Metric tensor for a uniformly accelerated observer

[PeroK]

So the takeaway is: a rotating observer in a flat space and a still observer in a gravity field are not equivalent (in fact the latter can detect space curvature with some experiment), but "equivalent enough" that their clock dilation obey the exactly same rules.

There is no local experiment to measure time dilation. Any local experiment, by definition, will use its own local proper time.

 

[Pyter]

There is no local experiment to measure time dilation. Any local experiment, by definition, will use its own local proper time.

I mean the time dilation with respect to an inertial observer in flat space.

Cit.:

[ 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 ϕ 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.

 

[pervect]

Yes, that would definitely tell a still observer in a gravity field apart from an accelerating observer in flat space, since the latter would experience a homogeneous acceleration field with no gradient whatsoever.

It's not that different for an accelerating observer than a stationary one, though it's harder to implement because the common-mode accelerations need to be compensated for if you are interested in the tidal components parallel to the direction of the acceleration. If you're only interested in the traverse tidal components, the acceleration doesn't matter at all.

The airplane mounted gravity gradiometers used in the oil surveys certainly have a non-zero proper acceleration, due to both gravity and turbulence. But the gravity gradiometer instruments can operate anyway, they are designed to reject the common-mode signal, basially by using matched acceleration sensors attached to a rotating wheel.

There are some minor theoretical relativistic effects that at practical accelerations are not measurable. At 1 g (10 m/s^2) acceleration, I make these effects out of the order of $10^{-15}$ (meters/second^2) / meter, i.e. for a 1 meter spaceship accelerating with a proper acceleration of 10 $m/s^2$ at the nose, the proper acceleration at the tail would be very slightly higher, 10 m^s^2 plus $10^{-15}$ m/s^2. This is well beyond the resolution of practical instruments, so rejecting the common mode signal is the only practical problem for using a gravity gradiometer on an accelerating spaceship if it's accelerating at reasonable accelerations.

If the hypothetical spaceship is trying to hold station near a black hole and pulling a few billion or trillion g's by doing so, things might be different.

It's also different if the spaceship is very long - for instance, if you had 1 light year (born rigid) spaceship accelerating at 1g at the nose, the proper acceleration at the tail would be infinite.

 

[Pyter]

No, that's not correct. There is still a gradient in the "acceleration field" for an observer inside an accelerating rocket in flat spacetime; the acceleration is slightly smaller at the top of the rocket than at the bottom.

Don't they experience the same proper acceleration at the top and the bottom?

 

[PeroK]

I mean the time dilation with respect to an inertial observer in flat space.

You need to be careful with that argument above. If we are considering a case where there is curved spacetime, then there is no "inertial observer in flat space". There are no global inertial frames.

With a centrifuge in flat spacetime, then there can be an inertial observer.

 

[Pyter]

You need to be careful with that argument above. If we are considering a case where there is curved spacetime, then there is no "inertial observer in flat space".

An observer at such a great distance from the gravity source that she experiences a nearly flat space.

 

[PeroK]

An observer at such a great distance from the gravity source that experiences a nearly flat space.

That inertial frame does not extend globally. To them, space is flat only locally.

The point I'm making is that you have already applied the equivalence principle beyond its range of applicability by assuming the "acceleration is equivalent to gravity" heuristic. You run the risk of over-simplifying the nature of curved spacetime by using the "gravity is equivalent to centrifugal motion" heuristic.

 

[vanhees71]

The naive equivalence principle is quite dangerous and has to be applied with great care. It's a heuristics to introduce the idea that inertial frames become a local concept, i.e., that the gravitational interaction can be reinterpreted as the description of spacetime as a pseudo-Riemannian manifold with a fundamental form of signature (1,3) (or (3,1), depending on your preference of the west- or east-coast convention). I think this latter more abstract mathematical formulation is the clearest formulation of what's meant by "equivalence principle" (or to be more precise the "strong equivalence principle").

What's for sure wrong is the assumption that gravity is equivalent to a reference frame accelerated against some global inertial frame. This holds only approximately locally, i.e., in a spacetime region, where the gravitational field can be considered homogeneous. True gravitational fields imply curvature of spacetime and thus cannot be equivalently described by an "accelerated reference frame".

 

[A.T.]

An observer at such a great distance from the gravity source that she experiences a nearly flat space.

Make sure to differentiate between a "reference frame" and an "observer". The common the use of "observer" as a synonym for "reference frame" leads to a lot of confusion.

 

[Pyter]

That inertial frame does not extend globally. To them, space is only flat locally.

As you can read in the Einstein citation, there's the formula for the frequency $\nu$ seen by a rotating/gravity-subjected observer as a function of the frequency $\nu_0$ of a clock/atom site at the center of the disc/infinite distance from the massive object surface.
It's not me who brought in this observer.

 

[jbriggs444]

Don't they experience the same proper acceleration at the top and the bottom?

No. They do not. That's Bell's spaceship paradox.

 

[vanhees71]

As you can read in the Einstein citation, there's the formula for the frequency ν seen by a rotating/gravity-subjected observer as a function of the frequency ν0 of a clock/atom site at the center of the disc/infinite distance from the massive object surface.
It's not me who brought in this observer.

As you can also read in this quote this can indeed be measured by local observations: Just put some light source (an atom radiating some light at the frequency $\nu_0$) at the center of the rotating disk. Then via the time dilation you get (in leading order in $v^2/c^2$) the formula for $\nu$ from the transverse Doppler effect. The measurement of the frequency is of course a local one of the observer sitting at the radius $r$.

 

[PeterDonis]

There is no local experiment to measure time dilation.

Careful. Einstein's original argument for gravitational time dilation, based on a rocket accelerating in a straight line in flat spacetime, is a local argument from the standpoint of the equivalence principle.

To put it another way, observing the time dilation between two observers close together who are at rest relative to each other and both have nonzero proper acceleration does not, in itself, tell you whether those observers are in a flat or a curved spacetime. Observing how time dilation varies with height over a sufficient range of heights can do it, but time dilation alone, between just two observers close together, can't. So observing time dilation between two observers close together is a local observation from the standpoint of the equivalence principle.

 

[PeterDonis]

So the takeaway is: a rotating observer in a flat space and a still observer in a gravity field are not equivalent

No. Local measurements cannot distinguish these two cases, any more than they can distinguish an observer in a rocket accelerating in a straight line in flat spacetime from an observer at rest in a gravitational field. All three of these cases are equivalent from the standpoint of the equivalence principle (provided the proper acceleration is the same in all three cases).

(in fact the latter can detect space curvature with some experiment)

Not with any local experiment, no.

Don't they experience the same proper acceleration at the top and the bottom?

No. Take some time and work your way through the math. The easiest way is to use Rindler coordinates.

 

[Pyter]

All three of these cases are equivalent from the standpoint of the equivalence principle (provided the proper acceleration is the same in all three cases).

Now I'm confused. Earlier you said Einstein wasn't using the equivalence principle in the excerpt, now you state they're equivalent.

Not with any local experiment, no.

What do you mean by "local experiment" or "measurement"? Aren't they all "local" to the observer?

 

[PeterDonis]

Earlier you said Einstein wasn't using the equivalence principle in the excerpt, now you state they're equivalent.

The fact that Einstein wasn't using that particular excerpt as an argument for the equivalence principle (as I said in post #28) does not mean the equivalence principle does not apply to that case. It does. The equivalence principle is completely general: it says that, locally, observers with the same proper acceleration in any spacetime whatever are equivalent. It is not restricted to just the particular example that Einstein used in his original argument, the one in his published paper (in 1907, IIRC).

What do you mean by "local experiment" or "measurement"?

"Local" means "confined to a region of spacetime that is small enough that the effects of tidal gravity cannot be observed".

As I've just stated it, of course what counts as "local" depends on how accurate your measurements of tidal gravity are. There is a way of formulating the equivalence principle using limits in order to avoid that technical difficulty, but in practice it's not an issue because there are plenty of cases in which the limits on the accuracy of our measurements of tidal gravity leave plenty of room for useful measurements to be local.

 

[Pyter]

"Local" means "confined to a region of spacetime that is small enough that the effects of tidal gravity cannot be observed".

Fine. Then I'll specify better my previous statement:

So the takeaway is: a rotating observer in a flat space and a still observer in a gravity field are "non-locally" not equivalent (in fact the latter can detect space curvature with some "non-local" experiment), but "non-locally" "equivalent enough" that their clock dilation obeys the exact same rules.

 

[PeterDonis]

Then I'll specify better my previous statement

The last part of your revised statement is false. Non-locally, the time dilation behaves differently (i.e., varies differently as a function of "height") in all three cases under discussion (at rest in the gravitational field of a massive object, in a rocket accelerating in a straight line in flat spacetime, on a rotating platform in flat spacetime).

 

[Pyter]

Non-locally, the time dilation behaves differently (i.e., varies differently as a function of "height") in all three cases under discussion

According to the Einstein's excerpt, time dilation is the same for the same potential $\phi$, which is a function of "height" (distance from the center of the disc) in one case, and of the surface gravity field in the other. If the rotating observer doesn't move radially, it remains the same also non-locally.

 

[PeterDonis]

According to the Einstein's excerpt, time dilation is the same for the same potential $\phi$,

You need to be careful. What the excerpt says is that given a specific scenario, observers with the same $\phi$ will have the same time dilation. So, for example, observers at any point on the rotating disc with the same $\phi$ (meaning the same value of $r$) will have the same time dilation (relative to the inertial observer at the center of the disc).

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 $\phi$. It doesn't say that because there's no way to even meaningfully compare values of $\phi$ in different spacetimes, or time dilations for that matter, so there's no way to even pick out where on Earth is "the same $\phi$" as some particular point on the rotating platform.

More generally, the excerpt says that time dilation in a gravitational field will depend on $\phi$. But that is a much weaker statement than the one you made that I quoted above.

 

[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.