# I Two orbiting clocks question

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1. May 11, 2017

### Bob Walance

Two clocks A and B are orbiting a non-rotating planet. Each clock periodically transmits its current time value via a radio signal and the other can receive that value.

Their orbits are in closely-spaced parallel planes and both orbits are the same distance from the planet. The two clocks are orbiting in opposite directions from each other. Twice every orbit, as they pass each other, they are very close together.

In this scenario, both clocks are inertial (according to Einstein's general theory) so their relative clock rates should only be affected by the special relativity factor "gamma". As I understand it, this gamma factor is not a function of whether two objects are getting closer or farther away from one another.

During most of the phases of each orbit, according to special relativity the A clock should see B's clock time between ticks increase (with respect to A's clock). The same holds for B's view of A.

On one orbit, just as they are passing each other, both clocks are reset to zero.

So, due to the symmetry of this scenario and due to common sense, as the two clocks are passing each other (twice per orbit), both clocks should receive a clock value from the other that is equal to their own clock's value.

How can this be if a given clock never sees the other clock rate increase to make up the amount that it has slowed down? Does gamma really not depend on whether two objects are approaching or receding?

Any help with this would sincerely be appreciated.

Bob

2. May 11, 2017

### Staff: Mentor

This is same problem as the symmetrical form of the twin paradox (both twins leave the earth in opposite directions, turn around, return to earth at the same time) about which we have several threads.

Search for these threads, and also check out the twin paradox FAQ at http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html - pay particular attention to the "Doppler Shift Analysis" section.

The easiest way to understand this problem is to imagine that both clocks carry strobe lights that flash once per second, accordng to an observer riding on the satellite along with the clock. We know how many times the clock on the other satellite ticked between flybys by counting the number of flashes of light that we receive from it between flybys.

3. May 11, 2017

### Ibix

If my friend is cross with me and declares that he never wants to see me again and stalks off, and doesn't realise that we're on a sphere not a Euclidean plane, there's an awkward moment coming down the line. You've made an analogous mistake. Special relativity does apply locally to both clocks in the same way that I use Euclidean geometry when tiling my kitchen floor. But it cannot apply to your whole scenario because special relativity provides no mechanism for something to move inertially in a circle.

In particular, what has bitten you is the assumption that the gamma factor is the only thing in play. That's a special relativity assumption that comes from the way to synchronise clocks used nearly universally in special relativistic scenarios. But general relativity provides many ways to answer the question "what time is it over there", with the only real constraints in this case being (a) later than the last time you asked; and (b) smoothly approaches the same time as you as we get closer together. @Nugatory has provided a very sensible strategy for answering what time it is, and if you work it out (I think you can treat light paths as straight lines, at least for a first approximation) you will find that "now, at the other clock" does not behave in the special relativity way you were citing. Rather, it will work as you correctly predicted from consistency requirements.

Last edited: May 11, 2017
4. May 11, 2017

### stevendaryl

Staff Emeritus
There are two different things that you might be getting mixed up:
1. The rate at which a receiver of signals gets signals from a sender.
2. The rate of a clock as measured using a coordinate system.
If you have a sender that is transmitting signals once per second (as measured by the sender's clock) to a receiver, then the rate at which those signals arrive at the receiver (as measured by the receiver's clocks) could be greater than once per second, or less than once per second, depending on whether the sender is moving toward the receiver or away from the receiver, and depending on the locations of the sender and the receiver. That's called the "Doppler shift", when there is no gravity involved. That's effect #1.

The second effect, which is what's called "relativistic time dilation" is not, in my opinion, best understood as a comparison of two clocks. Instead, it should be considered the rate of one clock as measured in a coordinate system in which that clock is not at rest. So it's a relationship between one clock and a coordinate system. It's not a relationship between two clocks. When there is no gravity involved, you can treat it as a relationship between two clocks, simply because for any clock that's moving inertially, there is an associated inertial coordinate system. When it comes to General Relativity, however, there is no unique coordinate system associated with a moving clock.

1. Pick some coordinate system. What are the clock rates for each clock in that coordinate system?
2. If the two clocks are continually sending signals saying their current time, what is the relationship between the time on one clock and the time shown on the signal from the other clock?
The answer to the first question is dependent on what coordinate system you use. The most convenient coordinate system is spherical polar coordinates centered on the planet (for example, the Schwarzschild coordinates). If the two clocks are moving in circular orbits of equal height, then their rates will be the same.

The answer to the second question is qualitatively the same as Doppler shift would suggest: the apparent clock rate for clock A as viewed by clock B seems greater than 1 when A is approaching B, but less than 1 when A is getting farther from B (and similarly with the apparent clock for B as viewed by A).

5. May 12, 2017

### Arkalius

Not quite (and I'm sure someone here will correct me if I get this a little wrong). They are in only approximately inertial frames within a small enough region around themselves where tidal gravity isn't noticeable. However, on the scale of their entire orbit, tidal gravity is clearly apparent, so they are not inertial on those scales.

The idea is that, if you have a circle, when you zoom in enough on part of it, it eventually looks approximately like a straight line. But the more you zoom out, the more you see the curvature of that line.

6. May 12, 2017

### DrGreg

Essentially yes. But to get the terminology right, I would say that the clocks are inertial, but they don't have inertial frames, only "approximately inertial frames" as you say.