I might be a little dense to this subject, but I would like to ask a question relating to this one

( Is gravitational time dilation different from other forms of time dilation? )

but asking if this can be extended from gravitational time dilation to acceleration-induced time dilation; that the time dilation under differing acceleration (between gravitation and propulsion) are not symmetric as they are in inertial frames under special relativity

The heart of what I want to know;

If a spaceship is accelerating Toward the center of the Earth at an acceleration greater than earth's gravitational "acceleration", does the spaceship (ignoring or correcting for doppler distortion) percieve clocks on earth to run fast in from their reference frame?

It seems to me that they should; the spaceship's acceleration should give rise to more-distorted spacetime than Earth's gravitation, no? By equivalence, would this be akin to observers watching the flow of time deeper in a gravitational well than the are?

  • 2
    $\begingroup$ I believe general relativity states that no experiment can distinguish between gravity and acceleration, so, yes, a higher acceleration would yield the same results as more gravity. And, accelerating (non-inertial) reference frames experience time slower than inertial ones. $\endgroup$
    – user854
    Jan 17, 2016 at 2:54
  • $\begingroup$ @Jules K Let me get this straight; are you asking whether time dilation due to acceleration could "cancel out", or "overcome" the time dilation due to the gravitational field? $\endgroup$
    – praeseo
    Jan 17, 2016 at 6:12
  • $\begingroup$ @HadrianEvan If I understand right then I think that's the gist of my question. I think more I want to know, if a ship's acceleration is greater than the acceleration of an object due to gravity, will the time dilation "go the other way". For a non-accelerating ship, anything in a potential well is perceived as having slowed-down time, and observers in the well perceive the ship to have sped-up time. I'm wondering if it's just always a matter of what has the higher acceleration that determines the direction of the time dilation $\endgroup$
    – Jules K
    Jan 17, 2016 at 6:21

1 Answer 1


The brief answer is 'yes’. Here is a thought experiment which I think makes it easy to see that the answer must be yes.

Consider the standard twin 'paradox':

  • twin a hangs around in free-fall;
  • twin b zooms off on their spaceship at some enormous speed with respect to twin a, turns around (in some smooth way, undergoing acceleration), and returns.

Well, we know that twin a's clock is fast with respect to twin b's, because twin a has followed a geodesic between the two meetings, and this has the maximal proper time of all smooth paths between the two events and both twins have followed smooth paths (in fact we can make stronger statements than this, but I only want to consider smooth paths, since I am making an argument from continuity and smooth paths are continuous (and much more)).

OK, so now let's modify the experiment: instead of being in free-fall, twin a is on a planet. We can make this planet as light as we like (really, we can make the gravitational acceleration experienced by twin a as low as we like).

Well, now it is obvious that for a sufficiently low gravitational acceleration the result of the experiment is unchanged: twin a still experiences more proper time than twin b. Equally it's clear that for a sufficiently high gravitational acceleration the result will go the other way: if twin a is hanging around near the event horizon of a black hole then they will experience less proper time than twin b unless twin b does something extraordinary. By continuity there's a setup where the proper times are the same.

Finally it remains to show that the twins can always observe each others' clocks, and so they will actually see the clocks running slow or fast. It's easy to convince yourself that this is true from continuity: consider versions of the experiment where twin b's path differs only a very small amount from a geodesic and/or gravity is ver small, then it's clear that they can always see each other's clocks because they can in the limit where the paths are the same. Now you can deform the path / increase gravity continuously and nothing goes wrong.

Sorry for the slightly informal nature of these arguments: all this can be made precise, just not on a touchscreen keyboard.

  • $\begingroup$ When you say "hanging around the event horizon of a black hole" do you mean they're in orbit or do you mean they're accelerating linearly away from the event horizon just fast enough to hover there? It seems important to distinguish those two cases. $\endgroup$
    – B T
    Nov 17, 2020 at 19:50
  • $\begingroup$ @BT: I think I meant accelerating away (so hovering). $\endgroup$
    – user107153
    Nov 18, 2020 at 0:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.