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So I am currently taking a course in Electrodynamics, and I was having a look at the Wikipedia article for the Electromagnetic Stress Energy Tensor (EMSET). I saw that the Maxwell stress tensor (MST) is simply the spacial part of the EMSET.

It appears that the EMSET tensor is of $(2,0)$ shape, $$T^{ij}=\frac{1}{\mu_0}\left(F^{ik}F^j{}_{k}-\frac{1}{4}\eta^{ij}F_{kl}F^{kl}\right)$$ Here $\mathbf{F}$ is the Electromagnetic stress tensor, which is most certainly contravariant, so I think it is safe to say that $\mathbf{T}$ is as well.

But, Wiki have defined the MST covariantly, $$\sigma_{ij}=\varepsilon_0E_iE_j+\frac{1}{\mu_0}B_iB_j-\frac{g_{ij}}{2}\left(\varepsilon_0|\boldsymbol E|^2+\frac{1}{\mu_0}|\boldsymbol B|^2\right).$$ And they seem to claim $$T^{ij}=-\sigma_{ij}\\ \text{when}~~i,j\neq 0.$$ But equating a covariant tensor to a contravariant one seems wrong to me. Is it instead supposed to be $$T^{ij}=-\sigma^{ij}\\ \text{when}~~i,j\neq 0$$ Where $$\sigma^{ij}=\varepsilon_0E^iE^j+\frac{1}{\mu_0}B^iB^j-\frac{g^{ij}}{2}\left(\varepsilon_0|\boldsymbol E|^2+\frac{1}{\mu_0}|\boldsymbol B|^2\right)~?$$

What's more interesting is that the Cauchy stress tensor in fluid mechanics is typically defined covariantly, see this post.

So yeah... basically I'm asking if the convention is to define the MST covariantly or contravariantly. The distinction really matters to me.

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  • $\begingroup$ In the (+---) metric $E^i = - E_i$ etc. so it is better to use the Minkowski metric to lower all Greek indices, then move to Cartesian components. Use $\alpha$ for 0,1,2,3 and $i,j$ for 1,2,3. $\endgroup$
    – DanielC
    Commented Oct 13, 2021 at 14:27
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    $\begingroup$ Well, since we have a metric tensor, we can always raise and lower indices, so it is really just a matter of convention. $\endgroup$
    – Qmechanic ♦
    Commented Oct 13, 2021 at 15:15
  • $\begingroup$ @K.defaoite By nature of geometry, the Faraday tensor is double-covariant. The stress-energy tensor is double covariant (think of its definition from GR en.wikipedia.org/wiki/…), but in the presence of a metric tensor, you can do whatever you want with the components. $\endgroup$
    – DanielC
    Commented Oct 13, 2021 at 15:22
  • $\begingroup$ @DanielC Right, but in general $\sigma_{ij}\neq \sigma^{ij}$. So is it $$T^{ij}=-\sigma_{ij}\\ \text{when}~~i,j\neq 0$$ Or is it $$T^{ij}=-\sigma^{ij}\\ \text{when}~~i,j\neq 0$$ $\endgroup$
    – K.defaoite
    Commented Oct 13, 2021 at 15:32
  • $\begingroup$ $\sigma_{ij}$ uses Cartesian components of the 3D electric and magnetic fields, therefore it only makes sense to consider $T_{ij}$ from $T_{\mu\nu}$. $\endgroup$
    – DanielC
    Commented Oct 13, 2021 at 15:34

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