B Metric tensor for a uniformly accelerated observer

[Pyter]

The excerpt does not say that, for example, an observer on the rotating disc with a particular value of $\phi$ has the same time dilation as an observer standing on the surface of the Earth with the same value of $\phi$ .

As I'm interpreting it, the same value of $\phi$ is exactly what links the two scenarios and make them equivalent.

[ibid, 157]
If we represent the difference of potential of the centrifugal force between the position of the clock and the centre of the disc by

, 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

From this it follows that

In the first place, we see from this expression that two clocks of identical construction will go at different rates when situated at different distances from the centre of the disc. This result is also valid from the standpoint of an observer who is rotating with the disc.

Now, as judged from the disc, the latter is in a gravitational field of potential ϕ, hence the result we have obtained will hold quite generally for gravitational fields. Furthermore, we can regard an atom which is emitting spectral lines as a clock, so that the following statement will hold:

An atom absorbs or emits light of a frequency which is dependent on the potential of the gravitational field in which it is situated.

[158]
The frequency of an atom situated on the surface of a heavenly body will be somewhat less than the frequency of an atom of the same element which is situated in free space (or on the surface of a smaller celestial body).

Now

,
where

is Newton's constant of gravitation, and

is the mass of the heavenly body. Thus a displacement towards the red ought to take place for spectral lines produced at the surface of stars as compared with the spectral lines of the same element produced at the surface of the earth, the amount of this displacement being

[...]

In the next "section" 158, Einstein equates the "centrifugal" potential of the rotating observer with the gravitational potential of the observer on the surface of the massive body and concludes that they are subject to the same frequency displacement with respect to $\nu_0$.

 

[A.T.]

As I'm interpreting it, the same value of $\phi$ is exactly what links the two scenarios and make them equivalent.

In the next "section" 158, Einstein equates the "centrifugal" potential of the rotating observer with the gravitational potential of the observer on the surface of the massive body and concludes that they are subject to the same frequency displacement with respect to $\nu_0$.

The potentials are not the same, but they are related to the respective frequency shift and gravitational time dilation in the same way.

 

[vanhees71]

One must be careful about this rotating disk example. It was not fully understood in the first discussions about it. Only a bit later with M. Noether and Herglotz it was understood that a Born-rigid body only allows for very limited motions, all of which were classified by these two authors. The Born-rigid body has only 3 degerees of freedom not 6 as the "Newtonian" rigid body.

One of the allowed motions is rotation at constant angular speed, but you cannot get there in a continuous way from the non-spinning disk. This implies that a realistic treatment must treat the situation, where the disk is brought from 0 to a finite spin, as an elastic body. Then all the paradoxes, including the one with the conserved particle number are easily solved.

As I already wrote above, the time-dilation effect of an observer rotating around a fixed axis measuring the frequency of light emitted from a source on the axis can be measured by a local experiment, and it's obeying the equivalence principle.

 

[A.T.]

One of the allowed motions is rotation at constant angular speed, but you cannot get there in a continuous way from the non-spinning disk. This implies that a realistic treatment must treat the situation, where the disk is brought from 0 to a finite spin, as an elastic body.

You could also construct the disc as already spinning at constant angular velocity, if you are only interested in that part.

 

[PeterDonis]

As I'm interpreting it, the same value of $\phi$ is exactly what links the two scenarios and make them equivalent.

You're interpreting it wrong. Einstein is relating the difference in $\phi$ between two observers to the time dilation between those two observers. (The observer at the center of the disc, and the observer at infinity in the massive object's gravitational field, both have $\phi = 1$.) That relationship is valid, but it is not what the equivalence principle is talking about.

The observer on the rotating disc with a given value of $\phi$ relative to the center of the disc will not, in general, have the same proper acceleration as the observer at rest in a massive object's gravitational field with the same value of $\phi$ relative to infinity. So the two observers, even though they both have the same $\phi$ and the same time dilation relative to some "standard" observer, are easily distinguishable by local measurements.

Conversely, an observer on a rotating disc with a given proper acceleration will not, in general, have the same value of $\phi$ (and thus time dilation) relative to the center of the disc as an observer at rest in a massive object's gravitational field with the same proper acceleration will have relative to infinity. But the value of $\phi$/time dilation is not observable by a local measurement, so these observers are not distinguishable by local measurements. And that lack of distinguishability by local measurements is what the equivalence principle is talking about.

 

[Pyter]

@PeterDonis sure. I was referring only to the equivalency of the time dilation and frequency difference in the two scenarios, we already agreed that it's always possible to tell a flat space from a non-flat one with non-local experiments.