Is there an absurdity in Einstein's equivalence principle? [closed]

A friend of mine, Amir Assarzadeh, upon noticing my thought experiments [1&2], asked me an intriguing question to which he also made an answer defending Einstein's equivalence principle (EEP). However, I thought that it is worth discussing here:

Assume a very massive planet and a shuttle at rest WRT the planet though keeping a large distance from it so that the effect of the planet's gravitational field is negligible on the shuttle. If we locate two similar clocks, one in the shuttle and the other on the surface of the planet, there would be a discrepancy between the clock rates for the very moment at which the shuttle starts up its engines undergoing a similar acceleration to the planet's gravitation from the viewpoint of the Schwarzschild observer.

In other words, we all know that the clock on the massive planet runs slower due to the gravitational time dilation as measured by a Schwarzschild observer, however, as the shuttle starts up its engines, the clock does not change its rate because the shuttle has still a zero relative velocity WRT both the planet and the Schwarzschild observer. Although the observer in the shuttle claims that EEP is held true at the moment of and all during the firing of the engines, and thus he anticipates that his clock must run slower similar to that located on the planet, the Schwarzchild observer claims that there is no time dilation for the shuttle's clock because it has no considerable velocity compared to that of light. Is this a paradox?

My friend, however, made a simple answer to this anomaly as to which I am not convinced. He claimed that EEP is applicable only for the observers inside the shuttle and on the planet. That is, a Schwarzschild observer is necessarily not able to use EEP for other noninertial observers from his own viewpoint.

Do you think his answer is valid?

• Maybe studying this will help plato.stanford.edu/entries/spacetime-iframes Jan 2, 2020 at 10:13
• You seem to post a lot of these questions that ask about general relativity in vague language, without mathematics. Although it's possible to learn a certain amount about GR without math, there is a limit to how far that will take you. You're getting correct feedback from a number of people, e.g., in Dale's comment above explaining that "the EEP discusses only local observations and gravitational time dilation is inherently non-local." I agree with anna v that your next step should be to read a textbook.
– user4552
Jan 2, 2020 at 17:12

The equivalence principle gets stated in all sorts of different ways, but it is basically the statement that the four-acceleration is given by the equation:

$$A^\alpha = \frac{\mathrm d^2x^\alpha}{\mathrm d\tau^2} + \Gamma^\alpha{}_{\mu\nu}U^\mu U^\nu$$

The first term on the right hand side is basically the coordinate acceleration, i.e. rate of change of position with time, while the second term is basically the gravitational acceleration, i.e. the gravitational force divided by the mass.

The reason the equivalence principle is embodied in this equation is that by changing our coordinate system we can make either term zero i.e. we can make the acceleration look purely kinetic or purely gravitational (more on this here). A change of coordinates is just a change in the way we measure positions in spacetime, not a physical change, so the conclusion is that the two types of acceleration must be physically equivalent.

However the equation I cite give the four-acceleration at a point in spacetime because the quantities that appear in the equation are functions of position in spacetime. It is vital to understand this because it means acceleration and gravity can be treated as the same at the same point in spacetime. The equivalence principle is a local principle not a global principle.

This is where your proposed contradiction falls down because you are comparing acceleration and gravity at different points in spacetime. The EEP does not claim they would be the same and there is no reason to suppose they would be.

EEP tells you that effect of gravitation is locally indistinguishable from that of an acceleration, so locally you can forget gravitational field and you can just use special theory of relativity (STR).

Locality is important though. For example, you cannot simulate radial field by acceleration. Moreover you can only consider area so small, that even tidal forces are neglectable. The observer in the shuttle has no "right" to compare his clock to any other clock other than those in the shuttle if he is to use only STR. If he wants to compare his clock anyway, STR+EEP is not enough, he will need additional assumptions, e.g. knowledge of Schwarzschild metric.

• What does "f.e." mean? Jan 2, 2020 at 12:27
• @KyleKanos for example Jan 2, 2020 at 13:10
• Ah. I imagine more people are familiar with the Latin abbreviate e.g. (for exempli gratia, which translates exactly to "for example"). Jan 2, 2020 at 13:19

My friend, however, made a simple answer to this anomaly as to which I am not convinced. He claimed that EEP is applicable only for the observers inside the shuttle and on the planet. That is, a Schwarzschild observer is necessarily not able to use EEP for other noninertial observers from his own viewpoint.

Do you think his answer is valid?

Yes, your friend's answer is correct. As described here the Einstein equivalence principle (EEP) is "The outcome of any local non-gravitational experiment in a freely falling laboratory is independent of the velocity of the laboratory and its location in spacetime" (emphasis added). The restriction to local experiments is crucial for applying the EEP.

According to the EEP any experiment conducted near the clock inside the shuttle will give the same result as any experiment conducted near the clock on the planet. But comparing either clock to an observer at infinity is clearly no longer a local experiment. The EEP makes no claim that such non-local experiments will give similar results.