# Derivation of the Lorentz Transformation

Let us consider a particle moving with the speed of light ${\displaystyle c}$ in the coordinate system ${\displaystyle K}$ and the coordinate system ${\displaystyle K'}$ moving with the velocity ${\displaystyle v}$ with respect to it. The trajectory equation in this system is ${\displaystyle x=ct}$ and in the coordinate system ${\displaystyle K}$ it is moving with the velocity ${\displaystyle c-v}$ with respect to the coordinate system ${\displaystyle K'}$. Assuming that is travels the same distance with respect to ${\displaystyle K'}$ in both systems and in both it travels with the speed of light it must be ${\displaystyle x'=(c-v)t=ct'}$. So there is a time dilation of its time at a given position in the system ${\displaystyle K}$:

${\displaystyle t'=\left(1-{\frac {v}{c}}\right)t=t-{\frac {v}{c^{2}}}ct=t-{\frac {v}{c^{2}}}x}$.

Analogically we can write

${\displaystyle x'=(c-v)t=x-vt}$.

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

${\displaystyle t'=t-{\frac {v}{c^{2}}}x}$,
${\displaystyle x'=x-vt}$.

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

Reverting the transformation for ${\displaystyle x,t}$ we obtain however

${\displaystyle t={\frac {t'+{\frac {v}{c^{2}}}x'}{1-{\frac {v^{2}}{c^{2}}}}}}$,
${\displaystyle x={\frac {x'+vt'}{1-{\frac {v^{2}}{c^{2}}}}}}$.

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

${\displaystyle t'=\left(t-{\frac {v}{c^{2}}}x\right)\kappa }$,
${\displaystyle x'=(x-vt)\kappa }$.

The reverse transformation in an obvious way becomes immediately:

${\displaystyle t={\frac {t'+{\frac {v}{c^{2}}}x'}{\left(1-{\frac {v^{2}}{c^{2}}}\right)\kappa }}}$,
${\displaystyle x={\frac {x'+vt'}{\left(1-{\frac {v^{2}}{c^{2}}}\right)\kappa }}}$.

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

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

or

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

that is

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

The obtained transformation is therefore the Lorentz transformation.