Derivation of the Lorentz Transformation

Let us consider a particle moving with the speed of light $c$ in the coordinate system $K$ and the coordinate system $K^{\prime }$ moving with the velocity $v$ with respect to it. The trajectory equation in this system is $x=ct$ and in the coordinate system $K$ it is moving with the velocity $c-v$ with respect to the coordinate system $K^{\prime }$. Assuming that is travels the same distance with respect to $K^{\prime }$ in both systems and in both it travels with the speed of light it must be $x^{\prime }=(c-v)t=c{t}^{\prime }$. So there is a time dilation of its time at a given position in the system $K$:

$t^{\prime }=(1-\frac{v}{c})t=t-\frac{v}{c^2}ct=t-\frac{v}{c^2}x$.

Analogically we can write

$x^{\prime }=(c-v)t=x-vt$.

The transformation sought in this way turns out to be a transformation of Galileo with time dilation:

$t^{\prime }=t-\frac{v}{c^2}x$,

$x^{\prime }=x-vt$.

We will now assume that it is valid for any event coordinates, not only for coordinates of the photon trajectory.

Reverting the transformation for $x,t$ we obtain however

$t=\frac{t^{\prime }+\frac{v}{c^2}{x}^{\prime }}{1-\frac{v^2}{c^2}}$,

$x=\frac{x^{\prime }+v{t}^{\prime }}{1-\frac{v^2}{c^2}}$.

However the physical situation seen from the coordinate system $K^{\prime }$ is identical as seen from $K$ but only the system $K$ is moving with respect to the system $K^{\prime }$ with the velocity $-v$. So we will try to improve the transformation with the scaling factor $\kappa$ which naturally preserves the speed of light:

$t^{\prime }=(t-\frac{v}{c^2}x)\kappa$,

$x^{\prime }=(x-vt)\kappa$.

The reverse transformation in an obvious way becomes immediately:

$t=\frac{t^{\prime }+\frac{v}{c^2}{x}^{\prime }}{(1-\frac{v^2}{c^2})\kappa }$,

$x=\frac{x^{\prime }+v{t}^{\prime }}{(1-\frac{v^2}{c^2})\kappa }$.

In order for the two transformations to be identical except for the physical change of the relative velocity sign it therefore must be:

$\frac{1}{(1-\frac{v^2}{c^2})\kappa }=\kappa$

or

${\kappa }^2=\frac{1}{1-\frac{v^2}{c^2}}$,

that is

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

The obtained transformation is therefore the Lorentz transformation.