# Momentarily co-moving inertial frames

1. Oct 31, 2015

### Geofleur

I wanted to check my understanding of momentarily co-moving inertial frames, so I came up with this example:

Consider an inertial frame (with unprimed coordinates), about whose origin a clock moves in a circular path with constant speed, $v$. What is the time elapsed on the moving clock after one period of the motion when the time elapsed on the stationary clock is $T$?

Construct a series of inertial frames, each one of which instantaneously spatially coincides with the moving clock at a different point along its trajectory. Moreover, each momentarily co-moving inertial clock reads the same time as the orbiting clock at the moment the two coincide. Then the elapsed time interval on any one of the co-moving inertial clocks and the stationary clock are related by

$\Delta t' = \Delta t \sqrt{1-\frac{v^2}{c^2}}$.

If $T$ is the period of the orbiting clock as viewed from the stationary frame, then the time elapsed on the orbiting clock after one period is therefore

$T' = \int_0^T \sqrt{1-\frac{v^2}{c^2}} dt = T\sqrt{1-\frac{v^2}{c^2}}$.

Is all this correct?

2. Oct 31, 2015

### TSny

According to the comoving frame, the stationary clock is moving while the orbiting clock is momentarily at rest. So, according to the comoving frame, the stationary clock is ticking slow due to time dilation while the orbiting clock is ticking at a normal rate. So, if primes denote time on the orbiting clock, you should have $dt= dt' \sqrt{1-\frac{v^2}{c^2}}$ according to the comoving frame.

EDIT: Your expression is valid according to the stationary frame. So, if that's what you had in mind, everything looks good.

Last edited: Oct 31, 2015
3. Oct 31, 2015

### Geofleur

Yes, it was the stationary frame's point of view I had in mind. Thanks!

4. Oct 31, 2015

### TSny

OK, good.

5. Oct 31, 2015

### Geofleur

I don't know whether this should go in a separate thread, but I've gotten confused. In Landau & Lifshitz, Fields, they say

"If we have two clocks, one of which describes a closed path returning to the starting point (the position of the clock which remained at rest), then clearly the moving clock appears to lag relative to the one at rest. The converse reasoning, in which the moving clock would be considered to be a rest (and vice versa) is now impossible, since the clock describing a closed trajectory does not carry out a uniform rectilinear motion, so that a coordinate system linked to it will not be inertial."

Does that mean that we cannot, after all, describe my example from the point of view of the orbiting clock?

6. Oct 31, 2015

### TSny

According to a particular comoving frame, when the orbiting clock increases its time by $dt’$ the stationary clock will increase its time by $dt = dt' \sqrt{1- \frac{v^2}{c^2}}$. If you integrate this you get $T = T' \sqrt{1- \frac{v^2}{c^2}}$. It is tempting to interpret this as saying that an observer who traveled with the orbiting clock would conclude that the stationary clock increases its reading by $T$ while the orbiting clock increased its reading by $T’$. However, it is not that simple. The orbiting clock is continually changing from one comoving frame to another and there is an additional effect due to the fact that different comoving frames do not agree on simultaneity of events.

It’s similar to the classic twin paradox. A stationary clock remains on the earth while a traveling clock moves away at constant speed and then returns at constant speed. The two clocks read zero at takeoff. Consider a frame comoving with the traveling clock on the way out. A comoving-frame observer would say that during the outbound trip the earth clock advances by some amount $\Delta T_1$ while the traveling clock advances by $\Delta T’_1$. Time dilation from the point of view of the comoving frame gives $\Delta T_1 = \Delta T’_1 \sqrt{1-\frac{v^2}{c^2}}$.

Likewise, on the way back, $\Delta T_2 = \Delta T’_2 \sqrt{1-\frac{v^2}{c^2}}$ according to the returning comoving frame.

At the instant that the traveling clock returns to earth, the traveling clock will read $T’_{final} = \Delta T’_1 + \Delta T’_2$. But the earth clock will not read $\Delta T_1 + \Delta T_2$.

That is $T_{final} \neq \Delta T_1 + \Delta T_2 = T’_{final} \sqrt{1-\frac{v^2}{c^2}}$. As we know, the correct relation is $T_{final} =T’_{final}/ \sqrt{1-\frac{v^2}{c^2}}$.

Adding $\Delta T_1$ and $\Delta T_2$ does not take into account the fact that switching from one comoving frame to another comoving frame (at the turnaround point) involves a sudden change in the reading of the stationary clock. This is due to the two different comoving frames disagreeing on which reading of the stationary clock is simultaneous with the turnaround event. Thus there is an extra $\Delta T_{extra}$ of the stationary clock associated with the switching of comoving frames. When this is taken into account, you find that the reading of the stationary clock at the instant of return to earth (using the comoving frames) is $T_{final} = \Delta T_1 + \Delta T_2 + \Delta T_{extra} = T’_{final}/ \sqrt{1-\frac{v^2}{c^2}}$. So the comoving-frame analysis agrees with the earth-frame analysis.

In the case of the orbiting clock, there is a continual switching of comoving frames. Roughly speaking, each switch involves an extra $dT_{extra}$ that needs to be included if you want to be able to deduce what the stationary clock reads from the point of view of the comoving frames of the orbiting clock.

Last edited: Nov 1, 2015
7. Nov 1, 2015

### Geofleur

Thank you for that very clear explanation!

8. Nov 1, 2015

### Staff: Mentor

You did fine in your example. When you are dealing with non inertial objects, whether the twins paradox or some rotational motion or whatever, you have several options for analysis:

1. You can analyze it using an inertial frame in which it is moving. The laws of physics take their standard form.

2. You can analyze it using the momentarily comoving inertial frames. In each of these frames it is at rest, but you can only use it for a moment. The laws of physics take their standard form, but you may have to do a bunch of transforms and integrals since the frame can be different at each moment.

3. You can analyze it using a non inertial frame in which it is at rest. You will have to make corrections to the laws of physics in that frame.

You used approach 2, which is fine. The quote is just pointing out that if you want to use a single frame and you want to use the frame where it is at rest, then you are using approach 3.