# Help with tensor formulation of special relativity

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1. Nov 6, 2016

### lonewolf219

1. The problem statement, all variables and given/known data

Hi, I can't seem to understand the following formula in my professor's lecture notes:

F_αβ = g_αγ*g_βδ*F^(γδ)

2. Relevant equations
Where g_αβ is the diagonal matrix in 4 dimensions with g_00 = 1 and g_11 = g_22 = g_33 = -1 and F^(γδ) is the electromagnetic tensor with c=1.

3. The attempt at a solution
I keep wanting to perform matrix multiplication, but g_αγ*g_bδ would just be the unit matrix if we did this, right? I don't understand how to perform the RHS of this equation...

2. Nov 6, 2016

### Fightfish

$g_{\alpha\gamma} g_{\beta \delta}$ is not matrix multiplication - it is an object with four free indices that are uncorrelated. In tensorial language, matrix multiplication corresponds to a contraction of two rank-two tensors. For example, $A_{i j} B^{j k}$ can be viewed as a matrix multiplication - try to compare this with the usual component-wise multiplication definition of matrix multiplication before reading on!

As for how to evaluate the RHS of the equation, well we have to perform a summation over the dummy indices. As an example, for $\alpha = \beta = 0$, we have
$$F_{0 0} = \sum_{\gamma = 0}^{3}\sum_{\delta = 0}^{3} g_{0\gamma} g_{0\delta} F^{\gamma \delta}$$
where I have put the summations in explicitly just to show things clearly.

Of course there is an easier way to do it since these are rank-two tensors - recall earlier that I mentioned that matrix multiplication involves contracting one index in a product of two such tensors. So, $C_{i k} = A_{i j} B^{j k}$ can be viewed as taking the i-th row of A and multiply it element-wise with the k-th column of B, which is what we do when we multiply matrices in the usual sense. If we rewrite the given equation as
$$F_{\alpha\beta} = g_{\alpha \gamma} F^{\gamma \delta} (g^{T})_{\delta \beta}$$
(note that I've taken the transpose of the second $g$ to reverse the indices so that we can interpret it as matrix multiplication)
then we can write
$$\mathbf{F}' = \mathbf{g}\,\mathbf{F}\,\mathbf{g}^{T}$$