# Tensor product notation [closed]

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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## closed as too localized by Qmechanic♦Jan 30 '13 at 18:21

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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. – Qmechanic Dec 1 '12 at 10:27

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.
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.
@Robert: there are different ways to contract tensors (i.e., make inner products). The second choice is written as $F_{\mu\nu} F^{\nu\mu}$. – Fabian Dec 1 '12 at 8:50