# Calculating electromagnetic invariant in matrix form

I'm kind of confused. I want to calculate the electromagnetic invariant $I := F^{\mu\nu}F_{\mu\nu}$, but I'm not sure what is the easiest way to do so. So, I was trying to do it in matrix form, i.e. defining $$\mathbf{F}:=\begin{pmatrix}0 & -E_{1} & -E_{2} & -E_{3}\\ E_{1} & 0 & -B_{3} & B_{2}\\ E_{2} & B_{3} & 0 & -B_{1}\\ E_{3} & -B_{2} & B_{1} & 0 \end{pmatrix}$$ and then calculate the quantity $I$, but I'm not sure how to obtain the matrix form for $I$ (By "matrix form" I mean expressed in terms of the matrix $\mathbf{F}$).

So, I have two questions, 1) what is the easiest way to calculate $I$?, and 2) how to obtain the matrix form for $I$ starting with $I := F^{\mu\nu}F_{\mu\nu}$?. I'm using the metric $\eta:=diag(1,-1,-1,-1)$.

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– Qmechanic Jan 31 '13 at 1:17
I isn't a matrix, its a scalar. – DJBunk Jan 31 '13 at 1:21
@DJBunk I know that, I'm looking for the scalar expression but in terms of the matrix F. – Ana Jan 31 '13 at 1:38
Actually I understand that $I$ can be putted as the trace of a matrix product, but I don't know how to obtain that product starting with the definition if $I$ that I've just gave. – Ana Jan 31 '13 at 1:40
@Qmechanic Yes, my 1st question is related with your link, but it's not my 2nd question. – Ana Jan 31 '13 at 1:46

Notice that if we define matrices $F = (F^{\mu\nu})$ and $\eta = (\eta_{\mu\nu})$, then notice that $$I = F_{\mu\nu}F^{\mu\nu} = \eta_{\mu\alpha}\eta_{\nu\beta}F^{\alpha\beta}F^{\mu\nu} = \eta_{\mu\alpha}F^{\alpha\beta}\eta_{\nu\beta}F^{\mu\nu} = -\eta_{\mu\alpha}F^{\alpha\beta}\eta_{\beta\nu}F^{\nu\mu} =-\mathrm {tr}(\eta F\eta F)$$ where $\mathrm{tr}$ denotes the trace and I have used antisymmetry and symmetry of $F$ and $\eta$ respectively. I'm guessing this is the type of matrix expression you're after?

Cheers!

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Yes, you are right. By the way, there are another equivalent expressions $I=F:(\eta F\eta)^{T}=Tr(F\eta F^{T}\eta)=-Tr(F\eta F\eta)$ where $A:B=\sum_{\mu,\nu,\sigma,\rho}e_{\mu}e_{\nu}:e_{\sigma}e_{\rho}A_{\mu\nu}B_{\si‌​gma \rho}=\sum_{\mu,\nu}A_{\mu\nu}B_{\nu\mu}$ is the dyadic product. – Ana Jan 31 '13 at 6:11
Hmm interesting hadn't seen that notation before. – joshphysics Jan 31 '13 at 18:40