# Question on tensor contraction

I have seen that the contraction of the electromagnetic field (emf) tensor between covariant and contravariant versions of the electromagnetic field tensor is $$-2E^2/c^2 +2B^2$$ but my confusion is that since the emf tensor is a matrix so when we contract we are multiplying two $$4\times 4$$ square matrices so how does it produce just $$-2E^2/c^2+2B^2$$,shouldn't it also be a matrix? Or am i missing something?

when we contract we are multiplying two 4×4 square matrices

No, matrix multiplication is not the same thing as matrix (or vector) contraction.

Matrix multiplication of two Faraday tensors would look like:

$$\mathbf{F}^2= \mathbf{F}\times \mathbf{F} \quad \leftrightarrow \quad \sum_\lambda F^{\mu\lambda} F_{\lambda\nu} = (F^2)^\mu_\nu,$$ whereas a full contraction is going a step further and setting, $$\mu = \nu$$ after the multiplication stage, so that:

$$F^{\mu\nu} F_{\nu\mu} = (F^2)^\mu_\nu |_{\mu = \nu} = \sum_\mu (F^2)^\mu_\mu,$$ i.e. the sum of the diagonal elements (the trace) of $$\mathbf{F}^2$$.

You are partly correct.

In tensor notation, we break your traditional notion of “matrix multiplication” into two separate parts.

The first part is called an outer product, the outer product of $$A$$ and $$B$$, both being “matrices” aka [1, 1]-tensors having one upper index and one lower index, is the outer product $$(A\otimes B)^{\alpha\phantom{\beta}\gamma}_{\phantom{\alpha}\beta\phantom{\gamma}\delta}= A^\alpha_{\phantom{\alpha}\beta} B^\gamma_{\phantom{\gamma}\delta}.$$ As you can see this tensor expression on the far right is kind of just a wiring diagram saying “that first top index (first overall) is wired into the top index of $$A$$, the second top index (third overall) is wired into the top index of $$B$$, etc.” The result has two places it can be wired on top and two places it can be wired on the bottom. So in 4D space this thing has 256 entries as a [2,2]-tensor.

After one contraction this can become a [1, 1] tensor again, another matrix, and this is matrix multiplication. It is what we write with a repeated index, $$(A\cdot B)^{\alpha}_{\phantom{\alpha}\delta}= (A\otimes B)^{\alpha\phantom{\beta}\beta}_{\phantom{\alpha}\beta\phantom{\gamma}\delta}= A^\alpha_{\phantom{\alpha}\beta} B^\beta_{\phantom{\beta}\delta}.$$in terms of wiring, we are shorting out a lower index to a higher index with a small wire. Note that there are actually four wirings we can choose from, $$A\cdot B$$ and $$B\cdot A$$ being the obvious ones. If I “short” a matrix in this way it generates a trace, so the other two are $$(\operatorname{tr} A)~B$$ and $$(\operatorname{tr} B)~A.$$

This generalization of trace is probably easiest to understand for a newcomer in terms of this idea of Einstein summation, which says that when I see an index repeated between an upper and a lower position, I quietly sum over the values of that index.

If we apply one more contraction we get $$\operatorname{tr}(A\cdot B)=\operatorname{tr}(B\cdot A)$$.

Now, technically, the “matrix” presentations of $$F^{\mu\nu}$$ or $$F_{\mu\nu}$$ are a sort of lie, those are I guess more properly tables of numbers rather than matrices. But, we use them because it doesn't actually mean anything for a [1,1]-tensor to have a symmetric or antisymmetric matrix in some coordinates, whereas [2,0] and [0,2] tensors can be symmetric or antisymmetric in a coordinate-independent geometrical meaning. Indeed some of our “wiring diagrams” that we can construct, $$\frac12 \delta^\alpha_\beta\delta^\mu_\nu \pm \frac12 \delta^\alpha_\nu\delta^\mu_\beta,$$can be used to express a [2, 0]-tensor, say, as the sum of a symmetric and an anti-symmetric tensor in a coordinate-independent fashion.

With that said the basic point still stands, you can use the metric tensor to convert a [2,0] $$F^{\mu\nu}$$ into a [1,1] $$F^\mu_{\phantom{\mu}\nu}$$ and then they are asking you for $$\operatorname{tr(F\cdot F)},$$ your interpretation is basically just missing the trace aspect coming from the continued contraction of the remaining indices.