electromagnetism - How much of Maxwell's equations is recoverable from the zero divergence of the stress-energy tensor?

As a motivating example, consider the static electromagnetic field defined by $\textbf{E}=(\text{const})x\hat{\textbf{y}}$, $\textbf{B}=0$. The stress-energy tensor for this field is $T=\operatorname{diag}(u,-u,u,-u)$, where $u$ is the energy density. The divergence of this stress-energy tensor is nonzero, since $\partial T^{xx}/\partial x\ne 0$. This field also violates Maxwell's equations, since the curl is nonzero but there are no time-varying magnetic fields present that could induce a curly electric field.

If we start from Maxwell's equations, we can prove that the divergence of $T$ is zero, which is a statement of conservation of energy-momentum. To what extent can we go the opposite way? I.e., can we start from

$\qquad(\operatorname{div} T=0$) and (other appealing principles)

and derive Maxwell's equations? (This is all assuming that the stress-energy tensor has the form we already know for the electromagnetic field, so it's symmetric, has zero trace, and so on.) If not, then what is a good counterexample that provides further insight? I would be happy with a discussion that was restricted to the vacuum field equations.

Answer

1. 
   

   Notation. The Lagrangian density without sources in E&M is

   $${\cal L}_0~=~ -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} \tag{1}$$ with $$F_{\mu\nu}~:=~A_{\nu,\mu}-A_{\mu,\nu}, \qquad \frac{\partial{\cal L}_0}{\partial A_{\mu,\nu}}~=~ F^{\mu\nu}.\tag{2}$$ Eqs. (1) & (2) are just to explain notation for later. We are not actually going to use eq. (1) to derive Maxwell's equation, cf. OP's title question.
   
   



2. 
   

   Stress-energy-momentum (SEM) tensor. In E&M, the canonical SEM tensor is$^1$

   $$\Theta^{\mu}{}_{\nu}~=~\delta^{\mu}_{\nu}{\cal L}_0+F^{\mu\alpha}A_{\alpha,\nu} \qquad\Rightarrow\qquad 0~\approx~d_{\mu} \Theta^{\mu}{}_{\nu}~=~d_{\mu}F^{\mu\alpha}~A_{\alpha, \nu},\tag{3}$$ while the symmetric SEM tensor is $$T^{\mu}{}_{\nu}~=~\delta^{\mu}_{\nu}{\cal L}_0+F^{\mu\alpha}F_{\nu\alpha}\qquad\Rightarrow\qquad 0~\approx~d_{\mu} T^{\mu}{}_{\nu}~=~d_{\mu}F^{\mu\alpha}~F_{\nu\alpha}.\tag{4}$$
   
   


3. 
   

   So the rhs. of eqs. (3) & (4) must be zero. If $A_{\alpha, \nu}$ or $F_{\nu\alpha}$ generically are invertible $4\times 4$ matrices, we can conclude Maxwell's equations (Gauss's law + Maxwell-Ampere's law)

   $$d_{\mu}F^{\mu\nu}~\approx~0.\tag{5}$$
   
   


4. 
   

   The other Maxwell equations (Faraday's law & no magnetic monopoles) are automatically satisfied since we assume that the 4-gauge potential $A_{\mu}$ exists, cf. e.g. this Phys.SE post.

   
   

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$^1$ Some references, e.g. Weinberg QFT, have the opposite notational conventions for $T$ and $\Theta$. Here we are using $(-,+,+,+)$ Minkowski sign convention, and work in units where $c=\epsilon_0=\mu_0=1$. The $\approx$ symbol means an on-shell equality, i.e. equality modulo EOM.