Lorentz-transformation

I don't understand how to derive the matrix representing the Lorentz-transformation given two systems S and S':

$$x' = \Lambda x$$

these transformations do not leave the differences $$\Delta x^\mu$$ unchanged, but multiply them also by the matrix $$\Lambda$$:

$$(\Delta s)^2 = (\Delta x)^T \eta (\Delta x) = (\Delta x')^T \eta (\Delta x') = (\Delta x)^T \Lambda^T \eta \Lambda (\Delta x) \tag{1.26}$$

and therefore

$$\eta = \Lambda^T \eta \Lambda$$

I don't understand the mathematical passages in eq. 1.26 particularly:

1. why is it needed to multiply by $$\Lambda$$?
2. The role of the transpose symbol

I know that the space-time interval is given by

$$(\Delta s)^2 = \eta_{\mu \nu}\Delta x^\mu\Delta x^\nu$$

and I understand that the metric given by $$\eta$$ should be the same in every reference system.

• I am not sure I understand as to which step you don't understand. – Dvij Mankad May 18 at 10:19

The role of the transpose symbol

The vectors are usually represented by the column matrix. $$\Delta x= \begin{bmatrix}c\Delta t \\ \Delta x \\\Delta y\\ \Delta z \end{bmatrix} .$$ Thus $$\Delta x^T= \begin{bmatrix}c\Delta t & \Delta x &\Delta y& \Delta z \end{bmatrix} .$$ The invariant interval in matrix representation is given by $$\Delta s^2=\Delta x^T\eta\Delta x=\begin{bmatrix}c\Delta t & \Delta x &\Delta y& \Delta z \end{bmatrix}\begin{bmatrix} 1& 0 & 0 & 0\\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0& 0 & 0 & -1 \end{bmatrix}\begin{bmatrix}c\Delta t \\ \Delta x \\\Delta y\\ \Delta z \end{bmatrix}$$ $$=(c\Delta t)^2 - \Delta x^2 -\Delta y^2 - \Delta z^2$$

Why is it needed to multiply by Λ?

Because the Lorentz transformation equation is given by $$\Delta x' = \Lambda \Delta x$$ substituting this in the interval $$(\Delta s)^2 = (\Delta x')^T \eta (\Delta x')$$ we get $$(\Delta s)^2= (\Lambda\Delta x)^T \eta (\Lambda\Delta x)=\Delta x^T\Lambda^T\eta \Lambda\Delta x$$
This is same as $$(\Delta x)^T \eta (\Delta x)$$ . Analyse it with the above equation, you will obtain $$\eta = \Lambda^T \eta \Lambda$$

1. If you look at $$\mathrm{(1.26)}$$ carefully, you will see that we have $$(\Delta s)^2 = (\Delta x )^T \eta(\Delta x) = (\Delta x')^T \eta(\Delta x')$$ This is equating the line element in the two different inertial frames. Notice the expression on the right hand side is a function of $$\Delta x'$$ not $$\Delta x$$. Using $$\Delta x' = \Lambda \Delta x$$ and $$(\Delta x')^T = (\Delta x)^T \Lambda^T$$ you can see where the both $$\Lambda$$ and $$\Lambda^T$$ have come from.
2. The transpose symbol is because the line element must be a scalar, so the "matrix" that represents both $$\Delta x$$ terms must have inverse dimensions to the "matrix" that represents the metric. I personally don't like this formalism, and if you go on to do general relativity you will be introduced to tensors and the Einstein summation convention. In this formalism it is a lot clearer because the line element is given as $$ds^2=g_{\mu \nu} dx^\mu dx^\nu$$ where $$g_{\mu \nu}$$ is the metric. In this notation you can clearly see that the line element is a scalar and that metric is a (0,2) tensor so $$dx^\mu dx^\nu$$, if treated like a single object like it is in your question, must be a rank (2, 0) tensor.
I must admit I personally do not like the form given in $$\mathrm{(1.26)}$$ since I feel it is confusing as to what $$\Delta x$$ is and why, as you asked, we want to take the transpose. So I will write the same derivation in the notation used in general relativity, hopefully you may find this clearer.
We have the line element in the two frames as $$ds^2 = \eta_{\mu \nu}dx^\mu dx^\nu = \eta'_{\mu' \nu'} dx^{\mu'} dx^{\nu'}$$ where we are working in Minkowski space (special relativity) so $$\eta = \eta'$$. The $$\mu'$$ and $$\nu'$$ are to denote that these are new indices in the new frame. These are given by the Lorentz transformation $$dx^{\mu'} = \Lambda^{\mu'}{}_\mu dx^\mu.$$ Notice that here the transmormation "matrix" is actually a (1,1) tensor. Einstein summation convention tells us we sum over the repeated $$\mu$$ so what the only free index on the right-hand side is $$\mu'$$, just like on the left-hand side. Substituting this into the above gives $$\eta_{\mu \nu}dx^\mu dx^\nu = \eta'_{\mu' \nu'} \Lambda^{\mu'}{}_\mu \Lambda^{\nu '}{}_\nu dx^\mu dx^\nu$$ So, $$\eta_{\mu \nu} = \Lambda^{\mu'}{}_\mu \Lambda^{\nu'}{}_{\nu} \eta'_{\mu' \nu'}$$ Which is the same as $$\eta'=\Lambda^T\eta\Lambda.$$