I never claimed to be weighing in on your discussion which I think is probably distracting from the OP’s question. I was just surprised how far out the Newtonian and relativistic accelerations matched and thought I would post it.

 

[PeterDonis]

It is sufficient to make the point that $h/c$ is a valid first order approximation

You have shown that it's a valid first order approximation for describing the worldline of a single light pulse (or wave crest). But I still think it's worth noting that any argument for gravitational redshift needs to look at the worldlines of two successive light pulses (or wave crests).

 

[Sagittarius A-Star]

Calculation of gravitational redshift in an accelerating elevator

The metric in Kottler–Møller coordinates of the Rindler frame is
$ds^2 = (1+ \frac{g\,x} {c^2})^2c^2dt^2 - (dx^2 + dy^2 + dz^2)$, from which follows the time-dilation formula:$$d\tau = dt\sqrt {(1+ \frac{g\,x} {c^2})^2 - \frac{v^2}{c^2}}$$
If two clocks are at rest in the accelerated rocket ($v_1=v_2=0$), the reference clock located at $x_1=0$ and the other clock with proper time $\tau$ at $x_2=h$, then from the time dilation formula follows:
$$\frac{d\tau}{dt} = 1+ \frac{g\,h} {c^2} $$

 

[externo]

I believe that I have understood.
From inside the cabin, the light takes t = h/c to overcome the distance. During this time, the cabin has moved away from g*h/c with respect to the source's emission point, and this produces redshift. There is a contradiction here, and this contradiction is removed by Einstein in a 1911 paper by saying that time passes more slowly in the back of the cabin than in the front and this is what explains the redshift. One cannot otherwise explain the redshift within the framework of the assumptions of Einstein because the source of the light and its receiver are the same object which is motionless compared to itself. So we cannot explain the redshift by case 1, ie by the fact that the wave is elongated because the source moves away.
On the other hand, assuming the anisotropy of the speed of light, the redshift can be explained by a loss of speed of light relative to the cabin (case 2). But this redshift would then cause no time lag between the top and the bottom.
More later.
https://journals.openedition.org/bibnum/1072

 

[Dale]

The metric in Kottler–Møller coordinates of the Rindler frame is
$ds^2 = (1+ \frac{g\,x} {c^2})^2c^2dt^2 - (dx^2 + dy^2 + dz^2)$, from which follows the time-dilation formula:$$d\tau = dt\sqrt {(1+ \frac{g\,x} {c^2})^2 - \frac{v^2}{c^2}}$$
If two clocks are at rest in the accelerated rocket ($v_1=v_2=0$), the reference clock located at $x_1=0$ and the other clock with proper time $\tau$ at $x_2=h$, then from the time dilation formula follows:
$$\frac{d\tau}{dt} = 1+ \frac{g\,h} {c^2} $$

This is, IMO, the best way to derive time dilation as it is both completely general and usually easier than any other approach.

 

[Dale]

On the other hand, assuming the anisotropy of the speed of light, the redshift can be explained by a loss of speed of light relative to the cabin (case 2).

No, simple anisotropy doesn’t lead to redshift here. You can use LET, but LET involves more than just anisotropy

 

[externo]

No, simple anisotropy doesn’t lead to redshift here. You can use LET, but LET involves more than just anisotropy

In the mathpage ou can read at the end :
"As we’ve seen, for classical particles in Galilean spacetime, the travel time for each successive particle is the same, so there is no frequency shift between emission and absorption, although the equivalence principle is perfectly satisfied. One might try to salvage the argument by postulating a classical wave theory of light in some stationary ether, but this not only conflicts with many experiments, it also restricts the argument to frequencies of electromagnetic waves, which would not account for the fact that time dilation entails changes in frequency for sequences of ballistic particles as well. Thus the only way that the equivalence principle leads to gravitational frequency shift is when it is combined with the postulate of local Lorentz invariance (equivalent to the postulate E = mc2)."

This sentence "but this not only conflicts with many experiments" is wrong, but for the rest I agree.
Classical theory does not explain the change in frequency of ballistic particles. If I don't mistaken this paper states that LET is falsified.
LET would be falsified, unless we find another explanation for the difference in the passage of time between the front and the back.

 

[Dale]

I don’t think that they are referring to LET in that statement. I think that they are referring to classical rigid or dragged aether theories. Those have been contradicted experimentally.

The second part which you highlighted is correct. The Einstein equivalence principle indeed says that locally physics reduces to special relativity. You cannot get the right equations (to high orders) by assuming Newtonian physics locally.

 

[Sagittarius A-Star]

This is, IMO, the best way to derive time dilation as it is both completely general and usually easier than any other approach.

I agree. However, the derivations of gravitational red-shift, linked by the OP, have a didactical advantage. They provide the insight, that the gravitational red/blue-shift is nothing else than a longitudinal Doppler-effect, if described with reference to an inertial frame.

 

[Dale]

I agree. However, the derivations of gravitational red-shift, linked by the OP, have a didactical advantage. They provide the insight, that the gravitational red/blue-shift is nothing else than a longitudinal Doppler-effect, if described with reference to an inertial frame.

I agree. If I were teaching a course on this material I would introduce the time dilation derivation from the metric. Then I would go through the cumbersome exercise of deriving the gravitational redshift from the Doppler shift in an accelerated frame in flat spacetime. And then I would have the students compare the two results to obtain the physical insight. But I would always present the metric-based approach as the definition of the time dilation and the other as a physically intuitive "gee whiz" exercise.