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?
Formula for the Maxwell Stress tensor in arbitrary coordinates
electromagnetismindex-notationtensors
Best Answer
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, $$\boxed{T^{ij}=\frac{1}{\mu_0}\left(-\frac{1}{c^2}E^iE^j-B^iB^j+\frac{\delta^i_j}{2}\left(|\boldsymbol B|^2+\frac{1}{c^2}|\boldsymbol E|^2\right)\right)}$$