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 !

 

 

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