Derivation of the Lorentz Transformation

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

t ′ = ( 1 − v c ) t = t − v c 2 c t = t − v c 2 x {\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

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

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

t ′ = t − v c 2 x {\displaystyle t'=t-{\frac {v}{c^{2}}}x} ,
x ′ = x − v t {\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  x , t {\displaystyle x,t} we obtain however

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

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

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

The reverse transformation in an obvious way becomes immediately:

t = t ′ + v c 2 x ′ ( 1 − v 2 c 2 ) κ {\displaystyle t={\frac {t'+{\frac {v}{c^{2}}}x'}{\left(1-{\frac {v^{2}}{c^{2}}}\right)\kappa }}} ,
x = x ′ + v t ′ ( 1 − v 2 c 2 ) κ {\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:

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

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

that is

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

The obtained transformation is therefore the Lorentz transformation.