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

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

Let us now envision two frames in ‘standard configuration’ with *K´* 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 *O´* be the spatial origins of the frames; *O* and *O´* cöıncide at time *t* = *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 *x´*/*γ* since *x´* is O′M as measured in *K´*. Hence a first relation is:

*x* = *vt* + *x´*/*γ* (equation 6)

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

*x*/*γ* = *vt´+ x´* (equation 7)

Relation (6a) yields immediately

*x´* = *γ*(*x − vt*) (equation 8a)

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

* x* = *γ*(*x´ *+ *vt´*) (equation 9a)

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

*t´* = *γ*(*t* − *vx*/*c*²) (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* = *γ*²(*x* − *vt*) + *γvt´*

*t´* = *x*/*γv* − *γ*(*x*/*v* − *t*) = *γ* [*x*/*γ*²*v* + *t* − *x*/*v*] = *γ* [(*x*/*v*)(*c*² − *v*²)/*c*² + *t* − *x*/*v*]

*t´* = *γ* [*c*²*x* − *xv*² + *c*²*vt* − *c*²*x*]/*c*²*v *= *γ*(*c*²*vt* − *xv*²) / *c*²*v* = *γ*(*t* − *xv*/*c*²).

Details on the inverse of equation 10a:

*x´* = *γ*²(*x´* + *vt´*) − *γvt*

*t* = *γ*(*x´*/*v* + *t´*) − *x´*/*γv* = *γ* [*x´*/*v* + *t´* − *x´*/*γ*²*v*] = *γ* [*t´* + *x´*/*v *− (*x´*/*v*)(*c*² − *v*²)/*c*²]

*t* = *γ* [*c*²*vt´* + *x´v*² + *c*²*x´* − *c*²*x´*]/*c*²*v* = *γ*(*c*²*vt´* + *x´v*²) / *c*²*v* = *γ*(*t´* + *x´v*/*c*²).

This derives the Lorentz transformation in *spacetime*.

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

Let us now envision two frames in ‘standard configuration’ with *K´* 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 *O´* be the *temporal* origins of the frames; *O* and *O´* cöıncide at *stance* *x* = *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*/*c*² and O*´*M seen from *K* is *t´*/*γ* since *t´* is O′M as measured in *K´*. Hence a first relation is:

*t* = *xv*/*c*² + *t´*/*γ* (equation 6b)

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

*t*/*γ* = *xv*/*c² + t´* (equation 7b)

Relation (6b) yields immediately

*t´* = *γ*(*t − xv*/*c²*) (equation 8b)

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

*t* = *γ*(*t´ *+ *xv*/*c*²) (equation 9b)

of this ‘time part’. Eliminating *t´* between (8b) and (b9) quickly leads to the formula for the transformed *space*:

*x´* = *γ*(*x* − *vt*) (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*) + *xc*²/*v*) = *γ*²(*t* − *xv*/c²) + *γx´v*/*c*²

*x´* = [*t* − *γ*²(*t* − *xv*/*c*²)] / (*γv*/*c*²) = [*t*(*c*² − *v*²) − *c*²*t* + *xv*] / γv(*c*² − *v*²)

*x´* =(*xv* − *v*²*t*) / *γv*(1 – *v*²/*c*²) = *γ*(*x* − *vt*).

Details on the inverse of equation 10b:

*t* = *γ*(*γ*(*t + xc²/v*) − *xc*²/*v*) = *γ*²(*t* + *xv*/c²) − *γx´v*/*c*²

*x´* = [−*t* + *γ*²(*t* + *xv*/*c*²)] / (*γv*/*c*²) = [−*t*(*c*² − *v*²) + *c*²*t* + *xv*] / γv(*c*² − *v*²)

*x´* =(*xv* + *v*²*t*) / *γv*(1 – *v*²/*c*²) = *γ*(*x* + *vt*).

This derives the Lorentz transformation in *timespace*.