# Time dilation in a centrifuge: effect of velocity and acceleration?

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?

• 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! Apr 19 at 6:09
• @JohnRennie thanks for pointing that out. You have a good answer there too. Apr 19 at 17:26