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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.

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  • $\begingroup$ How did you find out that stress-energy tensor? Did you assume the Poynting formulae for its components? Those are derived from Maxwell's equations, so if you assume EM field that is not Maxwell's, the Poynting formulae are not justified. $\endgroup$ – Ján Lalinský Nov 23 '18 at 16:15
  • $\begingroup$ @JánLalinský: One way to figure out the form of the stress-energy tensor is if you already know Maxwell's equations and derive the Poynting vector from it in the style you suggest. However, it's also possible to justify the form of the stress-energy tensor on more general grounds, without even knowing Maxwell's equations. For example, the momentum density has to be proportional to $\textbf{E}\times\textbf{B}$, because there is no other rotationally invariant vector-valued expression that is bilinear in the fields. $\endgroup$ – Ben Crowell Nov 23 '18 at 18:20
  • $\begingroup$ If you do not assume Maxwell's equations or the standard interpretation of the Poynting expressions, why would you assume that momentum density has to be bilinear in E,B? Is the question restricted to linear system of equations for E,B and those are the only admissible independent variables? Because if not, the Born-Infeld model is one possible counterexample - divergence of T is zero, but Maxwell's equations for E,B are not obeyed (one has to introduce other independent variables H,D to recover the EBHD variant of them). $\endgroup$ – Ján Lalinský Nov 23 '18 at 23:37
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  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.

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  • $\begingroup$ Thanks for adding more detail, that made it much easier to follow. This is just a side issue, but in the expression $F^{\mu\alpha}F_{\nu\alpha}$, what system of units is it that doesn't require a coupling constant or a $1/4\pi$? WP has a constant in both SI and cgs: en.wikipedia.org/wiki/… . (Of course this has no effect on your argument.) $\endgroup$ – Ben Crowell Nov 23 '18 at 14:27
  • $\begingroup$ I don't get it. This argument seems to assume what it proves. The second one assumes Lagrangian density is $L \propto F^{\mu\nu}F_{\mu\nu}$, one has already decided Maxwell's free field equations for $F$ are obeyed. $\endgroup$ – Ján Lalinský Nov 23 '18 at 16:04
  • $\begingroup$ I updated the answer. $\endgroup$ – Qmechanic Nov 23 '18 at 19:29
  • $\begingroup$ This is still circular. There is no reason I can see other than Maxwell's equations for those expressions for $\Theta$ and $F$. For example, in the Born-Infeld model, the stress energy tensor looks somewhat different, has zero divergence, and Maxwell's equations for E,B are not obeyed. Cf. formula 3.6A in their paper, rspa.royalsocietypublishing.org/content/royprsa/144/852/… $\endgroup$ – Ján Lalinský Nov 23 '18 at 23:40

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