Derivation of Lorentz Transformation's Equations

Derivation:

Let us consider two inertial frames $S$ and $S'$ in which frame $S'$ is moving with constant velocity $v$ along the positive x-axis direction relative to the frame $S$. Let $t$ and $t'$ be the time recorded in two frames. Let the origin $O$ and $O'$ of the two reference systems coincide at $t=t'=0$.

Now suppose, a source of light is situated at the origin $O$ in the frame $S$, from which a wavefront of light is emitted at time $t=0$. When the light reaches point $P$, the time required by a light signal in travelling the distance OP in the Frame $S$ is

$ t=\frac{OP}{c}$

$ t=\frac{\left (x^{2}+y^{2}+z^{2} \right )}{c}$

$ x^{2}+y^{2}+z^{2}=c^{2}t^{2}\qquad (1)$

The equation $(1)$ represents the equation of wavefront in frame $S$. According to the special theory of relativity, the velocity of light will be $c$ in the second frame $S'$. Hence in frame $S'$ the time required by the light signal in travelling the distance $O'P$ is given by

$ t'=\frac{O'P}{c}$

$ x'^{2}+y'^{2}+z'^{2}=c^{2}t^{2}\qquad (2)$

According to the Galilean transformation equation:

$ \begin{Bmatrix} x'=x-vt\\ y'=y\\ z'=z\\ t'=t \end{Bmatrix}$

Now substitute these values in equation $(2)$ then we get

$ (x-vt)^{2}+y^{2}+z^{2}=c^{2}t^{2}$

$ x^{2}+v^{2}t^{2}-2xvt+y^{2}+z^{2}-c^{2}t^{2}=0$

The above equation is certainly not same as the equation $(1)$ because it contains an extra term $(-2xvt+v^{2}t^{2})$. Thus the Galilean transformation fails.

Further $t=t'$ because $\left( t=\frac{OP}{c} \: and \: t'=\frac{O'P}{c} \right)$ which does not agree with Galilean transformation equations.

The extra term $(-2xvt+v^{2}t^{2})$ indicates that transformations in $x$ and $t$ should be modified so that this extra term is cancelled. So modification in transformation

$ x'=\alpha (x-vt) \quad for \: x'=0,\: x=vt$

$ t'=\alpha (t+fx)$

Where $α$, $α'$ and $f$ are constant to be determined for Galilean Transformations $α= α'=1$ and $f=0$. Now substituting these modified values in equation $(2)$ so

$ \alpha ^{2}(x-vt)^{2}+y^{2}+z^{2}=c^{2}\alpha'^{2}(t+fx)^{2}$

$ \alpha ^{2}(x^{2}+v^{2}t^{2}-2xvt)+y^{2}+ z^{2} =c^{2}\alpha{'}^{2}(t^{2}+f^{2}x^{2}+2fxt)$

$\alpha^{2}x^{2}+\alpha^{2}v^{2}t^{2}-2xvt\alpha^{2}+y^{2}+ z^{2} =c^{2} \alpha'^{2}t^{2}+c^{2} \alpha'^{2}f^{2}x^{2}+c^{2} \alpha'^{2}(2fxt)$

$ x^{2}(\alpha^{2}-c^{2} \alpha'^{2}f^{2})- 2xt(v\alpha^{2}+c^{2}\alpha '^{2}f)+y^{2}+ z^{2} =c^{2}t^{2}\left ( \alpha'^{2}-\frac{\alpha ^{2}v^{2}}{c^{2}} \right )\qquad(3)$

This result obtained from applying transformations from frame $S'$ to $S$ must be identical to equation $(1)$. Therefore

$ \alpha ^{2}-c^{2}\alpha '^{2}f^{2}=1\qquad (i)$

$ v\alpha ^{2}+c^{2}\alpha '^{2}f=0\qquad (ii)$

$ \alpha '^{2}-\frac{\alpha ^{2}v^{2}}{c^{2}}=1\qquad (iii)$

After solving equation $(i)$ equation $(ii)$ and equation $(iii)$, we get

$ \alpha =\alpha '=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$ $ f=-\frac{v}{c^{2}}$

Thus, The new transformation equation is invariant at the velocity of light $c$. These are:

$ \begin{Bmatrix} x'=\frac{x-vt}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\\ y'=y\\ z'=z\\ t'=\frac{(t-\frac{vx}{c^{2}})}{\sqrt{1-\frac{v^{2}}{c^{2}}}} \end{Bmatrix}$

These equations are called Lorentz Transformations because they were first obtained by Dutch Physicist H. Lorentz.

The above transformation equation shows that frame $S'$ is moving in positive x-direction with velocity $v$ relative to the frame $S$. But if we say that frame $S$ is moving with $v$ velocity relative to frame $S'$ along negative x-direction then the transformation is:

$ \begin{Bmatrix} x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\\ y=y'\\ z=z'\\ t=\frac{(t'+\frac{vx'}{c^{2}})}{\sqrt{1-\frac{v^{2}}{c^{2}}}} \end{Bmatrix}$

These are known as inverse Lorentz Transformations equations.