(x’ — ct’) = l (x — ct) . . . (3).

is fulfilled in general, where l indicates a constant ; for, according to (3), the disappearance of (x — ct) involves the disappearance of (x’ — ct’).

If we apply quite similar considerations to light rays which are being transmitted along the negative x-axis, we obtain the condition

(x’ + ct’) = µ(x + ct) . . . (4).

By adding (or subtracting) equations (3) and (4), and introducing for convenience the constants a and b in place of the constants l and µ, where

eq. 29:  file eq29.gif

and

eq. 30:  file eq30.gif

we obtain the equations

eq. 31:  file eq31.gif

We should thus have the solution of our problem, if the constants a and b were known.  These result from the following discussion.

For the origin of K1 we have permanently x’ = 0, and hence according to the first of the equations (5)

eq. 32:  file eq32.gif

If we call v the velocity with which the origin of K1 is moving relative to K, we then have

eq. 33:  file eq33.gif

The same value v can be obtained from equations (5), if we calculate the velocity of another point of K1 relative to K, or the velocity (directed towards the negative x-axis) of a point of K with respect to K’.  In short, we can designate v as the relative velocity of the two systems.

Furthermore, the principle of relativity teaches us that, as judged from K, the length of a unit measuring-rod which is at rest with reference to K1 must be exactly the same as the length, as judged from K’, of a unit measuring-rod which is at rest relative to K. In order to see how the points of the x-axis appear as viewed from K, we only require to take a " snapshot " of K1 from K; this means that we have to insert a particular value of t (time of K), e.g. t = 0.  For this value of t we then obtain from the first of the equations (5)

x’ = ax

Two points of the x’-axis which are separated by the distance Dx’ = I when measured in the K1 system are thus separated in our instantaneous photograph by the distance

eq. 34:  file eq34.gif

But if the snapshot be taken from K’(t’ = 0), and if we eliminate t from the equations (5), taking into account the expression (6), we obtain

eq. 35:  file eq35.gif

From this we conclude that two points on the x-axis separated by the distance I (relative to K) will be represented on our snapshot by the distance

eq. 36:  file eq36.gif

But from what has been said, the two snapshots must be identical; hence Dx in (7) must be equal to Dx’ in (7a), so that we obtain

eq. 37:  file eq37.gif

The equations (6) and (7b) determine the constants a and b.  By inserting the values of these constants in (5), we obtain the first and the fourth of the equations given in Section 11.