Consider point B in uniform circular motion around point A, i.e. observer B is undergoing centripetal acceleration, and observer A is at rest (inertial) at the center of the circle. I think that A will observe B's clocks running slow, i.e. the usual time dilation effect, while B (who is in a non-inertial frame) will observe A's clocks moving fast.

Is this true? My reasoning is as follows.

Consider a "twin paradox" style example where twin A stays on earth and twin B is an astronaut who goes to the international space station for a while. When B returns to earth, B will be slightly younger than A. (Since we can work in A's (inertial) frame and calculate B's proper time by the usual formula, and find that it is smaller than the time elapsed on earth.)

In the classic version of the twin paradox (where twin B goes to a distant star and back, rather than orbit the earth), the crucial point is when B turns around, changing reference frames. In typical versions of the narrative (where B is mostly cruising along at a constant speed, and the acceleration comes in quick bursts), what B "actually sees" is A suddenly aging rapidly at B's turn-around point.

In the "space station" variant however, no such turning-around happens, so it seems that what must have happened is that B observes A aging quickly all along.

Edit: thanks to Dale for specifying that it depends on whether gravity is involved or not. Originally, I was thinking of the non-gravitational case. So some of the details (e.g. space station/orbits) are wrong but the idea should be clear.

  • 3
    $\begingroup$ Is B actually orbiting with gravity or is this a non-gravitational scenario where B is circling A using some other force. This makes a difference $\endgroup$
    – Dale
    Apr 27, 2021 at 0:05
  • $\begingroup$ @Dale that's a good point. I was thinking of the pure-special relativity case but I'd be interested in either $\endgroup$
    – Aqualone
    Apr 27, 2021 at 0:08
  • $\begingroup$ You really need to make a choice because it makes a big difference. The special relativity case is less complicated. $\endgroup$
    – Dale
    Apr 27, 2021 at 0:10
  • 2
    $\begingroup$ @Dale if you mean by StackExchange rules/conventions I should choose one or the other, then I'll pick the non-gravity case for this question. $\endgroup$
    – Aqualone
    Apr 27, 2021 at 0:12
  • $\begingroup$ The resolution to the twin paradox is the acceleration of the turnaround point as you mention, and here that acceleration is essentially always happening. $\endgroup$
    – Señor O
    Apr 27, 2021 at 1:12

2 Answers 2


Yes, you are correct. A sees B’s clocks run slow and B sees A’s clocks run fast.

This is different than the usual case where each sees the other’s clocks running slow. In that usual case the situation is symmetric because both A and B are inertial. But in this case the two observers are not symmetric because A is inertial and B is non-inertial.

In B's reference frame the metric is given by $$d\tau^2= \left(1-r^2 \omega^2\right) dt^2-r^2 d\theta^2 -dr^2-2 r^2 \omega \ d\theta \ dt $$ so the time dilation is given by $$ \gamma = \frac{dt}{d\tau} = \left( (1-r^2 \omega^2)-r^2 v_\theta^2 - v_r^2-2r^2 \omega v_\theta\right)^{-1/2} $$ where $v_\theta=d\theta/dt$ and $v_r=dr/dt$.

Note, for objects at rest in this reference frame, like B and A we have $v_r=v_\theta = 0$ so $\gamma = (1-r^2\omega^2)^{-1/2}$. So indeed A at $r=0$ runs faster than B at $r>0$ in B’s frame.

  • $\begingroup$ Aqualone's question is about B in a circular motion with a constant orbital speed v=omega*r and A at rest at the center of the orbit. Consider another case, in which A is at rest in the orbital plane of B but somewhere outside the orbit. In this case, the speed v of B relative to A changes all the time. Is your answer dt/dtau=gamma=1/sqrt(1-v^2) also valid in this case (without gravity)? $\endgroup$
    – gamma1954
    Oct 16, 2021 at 19:27
  • $\begingroup$ @gamma1954 the answer remains valid. Also, the speed of B relative to A does not change in the altered scenario. What changes is the distance, not the speed. The speed is the magnitude of the velocity, not the time derivative of the distance. $\endgroup$
    – Dale
    Oct 16, 2021 at 21:17

You mention that you are also interested in the satellite-orbiting-a-celestial-body scenario.

There is a discussion written by Kevin Brown, in 1997, posted in the Usenet group sci.physics

Nowadays Googlegroups is the custodian of the Usenet archive.

The name of the thread is Relativistic time on satellites
Scroll to the answer posted by Kevin Brown.

(The content has suffered a bit; Usenet communication is written to be displayed with a fixed width font, such as New Courier. Formula's are sometimes formatted on three lines, using spaces for alignment. To read the Usenet message: copy the text, paste it in a text editor, and set the font to a fixed width font.)(later edit: I'm not sure, but I think content has been damaged; I think leading spaces have been removed.)

Kevin Brown starts with pointing out that in terms of General Relativity there is is only a single time dilation. In orbital motion there is a single time dilation. Distinction between velocity time dilation and gravitational time dilation is a matter of interpretation; it's not inherent in GR.

Separation in velocity time dilation component and gravitational time dilation component can be obtained for example as follows: compute the total time dilation for orbit at some altitude, then compare that to gravitational time dilation between surface of the planet and being stationary at that altitude.

For a satellite in low Earth orbit, altitude about 300 kilometers, the difference in altitude with the Earth surface is negligable. (Gravitational potential is with respect to the Earth center.) In low Earth orbit the orbital velocity is very large, one circumnavigation about every 90 minutes. So: for a satellite orbiting in low Earth orbit a smaller amount of proper time elapses than on the Earth's surface.

As we know, there is an altitude where one circumnavigation of the Earth takes a sidereal day, giving rise to geostationary orbit. Given that gravity is an inverse square force: the higher the altitude of the orbiting motion the slower the velocity. It's not just the angular velocity that is slower, the tangential velocity is slower too.

At geostationary orbit altitude the difference in gravitational potential is such that for a satellite in geostationary orbit a larger amount of proper time elapses than the amount of proper time that elapses at Earth surfacee.

The amount of proper time that elapses for a satellite in orbit goes from smaller in low Earth orbit to larger in geostationary orbit.

Kevin Brown derives a formula for relativistic time on a satellite, finding that the cross-over altitude is at about 2000 miles up. So if you would place a satellite at precisely that orbital altitude the amount of proper time elapsing for that satellite would be the same as at Earth surface.

Incidentally, In your question you phrased everything in terms of 'observer A sees' and 'observer B sees'.

In the case of orbiting motion the effect is cumulative, in which case it is meaningful (and in my opinion better) to phrase the time effect in terms of 'amount of proper time that elapses'

Proper time is a well defined concept. Muons have half-life of 1.56 micro seconds. But for Muons moving at relativistic velocity in a storage ring a smaller amount of proper time elapses per unit of laboratory proper time. So the Muons in the storage ring are available for the experiment for a longer time (laboratory proper time).

Proper time is well defined in the following sense:
You create a scenario with protagonists who are all experts in applying relativistic physics. Being experts they will all agree on the amount of proper time that wil elapse onboard a satellite orbiting at a particular altitude. They will agree on the ratio of proper time onboard the satellite compared to the proper time at Earth surface.


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