Lorentz for space and time

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/c21/v2)

so that

ρ = (1 – c2/v2) -1/2.

which completes the temporal Lorentz transformation.