Proper Time and Time Dilation

Suppose we have a particle moving with a velocity v in a given coordinate system $K$. In a time $dt$ (in that system) it moves $d{\bf x} = {\bf v}dt$. Then its invariant infinitesimal interval is

$$(ds)^2 = (c dt)^2 - \left\vert d{\bf x} \right\vert^2 = c^2 dt^2 (1 - \beta^2).$$ (15.59)

In the particular frame where the particle is at rest ($dx' = 0$) we define the proper time to be

$$d\tau = dt'$$ (15.60)

so that

$$(ds)^2 = c^2 (d\tau)^2 .$$ (15.61)

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

$$d \tau = dt \sqrt{1 - \beta^2(t)} = \frac{dt}{\gamma(t)}$$ (15.62)

and to find the interval between two events on some world line it is necessary to integrate:

$$t_2 - t_1 = \int_{\tau_1}^{\tau_2} \frac{d\tau}{\sqrt{1 - \beta^2(\tau)}}$$

$$= \int_{\tau_1}^{\tau_2} \gamma(\tau) d\tau .$$ (15.63)

If $\beta$ is constant (so the frames are inertial) then we get the usual time dilation

$$\Delta t = \gamma \Delta \tau$$ (15.64)

or

$$\Delta \tau = \frac{\Delta t}{\gamma}$$ (15.65)

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 $\beta(\tau)$). 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.