# Lorentz transformation

## Concise derivation of the Lorentz transformation

The problem is usually restricted to two dimensions by using a velocity along the x axis such that the y and z coordinates do not intervene. It is similar to that of Einstein Albert Einstein, Relativity: The Special and General Theory. More details may be found in Bernard Schaeffer, Relativités et quanta clarifiés As in the Galilean transformation, the Lorentz transformation is linear : the relative velocity of the of the reference frames is constant. They are called inertial or Galilean reference frames. According to relativity no Galilean (or none, according to Mach) reference frame is priviledged. Another condition is that the speed of light must be independent of the reference frame, in practice of the velocity of the light source.

### Galilean reference frames

In classical kinematics, the total displacement x in the R frame is the sum of the relative displacement x' in frame R' and of the displacement x in frame R. If v is the relative velocity of R' relative to R, we have v : x = x’+vt or x’=x-vt. This relationship is linear for a constant v, that is when R and R' are Galilean frames of reference.

In Einstein's relativity, the main difference is that space is a function of time and vice-versa: t ≠ t’. The most general linear relationship is obtained with four constant coefficients, α, β, γ and v:

$x'=\gamma\left(x-vt\right)$
$t'=\beta\left(t+\alpha x\right)$

The Lorentz transformation becomes the Galilan transformation when β = γ = 1 et α = 0.

### Speed of light independent of the velocity of the source

Light being independent of the reference frame as was shown by Michelson, we need to have x = ct if x’ = ct’. Replacing x and x' in the preceding equations, one has:

$ct'=\gamma\left(c-v\right)t$
$t'=\beta\left(1+\alpha c\right)t$

Replacing t’ with the help of the second equation, the first one writes:

$c\beta\left(1+\alpha c\right)t=\gamma\left(c-v\right)t$

After simplification by t and dividing by cβ, one obtains:

$1+\alpha c=\frac{\gamma}{\beta}(1-\frac{v}{c})$

### Principle of relativity

According to the principle of relativity, there is no priveledged galilean frame of reference. One has to find the same Lorentz transformation from frame R to R' or from R' to R. As in the Galilean transformation, the sign of the transport velocity v has to be changed when passing from one frame to the other.

The following derivation uses only the principle of relativity which is independent of light velocity. The inverse transformation of

$x'=\gamma\left(x-vt\right)$
$t'=\beta\left(t+\alpha x\right)$

is :

$x=\frac{1}{1-\alpha v}\left(\frac{x'}{\gamma}-\frac{vt'}{\beta}\right)$
$t=\frac{1}{1-\alpha v}\left(\frac{t'}{\beta}-\frac{\alpha x'}{\gamma}\right)$

In accordance with the principle of relativity, the expressions of x and t are:

$x=\gamma\left(x'+vt'\right)$
$t=\left(t'+\alpha x'\right)$

They have to be identical to those obtained by inverting the transformation except for the sign of the velocity of transport v:

$x=\frac{1}{1+\alpha v}\left(\frac{x'}{\gamma}+\frac{vt'}{\beta}\right)$
$t=\frac{1}{1+\alpha v}\left(\frac{t'}{\beta}-\frac{\alpha x'}{\gamma}\right)$

We thus have the identities, verified for any x’ and t’ :

$x=\gamma\left(x'+vt'\right)=\frac{1}{1+\alpha v}\left(\frac{x'}{\gamma}+\frac{vt'}{\beta}\right)$
$t=\left(t'+\alpha x'\right)=\frac{1}{1+\alpha v}\left(\frac{t'}{\beta}-\frac{\alpha x'}{\gamma}\right)$

Finally we have the equalities :

$\beta =\gamma=\frac{1}{\sqrt{1+\alpha v}}$

### Expression of the Lorentz transformation

Using the relation

$1+\alpha c=\frac{\gamma}{\beta}(1-\frac{v}{c})$

obtained earlier, one has :

$\alpha =-\frac{v}{c^2}$

and, finally:

$\beta =\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$

We have now all the coefficients needed and, therefore, the Lorentz transformation :

$\mathbf{ x=\frac{x' + vt'}{ \sqrt[]{1 -\frac{v^2}{c^2}} }}$
$t=\mathbf{\frac{t' + \frac{vx'}{c^2}}{ \sqrt[]{1 -\frac{v^2}{c^2}}}}$

The inverse Lorentz transformation writes, using the Lorentz factor γ:

$\mathbf{x'= \gamma\left(x - vt\right)}$
$\mathbf{t'=\gamma\left(t - \frac{vx}{c^2}\right)}$