Formula for the Maxwell Stress tensor in arbitrary coordinates

Question

This question is nearly identical to my last, except this time its the Maxwell stress tensor, not the Cauchy stress tensor. I often see its components written as $$\sigma_{ij}=\varepsilon_0E_iE_j+\frac{1}{\mu_0}B_iB_j+\frac{\delta_{ij}}{2}\left(\varepsilon_0|\boldsymbol E|^2+\frac{1}{\mu_0}|\boldsymbol{B}|^2\right)$$ With $E_i$ being understood as the components of the vector $\boldsymbol E$. But I thought, "hang on", $\boldsymbol E$ is a vector, and thus is contravariant, and its components should be written $E^i$, and similarly for $\boldsymbol B$. But, I know that the stress tensor should be fully covariant, since it measures force (a vector) per unit area (which can be represented as a normal vector). I.e, it takes in two vector inputs and outputs a scalar, so it should be second order covariant. So I thought, we should replace $E_i$ with $g_{ki}E^k$. Similarly, the Kronecker delta bugs me as well - it is defined as a $(1,1)$ tensor, with components $$\delta^i_j=1 \text{ if }i=j, ~\delta^i_j=0\text{ if }i\neq j$$ So $$\delta_{ij}=g_{ki}\delta^k_j=g_{ij}$$ So, the "correct" formula, written out in all its glory, should really be $$\sigma_{ij}=\varepsilon_0(g_{ki}E^k)(g_{lj}E^j)+\frac{1}{\mu_0}(g_{ki}B^k)(g_{lj}B^j)+\frac{g_{ij}}{2}\left(\varepsilon_0|\boldsymbol E|^2+\frac{1}{\mu_0}|\boldsymbol{B}|^2\right)$$ Or of course, in shorter form $$\sigma_{ij}=\varepsilon_0E_iE_j+\frac{1}{\mu_0}B_iB_j+\frac{g_{ij}}{2}\left(\varepsilon_0|\boldsymbol E|^2+\frac{1}{\mu_0}|\boldsymbol{B}|^2\right)$$ Where $E_i$ are recognized not as the components of $\boldsymbol E$, but rather as the components of $\boldsymbol E^{\flat}$, its dual. And of course $|\boldsymbol E|^2=g_{ab}E^aE^b$. Am I right?

Answer 1

It is actually better if we define the Lorentz invariant Maxwell stress tensor. It is a second order contravariant tensor: $$T^{\alpha\beta}=\frac{1}{\mu_0}\left(F^{\alpha\gamma}F^\beta{}_{\gamma}-\frac{1}{4}\eta^{\alpha\beta}F^{\gamma\delta}F_{\gamma\delta}\right)$$

Here $F^{\mu\nu}$ is the $\mu,\nu$ component of the field strength tensor and $\eta^{\mu\nu}$ is the $\mu,\nu$ component of the inverse spacetime metric. This formula works no matter which coordinate system we use (but, the components of $\boldsymbol{\eta}$, $\mathbf{F}$ might be very complicated). We use the well known Lorentz scalar $$F^{\mu\nu}F_{\mu\nu}=2|\boldsymbol B|^2-2|\boldsymbol E^2|/c^2$$ Let $i,j,k,l,m\in\{1,2,3\}$. Let's compute the spacial components of the MST:

$$T^{ij}=\frac{1}{\mu_0}\left(F^{i\gamma}F^j{}_{\gamma}-\frac{\eta^{ij}}{2}(|\boldsymbol B|^2-|\boldsymbol E^2|/c^2)\right)$$

Working for example in Cartesian spacial coordinates (i.e $\boldsymbol\eta=\operatorname{diag}(-1,1,1,1)$), we have the following identities: $$F^{i~0}=\frac{1}{c}E^i~~;~~F^{ij}=\varepsilon_{ijk}B^k \\ F^j{}_\gamma=\eta_{\gamma\lambda}F^{j\lambda} \\ \implies F^j{}_0=-F^{j~0}=\frac{-1}{c}E^j \\ \text{and}~~F^j{}_k=\eta_{k\lambda}F^{j\lambda}=\eta_{kk}F^{jk}=\varepsilon_{jkl}B^l$$ Hence $$T^{ij}=\frac{1}{\mu_0}\left(F^{i~0}F^j{}_0+F^{ik}F^j{}_k-\frac{\delta^i_j}{2}(|\boldsymbol B|^2-|\boldsymbol E^2|/c^2)\right) \\ =\frac{1}{\mu_0}\left(-\frac{1}{c^2}E^iE^j+\varepsilon_{ikl}B^l\varepsilon_{jkm}B^m-\frac{\delta^i_j}{2}(|\boldsymbol B|^2-|\boldsymbol E^2|/c^2)\right)$$ Now, $$\varepsilon_{ikl}\varepsilon_{jkm}=\varepsilon_{ilk}\varepsilon_{jmk}=\delta^i_j\delta^l_m-\delta^i_m\delta^l_j$$ So $$\varepsilon_{ikl}B^l\varepsilon_{jkm}B^m=(\delta^i_j\delta^l_m-\delta^i_m\delta^l_j)B^lB^m=\delta^i_jB^lB^l-B^iB^j$$ Hence, in Cartesian coords, $$\boxed{T^{ij}=\frac{1}{\mu_0}\left(-\frac{1}{c^2}E^iE^j-B^iB^j+\frac{\delta^i_j}{2}\left(|\boldsymbol B|^2+\frac{1}{c^2}|\boldsymbol E|^2\right)\right)}$$