JesseM
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You can also forego accelerating coordinate systems, and just analyze the time elapsed on an accelerating clock using the coordinates of some inertial frame in which you know the clock's coordinate position and velocity as a function of coordinate time. The trick is to approximate a smoothly-varying path by a polygonal path made up of a bunch of short inertial segments lasting a coordinate time \Delta t, that way the time elapsed on the clock on each segment will be \sqrt{1 - v^2/c^2} \Delta t, and then you can just add up the clock times on all the segments (using the appropriate v for each segment, which can vary from one to another) to get the total time elapsed on the polygonal path. Then you let the time of each segment become infinitesimal, so the sum becomes an integral and the total time elapsed on a clock with velocity as a function of time v(t) is just \int \sqrt{1 - v(t)^2/c^2} dt.Nugatory said:There's a widespread misconception that you need general relativity in situations involving acceleration, but it's just not true; special relativity handles acceleration just fine. You can google for "Rindler coordinates" for one example, and you'll find another example (a clock experiencing uniform circular motion due to the Earth's rotation) in Einstein's original 1905 paper to which ghwellsjr gave you a link above.
Einstein doesn't go into detail, but he does allude to this method at the end of section 4 of the 1905 paper, when he writes:
(note that the factor he gives is the result of a first-order approximation to the fully accurate formula t\sqrt{1 - v^2/c^2})It is at once apparent that this result still holds good if the clock moves from A to B in any polygonal line, and also when the points A and B coincide.
If we assume that the result proved for a polygonal line is also valid for a continuously curved line, we arrive at this result: If one of two synchronous clocks at A is moved in a closed curve with constant velocity until it returns to A, the journey lasting t seconds, then by the clock which has remained at rest the traveled clock on its arrival at A will be \frac{1}{2} tv^2/c^2 second slow.