Imagine three observers - one (A) stationary on the surface of Earth (latitude 0 deg) and two others orbiting the planet in the same circular equatorial orbit just in the opposite direction. When the orbiting observers B and C meet each other just above the observer A all theisr clocks are synchronized. When their meet again what will their clock show? They move relative to each other, but experience equal acceleration and gravitational potential. Will the orbiting observers have their clock synchronized all the time? If the time intervals differ who measures greater time and why? Suppose Earth in non-rotating itself and with perfect spherical symmetry.
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4 Answers
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Interesting question... with the assumptions you've specified, obviously observers B and C will have the same time on their clocks when they meet again because their situations are identical. Observer A, on the other hand, has to hold itself in place with a rocket or something, so it's not in an inertial reference frame. Based on that, A will have a different time on its clock when the three meet up again.
Let's see what the math says. Given a spherical, nonrotating Earth, we can use the Schwarzschild metric to describe spacetime outside it.
$$c^2\mathrm{d}\tau^2 = \biggl(1 - \frac{r_s}{r}\biggr)c^2\mathrm{d}t^2 - \biggl(1 - \frac{r_s}{r}\biggr)^{-1}\mathrm{d}r^2 - r^2(\mathrm{d}\theta^2 + \sin^2\theta\,\mathrm{d}\phi^2)$$
For the spacetime trajectories of the three observers, $\mathrm{d}r = 0$ (because they stay at a constant radius) and $\mathrm{d}\phi = 0$ (because they orbit in a single plane). So a bit of algebra gets you to
$$\frac{\mathrm{d}\tau}{\mathrm{d}t} = \sqrt{\biggl(1 - \frac{r_s}{r}\biggr) - \frac{r^2}{c^2}\biggl(\frac{\mathrm{d}\theta}{\mathrm{d}t}\biggr)^2}$$
In this formula, $r$ is equal to the radial coordinate of the three observers, $c$ is the speed of light, and $r_s$ is the Schwarzschild radius of the Earth. All three of those are constants. The only thing that differs from one observer to another is the coordinate angular velocity $\frac{\mathrm{d}\theta}{\mathrm{d}t}$, which is something like the angular velocity as measured by a distant observer. For observer B, this will be equal to some constant $\omega$, for C it'll be equal to $-\omega$, and for A it will be zero. This means that $\frac{\mathrm{d}\tau}{\mathrm{d}t}$ is equal for B and C, and slightly greater for A.
Now, this quantity $\frac{\mathrm{d}\tau}{\mathrm{d}t}$ is the rate at which proper time ($\tau$) elapses relative to coordinate time ($t$). The coordinate time is, again, basically what would be measured by a distant observer. So each time the three observers A,B,C meet up, the meeting takes place at the same coordinate time for all three of them. However, the proper time $\tau$, which is the time each observer measures internally, is not the same for all three. The fact that $\frac{\mathrm{d}\tau}{\mathrm{d}t}$ is larger for observer A means that for a given amount of coordinate time (like, say, the interval between two successive meetings of the three observers), A will experience more time than B or C. So if the observers start with synchronized clocks, when they next meet up, A will find that its clock is a little bit ahead of B's and C's clocks.
If you're curious about the numbers: we may not have enough precision to get an accurate result, but I can do this just to show how the calculation would work. Let's plug in the Earth's Schwarzschild radius of $r_s = 8.9\text{ mm}$ and the orbital radius of, say, the International Space Station at $r = 6750\text{ km}$ (rough average). We can also use the ISS's orbital speed of $r\frac{\mathrm{d}\theta}{\mathrm{d}t} = 7.68\ \mathrm{\frac{km}{s}}$ for observers B and C. That gives the following rates:
$$\begin{align}\frac{\mathrm{d}\tau_A}{\mathrm{d}t} &= 1 - 1.2027\times 10^{-8} & \frac{\mathrm{d}\tau_{B,C}}{\mathrm{d}t} &= 1 - 1.2355\times 10^{-8}\end{align}$$
The difference works out to $3\times 10^{-10}$. So over a 90-minute orbital period, the clock on A would come out ahead of that on B or C by 1.7 microseconds. But again, I'm not sure this number is necessarily trustworthy because we are talking about very small numbers here, and some of the GR effects I've neglected may contribute.
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$\begingroup$ Thanks for detailed answer. How would be the entire situation seen or calculated by one of the orbiting observers (rather than Schwarzschild far away from Earth)? I assume that for him/her (B) the clock of C will has different rate but that the dilatation due to relative velocity average out during the orbit? $\endgroup$ Commented Apr 7, 2012 at 10:22
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$\begingroup$ All I've calculated here is the difference between the various observers' clocks that accumulates over one orbit. Since all the observers meet at the same position after the orbit, this quantity is the same no matter who is observing it - in other words, there is no distant observer actually involved. The calculation tells you what A,B,C themselves would actually measure. Of course a distant observer would see the same thing as well. $\endgroup$– David ZCommented Apr 7, 2012 at 18:56
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$\begingroup$ Now, if you wanted to calculate the rate at which time passes for C as observed by B, you would have to do a more complicated calculation using the Kerr metric, because B observes the Earth to be rotating relative to its own inertial frame. I could try to add that in (when I have time) if you want. $\endgroup$– David ZCommented Apr 7, 2012 at 19:03
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$\begingroup$ Thanks for the additional comments. If you have time later to add the Kerr metric solution I would much appreciate it. $\endgroup$ Commented Apr 7, 2012 at 20:23
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$\begingroup$ @Leos I will do that when I have a chance but it might have to wait for a day or so. $\endgroup$– David ZCommented Apr 7, 2012 at 22:09
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As soon as there is any gravitation or acceleration you need to use general relativity and the simple time-dilation effects you get in special relativity are not present (that is they are no longer as simple).
In fact the scenario you describe is equivalent to the standard twin paradox with the exception that your twins are smoothly accelerating with respect to each other and that there is symmetry between twins (triplets?) B and C. The resolution to the twin paradox is that for the travelling twin to return she must accelerate and that is counter to the assumptions made in special relativity.
The nice thing about your example is that it shows that the problem with the twin paradox is not that there is symmetry breaking between the twins, but that acceleration or gravitation is present.
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B and C will have the same time, but A time will differ. It will either be faster, if B and C are in a low enough orbit, or slower if they are in a high enough orbit. You must use GTR if gravity is involved
answered Jun 6, 2019 at 14:30
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Interesting problem. The ref A, the static one can be ommited. The others that are orbitng the Earth, will experience the same gravitational force, and both clocks are equally affected and so GR can not contribute to any difference.
The orbit is circular and, at first sight, it calls for an acceleration.
Is it so? Nope, imo. Although they are subject to forces, the gravitation and the centripetal force, they cancel each other and the net force is zero in the radial direction. If the engines are off then the tangencial acceleration is also zero.
They are simply following the 'geodesic', in free fall.
The problem is clearly symetrical at all levels, and SR is enough.
The moment they cross each other thay can send a message with the clock time and they will see that they mark the same time.
Mr David in his answer already told what happens to the clock of A that remained in the surface of the Earth.
I'will change its position to the same altitude of the others that are orbiting. If it goes to a stationary position wrt the non-rotating Earth surface it must apply engines to counter the effect of the gravitation, by the same amount of force. But, inspite of this, is the observer accelerating ? Nope, imo, by the same reason that he suffer zero net force in all directions.
SR is enough? Make your own conclusions.
