# Light invariance principle

## Sommaire

### Galilean reference frames

In classical kinematics, the total displacement x in reference frame R is the sum of the relative displacement x’ in R’ and of the displacement vt of R’ relative to R at a velocity v : x = x’+vt or, equivalently, x’=x-vt. This relation is linear when the velocity v is constant, that is when the frames R and R' are galilean. Time t is the same in R and R’, which is no more valid in special relativity, where t ≠ t’. The more general relationship, with four constants α, β, γ and v is :

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

The Lorentz transformation becomes the Galilean one for β = γ = 1 et α = 0.

### Light invariance principle

The velocity of light is independent of the velocity of the source, as was shown by Michelson. We thus need to have x = ct if x’ = ct’. Replacing x and x' in these two equations, we have

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

Replacing t' from the second equation, the first one becames

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

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

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

#### Relativity principle

This derivation does not use the speed of light and allows therefore to separate it from the principle of relativity. 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 accord with the principle of relativity, the expressions of x and t should write :

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

They should be identical to the original expressions except for the sign of the velocity :

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

We should then have the following identities, verified independently of x’ and t’ :

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

This gives the following equalities :

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

### Expression of the Lorentz transformation

Using the above relationship

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

we get :

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

and, finally:

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

We have now all the four coefficients needed for the Lorentz transformation which writes in two dimensions :

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

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

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

These four equations are used according to the needs.