How to show invariance using the Maxwell tensor?

Question

I want to show the invariance of $E^2-c^2B^2$ under the Lorentz transformations. The obvious way to do this is to show that $$E^2-c^2B^2=E'^2-c^2B'^2,$$ where $E'$ and $B'$ are the Lorentz transformations of the electric and magnetic fields, respectively. This is quite a simple yet very inelegant calculation (one applies the Lorentz transformation to each of $E_x, E_y, E_z, B_x, B_y, B_z$ and the invariance follows).

How can I do this using the Maxwell tensor? I believe that it is possible to show invariance by doing the matrix multiplication of $$F_{\mu\nu}F^{\mu\nu}$$ but I am unsure of how to actually do this (I am confused about what what $F_{\mu\nu}$ actually is), and I am confused how this could stretch to showing the invariance of other things, because surely this matrix multiplication would always give the same answer...

Answer 1

How does a scalar quantity transforms under a Lorentz transformation?

I will show you here how this works for a vector (contravariant) and a covector.

The transformation rule for these objects are

$$u^a = { \partial x^a \over \partial x' ^b} u' ^b $$

$$u_a = { \partial x'^b \over \partial x^a}u' _b $$

Multiplying the two:

$$ u^{a} u_{a} = { \partial x^a \over \partial x'^b} {\partial x'^b \over \partial x^a} u' ^{b} u' _{b} .$$

So, by eliminating the terms that are the same we have

$$u^a u_a =u' _b u' ^b =u' ^a u' _a,$$ where we rename the indexes b to a since b is a mute index.

Since this quantity is a scalar, and thus the same in both reference frames. By taking the general transformation rule for the tensors you have you can show the same.

Hope this helps

Answer 2

I believe $F^{\mu\nu}$ is better known as the Faraday tensor; Maxwell tensor is only the spatial part of it, 3x3 matrix.

The argument goes like this: $F$ is a four-tensor, so if any quantity is constructed from it by multiplication of its components, this new quantity also has to transform as a tensor. The construct $$ F^{\mu\nu}(g_{\mu\rho}g_{\nu\sigma}F^{\rho\sigma}) $$ results in a single number. It has no index left, because there is summation over all of them.

The only way a single number could transform as a tensor is if it is the same number in all coordinate systems. Hence $F^{\mu\nu}F_{\mu\nu}$ is invariant. When components of $F$ are expressed through components of $\mathbf E,\mathbf B$, it turns out this number is $2c^2(B_x^2+B_x^2+B_x^2)-2(E_x^2+E_y^2+E_z^2)$ (in SI units).