Lorentz-transformation

Question

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.

Answer 1

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

Answer 2

To answer your question:

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 $$\eta'=\Lambda^T\eta\Lambda.$$