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I'm working through relativistic electrodynamics, but I'm stuck on determining the $00$-th component of the energy-momentun tensor. The (contravariant) 2-tensor (in Heaviside-Lorentz units) reads:

$$T^{\mu\nu}=F^{\mu\alpha}F_{\alpha}^{\:\nu}+\frac{1}{4}\eta^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}$$

It's the first term that's causing me pain (the second is just a standard field invariant), so if we just look at that:

$$T^{00}=F^{0\alpha}F_{\alpha}^{\:0}$$

I want to show this to be the square of the electric field, but I'm confused about the indices. I would have thought that

$$F_{\alpha0}=\eta_{0 \sigma}F_{\alpha}^{\:\sigma}=\eta_{00}F_{\alpha}^{\:0}=F_{\alpha}^{\:0}$$

with no sign change as I'm lowering a temporal index. But when I swap the indices for $F^{0\alpha}$, I pick up a minus sign due to the antisymmetry of $F$. Since I want the whole thing to just be the EM energy density, I need to get rid of this minus sign. I presume that I must compensate for it in the index lowering process that I'm messing up somehow.

I presume I'm just making some novice mistake, so if someone could point it out and set me straight that would be much appreciated!

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The term is$$F^{0\alpha}F_\alpha{}^0=F^{0i}F_i{}^0=F^{0i}\eta_{i\mu}F^{\mu0}=F^{0i}\eta_{ij}F^{j0}\stackrel{\ast}{=}\pm F^{0i}\delta_{ij}F^{j0}\stackrel{\ast}{=}\pm F^{0i}F^{i0}=\mp E^2$$if you work in $\mp\pm\pm\pm$, where each $\ast$ breaks the usual contraction rules for indices to emphasize the value.

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  • $\begingroup$ Oh brilliant, that cleared it up! I prefer -+++ signature as I'm just learning this and differential geometry (with GR in view) for the first time. $\endgroup$ Dec 8, 2021 at 13:56

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