§ 3. Proper time

Suppose that in a certain inertial reference system we observe clocks which are moving relative to us in an arbitrary manner. At each different moment of time this motion can be considered as uniform. Thus at each moment of time we can introduce a coordinate system

rigidly linked to the moving clocks, which with the clocks constitutes an inertial reference system.

In the course of an infinitesimal time interval $dt$ (as read by a clock in our rest frame) the moving clocks go a distance $\sqrt{dx^2 + dy^2 + dz^2}$ . Let us ask what time interval $dt'$ is indicated for this period by the moving clocks. In a system of coordinates linked to the moving clocks, the latter are at rest, i.e., $dx' = dy' = dz' = 0$ . Because of the invariance of intervals

from which

But

where $v$ is the velocity of the moving clocks; therefore

Integrating this expression, we can obtain the time interval indicated by the moving clocks when the elapsed time according to a clock at rest is $t_2 - t_1$ :

The time read by a clock moving with a given object is called the proper time for this object. Formulas (3.1) and (3.2) express the proper time in terms of the time for a system of reference from which the motion is observed.

As we see from (3.1) or (3.2), the proper time of a moving object is always less than the corresponding interval in the rest system. In other words, moving clocks go more slowly than those at rest.

Suppose some clocks are moving in uniform rectilinear motion relative to an inertial system $K$ . A reference frame $K'$ linked to the latter is also inertial. Then from the point of view of an observer in the $K$ system the clocks in the $K'$ system fall behind. And conversely, from the point of view of the $K'$ system, the clocks in $K$ lag. To convince ourselves that there is no contradiction, let us note the following. In order to establish that the clocks in the $K'$ system lag behind those in the $K$ system, we must proceed in the following fashion. Suppose that at a certain moment the clock in $K'$ passes by the clock in $K$ , and at that moment the readings of the two clocks coincide. To compare the rates of the two clocks in $K$ and $K'$ we must once more compare the readings of the same moving clock in $K'$ with the clocks in $K$ . But now we compare this clock with different clocks in $K$ —with those past which the clock in $K'$ goes at this new time. Then we find that the clock in $K'$ lags behind the clocks in $K$ with which it is being compared. We see that to compare the rates of clocks in two reference frames we require several clocks in one frame and one in the other, and that therefore this process is not symmetric with respect to the two systems. The clock that appears to lag is always the one which is being compared with different clocks in the other system.

If we have two clocks, one of which describes a closed path returning to the starting point (the position of the clock which remained at rest), then clearly the moving clock appears to lag relative to the one at rest. The converse reasoning, in which the moving clock would be considered to be at rest (and vice versa) is now impossible, since the clock describing a closed trajectory does not carry out a uniform rectilinear motion, so that a coordinate system linked to it will not be inertial.

Since the laws of nature are the same only for inertial reference frames, the frames linked to the clock at rest (inertial frame) and to the moving clock (non-inertial) have different properties, and the argument which leads to the result that the clock at rest must lag is not valid.

The time interval read by a clock is equal to the integral

taken along the world line of the clock. If the clock is at rest then its world line is clearly a line parallel to the $t$ axis; if the clock carries out a nonuniform motion in a closed path and returns to its starting point, then its world line will be a curve passing through the two points, on the straight world line of a clock at rest, corresponding to the beginning and end of the motion. On the other hand, we saw that the clock at rest always indicates a greater time interval than the moving one. Thus we arrive at the result that the integral

taken between a given pair of world points, has its maximum value if it is taken along the straight world line joining these two points.†