2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 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. $\endgroup$ – Matt Dickau Mar 4 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 at 22:22
2
$\begingroup$
  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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Matt Dickau Mar 5 at 18:21
  • 1
    $\begingroup$ $\uparrow$ Yes. $\endgroup$ – Qmechanic Mar 5 at 18:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.