I am trying to work out the symmetric stress-energy tensor for a free massive vector field and show that it is conserved. The Lagrangian density and resulting EoM are:

$$L=-\frac{1}{4}F^{\alpha\beta}F_{\alpha\beta}+\frac{1}{2}m^2 A^{\beta}A_{\beta}$$ $$\partial_\mu F^{\mu\nu}+m^2 A^\nu=0$$

Where I am using the metric $\eta_{\mu\nu}=diag(+1,-1,-1,-1)$. Evaluating the usual expression for the canonical stress energy tensor:

$$T_C^{\mu \nu}=\frac{\partial L}{\partial \left(\partial_{\mu} A_{\lambda}\right)}\eta^{\nu\beta}\partial_{\beta} A_{\lambda}-\eta^{\mu \nu}L$$ $$\frac{\partial L}{\partial \left(\partial_{\mu} A_{\lambda}\right)}=F^{\lambda\mu}$$ $$T_C^{\mu \nu}=F^{\lambda\mu}\eta^{\nu\beta}\partial_{\beta} A_{\lambda}+\frac{1}{4}F^{\alpha\beta}F_{\alpha\beta}\eta^{\mu \nu}-\frac{1}{2}m^2 A^{\beta}A_{\beta}\eta^{\mu \nu}$$

According to what I can find on this site and elsewhere, the symmetric stress energy should be (adding the mass term to the form found here):

$$T^{\mu\nu}=\frac{1}{4}F^{\alpha\beta}F_{\alpha\beta}\eta^{\mu \nu}+F^{\mu\lambda}\eta_{\lambda\kappa}F^{\kappa\nu}-\frac{1}{2}m^2 A^{\beta}A_{\beta}\eta^{\mu \nu}$$

When I calculate the difference between these I get:

$$T^{\mu\nu}-T_C^{\mu \nu}=S^{\mu\nu}=-F^{\beta\nu}\partial_\beta A^\mu$$

It should be that $\partial_\mu S^{\mu\nu}=0$ so that $T^{\mu\nu}$ is conserved just as $T_C^{\mu\nu}$ is. But I calculate:

$$\partial_\mu S^{\mu\nu}=-\left( \partial_\mu F^{\beta\nu}\right) \partial_\beta A^\mu-F^{\beta\nu}\partial_\beta \left( \partial_\mu A^\mu\right)$$

The second term is zero since $\partial_\mu A^\mu=0$ by the EoM. But I can't figure out how to show that the first term is zero, nor can I see that $\partial_\mu T^{\mu\nu}=0$ when I try to calculate it directly. Where am I going wrong?

  • $\begingroup$ Minimally couple the Lagrangian and consider taking the derivative with respect to the metric, and then finally substituting the Minkowski metric. $\endgroup$
    – JamalS
    Mar 4, 2020 at 17:41
  • $\begingroup$ @JamalS Could you provide an answer showing how to do that? I am not familiar with how to take the derivative with respect to the metric. $\endgroup$ Mar 4, 2020 at 17:57
  • $\begingroup$ The conservation law is $\partial_\nu T^{\mu\nu} = 0$, not $\partial_\mu T^{\mu\nu} = 0$. $\endgroup$
    – Prahar
    Mar 5, 2020 at 22:22

1 Answer 1

  1. The Lagrangian density for massive E&M is $${\cal L}~=~\sqrt{|g|}L, \qquad L~=~-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}\mp \frac{1}{2}m^2 A^2, \tag{1}$$ for Minkowski sign convention $(\mp,\pm,\pm,\pm)$. The Euler-Lagrange (EL) equations read $$ d_{\mu}F^{\mu\nu}~\stackrel{(1)}{\approx}~\pm m^2A^{\nu} .\tag{2}$$

  2. The mass term breaks gauge-invariance but not translational invariance, so that the canonical stress-energy-momentum (SEM) tensor $$\mp T_C^{\mu}{}_{\nu}~:=~\frac{\partial L}{\partial(\partial_{\mu}A_{\lambda})}\partial_{\nu}A_{\lambda}-\delta^{\mu}_{\nu}L ~\stackrel{(1)}{=}~F^{\lambda\mu}\partial_{\nu}A_{\lambda}-\delta^{\mu}_{\nu}L\tag{3}$$ is conserved cf. Noether's first theorem

  3. The Hilbert/metric SEM tensor is $$ \mp T_{\mu\nu}~:=~\frac{2}{\sqrt{|g|}}\frac{\delta S}{\delta g^{\mu\nu}} ~\stackrel{(1)}{=}~- F_{\mu}{}^{\lambda}F_{\nu\lambda}\mp m^2A_{\mu}A_{\nu} - g_{\mu\nu}L.\tag{4}$$ In eq. (4) it is important that we think of $A$ as a co-vector/one-form rather than a vector.

  4. The difference becomes $$S^{\mu}{}_{\nu}~:=~T^{\mu}{}_{\nu}-T_C^{\mu}{}_{\nu} ~\stackrel{(3(+(4)}{=}~\mp F^{\mu\lambda}d_{\lambda}A_{\nu}+ m^2A^{\mu}A_{\nu} ~\stackrel{(2)}{\approx}~\mp d_{\lambda}(F^{\mu\lambda}A_{\nu}),\tag{5}$$ so that $$d_{\mu}S^{\mu}{}_{\nu}~\stackrel{(5)}{\approx}~\mp d_{\mu}d_{\lambda}(F^{\mu\lambda}A_{\nu})~=~0,\tag{6}$$ by (anti)symmetry of indices $\mu\leftrightarrow\lambda$. Hence the conservation law for the two SEM tensors (3) & (4) are equivalent, cf. OP's question.

  • $\begingroup$ Thanks! To take the derivative with respect to the metric in finding the Hilbert SEM tensor, do you just have to lower all the indicies on the A field so that there is an instance of the metric for each contraction in the Lagrangian? $\endgroup$ Mar 5, 2020 at 18:21
  • 1
    $\begingroup$ $\uparrow$ Yes. $\endgroup$
    – Qmechanic
    Mar 5, 2020 at 18:51

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