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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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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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