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In the image there is a tensor product:

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

It's about how this operation on the co- and contravariant field strength tensors can give one of the invariants of the electromagnetic field.

I've tried it and it's actually the double inner product, F_lower(row,column)*F^upper(column,row) summed over all rows and columns

$$F_{\mu \nu}F^{\nu \mu}$$

Is this the way I would write it with subscript summation?

How did it come?

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    $\begingroup$ The difference between $F_{\mu \nu}F^{\mu \nu}$ and $F_{\mu \nu}F^{\nu \mu}$ is a sign because $F_{\mu \nu}$ is antisymmetric. $\endgroup$
    – Qmechanic ♦
    Commented Dec 1, 2012 at 10:27

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You should be contracting the following two objects $$ F_{\mu \nu}= \begin{pmatrix} 0 & E_x & E_y & E_z \\ -E_x & 0 & -B_z & B_y \\ -E_y & B_z & 0 & -B_x \\ -E_z & -B_y & B_x & 0 \end{pmatrix} \quad \text{and}\quad F^{\mu \nu} = \begin{pmatrix} 0 & -E_x & -E_y & -E_z \\ E_x & 0 & -B_z & B_y \\ E_y & B_z & 0 & -B_x \\ E_z & -B_y & B_x & 0 \end{pmatrix} $$ Some of the tensor components change signs when you move the indices around. Now you should be doing what was mentioned by Fabian $$ \sum_{\mu=0}^{3}\sum_{\nu=0}^{3}F_{\mu \nu}F^{\mu \nu}=F_{00}F^{00}+F_{01}F^{01}+...+F_{33}F^{33} $$ As you can see, the electric field multiplication will come out with an overall minus, and the magnetic field will come out positive.

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You can write it as $$\sum_{\mu=0}^3 \sum_{\nu=0}^3 F_{\mu\nu} F^{\mu\nu}.$$ It is usually not written because of lazyness and shortness of notation (this notation is called Einstein summation convention.

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    $\begingroup$ I just realized that two sources I was looking at have different definitions of the double inner product. For example, foamcfd.org/Nabla/guides/ProgrammersGuidese3.html and people.rit.edu/pnveme/EMEM851n/constitutive/tensors_rect.html Which one is it supposed to be? $\endgroup$
    – Robert
    Commented Dec 1, 2012 at 8:09
  • $\begingroup$ I don't see a difference, if I compare (1.15) with `Double Dot Product'. In any case, it should have the form in my answer. $\endgroup$
    – Fabian
    Commented Dec 1, 2012 at 8:25
  • $\begingroup$ I believe you, but on this picture people.rit.edu/pnveme/EMEM851n/constitutive/tensor_rect11.gif the formula flips the row and column on the second tensor. Do you know where that derivation comes from? $\endgroup$
    – Robert
    Commented Dec 1, 2012 at 8:39
  • $\begingroup$ If the matrices are [ 1 -1; 1 1] and [1 -1; 1 1] then your equation gives A00B00+A01B01+A10B10+A11B11 = 4. The equation A00B00 + A01B10 + A10B01 + A11B11 = 1 -1 -1 +1 =0. The second equation is what's given in people.rit.edu/pnveme/EMEM851n/constitutive/tensor_rect11.gif $\endgroup$
    – Robert
    Commented Dec 1, 2012 at 8:45
  • $\begingroup$ @Robert: there are different ways to contract tensors (i.e., make inner products). The second choice is written as $F_{\mu\nu} F^{\nu\mu}$. $\endgroup$
    – Fabian
    Commented Dec 1, 2012 at 8:50

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