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Time Dilation

  

Figure: A "light clock" is constructed aboard a glass spaceship (reference frame O') as follows: the "tick" of the clock is defined by one half the time interval  t'  required for the light from a strobe light to traverse the width of the ship (a height h), bounce off a mirror and come back, a total distance of 2h. In the reference frame of a ground-based observer O (with respect to whom the ship is travelling at a velocity u), the light is emitted a distance  2ut  behind the place where it is detected a time  2t  later. Since the light has further to go in the O frame (a distance $\ell = \sqrt{h^2 + u^2t^2}$), but it travels at  c  in both frames,  t  must be longer than  t'. This effect is known as TIME DILATION.

Fig. 23.3 pictures a device used by R.P. Feynman, among others, to illustrate the phenomenon of TIME DILATION: a clock aboard a fast-moving vessel (even a normal clock) appears^23.7 to run slower when observed from the "rest frame" -- the name we give to the reference frame arbitrarily chosen to be at rest. Now, if we choose to regard the ship's frame as "at rest" (as is the wont of those aboard) and the Earth as "moving," a clock on Earth will appear to be running slowly when observed from the ship! Who is right? The correct answer is "both," in utter disregard for common sense. This seems to create a logical paradox, which we will discuss momentarily. But first let's go beyond the qualitative statement, "The clock runs slower," and ask how much slower.

For this we need only a little algebra and geometry; nevertheless, the derivation is perilous, so watch carefully. For O', the time interval described in Fig. 23.3 is simply

$$t' \; = \; {h \over c} \qquad \qquad \hbox{\rm so that} \qquad \qquad h \; = \; ct'$$

whereas for O the time interval is given by

$$t \; = \; {\ell \over c} \qquad \qquad \hbox{\rm where} \qquad \qquad \ell^2 \; = \; h^2 \; + \; u^2 t^2$$

by the Pythagorean theorem. Expanding the latter equation gives

$$t \; = \; {\sqrt{ h^2 + u^2 t^2 } \over c} \qquad \qquad \hbox{\rm or} \qquad \qquad c^2 t^2 \; = \; h^2 \; + \; u^2 t^2$$

which is not a solution yet because it does not relate  t  to  t'. We need to "plug in"   h² = c² t'²  from earlier, to get

$$\begin{eqnarray*}c^2 t^2 &=& c^2 t'^2 \; + \; u^2 t^2 \cr \hbox{\rm or} \qquad . . . . . . , t^2 \cr \hbox{\rm or} \qquad t^2 \, (1 - \beta^2) &=& t'^2 \end{eqnarray*}$$

where we have recalled the definition   $\beta \equiv u/c$. In one last step we obtain

$$t \; = \; { t' \over \sqrt{ 1 - \beta^2 } } \qquad \qquad \hbox{\rm or} \qquad \qquad t \; = \; \gamma \; t'$$

where  $\gamma$  is defined as before: $\gamma \equiv 1/\sqrt{1 - \beta^2}$.

This derivation is a little crude, but it shows where  $\gamma$  comes from.