-3

In his book, Relativity: The Special and General Theory, Einstein claimed that the clocks in a gravitational field, as well as those located on a rotating disc, run slower solely due to the gravitational potential, no matter how much acceleration they undergo. He then replaces potential per unit mass with velocity square $(r^2\omega^2)$ of a clock located at a radius $r$ on the disc:

... If we represent the difference of potential of the centrifugal force between the position of the clock and the centre of the disc by $\phi$ , i.e. the work, considered negatively, which must be performed on the unit of mass against the centrifugal force in order to transport it from the position of the clock on the rotating disc to the centre of the disc, then we have

$$\phi=\frac{\omega^2r^2}{2}$$

... Now, as judged from the disc, the latter is in a gravitational field of potential $\phi$, hence the result we have obtained will hold quite generally for gravitational fields...

... Now $\phi=-K\frac{M}{r}$, where $K$ is Newton’s constant of gravitation, and $M$ is the mass of the heavenly body.

Assume that we have a massive spherical planet with a hollow core. At the first approximation, the gravitational acceleration is zero everywhere inside the hollow core as predicted by classical mechanics. However, the gravitational potential is not zero in the core. Now, consider two observers, one has been trapped and is floating inside the core of this massive planet, and the other floating inside a massless shell in an interstellar space away from the gravitation of the planet, yet at rest with respect to the planet. (A Schwarzschild observer)

If these observers hold similar light clocks, GR predicts that time dilates for the clock within the planet's core because of non-zero gravitational potential compared to the clock held by the observer inside the massless shell. Nonetheless, do you think that this result is logically tenable, or is it compatible with EEP (Einstein equivalence principle)?

Indeed, using EEP, each of these observers can by no means detect if he is inside a massive planet or a massless shell floating in interstellar space. That is, observers' feelings are the same, both undergoing similar weightlessness, and all experiments have similar outcomes from the viewpoint of these observers.

According to the fact that the relative velocity of these observers is zero, it is logically anticipated that their clocks run at the same rate but this is not the case! Where is the problem?

Community
  • 1
Mohammad Javanshiry
  • 2,513
  • Why do you believe it is "logically anticipated" that the clocks run at the same rate if the relative velocity of the clocks is zero? (The whole point of gravitational time dilation is that this not the case.) – TimRias Nov 26 '19 at 08:58
  • @mmeent Because no experiment carried out by the observers can distinguish these two cases from each other since the observers experience a weightlessness situation. If the clock in the planet's core runs slower, this can be considered as a distinguishing experiment, which violates my first sentence. – Mohammad Javanshiry Nov 26 '19 at 09:10
  • 1
    That is not what the EEP says. The EEP says that the observers individually cannot do (local) experiments that distinguish the two cases. An observer in a freely falling elevator cannot distinguish his situation from floating in free space, but an external observer can easily tell the two situations apart. – TimRias Nov 26 '19 at 09:35
  • As an aside, note that the relative velocity between to distant observers is not a well defined observable in the context of general relativity. Hence there is (typically) no way to establish that the relative velocity between two observers is zero! – TimRias Nov 26 '19 at 09:37
  • @mmeent There can be different versions and interpretations of EEP. (A version is about free fall that mentioned by you. The other is about the equivalence between an observer in an accelerated craft with that in a uniform G-field, and mine can be considered as another one.) But the cornerstone of all is that if the feelings of the observers or the outcomes of their experiments are assumed to be the same, these observers are equal. You can forget about EEP as well. Investigate the question from the viewpoint of logic. – Mohammad Javanshiry Nov 26 '19 at 09:48
  • @mmeent You can consider the velocity negligible compared to the great effect of the G-potential inside the core. Therefore, you can assume that the observer in the massless shell is far enough from the planet that his acceleration and instantaneous velocity towards the planet are very small. – Mohammad Javanshiry Nov 26 '19 at 09:55
  • Your "version" of the EEP is simply not satisfied by nature. (And may not be possible to realize in a logically consistent way.) – TimRias Nov 26 '19 at 09:57
  • @mmeent I'm not following. Are you saying that you are nature's legal representative?! – Mohammad Javanshiry Nov 26 '19 at 10:07
  • I think it's more accurate to say that our current model of nature (specifically General Relativity) doesn't include the version of the EEP you're proposing. You're welcome to try to come up with a model of nature that does embody your version. If you're successful, you can then see if it does a better job of predicting physical phenomena than GR does. – Michael Seifert Nov 26 '19 at 12:33
  • 1
    Why the downvote? It seems a finely formulated question – lcv Dec 15 '19 at 12:35

1 Answers1

9

The Einstein Equivalence Principle states (roughly) that all local experiments in a freely falling observatory are indistinguishable from those in flat spacetime. If these two observers sent radar pulses to each other and compare the rates at which their clocks are ticking, this is a non-local experiment. And so long as your two observers do not communicate with each other, they have no way of knowing that there is another observer out there whose clocks run differently from theirs.

Thus, there's no inconsistency with the Einstein Equivalence Principle.

Michael Seifert
  • 48,772
  • 1
    @MohammadJavanshiry: You seem to have interpreted my response as saying that the observers cannot possibly communicate with each other. This is not what I'm saying. I'm saying that when they do communicate with each other, their communication does not qualify as a "local experiment" for the purposes of the EEP. – Michael Seifert Nov 25 '19 at 20:50
  • @J. Murray This answer is misleading, and it is not correct. Observers' communication can usually upset local results when they are transferred via non-local arrangments. This is not my question's problem. Many other experiments do have this deficiency. – Mohammad Javanshiry Nov 25 '19 at 21:14
  • @Michael Seifert Even in SR, when a lab observer receives the increased frequency of an approaching light source, he is not allowed to deduce that the source clock runs faster. He needs to consider the relativistic Doppler effect to obtain the correct time dilation. It is obvious in this example that communication changes the true results. But these changes cannot intrinsically affect the local results. For any observer, it suffices to imagine any experiment, rather than to receive the results by communication. – Mohammad Javanshiry Nov 25 '19 at 21:28
  • @MohammadJavanshiry: What are the "local results" you're referring to? What experiment can the observer do inside the shell that allows them to say something about time dilation relative to observers outside the shell? – Michael Seifert Nov 25 '19 at 21:37
  • @Michael Seifert I'm referring to two light clocks. I think you are mixing apples and oranges. This is a relativity question, it is not Schrödinger's cat! – Mohammad Javanshiry Nov 25 '19 at 21:55
  • @Mohammad Javanshiry this answer is correct. The EEP is not violated. All local measurements by both observers follow the EEP and the EEP simply does not apply for non local measurements like gravitational time dilation. The fact that measurements of gravitational time dilation are non local is indeed the key to resolving the issue raised in your question – Dale Nov 27 '19 at 13:51
  • @Dale I do not understand why for outside a non-rotating sphere the time dilation has easily been calculated WRT the Schwarzschild observer, whereas for its inside, as reflected in my example, everything becomes non-local. https://en.wikipedia.org/wiki/Gravitational_time_dilation#Outside_a_non-rotating_sphere – Mohammad Javanshiry Nov 27 '19 at 14:03
  • Between the outside and the inside of the shell the spacetime is substantially curved due to the shell itself. Local means that the measurements are restricted to a region of spacetime over which the curvature is negligible. By construction that cannot include anything crossing the shell. – Dale Nov 27 '19 at 14:12
  • @Dale Is it correct if I say: Because the observer on the planet can never make any local measurements for time dilation WRT a Schwarzschild observer in that he has to communicate with the Schwarzschild observer non-locally, and thus Wikipedia is flawed along with many textbooks? – Mohammad Javanshiry Nov 27 '19 at 14:15
  • I am not sure what flaw you are referring to (probably there are many), but yes the non-local communication is necessary for establishing gravitational time dilation. Is the flaw a claim that gravitational time dilation can be measured locally? If so, that is indeed a big flaw. – Dale Nov 27 '19 at 14:22
  • @Dale I said the hollow core, there is no material in there. It is possible that GR fails inside matter, but it is applicable for all vacuum regions or those with very low mass densities such as dust. – Mohammad Javanshiry Nov 27 '19 at 14:22
  • @Dale Is the flaw a claim that gravitational time dilation can be measured locally? What do you interpret from the Wiki article? Isn't it as if the gravitational time dilation is measured locally? I want to say whatever measurements that are made on a planet and are compared to those made by a Schwarzschild observer are mostly considered to be local, yet I think it's not problematic at all. – Mohammad Javanshiry Nov 27 '19 at 14:37
  • Let us continue this discussion in chat. – Dale Nov 27 '19 at 14:40