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Consider a circular orbit whereby a spaceship travels around near the speed of light.

Say the radius of this orbit is such that the angular velocity is low.

An observer is placed at the center of the orbit, which has a large enough telescope to see this spaceship. He uses two telescopes facing back to back that can each see 180 degrees of view. He then observes the spaceship for a year before the spaceship slows down to a stop which they then compare times.

I assume that the times must be the same. This is because if instead of observing the rotating spaceship, one of the telescopes tracked the ship, such that to the telescope, the spaceship is stationary, then to an observer looking through the telescope the times between the ship and the observer would be the same (as the telescope is now rotating at a low angular velocity). But the ship cannot have aged more slowly just because the telescope decided not to track the ship but to just observe it.

Qmechanic
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terminate
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2 Answers2

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These sorts of questions can be solved by checking world lines. Let $S$ be the frame of the stationary observer at the center of the orbit. His/her worldline in that frame is $x^{\mu}=\left(ct,\,\mathbf{0}\right)^{\mu}$ (Cartesian coordinates, flat space-time, $c$ is the speed of light, $t$ is time in $S$ - stationary observer).

Let the world-line of the rotating observer be: $y^{\mu}=\left(ct,R\cos\omega t,R\sin\omega t, 0\right)^\mu$ where $R$ is the radius of the orbit and $\omega$ is the angular velocity.

The four-velocity of the rotating observer is then:

$$ \frac{dy^\mu}{d\tau}=\gamma \left(c, - R\omega\sin\omega t, R\omega\cos\omega t, 0\right)^\mu $$

Where $\tau$ is proper time of the rotating observer and $\gamma=dt/d\tau$ is the Lorentz factor. One can then use normalization of the four-velocity:

$$ \frac{dy^\mu}{d\tau}\,\frac{dy_\mu}{d\tau}=c^2=\gamma^2 \cdot\left(c^2-R^2\omega^2\right) $$

So

$$ \frac{dt}{d\tau}=\gamma=\frac{1}{\sqrt{1-\left(R\omega/c\right)^2}} $$

For non-zero rotation, $dt/d\tau>1$ so time in $S$ (stationary observer) is flowing faster than for the rotating observer. Rotating observer will age slower.

Cryo
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  • But what about for the other case where the telescope tracks the ship such that both now have matching angular velocities? Surely then when the telescope tracks it, it should see the ship's time as being slower, even though the ship and the telescope are both in a stationary frame. Is this right? – terminate Sep 15 '22 at 16:45
  • @terminate, what do you mean with stationary frame? Both telescope and ship change permanently velocity, however, not its magnitude but the direction. The answer of Cryo is perfectly correct. In four dimensional space the world-line of ship is a helix and of central observer a line of the same length. – JanG Sep 15 '22 at 18:22
  • @Jan Gogolin, I mean that as both telescope and ship have matching angular velocity, then I can trace an imaginary line from the telescope to the ship. Both the ship and the telescope will not move off this line and so both will always be aligned with each other and therefore stationary with respect to each other. – terminate Sep 15 '22 at 19:42
  • @terminate, what matters here is linear velocity, not angular velocity. We are describing stationary observer and one rotating around it. Both are described by single point. There is no concept of telescope here. If your telescope is in the stationary reference frame then the analysis above applies. Perhaps you want to know about the rate of time flow for two observers rotating with the same angular velocity (same plane and axis of rotation), but with different radii. This is a slightly different question, but, I think the analysis done above is easily extended to cover that case – Cryo Sep 15 '22 at 21:03
  • You will get a ratio of square roots – Cryo Sep 15 '22 at 21:03
  • I would suggest to work on world-lines of objects and maths. Hand-wavy discussions will only make things complicated – Cryo Sep 15 '22 at 21:05
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The person in the rocket will age more slowly than the observer at rest in the center. The observer in the rocket experiences an acceleration and that's equivalent to staying in a gravity field, in which time dilates. The relative velocity between both observers is constantly changing because the situation is asymmetrical. It resembles the twin paradox in the sense that the rocket experiences a constant acceleration. If de rocket meets the observer in the center they will show different clocktimes (assuming their clocks were synchronized previously).

MatterGauge
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  • But if that's the case, what about the example where the telescope tracks the ship? Both the telescope and the ship now have matching angular velocities. But the linear velocity of the ship is much greater than that of the telescope and from the telescope's perspective, the ship is stationary with respect to it. So in this situation, the times should be the same. – terminate Sep 15 '22 at 16:23
  • @terminate the problem isn't that simple, you have to remember that tracking the ship causes you to see fictitious forces that depend on distance to the ship. You have to carry out this analysis in a rotating reference frame. – Triatticus Sep 15 '22 at 16:28