two-fold application of the transformation-equations, we obtain
t´ = φ(-v)β(-v){τ + (v/c^2)ξ} = φ(v)φ(-v)t,
x´ = φ(v)β(v)(ξ + vτ) = 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.
[therefore] phi(v)phi(-v) = 1,
Let us now turn our attention to the part of the y-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
x_{1} = vt, y = l/φ(v), z_{1} = 0 }
x_{2} = vt, y_{2} = 0, z_{2} = 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/φ(v) = l/φ(-v)
[therefore] φ(v) = φ(-v), [therefore] φ(v) = 1.
