we notate the 4 – coordinate as x in S frame and y is S’ frame.

let the Lorentz transform be

$y_i = L_{ij} x_j$

notice that when writing like this, it is equivalent as a matrix equation on column vector:

$\begin {pmatrix} y_0 \\ y_1 \\ y_2 \\ y_3 \end {pmatrix} = L \cdot x$

For row vector, the Einstein notation read:

$y_i = x_j L_{ji}$

be careful the order of the subscript of L.

to derive the Lorentz transform, we only need to know 3 things,

1. the wave equation does not change ( i.e. the speed of light does not change )
2. the Lorentz transform has inverse
3. the relative motion of 2 frame is given by a speed $\beta$

we need Jacobian when relate the derivative between 2 frames.

$\frac { \partial }{\partial x_i} = \frac { \partial } { \partial y_j} \frac { d y_j }{ d x_i }$

Thus,

$\frac { \partial }{\partial x_i} = \frac { \partial } { \partial y_j} L_{ji}$

(this is a row vector equation )

The wave equation is :

$\left( \frac{\partial^2}{\partial x_0^2} - \nabla^2 \right) \Psi = 0$

this wave equation can be viewed by using this:

$a^2-b^2 = (a+b) (a - b) = ( a, b ) \cdot ( a, -b) = (a,b) \cdot \begin {pmatrix} 1 & 0 \\ 0 & -1 \end {pmatrix} \cdot ( a, b)$

There, we can write the wave equation as :

$\frac { \partial } {\partial x_i } g_{ij}\frac {\partial}{\partial x_j} \Psi = 0$

where the g is a matrix:

$g = \begin {pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end {pmatrix}$

Subsitute the Jacobian, we have

$\frac {\partial }{ \partial y_{\alpha} }L_{\alpha i}g_{ij}L_{j \beta}\frac {\partial}{\partial y_\beta} \Psi = 0$

the wave equation does not change with frame. there fore, this must be hold:

$L_{\alpha i} g_{ij} L_{j \beta} = g_{\alpha \beta }$

ntoics that the left Lorentz transform is transpose, and the right one is normal one. For simplicity, for a 1-D spatial dimension. we can let L be:

$\begin {pmatrix} a & b \\ c & d \end {pmatrix}$

by solving, we have 3 indenpent equations:

$a^2 - c^2 = 1$

$ab-cd = 0$

$b^2 - d^2 = -1$

Using the inverse of Lorentz transform, we have another 3 equations:

$a^2 - bc = 1$

$b(a - d) = 0$

$d^2 - bc = 1$

Solving this 6 equations for non-trivial solution.

$L = \begin {pmatrix} a & b \\ b & a \end {pmatrix}$

$a^2 - b^2 = 1$

Now, using the original of frame S ( $x_1 = 0$ ) is moving with speed $\beta$ in the S’-frame.

$y_1' / y_0' = b/a = \beta$

then we have the Lorentz Transform !