# Will observers moving on a sphere experience time dilation?

A single source of light exists at a fixed point in space relative to two observers. The two observers move on the surface of a shell with a fixed radius with the light source at its centre. They move along a circular path on the sphere. They move at different velocities u and v. Will they experience time dilation relative to each other?

If relativity says that there will be time dilation on the shell of a sphere with a fixed radius with a light source at its centre, then there is a problem. Let's say the light blinks once every 10 secs. Then that light will reach all observers on the shell at the same time no matter what their relative velocity to each other. All observers would agree the light blinks once every 10 secs, no matter what their velocity is as long as they are on the shell.

• What does this have to do with general relativity? Why does the light source matter? Anytime there is relative motion between observers, there will be time dilation. – DilithiumMatrix Mar 20 '13 at 0:27
• Hi Physics Man - this is a site for conceptual questions about physics, not general homework help (or for people to answer homework-like questions for you). If you can edit your question to ask about the specific physics concept that is giving you trouble, I'll be happy to reopen it. See our FAQ and homework policy for more information. – David Z Mar 20 '13 at 2:42
• It is actually a conceptual question. Because if relativity says that there will be time dilation on the shell of a sphere with a fixed radius with a light source at its centre, then there is a problem. Let's say the light blinks once every 10 secs. Then that light will reach all observers on the shaell at the same time no matter what their relative velocity to each other. That is why I asked this question. It's not a homework question. All observers would agree the light blinks once every 10 secs, no matter what their velocity is as long as they are on the shell. – Physics n00b Mar 20 '13 at 4:43
• I think this should be reopened. It strikes me as quite a subtle question and I'm not sure I know the answer to it. – John Rennie Mar 20 '13 at 10:56
• @PhysicsMan: assuming we can ignore the curvature due to the small masses present, I think the time dilation on the circling rockets is just due to their speed. So the time dilation on the two rockets will be different. However the flashes won't reach the rockets at the same time in the frame of the rockets. I must admit I'm not sure from a quick look at the problem exactly what would be reported by the various observers. – John Rennie Mar 20 '13 at 10:59

This is only a qualitative answer because a proper treatment will take me much head scratching if I could manage it at all. Anyhow this is what I think happens.

The question is a variant of the twin paradox. At some point the positions of two spaceships, $u$ and $v$, will coincide and at that point they set their clocks to read the same. If their speeds (measured in the frame of the light source) have some rational ratio then at some future time their positions will again coincide and they can compare clocks (which will differ). During this time all observers must agree on the number of light pulses received.

In the absence of any significant amount of mass the time dilation on the ships, as measured by the light source, is simply the usual factor of $\gamma$ that depends on their speed, so the light source sees the clocks on the two spaceships slowed by different amounts. However by symmetry the observers on the ships see the light source clock slowed by the (two different) factors of $\gamma$ so the faster ship will see a lower light pulse frequency than the slower ship.

The trouble is that faster ship sees the slower light pulse frequency and it records less elapsed time between the two meetings. So it must have counted fewer pulses than the slower ship. How can this be reconciled with the requirement that all observers count the same number of pulses?

The solution is that the spaceships are not in inertial frames. If you have two inertial frames, $S$ and $S'$, moving at some non-zero relative velocity then it's true that both frames see clocks in the other frame slowed by the same factor of $\gamma$. However if you are accelerating you are constantly changing inertial frames and the symmetry argument does not apply. You can see this in the simple example of the relativistic rocket. The time dilation between the accelerating rocket and the static observer is not symmetrical (which is the solution to the twin paradox).

I don't know how to calculate the time dilation for an object in a circular motion, so I can't give a quantitative answer. However, qualitatively, the spaceships circling the light source see the light source clock run fast not slow, and the faster the ships travel the faster they will see the light source clock run.

So in the example I started with, in between their two meetings the faster moving ship $v$ will record less elapsed time than the slower moving ship $u$, but it will have measured a higher light pulse frequency than $u$ does, and $u$ in turn measures a higher pulse frequency than the static observer. The end result is that both ships and the static observer count the same number of light pulses over the period between the ships meetings.

Response to comment:

You're concerned that because the circling spaceships have zero velocity in the direction of the light source there can't be any time dilation. This isn't true, and you can see this very simply from the following:

This diagram shows two inertial frames $S$ and $S'$ with a relative velocity $v$. The light source $L$ is in $S$ and the two points $A$ and $B$ are in $S'$. Ignore point $B$ for now, then $A$ is travelling towards $L$ at velocity $v$ and the time dilation is the usual $\gamma$.

But $A$ and $B$ are stationary wrt each other, therefore their clocks tick at the same rate. So the time dilation for $B$ is the same as for $A$. Now consider the moment when $A$ and $L$ coincide. At this point the component of $B$'s velocity in the direction of $L$ is zero. However we know that $A$ and $B$'s clocks are in sync, and we know that $A$ is time dilated by $\gamma$ as viewed from $L$, so the conclusion is that $B$'s clock must also be time dilated by $\gamma$ as viewed from $L$.

At this moment my example of the two frames looks like your example of circular motion, where the distance between $A$ and $B$, $r$, is the radius of the circular motion and $B$ is the position of the spaceship. That's why the light source sees the circling ships time dilated by a factor of $\gamma$.

Your setup is really a variant of the relativistic rotating disk. Unusually Wikipedia has no good article on this, though there is some discussion of it in the articles on the Ehrenfest Paradox and Gravitational Time Dilation. However if you Google for relativistic rotating disk you'll find lots of web sites discussing it.

• My question then is this: On any circle on the surface of the sphere, there is no length contraction in the direction of the light source (at least I think so). Hence the observers on the space ships should measure the "actual" distance to the light source. Now if both observers measure actual differences in length between the light source and their position (the length observed by the two observers would be the same). Now if length they measure is the same, and the speed of light they measure is the same, then time must also be the same. – Physics n00b Mar 20 '13 at 22:30
• The distance measured by both observers should be the radius of the sphere (no length contraction) since they are at any point on a circle on the surface of the sphere with the blinking light source at the centre. – Physics n00b Mar 20 '13 at 22:54
• The velocity at any given point on the path of a circle on the surface of the sphere will be a tangent to the circle and that velocity will have no component towards the light source. – Physics n00b Mar 20 '13 at 23:34
• Thank you so much for taking the time to answer this. I will read up on this topic and get back if I have more questions, if I don't I will accept your answer. – Physics n00b Mar 21 '13 at 9:47

You write: "All observers would agree the light blinks once every 10 secs, no matter what their velocity is as long as they are on the shell."

Using their own stationary clock, no they wouldn't, because the light source is moving relative to them. Each would record the light blinking at a rate slower by a factor $\gamma$