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An object in a centrifuge has a tangential velocity of $v = \omega r$ and a radial acceleration of $g = \omega^2 r$. By itself, the velocity would cause a time dilation of $1/\sqrt{1-(v/c)^2}$ = $1/\sqrt{1-(\omega r/c)^2}$.

The potential associated with the radial acceleration is $U = -(\omega r)^2/2$, so the time dilation due to the potential alone would be $1/\sqrt{1+2U/c^2}$ = $1/\sqrt{1-(\omega r/c)^2}$, which interestingly has the same value.

I feel like I ought to be able to multiply these together to get $1/(1-(\omega r/c)^2)$ for the combined time dilation, but I doubt it's that simple.

I see a comprehensive answer from Crowell to a related question but I don't really understand the answer. It seems like a well-defined question with just two parameters, $\omega$ and $r$, so I assume the answer is a relatively simple expression?

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    $\begingroup$ This is discussed in Can a ultracentrifuge be used to test general relativity? though I'm reluctant to close this as a duplicate because I think Dale's answer here is spot on! $\endgroup$
    – John Rennie
    Commented Apr 19, 2022 at 6:09
  • $\begingroup$ @JohnRennie thanks for pointing that out. You have a good answer there too. $\endgroup$
    – Roger Wood
    Commented Apr 19, 2022 at 17:26

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Acceleration, in and of itself, does not cause time dilation. This is the clock postulate and is well confirmed by experiment e.g. in particle accelerators. So only the speed (relative to the observer) causes time dilation. If that speed is constant (even though the velocity is changing) then you can just use the ordinary SR time dilation formula. In the case of a centrifuge, you could instead use coordinates which are co-rotating, in which case the speed is 0 but you'd get the same value for time dilation by introducing a "gravitational" potential. But you shouldn't do both at the same time, because that would be mixing coordinate systems.

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  • $\begingroup$ that link is very helpful, thanks! $\endgroup$
    – Roger Wood
    Commented Jun 1, 2021 at 20:32
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which interestingly has the same value.

I feel like I ought to be able to multipy these together

It is not a coincidence that they have the same value. They are both the same time dilation from different perspectives. The velocity formula is the time dilation from the perspective of the inertial frame. The potential formula is the time dilation from the perspective of the rotating frame.

You do not multiply them. The potential doesn’t exist in the inertial frame. The velocity doesn’t exist in the rotating frame. So they are different ways of calculating the same thing, and cannot be combined

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  • $\begingroup$ I hear what you are saying, but still find it confusing. A time-dilation is always relating times measured in two systems, not just one. Otherwise what would it be dilated to. So for example, a clock in the centrifuge would according to your explanation "feel" the gravitational effect. But if I want to know the dilation w.r.t. the clock at rest in the lab, then from the point of view from the centrifuge, this lab clock would be moving with a certain speed, and I again have a special relativistic effect. So again: why not both effects? (1/2) $\endgroup$
    – Britzel
    Commented Oct 20, 2023 at 9:25
  • $\begingroup$ I do find that very confusing, still, but I think it is due to that these explanations are all using special relativity in combination with accelerated frames, whereas the full theory is in fact general relativity, in which masses create curvature and hence a gravitational field, but accelerations do not. (2/2) $\endgroup$
    – Britzel
    Commented Oct 20, 2023 at 9:26
  • $\begingroup$ @Britzel said “time-dilation is always relating times measured in two systems”. More correctly stated: time dilation is always relating a proper time to a coordinate time. $d\tau/dt=1/\gamma$. So in this question there is one clock’s proper time that is compared to coordinate time in an inertial frame where it is moving at constant speed but there is no potential, and coordinate time in a non-inertial frame where it is at rest but there is a potential. In neither frame is there both motion and a potential. GR is not needed here $\endgroup$
    – Dale
    Commented Oct 20, 2023 at 11:42
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Stop debating and do the experiment. Put a radioactive substance in an ultracentrifuge and see to what degree the decay rate changes.

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  • $\begingroup$ The experiment is done every day in particle accelerators around the world. The decay rate changes according to the speed, not the acceleration. Bailey et al did a particularly detailed study of this published in Nature 268 in July 28, 1977 pg 301. $\endgroup$
    – Eric Smith
    Commented Jul 11 at 17:30
  • $\begingroup$ Thank you. But speed only doesn't seem compatible with the equivalence principle. A particle has no way of knowing whether it is accelerating in a circle or a gravitational field. Did Bailey use experimental data or just theory? $\endgroup$
    – jamesraymond
    Commented Jul 12 at 18:20
  • $\begingroup$ They actually did the experiment, with muons. See math.ucr.edu/home/baez/physics/Relativity/SR/clock.html for more details. The equivalence principle is a useful rule of thumb more than an inviolable law of physics. In this particular case if you want to apply it you have to use a frame co-moving with the particles, in which case it will (correctly) suggest that the lab occupants, at a "higher" potential, will age more quickly. $\endgroup$
    – Eric Smith
    Commented Jul 12 at 20:18

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