Consider again the now-classic scenario in which observer K is at rest and observer K′ is moving in the positive direction of the x axis with constant velocity, v. This time there is a characteristic constant speed, c. The basic problem is that if they both observe a point event E, how should one convert the coordinates of E from one reference frame to the other?
We return first to the Galilean transformation and include a factor, γ, in the transformation equation for the positive direction of the x axis:
rx′ = γ (rx − vtx)
where rx is the spatial coordinate and tx is the temporal coordinate. Only the coordinates of the x axis are affected; the other coordinates do not change.
The inverse spatial transformation is then:
rx = γ (rx′ + vtx′).
The trajectory of a reference particle or probe vehicle that travels at the characteristic speed in the positive direction of the x axis will follow the equations:
rx = ctx and rx′ = ctx′.
With the spatial transformations we conclude that
rx′ = ctx′ = γ (rx − vtx) = γ tx (c − v),
rx = ctx = γ (rx′ + vtx′) = γ tx′ (c + v).
Multiplying these together and cancelling tx tx′ leads to:
c2 = γ2 (c − v) (c + v) = γ2 (c2 − v2)
so that
γ = (1 – v2/c2) -1/2.
which completes the spatial Lorentz transformation.
We return next to the dual Galilean transformation and include a factor, ρ, in the transformation equation for the positive direction of the x axis:
tx′ = ρ (tx − rx/v)
where rx is the spatial coordinate and tx is the temporal coordinate. Only the coordinates of the x axis are affected; the other coordinates do not change.
The inverse temporal transformation is then:
tx = ρ (tx′ + rx′/v).
The trajectory of a reference particle or probe vehicle that travels at the characteristic speed from the origin will follow the equations:
rx/c = tx and rx′/c = tx′.
With the temporal transformations we conclude that
tx′ = rx′/c = ρ (tx − rx/v) = ρ rx (1/c − 1/v),
tx = rx/c = ρ (tx′ + rx′/v) = ρ rx′ (1/c + 1/v).
Multiplying these together and cancelling rx rx′ leads to:
1/c2 = ρ2 (1/c − 1/v) (1/c + 1/v) = ρ2 (1/c2 − 1/v2)
so that
ρ = (1 – c2/v2) -1/2.
which completes the temporal Lorentz transformation.