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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?

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A clock near the surface of the earth will run slower than one on the top of the mountain.

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").

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

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.

the one near the earth will have a greater acceleration than the one on the mountain

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.

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The fact that the clock near the earth would run continually slower - i.e. the difference between the two would grow the more time they are seperated - is enough to be equivalent to different rates of acceleration. It is not like it runs more slowly for at bit and then runs at the same rate as the other one, but slightly behind.

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