I want to compute the stress-energy tensor for the following Lagrangian: $$\mathcal{L} = -\frac{1}{4} F_{\mu \nu} F^{\mu \nu} - \frac{1}{2\xi} (\nabla_\mu A^\mu)^2$$

but I'm struggling with the gauge-fixing term. When taking variations with respect to the metric, it seems that it will involve many terms, some of then not clearly covariant. I am only interested in its value in Minkowski, but that is also hard to find.

By stress-energy tensor I mean the Hilbert one, defined as:

$$T_{\mu \nu} = \frac{-2}{\sqrt{-g}} \frac{\delta S}{\delta g^{\mu \nu}}$$

My attempt

I have attempted to include an auxiliary field, such that

$$\mathcal{L}' = -\frac{1}{4} F_{\mu \nu} F^{\mu \nu} + B \nabla_\mu A^\mu + \frac{\xi}{2} B^2$$

which becomes the original one after finding and imposing the equations of motion for $B$. But is it true that the two stress energy tensors are going to be the same? What I find is that

$$T_{\mu \nu} = \eta_{\mu \nu} \mathcal{L} + F_{\alpha \mu} F_{\beta \nu} g^{\alpha \beta} + \frac{4}{\xi} \partial_\alpha A^\alpha \partial_{(\mu} A_{\nu)} + \frac{2}{\xi} \partial_{(\mu} \left(\partial_\alpha A^\alpha \right) A_{\nu)} - \frac{1}{\xi} \eta_{\mu \nu} \partial_\rho(A^\rho \partial_\alpha A^\alpha)$$

in Minkowski, and after getting rid of the auxiliary field. Can this be correct? What if I integrate by parts the first term involving the auxiliary field, so that I get:

$$\mathcal{L}' = -\frac{1}{4} F_{\mu \nu} F^{\mu \nu} - \nabla_\mu B A^\mu + \frac{\xi}{2} B^2 = -\frac{1}{4} F_{\mu \nu} F^{\mu \nu} - \partial_\mu B A^\mu + \frac{\xi}{2} B^2 $$

given that $B$ is a scalar field. This seems to be giving me a different stress-energy tensor as well.

  • 1
    $\begingroup$ Did you try replacing $\nabla_\mu A^\mu$ by $\frac{1}{\sqrt{|g|}}\partial_\mu(\sqrt{|g|} A^\mu)$? I am almost certain it should work, but I don't currently have time to work out the details $\endgroup$ Mar 30, 2019 at 16:19
  • $\begingroup$ No, I haven't tried that. What I tried is, given $\mathcal{L}$, use the definition of the covariant derivative and use the variation of the Christoffel symbols. That involves many terms that are not clearly covariant, I believe (i.e. $\nabla_\mu (\partial_\mu A^\nu A^\mu)$ [I am not saying that you get this term exactly, just that it involves this kind of terms]) $\endgroup$
    – exforest
    Mar 30, 2019 at 16:26
  • 1
    $\begingroup$ Yup, that really sounds messy, but my approach is clearly coordinate independent so it shouldn't be as messy $\endgroup$ Mar 30, 2019 at 16:33
  • 2
    $\begingroup$ Notice that the energy-momentum tensor is not unique but can be modified by terms that vanish by the equations of motion. See e.g. max2.physics.sunysb.edu/~rastelli/HW2Solutions.pdf. The concept of these so-called improvement terms is widely known, but I am not sure I can point to a text book that discusses them. So it is perfectly conceivable that you get two seemingly different results as long as the difference is such an "improvement term". $\endgroup$
    – user178876
    Mar 30, 2019 at 19:01

1 Answer 1


Well, it has been a week and I think I have convinced myself of the answer. Basically the solution was in the comments, but took me some time to understand. Indeed, as @marmot has said, the stress-energy tensor is an intrinsically off-shell object. Therefore, obtaining two seemingly different answer should not be a problem. Only when you project them on-shell (impose the equations of motion), you can expect to obtain a single, unique answer. .

The "two different" ways I propose of computing the tensor (i.e. with the Lagrangian $\mathcal{L'}$ containing $B \nabla_\mu A^\mu$ or, integrating by parts and using $\partial_\mu B A^\mu$), and the way that @Filipe Miguel explains (i.e. writing down the exact expression for the covariant derivative of $A^\mu$), end up giving the same answer after using the EOM.

Thank you very much everyone!


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