Page:CarmichealPostulates.djvu/20

and time coordinates. Let two systems of reference S and $S'$ have the relative velocity v in the line l. Let systems of rectangular coordinates be attached to the systems of reference S and $S'$ in such way that the x-axis of each system is in the line l, and let the y-axis and the z-axis of one system be parallel to the y-axis and the z-axis respectively of the other system. Let the origins of the two systems coincide at the time t = 0. Furthermore, for the sake of distinction, denote the coordinates on S by x, y, z, t and those on $S'$ by x', y', z', t'. We require to find the value of the latter coordinates in terms of the former.

From postulate L it follows at once that y' = y and z' = z. Let an observer on S consider a point which at the time t = 0 appears to him to be at distance^[1] x from the $y'z'$-plane; at time t = t it will appear to him to be at the distance x - vt from the $y'z'$-plane. Now, by an observer on $S'$ this distance is denoted by $x'$. Then from theorem, VI. we have

$x'{\sqrt {1-\beta ^{2}}}=x-vt$

Now consider a point at the distance x from the yz-plane at time t = t in units of system S. From theorem VII. it follows that to an observer on S the clock on $S'$ at the same distance x from the yz-plane will appear behind by the amount

${\frac {v}{c^{2}}}x$,

where c is the velocity of light. That is, in units of S this clock would register the time

$t-{\frac {v}{c^{2}}}x$.

Hence, by means of theorem IV., we have at once the result

$t'{\sqrt {1-\beta ^{2}}}=t-{\frac {v}{c^{2}}}x$.

Solving the two equations involving $x'$ and $t'$ and collecting results, we have

(A)	${\begin{array}{llc}t'=&{\frac {1}{\sqrt {1-\beta ^{2}}}}\left(t-{\frac {v}{c^{2}}}x\right),\\\\x'=&{\frac {1}{\sqrt {1-\beta ^{2}}}}(x-vt),&(MVLR)\\\\y'=&y\\\\z'=&z,\end{array}}$

where $\beta =v/c$ and c is the velocity of light.

In the same way we may obtain the equations which express t, x, y, z

1. The algebraic sign of the distance is supposed to be taken into account in the value of x.