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I want to express the tensor invariant ##^{\dagger}F^{\mu \nu} F_{\mu \nu}## in function of both ##\vec E## and ##\vec B##.

The Field tensor is:

$$F^{\mu\nu} = \begin{bmatrix} 0 & E_x & E_y & E_z \\ -E_x & 0 & -B_z & B_y \\ -E_y & B_z & 0 & -B_x \\ -E_z & -B_y & B_x & 0 \end{bmatrix}$$

Note that the Field tensor is antisymmetric:

$$F^{\mu \nu} = - F^{\nu \mu}$$

If we upper an index (or lower it; in this case I am going from covariant to contravariant) we have to change the sign if one of the indices is zero.

I am using the convention ##(- + + +)##.

##^{\dagger}F^{\mu \nu} F_{\mu \nu} = F^{0 0} F^{0 0} - F^{0 1} F^{0 1} - F^{0 2} F^{0 2} - F^{0 3} F^{0 3} - F^{1 0} F^{1 0} - F^{2 0} F^{2 0} - F^{3 0} F^{3 0} + F^{1 1} F^{1 1} + F^{2 2} F^{2 2} + F^{3 3} F^{3 3} + F^{1 2} F^{1 2} + F^{1 3} F^{1 3} + F^{2 1} F^{2 1} + F^{2 3} F^{2 3} + F^{3 1} F^{3 1} + F^{3 2} F^{3 2}##

Let's explain one case: ##-F^{0 1} F^{0 1}##

##F^{0 1} = \frac{E_x}{c}## by itself. But now I have to multiply it by the second ##F^{0 1}##. I have brought the indices up, so that means changing sign. Thus the second ##F^{0 1} = -\frac{E_x}{c}##, getting:

$$F^{0 1} F^{0 1} = -(\frac{E_x}{c})^2$$

The rest follows:

$$^{\dagger}F^{\mu \nu} F_{\mu \nu} = -\frac{2\vec E^2}{c^2} + 2 \vec B^2 = 2(\vec B^2 - \frac{\vec E^2}{c^2})$$

Please let me know if more details are required.