# Lorentz for spacetime and timespace

This post builds on the previous posts here and here, and follows the approach of J.-M. Levy here.

First, consider the Lorentz transformation for spacetime along the x axis per Levy’s section III:

Let us now envision two frames in ‘standard configuration’ with having velocity v with respect to K and let x, t (resp. x´, t´) be the coordinates of event M in the two frames. Let O and be the spatial origins of the frames; O and cöıncide at time t = = 0.

Here comes the pretty argument: all we have to do is to express the relation

OM = OO´ + O´M (equation 5)

between vectors (which here reduce to oriented segments) in both frames.

In K, OM = x, OO´ = vt and O´M seen from K is /γ since is O′M as measured in . Hence a first relation is:

x = vt + /γ (equation 6)

In , OM = x/γ since x is OM as measured in K, OO′ = vt´ and O′M = . Hence a second relation:

x/γ = vt´+ x´ (equation 7)

Relation (6a) yields immediately

= γ(x − vt) (equation 8a)

which is the x-axis ‘space part’ of the LT and relation (equation 7a) yields the inverse

x = γ(+ vt´) (equation 9a)

of this ‘space part’. Eliminating between (8a) and (9a) quickly leads to the formula for the transformed time:

= γ(tvx/c²) (equation 10a)

the inverse of which could easily be found by a similar elimination of x. Coordinates on the y and z axes are unchanged for the already stated reason that distances do not vary in the directions perpendicular to the velocity.

Further details on equation 10a:

x = γ²(xvt) + γvt´

= x/γvγ(x/vt) = γ [x/γ²v + tx/v] = γ [(x/v)(c² − v²)/c² + tx/v]

= γ [c²xxv² + c²vtc²x]/c²v = γ(c²vtxv²) / c²v = γ(txv/c²).

Details on the inverse of equation 10a:

= γ²( + vt´) − γvt

t = γ(/v + ) − /γv = γ [/v + /γ²v] = γ [ + /v − (/v)(c² − v²)/c²]

t = γ [c²vt´ + x´v² + c²c²]/c²v = γ(c²vt´ + x´v²) / c²v = γ( + x´v/c²).

This derives the Lorentz transformation for spacetime.

Next, similarly consider the Lorentz transformation for timespace along the t axis:

Let us now envision two frames in ‘standard configuration’ with having lenticity w ≡ 1/v with respect to K and let x, t (resp. x´, t´) be the coordinates of event M in the two frames. Let O and be the temporal origins of the frames; O and cöıncide at stance x = = 0.

Here comes the pretty argument: all we have to do is to express the relation

OM = OO´ + O´M (equation 5b)

between vectors (which here reduce to oriented segments) in both frames.

In K, OM = t, OO´ = xv/c² and O´M seen from K is /γ since is O′M as measured in . Hence a first relation is:

t = xv/c² + /γ (equation 6b)

In , OM = t/γ since t is OM as measured in K, OO′ = xv/ and O′M = . Hence a second relation:

t/γ = xv/c² + t´ (equation 7b)

Relation (6b) yields immediately

= γ(t − xv/) (equation 8b)

which is the t-axis ‘time part’ of the LT and relation (equation 7b) yields the inverse

t = γ(+ xv/c²) (equation 9b)

of this ‘time part’. Eliminating between (8b) and (b9) quickly leads to the formula for the transformed space:

= γ(xvt) (equation 10b)

the inverse of which could easily be found by a similar elimination of t. Coordinates on the other time axes are unchanged for the already stated reason that distances do not vary in the directions perpendicular to the lenticity.

Further details on equation 10b:

t = γ(γ(t − xc²/v) + xc²/v) = γ²(txv/c²) + γx´v/c²

= [tγ²(txv/c²)] / (γv/c²) = [t(c² − v²) − c²t + xv] / γv(c² − v²)

=(xvv²t) / γv(1 – v²/c²) = γ(xvt).

Details on the inverse of equation 10b:

t = γ(γ(t + xc²/v) − xc²/v) = γ²(t + xv/c²) − γx´v/c²

= [−t + γ²(t + xv/c²)] / (γv/c²) = [−t(c² − v²) + c²t + xv] / γv(c² − v²)

=(xv + v²t) / γv(1 – v²/c²) = γ(x + vt).

This derives the Lorentz transformation for timespace.