Question: Show that equation (2-36) follows from the arbitrary four-vector Lorentz transformation equations (2-35).

The Lorentz transformation is described as a set of some equations which mainly relate to the time and space coordinates. This transformation is mainly done with the help of an inertial frame.

The equation of the invariant quantity in the plane is expressed as:

$$\begin{gathered} {A\text{'}}_x={\gamma }_v\left(A_x-\frac{v}{c}{A}_t\right) \\ A{\text{'}}_x^2={\gamma }_{\text{v}}^{\text{2}}\left({\text{A}}_{\text{x}}^{\text{2}}{\text{- 2A}}_{\text{x}}\frac{\text{v}}{\text{c}}{\text{A}}_{\text{t}}\text{+}\frac{{\text{v}}^{\text{2}}}{{\text{c}}^{\text{2}}}{\text{A}}_{\text{t}}^{\text{2}}\right) \end{gathered}$$ (1)

The equation of the invariant quantity in the plane is expressed as:

$$\begin{gathered} {A\text{'}}_t={\gamma }_v\left(A_t-\frac{v}{c}{A}_x\right) \\ A{\text{'}}_t^2={\gamma }_v^2\left({\text{A}}_{\text{t}}^{\text{2}}{\text{- 2A}}_{\text{t}}\frac{\text{v}}{\text{c}}{\text{A}}_{\text{x}}\text{+}\frac{{\text{v}}^{\text{2}}}{{\text{c}}^{\text{2}}}{\text{A}}_{\text{x}}^{\text{2}}\right) \end{gathered}$$ (2)

Subtracting equation (1) from equation (2).

$A{\text{'}}_x^2-A{\text{'}}_t^2={\gamma }_v^2\left(\left(1-\frac{v^2}{c^2}\right)A_x^2-\left(1-\frac{v^2}{c^2}\right)A_t^2\right)$

As , localid="1658808288943" ${\gamma }_v^2=\frac{1}{1-\frac{v^2}{c^2}}$then $A{\text{'}}_x^2-A{\text{'}}_t^2=A_x^2-A_t^2$. Moreover,$A{\text{'}}_y^2=A_y^2$ and$A{\text{'}}_z^2=A_z^2$

Finally, the above equation will become:

$A{\text{'}}_t^2-\left(A{\text{'}}_x^2+A{\text{'}}_y^2+A{\text{'}}_z^2\right)=A_t^2-\left(A_x^2+A_y^2+A_z^2\right)$

The above equation is the equation (2-36) which is in the time-space context.

Thus, equation (2-36) follows from the arbitrary four-vector Lorentz transformation equations (2-35).