Time Dilation on Rotating Disk: Clocks on Disk Perspective

In summary: Nearby Stars" of the General Theory of Relativity. The motion of spectral lines is affected by the gravitational time-dilation of clocks on a rotating body.
  • #1
Thadriel
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TL;DR Summary
(no "master clock" at rest with the center of the disk)

Sorry I know this is a common one. I'm sure you've all had to answer this a thousand times, so I apologize, but I'm hoping to find a comment that lights the spark to get my brain to grasp what seems to not make sense.
Obviously, a third observer who is at rest with respect to the disk will see that the clock on the outside has a much faster velocity than a clock on the interior of the disk, so clearly the outside clock will show that it has measured less time.

But that's one question. What about looking at it from the perspective of the two clocks on the disk as they look at each other, and not bringing the "at rest" third clock into the conversation at all?*

These two clocks do not move with respect to each other during the rotation, so presumably only light signal delay would cause differences in what they see of each other's clocks during the rotation. Which means neither should measure the other as having slow or fast clock.

For simplicity, you could just put clock A at the center and clock B at the outside. As they the disk rotates, the two clocks remain at rest with respect to each other. I believe that, as the disk is moving, they should see no difference unrelated to light signal delay of their clocks. (not discussion differential aging yet).
.

If that is the case, how to we reconcile that the outer clock will have experience less time when the two clocks are brought together and compared locally? The only thing I can think of is that they will disagree on when the disk stops rotating. But how? They are both at rest with respect to each other according to each other's perspective. It's really easy to see the symmetry break in the standard twin paradox, because one must return. But with this example, we're free to move either clock and leave the other alone.

Somewhere in this example there has to be something that breaks the apparent symmetry between these two clocks according to both reference frames of the two clocks (as opposed to a third "master clock."). Because otherwise, if you remove all the matter in the universe except these two clocks, they have no way to distinguish each other from one another other than accelerometers. And acceleration is not the cause of time dilation (unless maybe it's a "gravitational field" caused by the acceleration, and then we're bringing GR into the problem).

*Or, is this what they mean when they say in special relativity you have to treat acceleration "very carefully?" As in, "don't you go examining acceleration without using an outside clock and measuring rod that are not accelerating." Which would mean, special relativity can deal with acceleration, but only if the reference frame used for examination is inertial.
 
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  • #2
Thadriel said:
But that's one question. What about looking at it from the perspective of the two clocks on the disk as they look at each other, and not bringing the "at rest" third clock into the conversation at all?*

These two clocks do not move with respect to each other during the rotation, ... Which means neither should measure the other as having slow or fast clock.
The analysis of relative motion and time dilation that you will have seen assumes an inertial reference frame. The rest frame of neither clock is inertial and so that analysis cannot be used.

Thadriel said:
Somewhere in this example there has to be something that breaks the apparent symmetry between these two clocks according to both reference frames of the two clocks (as opposed to a third "master clock."). Because otherwise, if you remove all the matter in the universe except these two clocks, they have no way to distinguish each other from one another other than accelerometers.
That the clocks experience proper acceleration is very relevant.
Thadriel said:
And acceleration is not the cause of time dilation (unless maybe it's a "gravitational field" caused by the acceleration, and then we're bringing GR into the problem).
There's no need for GR.
Thadriel said:
*Or, is this what they mean when they say in special relativity you have to treat acceleration "very carefully?"
You have to treat an accelerating reference frame carefully. You can study an accelerating particle using an inertial reference frame without any complications.
Thadriel said:
Which would mean, special relativity can deal with acceleration, but only if the reference frame used for examination is inertial.
It's harder to use an accelerating reference frame, but it can be done in some cases. E.g.

https://en.wikipedia.org/wiki/Rindler_coordinates
 
  • #3
Thadriel said:
For simplicity, you could just put clock A at the center and clock B at the outside. As they the disk rotates, the two clocks remain at rest with respect to each other. I believe that, as the disk is moving, they should see no difference unrelated to light signal delay of their clocks. (not discussion differential aging yet).

Consider a large spaceship. You can create artificial gravity by centrifugal force in there, if it rotates. In the rotating frame, this pseudo-gravity comes along with a gravitational time-dilation.

Einstein calculated this in the chapter "(c) Displacement of Spectral Lines Towards the Red" of "The Special and General Theory/Appendix":

https://en.wikisource.org/wiki/Rela...isplacement_of_Spectral_Lines_Towards_the_Red
 
  • #4
PeroK said:
The analysis of relative motion and time dilation that you will have seen assumes an inertial reference frame. The rest frame of neither clock is inertial and so that analysis cannot be used.That the clocks experience proper acceleration is very relevant.

There's no need for GR.

You have to treat an accelerating reference frame carefully. You can study an accelerating particle using an inertial reference frame without any complications.

It's harder to use an accelerating reference frame, but it can be done in some cases. E.g.

https://en.wikipedia.org/wiki/Rindler_coordinates
So it's almost what I suspected at the end of my post, but not quite. You say here that "it can be done in some cases." Do this mean Rindler coordinates can be used in this particular case to look at the problem from the perspective of the clocks on the disk, and if so, could these coordinates be able to explain from the perspective of the clocks on the disk why one of them ages more than the other?

If you'd have to go through tedious calculations to find out, I won't ask you to do so. Because regardless, it seems in order to answer this question as it is asked and understand it, I'll have to learn Rindler coordinates and how to use them.

One thing is for sure: there has to be a physical result agreed upon by everyone when the disk is stopped and clocks are brought together and compared. It's really easy to see the cause from an outside clock. I'm hoping Rindler coordinates will illuminate the explanation from the persepective of the clocks on the disk.

Regardless, thanks for the reply and pointing me in the direction to get better understanding.
Sagittarius A-Star said:
Consider a spaceship. You can create artificial gravity by centrifugal force in there, if it rotates. In the rotating frame, this pseudo-gravity comes along with a gravitational time-dilation.

Einstein calculated this in the chapter "(c) Displacement of Spectral Lines Towards the Red" of "The Special and General Theory/Appendix":

https://en.wikisource.org/wiki/Rela...isplacement_of_Spectral_Lines_Towards_the_Red
Thanks for this. While I'm aware of this phenomenon, I'm certainly not at a math level to examine it. I am sure this has a non-zero contribution to the differential aging between the clocks, but I do not suspect it fully explains it. Am I wrong in believing part of it must come from the velocity?
 
  • #5
Thadriel said:
Am I wrong in believing part of it must come from the velocity?
Velocity and time-dilation are frame-dependent quantities. In the non-rotating inertial frame, your clock B has a velocity, in the rotating frame not.
 
  • #6
Thadriel said:
One thing is for sure: there has to be a physical result agreed upon by everyone when the disk is stopped and clocks are brought together and compared. It's really easy to see the cause from an outside clock. I'm hoping Rindler coordinates will illuminate the explanation from the persepective of the clocks on the disk.

Regardless, thanks for the reply and pointing me in the direction to get better understanding.
I did an analysis here a few years ago. This breaks down a circular orbit into a sequence of inertial stages.

https://www.physicsforums.com/threa...ther-in-a-circular-orbit.896607/#post-5660815
 
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  • #7
Thadriel said:
One thing is for sure: there has to be a physical result agreed upon by everyone when the disk is stopped and clocks are brought together and compared. It's really easy to see the cause from an outside clock. I'm hoping Rindler coordinates will illuminate the explanation from the persepective of the clocks on the disk.
We also have to specify the procedure used to set the clocks to their initial values, but once having done that there is indeed a way of calculating the elapsed time on each clock when the disk is stopped and the clocks are brought together. This is basically a variation of the twin paradox, and the resolution is the same: the elapsed time on each clock is the spacetime interval along that clock's worldline.

This calculation is most easily done using coordinates from the inertial frame in which the center of the disk is at rest. Rindler coordinates are irrelevant here (unless you are a total masochist and want to analyze the problem from the point of view of an observer in an accelerating spaceship zooming past the disk, and you want to do so in the hardest possible way).
 
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  • #8
Thadriel said:
Do this mean Rindler coordinates can be used in this particular case to look at the problem from the perspective of the clocks on the disk, and if so, could these coordinates be able to explain from the perspective of the clocks on the disk why one of them ages more than the other?
No. Rindler coordinates are useful mainly for describing an observer under constant proper acceleration in the same direction, which is not the case for a rotation. What you can do is to simply introduce a rotating set of coordinates ##\xi = x \cos(\omega t) + y \sin(\omega t)## and ##\eta = -x\sin(\omega t) + y \cos(\omega t)##. This can be inverted to
$$
x = \xi \cos(\omega t) - \eta \sin(\omega t), \quad y = \xi \sin(\omega t) + \eta \cos(\omega t)
$$
and therefore
$$
dx = \cos(\omega t) d\xi - \sin(\omega t) d\eta - \omega [\xi \sin(\omega t) + \eta \cos(\omega t)]dt, \quad
dy = \sin(\omega t) d\xi + \cos(\omega t) d\eta + \omega [\xi \cos(\omega t) - \eta \sin(\omega t)] dt.
$$
The Minkowski metric expressed in this coordinate system becomes
$$
ds^2 = dt^2 - dx^2 - dy^2 - dz^2 = dt^2 [1 - \omega^2 (\xi^2 + \eta^2)] - [\ldots d\xi + \ldots d\eta] \omega\, dt - d\xi^2 - d\eta^2 - dz^2.
$$
For observers stationary in the rotating frame, we have ##d\xi = d\eta = dz = 0## and therefore ##ds^2 = dt^2 [1-\omega^2 r^2]## where ##r^2 = \xi^2 + \eta^2 = x^2 + y^2##. The time elapsed between two surfaces of constant coordinate time ##t## therefore decreases with radius ##r##. Just as expected.
 
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  • #9
PeroK said:
I did an analysis here a few years ago. This breaks down a circular orbit into a sequence of inertial stages.

https://www.physicsforums.com/threa...ther-in-a-circular-orbit.896607/#post-5660815
Thanks. I'll look at this shortly.

Nugatory said:
We also have to specify the procedure used to set the clocks to their initial values, but once having done that there is indeed a way of calculating the elapsed time on each clock when the disk is stopped and the clocks are brought together. This is basically a variation of the twin paradox, and the resolution is the same: the elapsed time on each clock is the spacetime interval along that clock's worldline.

This calculation is most easily done using coordinates from the inertial frame in which the center of the disk is at rest. Rindler coordinates are irrelevant here (unless you are a total masochist and want to analyze the problem from the point of view of an observer in an accelerating spaceship zooming past the disk, and you want to do so in the hardest possible way).
I would like to see how this would be done. I'll be looking at PeroK's post first, of course.

But just thinking of it intuitively, if I'm at the center of the disk (I'm clock A), while clock B is at the outside of the disk, I can think of at least two things which seem to be true: (1) there is no relative motion between me at clock A and clock B, and (2) I can definitely feel an acceleration, since I'm not a point (I don't know if a point would feel acceleration or not, but I am spatially extended, so I suspect I'd be spinning along with the disk). I'm sure that adds painful complication, but I'd like to think an accelerometer attached to my clock would measure the acceleration. The only obvious differences I can think of between me at clock A and clock B is that block B ought to feel like its being pushed outward (like when you ride a marry-go-round), and clock B travels more radial distance in the same time interval. I could naively interpret that as a "velocity" difference, since clock B travels more distance in the same amount of time than clock A, and then calculate time dilation that way, but that seems suspect to me, since relative to clock A, there is no obvious velocity.EDIT: It appears to me that if the clock at the center is treated as NOT rotating, this may completely change the problem from the case in which the clock at the center IS rotating (which seems more realistic to me).
 
  • #10
Thadriel said:
1) there is no relative motion between me at clock A and clock B
There is no change in distance between the two clocks.
That is not the same as saying that there is no motion of clock B relative to an inertial frame anchored at A.

You need to do a better job of defining "relative motion".
 
  • #11
Orodruin said:
No. Rindler coordinates are useful mainly for describing an observer under constant proper acceleration in the same direction, which is not the case for a rotation. What you can do is to simply introduce a rotating set of coordinates ##\xi = x \cos(\omega t) + y \sin(\omega t)## and ##\eta = -x\sin(\omega t) + y \cos(\omega t)##. This can be inverted to
$$
x = \xi \cos(\omega t) - \eta \sin(\omega t), \quad y = \xi \sin(\omega t) + \eta \cos(\omega t)
$$
and therefore
$$
dx = \cos(\omega t) d\xi - \sin(\omega t) d\eta - \omega [\xi \sin(\omega t) + \eta \cos(\omega t)]dt, \quad
dy = \sin(\omega t) d\xi + \cos(\omega t) d\eta + \omega [\xi \cos(\omega t) - \eta \sin(\omega t)] dt.
$$
The Minkowski metric expressed in this coordinate system becomes
$$
ds^2 = dt^2 - dx^2 - dy^2 - dz^2 = dt^2 [1 - \omega^2 (\xi^2 + \eta^2)] - [\ldots d\xi + \ldots d\eta] \omega\, dt - d\xi^2 - d\eta^2 - dz^2.
$$
For observers stationary in the rotating frame, we have ##d\xi = d\eta = dz = 0## and therefore ##ds^2 = dt^2 [1-\omega^2 r^2]## where ##r^2 = \xi^2 + \eta^2 = x^2 + y^2##. The time elapsed between two surfaces of constant coordinate time ##t## therefore decreases with radius ##r##. Just as expected.
So, if I understand this correctly, since we can choose a frame where either is stationary in the frame, and since r doesn't disappear when we do this, but it does vary between these observers, then we'll still get the expected time dilation?

Unless I'm mistaken, this also seems to imply that at the exact center (where r = 0), there is no motion at all through space and all "motion" is in the time dimension, right? Which means maximal aging? (I don't mean literal "motion" through time or spacetime; I don't want to bring in the confusion from Brian Greene's apparently annoying examples :wink:)

Out of the blue thought: r is one dimensional, much like a distance between the top and bottom of a spaceship which is accelerating (upward) would be. How different is this scenario from the one in which an accelerating rocket ends up with a time dilation component depending on how far away a clock is from the "floor" to the "ceiling" of the rocket?I'm sure these situations can't be directly compared, because rotational motion is different. But it's interesting in both cases, it seems that only distance determines how much time clocks measure.
jbriggs444 said:
There is no change in distance between the two clocks.
That is not the same as saying that there is no motion of clock B relative to an inertial frame anchored at A.

You need to do a better job of defining "relative motion".
Well, if we consider the case where the clock at the center is also rotating, then what is the difference between them with respect to their relative motion? If the clocks are set to be facing each other, and both are rotating, could they not always face each other? In that case, A would not be in an inertial frame, right (since it's rotating)?
 
  • #12
Thadriel said:
Well, if we consider the case where the clock at the center is also rotating, then what is the difference between them with respect to their relative motion? If the clocks are set to be facing each other, and both are rotating, could they not always face each other? In that case, A would not be in an inertial frame, right (since it's rotating)?
Again, what do you mean by "relative motion"?
And what do you mean by "in an inertial frame"?

An ideal clock is a point-like object. It has no associated orientation and, hence, no notion of rotation. It just measures time.

The normal formulation of special relativity is in terms of inertial frames, not Born-rigid objects. Two clocks can be motionless relative to the same born rigid frame without ticking at the same rate.
 
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  • #13
Thadriel said:
Well, if we consider the case where the clock at the center is also rotating, then what is the difference between them with respect to their relative motion? If the clocks are set to be facing each other, and both are rotating, could they not always face each other? In that case, A would not be in an inertial frame, right (since it's rotating)?
Unless parts of the clock are moving at significant speed, the clock is FAPP inertial. In terms of a central observer rotating or not, it's all about coordinates. If you spin once per second, say, then you have a choice regarding distant objects. You can establish an inertial coordinate system. Or, you can use a rotating coordinate system, with you at the centre. The latter case is not an inertial coordinate system, regardless of whether a central object is considered to be inertial or not. It's the coordinates of the distant object that are important.

Note that a distant object that was orbiting might be at fixed coordinates in one system and not in the other. And vice versa for a inertial distant object.

It's your (global) coordinate system that matters; not your state of motion directly.
 
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  • #14
jbriggs444 said:
Again, what do you mean by "relative motion"?
And what do you mean by "in an inertial frame"?

An ideal clock is a point-like object. It has no associated orientation and, hence, no notion of rotation. It just measures time.

The normal formulation of special relativity is in terms of inertial frames, not Born-rigid objects. Two clocks can be motionless relative to the same born rigid frame without ticking at the same rate.
Okay, by relative motion I mean if you remove all other things from the universe, how do the locations and orientations of the objects with respect to each other change over time. By inertial frame, I mean one in which Newton's law of inertia holds good, I guess.

It seems that clocks that are extended objects are not really relevant to the discussion based on what you said here, however. And if they were, it would certainly be way over my head in terms of physics. I once saw the difference in a point object and spatially extended object in a Newtonian physics examination. It was a dramatic jump in mathematical complexity.
 
  • #15
PeroK said:
Unless parts of the clock are moving at significant speed, the clock is FAPP inertial. In terms of a central observer rotating or not, it's all about coordinates. If you spin once per second, say, then you have a choice regarding distant objects. You can establish an inertial coordinate system. Or, you can use a rotating coordinate system, with you at the centre. The latter case is not an inertial coordinate system, regardless of whether a central object is considered to be inertial or not. It's the coordinates of the distant object that are important.

Note that a distant object that was orbiting might be at fixed coordinates in one system and not in the other. And vice versa for a inertial distant object.

It's your (global) coordinate system that matters; not your state of motion directly.
In the case where we choose a stationary frame, but imagine a scenario in which the clock at the middle is rotating at the same rate as the clock on the outer edges of the disk, does that change what happens physically in terms of what times each measure?

I see that according to Orodruin's analysis, he sets up a rotating coordinate system, and then shows that stationary objects in that system will have their time measured depend upon the distance r from the center (if I understood it correctly). Since these are coordinates, I am assuming that any spot in that system can be deemed "at rest," and if that's the case, then only r would effect how much time a clock measures. Is that correct?By the way, I did read through your post you linked. I just don't know if I'm quite connecting the dots, since the original post in that thread seems to be about two observers both of whom are orbiting, which seems more complex than the simple two observers on a disc scenario. I'm still trying to wrap my head around what's similar and different, but I'm afraid that, unlike Scar, when it comes to the brains portion of the gene pool, I didn't get the lion's share 🦁 . Didn't get the lion's share when it comes to brute strength either, though (sorry I just watched a bunch of Lion King memes).
 
  • #16
Thadriel said:
Okay, by relative motion I mean if you remove all other things from the universe, how do the locations and orientations of the objects with respect to each other change over time.
So a born rigid set of objects.

Thadriel said:
By inertial frame, I mean one in which Newton's law of inertia holds good, I guess.
Sounds good.

So a uniformly rotating disk with one clock at the center and one at the rim is born rigid but is not at rest in any inertial frame.

"in an inertial frame", I take to mean continually "at rest in an inertial frame".
 
  • #17
Thadriel said:
In the case where we choose a stationary frame, but imagine a scenario in which the clock at the middle is rotating at the same rate as the clock on the outer edges of the disk, does that change what happens physically in terms of what times each measure?
If it did, you could test the difference with a simple experiment. Local clocks on Earth differ by only tiny fractions of a second based on their state of motion.
Thadriel said:
I see that according to Orodruin's analysis, he sets up a rotating coordinate system, and then shows that stationary objects in that system will have their time measured depend upon the distance r from the center (if I understood it correctly). Since these are coordinates, I am assuming that any spot in that system can be deemed "at rest," and if that's the case, then only r would effect how much time a clock measures. Is that correct?
"At rest" just means has time-independent (constant) spatial coordinates. So, yes, a clock at rest at constant ##\xi## and ##\eta## in that system is time dilated compared to a clock orbiting at the same ##r##.
 
  • #18
jbriggs444 said:
So a born rigid object.Sounds good.

So a uniformly rotating disk with one clock at the center and one at the rim is born-rigid but is not at rest in any inertial frame.

"in an inertial frame", I take to mean continually "at rest in an inertial frame". All objects are "in" all frames at all times.
I don't know how much I want to get into spatially extended objects, but unfortunately I don't know if it's completely avoidable, since different parts of an object have their own little clocks (whatever repeating, clock-like biological processes may be going on in the cell on my scalp is at a different altitude than similar processes at my feet).

PeroK said:
If it did, you could test the difference with a simple experiment. Local clocks on Earth differ by only tiny fractions of a second based on their state of motion.
Oh man, this one is so obvious. Should have realized. :sorry::-p

Only question is, how much of that is due to the mass/energy/pressure/etc. of Earth (gravitational time dilation)? And how would you separate those results for experimental purposes? My guess is the normal procedure would be to calculate what those two scenarios would be with point masses and see how the sum differs. But I feel like that could still result in murkiness about what percentage of the total time difference each causes.

Not to mention, I suspect that moving bodies would interfere with the "pure" results of the gravitational time dilation calculation, since, unless I'm mistaken, that's all about energy, and moving bodies have energy (and whatever else they add to that). It seems like you could consider the time dilation due to velocity on its own, but it wouldn't make sense to look at the time dilation due to gravity on its own, since the moving body adds energy to the system (and probably interferes with all the other things that go into that calculation).

I'm sure this is mistaken at least in part.

PeroK said:
"At rest" just means has time-independent (constant) spatial coordinates. So, yes, a clock at rest at constant ##\xi## and ##\eta## in that system is time dilated compared to a clock orbiting at the same ##r##.

But you can choose any reference frame in which the the spatial coordinates of an observer/clock/whatever are time-independent, correct? Since everything is at rest in its own reference frame?
 
  • #19
Thadriel said:
Only question is, how much of that is due to the mass/energy/pressure/etc. of Earth (gravitational time dilation)?
You can look up the Hafele-Keating experiment.
Thadriel said:
But you can choose any reference frame in which the the spatial coordinates of an observer/clock/whatever are time-independent, correct? Since everything is at rest in its own reference frame?
There's no state of motion that defies the mathematics of defining a system of coordinates with the object in question constantly at the origin.
 
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  • #20
Thadriel said:
But you can choose any reference frame in which the the spatial coordinates of an observer/clock/whatever are time-independent, correct? Since everything is at rest in its own reference frame?
But not all rest frames are inertial.
 
  • #21
jbriggs444 said:
But not all rest frames are inertial.
In those cases, you'd be using the equivalence principle for small portions of spacetime, right?

If so, what about in the cases where instead of mere acceleration, you have lots of jerk?
 
  • #22
Also let me say thanks for the answers, everyone. I think the original question is solved.
 

1. What is time dilation on a rotating disk?

Time dilation on a rotating disk refers to the phenomenon where time appears to pass at a slower rate for objects that are rotating compared to objects that are stationary.

2. How does time dilation on a rotating disk occur?

Time dilation on a rotating disk occurs because of the effects of gravity and acceleration on time. As an object rotates, it experiences a centripetal acceleration, which causes a distortion in the fabric of space-time, resulting in a slower passage of time.

3. What is the equation for time dilation on a rotating disk?

The equation for time dilation on a rotating disk is t' = t / √(1 - r²ω²/c²), where t' is the time measured on the rotating disk, t is the time measured by a stationary observer, r is the distance from the center of rotation, ω is the angular velocity, and c is the speed of light.

4. How does the speed of rotation affect time dilation on a rotating disk?

The speed of rotation has a direct impact on the amount of time dilation on a rotating disk. The faster the disk rotates, the greater the centripetal acceleration and the stronger the time dilation effect.

5. Can time dilation on a rotating disk be observed in real life?

Yes, time dilation on a rotating disk has been observed in experiments using atomic clocks. The clocks on the rotating disk were found to measure time at a slower rate compared to the clocks on a stationary disk, confirming the predictions of Einstein's theory of relativity.

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