= φ(v)φ(-v)x, etc.
Since the relations between (x′, y′, z′, t′), and (x, y, z, t) do not contain time explicitly, therefore K and k′ are relatively at rest.
It appears that the systems K and k′ are identical.
∴ φ(v)φ(-v) = 1.
Let us now turn our attention to the part of the ξ-axis between (ξ = 0, η = 0, ζ = 0), and (ξ = 0, η = 1, ζ = 0). Let this piece of the y-axis be covered with a rod moving with the velocity v relative to the system K and perpendicular to its axis;—the ends of the rod having therefore the co-ordinates
x1 = vt, y1 = l / φ(v), z1 = 0
x2 = vt, y2 = 0, z2 = 0
Therefore the length of the rod measured in the system K is l/φ(v). For the system moving with velocity (-v), we have on grounds of symmetry,
l l
------ = ------