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# Proper Time and Time Dilation

Suppose we have a particle moving with a velocity v in a given coordinate system . In a time (in that system) it moves . Then its invariant infinitesimal interval is (ds)^2 = (c dt)^2 - dx^2 = c^2 dt^2 (1 - &beta#beta;^2). In the particular frame where the particle is at rest ( ) we define the proper time to be d&tau#tau;= dt' so that (ds)^2 = c^2 (d&tau#tau;)^2 . Thus the proper time is just the time experienced by the particle in its own rest frame.

From the relations above, it is easy to see that d &tau#tau;= dt 1 - &beta#beta;^2(t) = dt&gamma#gamma;(t) and to find the interval between two events on some world line it is necessary to integrate: t_2 - t_1 & = & &int#int;_&tau#tau;_1^&tau#tau;_2 d&tau#tau;1 - &beta#beta;^2(&tau#tau;)
& = & &int#int;_&tau#tau;_1^&tau#tau;_2 &gamma#gamma;(&tau#tau;) d&tau#tau;. If is constant (so the frames are inertial) then we get the usual time dilation &Delta#Delta;t = &gamma#gamma;&Delta#Delta;&tau#tau; or &Delta#Delta;&tau#tau;= &Delta#Delta;t&gamma#gamma; but this is not true if the particle is accelerating. Applying it without thought leads to the twin paradox''. However, the full integral relations will be valid even if the two particles are accelerating (so that ). You will need to evaluate these relations to solve the twin paradox for one of your homework problems.

Finally, I want to note (without discussing it further at this time) that proper time dilation leads to a relativistic correction to the usual doppler shift. Or should I say that the non-relativistic doppler shift is just a low velocity limit of the correct, time dilated result.

Now that we have some notion of what an infinitesimal time interval is, we could go ahead and try to defince 4-dimensional generalizations of momentum and energy. First, however, we will learn how velocities Lorentz transform.    Next: Addition of Velocities Up: Special Relativity Previous: 4-Vectors   Contents
Robert G. Brown 2017-07-11