# SR and circular motion

1. Mar 1, 2014

### rede96

Sorry if this has been asked a lot before but I did try a quick search for this but could find a simple answer.

If I am at a fixed point on the equator and a friend is in a space station in a geostationary orbit, and ignoring GR, will their be time dilation between us? Or can we be considered to be at rest?

2. Mar 1, 2014

### ghwellsjr

The simple answer is to pick an Inertial Reference Frame (IRF) in which you both are rotating around on a common radius with the earth at the center. Then each or your Time Dilations will be determined by your speeds which are proportional to your distances from the center of the earth. Since you are closer, your Time Dilation will be less than that of the space station. You will be able to observe that the clocks on the space station are ticking slower than yours and the space station will be able to observe that yours is ticking faster.

EDIT: You both are not really observing your individual Time Dilations according to the IRF but you are observing the difference between them.

3. Mar 1, 2014

### WannabeNewton

Just because two standard clocks are at rest with respect to one another doesn't mean they tick at the same rate. It's true for the special case of inertial clocks but not in general.

4. Mar 1, 2014

### Staff: Mentor

OK, so ignoring GR means there is no gravity, so you are both travelling in uniform circular motion in an inertial frame. There will be time dilation between you.

You can be considered at rest in a rotating frame. That is a non inertial frame, so there will still be the same time dilation.

5. Mar 1, 2014

### rede96

Thanks for the replies.

So assuming no gravity and that we are rotating on a common radius, then is it fair to say that unlike the situation of two people moving relative to each other who are not in circular motion, the time dilation is not reciprocal? I.e. I can say that my friends clock is running slower but he can not say the same about mine? He must see my clock running faster? Also does this mean we will be ageing at different rates?

6. Mar 1, 2014

### PAllen

yes, yes, and yes.

7. Mar 1, 2014

### ghwellsjr

Even in the case of two inertial people moving relative to each other, they are not seeing Time Dilation, they are seeing Doppler shifts and they can see each others clock going either slower or faster depending on their relative motion. If they are approaching, they see the others clock going faster, if retreating, then slower. But in either case, what they see is symmetrical.

Time Dilation is a relationship between the progress of time on a single clock and the progress of time in a frame (not necessarily another clock). It is well defined for Inertial Reference Frames but not for non inertial reference frames.

The cleanest way for you on earth and your friends on the space ship to talk about Time Dilation is with respect to the IRF that I mentioned in my first post. It's very easy to analyze and you can easily determine the Doppler shifts that you will each see and you can determine your relative aging according to the IRF. But there are other IRF's that will determine different Time Dilations (that change with time) and different aging functions (that also change with time).

8. Mar 1, 2014

### rede96

OK great. Again, thanks for the replies.

Ah ok. that makes sense. Thanks.

If it isn't too much to ask, would you mind doing an example of the math to work out the ageing of two people in rotational motion wrt the IRF?

9. Mar 1, 2014

### bcrowell

Staff Emeritus

The way I would put it is that when objects are not moving inertially, it isn't necessarily well defined whether or not they're at rest relative to one another: https://www.physicsforums.com/showthread.php?t=738656#post4664349

10. Mar 1, 2014

### WannabeNewton

That's a very good point. I was thinking of it in the second way you described (rotating frame fixed to the Earth) but as you say the same wouldn't be true if we instead used sequences of momentarily comoving inertial frames. A better way for me to have phrased it would have been in terms of the constant spatial coordinates of the orbiting clocks in the rotating frame because in a momentarily comoving inertial frame of any one of the clocks there will be a local circulation of neighboring clocks.

11. Mar 2, 2014

### yuiop

In the case of a Born rigid accelerating rocket, (non inertial) is there a well defined notion of the Rindler observers on board the rocket being at rest with respect to each other? Certainly, their mutual radar distances from each other remain constant over time and there are no complications due to rotation. If they can be considered to at rest with respect to each other (a rigid congruence) then this is an example of non inertial observers at rest wrt each other, that have relative time dilation without involving gravity.

For circular motion things are not so straight forward. Consider the case of a non rotating observer (A) watching another observer (B) circulating around him. From A's point of view, B has relative motion due to B's tangential velocity. Now if A spins around his own axis so that he keeps B in his direct line of sight, B appears to be stationary. In either case, the relative time dilation remains the same.

Last edited: Mar 2, 2014
12. Mar 2, 2014

### yuiop

The ratio of the clock rates between the observers at radii $r_1$ and $r_2$ circulating around a common centre is given by:

$\frac{\sqrt{1-v_1^2/c^2}}{\sqrt{1-v_2^2/c^2}}$

where v1 and v2 are the respective tangential velocities of the two observers. If they have the same angular velocity w, (which is a requirement if we want a notion of them being at rest wrt each other as far as constant spatial separation is concerned), then the ratio can be given by:

$\frac{\sqrt{1-w^2 r_1^2/c^2}}{\sqrt{1-w^2 r_2^2/c^2}}$

Last edited: Mar 2, 2014
13. Mar 3, 2014

### rede96

Thanks I think this is what I was thinking of. For example where person 1 may be 3500 miles from the centre of the earth and person 2 may my be 4000 miles from the centre but on the same radius.

Ah ok. So as the two people given in the above example are at different points along the radius they will have different angular velocities and therefore not at rest wrt each other?

14. Mar 4, 2014

### yuiop

It's the other way around. The second equation is for when they have the same angular velocity and constant spatial separation, so that if they are on the same radial they will remain on the same radial . The first equation if for arbitrary tangential velocity so they will not necessarily stay on the same radial or have constant spatial separtation.

15. Mar 4, 2014

### WannabeNewton

That's only because relative time dilation is measured with respect to the instantaneous inertial frame which is by definition Fermi-transported so you're forced to use case A. In other words we never consider case B for relative time dilation since case B involves a frame that isn't Fermi-transported.

However the exchange between Ben and I was regarding two slightly different cases from yours. Here we have A as the central observer whose frame is corotating with the uniform angular velocity of the circular orbits. In this frame all the observers in the circular orbit are at rest since it's corotating and the relative time dilation manifests itself as a pseudo gravitational time dilation. In other words the observers all have constant spatial coordinates in this frame.

Now consider an observer B in a circular orbit. We have two relevant choices of instantaneous rest frames for B. There's the instantaneous inertial frame and there's the "natural" rest frame of B adapted to the symmetries of the circular orbit. The first corresponds to a Fermi-transported frame and the second corresponds to a Lie transported frame. In the Lie transported frame of B, all neighboring observers in the family will be fixed in space at each instant because the entire family of circular orbits is described by a twisting time-like Killing field i.e. Born rigid rotation. Because of the twisting and the rigidity we can conclude that the Lie transported frame rotates relative to each instantaneous inertial frame (Fermi-transported frame) with some angular velocity (if we ignore gravitation then it's just the Thomas precession rate). So in the instantaneous inertial frame of B the neighboring observers circulate around B (more precisely, relative to the gyroscope axes of the inertial frame) at that instant with the Thomas precession rate.

In the case of Born rigid (uniform) linear acceleration, the "natural" rest frame and instantaneous inertial frames coincide; in principle I can equip each observer with a rotating frame instead, wherein neighboring observers in this Born rigidly accelerated family do not have fixed spatial positions anymore, just for the heck of it but, as noted above, the relative time dilation factor is always calculated in the instantaneous inertial frame for obvious reasons.

16. Mar 6, 2014

### rede96

Sorry for the late reply, been away for a couple of days.

So just for clarification, if the spatial separation of the two points remains constant and lay on the same radius then we use this calculation:

$\frac{\sqrt{1-w^2 r_1^2/c^2}}{\sqrt{1-w^2 r_2^2/c^2}}$

And if the special separation is not constant then we use this calculation:

$\frac{\sqrt{1-v_1^2/c^2}}{\sqrt{1-v_2^2/c^2}}$

(Oh and I assume by 'same angular velocity' you mean same number of revolutions per time?)

But what if I was in the middle of a big rotating disk that had a hole cut in the middle so I wasn't connected to the spinning at all. In fact I'd be in an inertial reference frame. Does the second calculation still apply?

17. Mar 10, 2014

### yuiop

This equation applies if the clocks are rotating about a common centre with the same angular velocity (w), which is what happens if the clocks are both on a rotating disc. The clocks can be anywhere on the disc and not necessarily on the same radius. To observers on the disc, all other points on the disc are stationary.

Yes, with the caveat that v1 and v2 are measured locally by inertial observers that are at rest in a common rest frame.

Yes, as in the clocks on the same rotating disc as described above.

Yes, but in this special case, the spatial separation remains constant (with the caution that you consider the issues mentioned by WBN) and the velocity of the clock at the centre is zero.