A clock near the surface of the earth will run slower than one on the top of the mountain. If the equivalence principal tells us that being at rest in a gravitational field is equivalent to being in an accelerated frame of reference in free space, shouldn't the clock near the earth run increasingly slower than the other clock over time? If those two clocks are considered to be in two different frames of acceleration, the one near the earth will have a greater acceleration than the one on the mountain, and thus over time, their relative velocity will increase over time which will increase the time dilation. Or is this not a proper use of the equivalence principle?
General Relativity – Time Dilation in a Gravitational Field and the Equivalence Principle
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But doesn't a clock in an accelerating spaceship run at the same rate no matter where in the ship you put it?
Remarkably, the answer is, even in the context of SR, no.
It turns out that acceleration of an extended object is quite subtle.
That is to say, we can't meaningfully speak of the acceleration of an extended object.
Essentially, the 'front' (top?) of the spacecraft accelerates less than the 'back' (bottom?) of the spacecraft if the spacecraft is not to stretch and eventually fail structurally.
Thus, the clocks at the front (top) run faster than the clocks at the back (bottom) as would be the case for clocks at rest at different heights in a gravitational potential.
This is actually well known and best understood in the context of Rindler observers.
Note that Rindler observers with smaller constant x coordinate are accelerating harder to keep up! This may seem surprising because in Newtonian physics, observers who maintain constant relative distance must share the same acceleration. But in relativistic physics, we see that the trailing endpoint of a rod which is accelerated by some external force (parallel to its symmetry axis) must accelerate a bit harder than the leading endpoint, or else it must ultimately break.
Now, this isn't meant to answer your general question but, rather, to address the particular question quoted at the top.
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.
Best Answer
Rather: the geometric (and kinematic) relations between two (or more) given, distinct, separated clocks must be determined and taken into consideration in order to compare intervals (from any one indication to any other indication) of each clock to each other, on this basis to compare the "proper rates" (or frequencies) at which these given clocks "ran" (or "ticked"), and thus to find out whether they "ran equally", or which "ran slower" than the other (in terms of their individual "proper rates").
Certainly it is possible to consider participants in a flat region which move with constant proper acceleration;
and certainly it is possible to consider pairs of such participants which are rigid to each other (i.e. either one of them finding constant ping duration to the other), just as "the surface of the earth" and "the top of the mountain" are rigid to each other.
Some relevant calculations concerning such geometric/kinematic relations between two participants in a flat region are shown in my answer there.
Yes; as shown in the indicated answer, the two participants which remain rigid to each other in a flat region move at ("slightly") different constant proper accelerations, too;
namely "the top" moving with proper acceleration $ k \, e^{(\frac{-k}{c^2} L )} $,
compared to "the bottom" moving with proper acceleration $k$,
where $\frac{2 L}{c}$ is the constant ping duration of "the bottom" to "the top" and back.