# Rotating Disk and Time Dilation

1. Feb 20, 2015

### ionm

My question is about comparing the time dilation of a clock on a spinning disk versus a clock in the vicinity of a massive object. It seems there should be a connection between the two, because of the equivalence principle, but I'm missing something because I don't quite get the answer I would expect.

These are my reasoning steps:

1. The relation between the proper time of a clock rotating uniformly with $\omega$ at a distance $r$ from the center, and a clock at rest at the center is:

$$\Delta t_r = \sqrt{1-r^2\omega^2/c^2}\Delta t_\text{center}$$

2. The relation between the proper time of a clock inside a spherical gravitational field and the proper time of a clock far away is:

$$\Delta t_r = \sqrt{1-\dfrac{2GM}{rc^2}}\Delta t_0$$

3. Now lets say we want to rotate our clock such that the centrifugal acceleration reproduces the effect of the gravitational acceleration, so we impose

$$r\omega^2 = \dfrac{GM}{r^2}$$

but then we get two different relations for the time. Why?

$$\Delta t_r = \sqrt{1-r^2\omega^2/c^2}\Delta t_\text{center}$$
$$\Delta t_r = \sqrt{1-2r^2\omega^2/c^2}\Delta t_0$$

4. Now of course, there is no reason why the radius should be the same, so let me rephrase the question. We can always pick $r$ and $\omega$ such that the centrifugal acceleration reproduces the effect of gravity:

$$r\omega^2 = \dfrac{GM}{R^2}$$

From the equivalence principle we would expect that these two cases should be equivalent, so the proper times of the two clocks should be the same. We can fix $\Delta t_0=\Delta t_\text{center}$ if they are not moving with respect to each other. However, the relations we get are not quite the same, but still depend on the radius:

$$\Delta t_r = \sqrt{1-r^2\omega^2/c^2}\Delta t_\text{center}$$
$$\Delta t_r = \sqrt{1-2rR\omega^2/c^2}\Delta t_0$$

Why is that? Isn't this result contradicting the principle of equivalence? Or is there a mistake in my reasoning?

2. Feb 20, 2015

### PeroK

I can see two mistakes in your reasoning:

1) You are comparing the time dilation due to relative velocity with the time dilation due to gravity/acceleration. The equation you used for gravitational time dilation is for two observers at rest with respect to each other, but one is in a gravitational field (or accelerating). The equation you used for time dilation is for two observers with a relative velocity but not accelerating.

2) In the spinning case, you have both relative velocity and relative acceleration. So, you would need two components of time dilation: one for the relative velocity and one for the relative acceleration.

If you have linear acceleration and linear velocity, then likewise you have two components of time dilation.

3. Feb 20, 2015

### PeroK

This is for relative velocity only and doesn't take account of the relative acceleration.

This takes account of gravity, but not any relative velocity.

4. Feb 20, 2015

### A.T.

You cannot reproduce the gravitional field of a spherrical mass between two distant two clocks by rotation. Graviational time dialtion is a function of the space-time geometry along the entrie path between the two clocks, not just of local accellaration of the clocks.

For a roating frame you have to use the apporpriate space-time metric (not the one for a spherical mass):

http://www.projects.science.uu.nl/igg/dieks/rotation.pdf [Broken]

If you intergrate the rotating frame metric along the radial coordiante, you should get the same factor for gravitational time dialtion, as in the inertial frame for kinetic time dialtion .

Last edited by a moderator: May 7, 2017
5. Feb 20, 2015

### Staff: Mentor

What you say about "time dilation due to acceleration" is incorrect. Acceleration does not cause time dilation.

In the case of gravity, time dilation is caused by a difference in gravitional potential. That is what the $\sqrt{1 - 2 G M / c^2 r}$ factor signifies. There is no acceleration in there.

In the case of relative motion, only velocity affects time dilation; acceleration does not. The factor $\sqrt{1 - v^2 / c^2}$ has no acceleration in there either. (So in the case of linear acceleration and linear velocity, only the velocity affects time dilation; there is no "acceleration component".

6. Feb 20, 2015

### Staff: Mentor

There is a mistake in your reasoning. The mistake is that you are assuming that time dilation is a function of acceleration. It isn't. (See my response to PeroK.) That's why your time dilation formulas (which are correct) are not the same, even though the accelerations are the same.

Note also that the equivalence principle only applies within a single local inertial frame. Comparing time dilation between points with different accelerations requires going outside a single local inertial frame, so the EP does not apply.

7. Feb 20, 2015

### harrylin

Are you sure? Einstein seems to suggest that the EEP must apply even for different accelerations, here: https://en.wikisource.org/wiki/Rela...easuring-Rods_on_a_Rotating_Body_of_Reference

8. Feb 20, 2015

### Staff: Mentor

I would not interpret what he says here that way. He says that the acceleration felt by an observer on the rotating disk can be regarded as the effect of a gravitational field; that's the EP, yes. But that's just local to that single observer.

When he talks about the "spatial distribution" of this field, he's going beyond the EP; he's talking about how to formulate a global description of this field in the non-inertial coordinates of the observer at rest on the rotating disk, a description that covers, not just the acceleration felt by that single observer, but the (different, in many cases) accelerations that would be felt by observers placed at different locations, all over the disk. There is no way to cover all those observers in a single local inertial frame--or, to put it another way, there is no single inertial frame that is comoving with all of those observers.

9. Feb 20, 2015

### ionm

Thank you! That's the explanation I was looking for. So what you are saying is this: If you have a rocket accelerating in empty space with $a=g$, then all the experiments inside the rocket will match the experiments on Earth (locally at least). A clock at the end of the rocket and one at front of the rocket (at some distance $d$) will have the same time dilation relation with two clocks on Earth placed at a height difference $d$ from each other.

However, you cannot simply compare the time dilation for a clock on Earth and an observer in empty space, vs a clock in the rocket and an observer at rest in empty space. The two time dilations would be:

$$\Delta t_\text{Earth} = \sqrt{1-\dfrac{2gR}{c^2}}\Delta t_\text{space}$$

$$\Delta t_\text{space ship} = \int^{\Delta t_\text{space}}_{0} \sqrt{1-\dfrac{(v_0+gt)^2}{c^2}} dt$$

So the only way the acceleration affects the time dilation of the clock in the rocket is by increasing the velocity. If $v_0=0$, in the first moments the time dilation effect is almost zero despite the acceleration.

And to finish the parallel, there is a difference between what the observers will conclude:

1. For the Observer on Earth his clock is running slow, with the above relation

2. For the Rocker Observer, the clock of the outer space observer is running slow

And this illustrates that acceleration and gravity are not equivalently overall, but only inside the rocket. Once the rocket observer interacts with observers at rest there will be differences between the Earth and the rocket.

Let me know if this is correct, and if you agree with it.

10. Feb 20, 2015

### Staff: Mentor

Since it's correct, yes, I agree with it. I always agree with what's correct.

11. Feb 20, 2015

### ionm

Thank you! This clarifies the equivalence principle. For some reason it encourages many people to equate the effect of gravity with the effect of acceleration.

12. Feb 21, 2015

### A.T.

The equivalence principle equates local effects of gravity with local effects of acceleration. Gravitational time dilation is not local effect, but if you somehow create globally equivalent potential fields, then gravitational time dilation will be the same.

13. Feb 24, 2015

### harrylin

I fully agree that there is no single inertial frame that is comoving with all of those observers. You seem to reason here 'that the equivalence principle only applies within a single local inertial frame' because you hold that a field with a "spatial distribution" is going beyond the EP, for that does not correspond to a single local inertial frame. That sounds perfectly circular to me.

Einstein's theory accepts no different laws between fictive and real gravitational fields, no matter the spatial distribution. He explained the EEP perhaps clearer a few years later:

"perhaps Newton's law of field could be replaced by another that fits in with the field which holds with respect to a "rotating" system of co-ordinates? My conviction of the identity of inertial and gravitational mass aroused within me the feeling of absolute confidence in the correctness of this interpretation. [..] We are familiar with the "apparent" fields which are valid relatively to systems of co-ordinates possessing arbitrary motion with respect to an inertial system. With the aid of these special fields we should be able to study the law which is satisfied in general by gravitational fields."
- https://en.wikisource.org/wiki/A_Brief_Outline_of_the_Development_of_the_Theory_of_Relativity

14. Feb 24, 2015

### Staff: Mentor

I wasn't offering that statement as an argument for the EP, only as my interpretation of what Einstein was saying. See below.

But those "no different laws" include a lot more than the EP. They include the full Einstein Field Equation, whose set of solutions includes both flat Minkowski spacetime and various curved spacetimes. The same "law", the EFE, covers all these possibilities, which include anything that might be called either a "fictive" or a "real" gravitational field. But that doesn't mean that the EP is just another name for the EFE. Even though the same law covers all these different spacetimes, that doesn't mean there is no way at all to experimentally determine which spacetime you're in; of course you can (just measure tidal gravity). So all these different spacetimes are certainly not "equivalent" in any global sense, even though the same law (the EFE) applies to all of them. They're only equivalent (i.e., experimentally indistinguishable) in a local sense, if you restrict the experiments to a small enough region of spacetime. That's what "the EP only applies within a single local inertial frame" means.

15. Feb 25, 2015

### harrylin

In other words, we simply differ about what we call the Einstein equivalence principle; I understand the EEP to correspond to the last two sentences as cited from Einstein in my post #13. But I thought that I was replying in the thread about the Equivalence principle, and now I notice that in fact this another thread - here it's off-topic. Sorry about that!

16. Feb 25, 2015

### Staff: Mentor

I understand that "Einstein Equivalence Principle" is a reasonable term to describe what Einstein was saying in that passage, but it's not the standard term for it. The standard term for what Einstein was describing in that passage is "general covariance". The term "Einstein Equivalence Principle" is standardly used in the more restrictive sense I described. But that's a matter of terminology, not physics; I agree that you have describe the physics correctly.