# Stress-Energy Tensor from Lagrangian: Technical Question II

1. Dec 21, 2013

### maykot

This thread is supposed to be a continuation of the discussion of this thread: (1) https://www.physicsforums.com/showthread.php?t=88570.
The previous thread was closed but there was a lot of things I did not understand.
This is also somewhat related to a recent thread I created: (2) https://www.physicsforums.com/showthread.php?t=729135.

So, starting with the discussion in thread (1):
Physics Monkey wrote,
Then, from what I understood, if the electromagnetic potential is "most naturally" a 1-form and the most natural way to define a derivative is also through a 1-form, it is the $F_{\mu\nu}$ part of the lagrangian that is constant with respect to the metric, therefore:
$$t^{\alpha\beta} \equiv \frac{\partial \left( F_{\mu\nu}F^{\mu\nu} \right)}{\partial g_{\alpha\beta}} = \frac{\partial}{g_{\alpha\beta}} \left( g^{\mu\lambda} g^{\nu\kappa} F_{\mu\nu} F_{\lambda\kappa} \right) = F_{\mu\nu} F_{\lambda\kappa} \left[ \frac{\partial g^{\mu\lambda}}{\partial g_{\alpha\beta}} g^{\nu\kappa} + g^{\mu\lambda} \frac{\partial g^{\nu\kappa}}{\partial g_{\alpha\beta}} \right] =$$
$$=-F_{\mu\nu} F_{\lambda\kappa} \left[ g^{\mu\alpha} g^{\lambda\beta} g^{\nu\kappa} + g^{\mu\lambda} g^{\nu\alpha} g^{\kappa\beta} \right] = -2 g_{\mu\nu} F^{\alpha\mu}F^{\beta\nu}$$
And:
$$t_{\alpha\beta} \equiv \frac{\partial \left( F_{\mu\nu}F^{\mu\nu} \right)}{\partial g^{\alpha\beta}} = \frac{\partial}{g^{\alpha\beta}} \left( g^{\mu\lambda} g^{\nu\kappa} F_{\mu\nu} F_{\lambda\kappa} \right) = F_{\mu\nu} F_{\lambda\kappa} \left[ \frac{\partial g^{\mu\lambda}}{\partial g^{\alpha\beta}} g^{\nu\kappa} + g^{\mu\lambda} \frac{\partial g^{\nu\kappa}}{\partial g^{\alpha\beta}} \right] =$$
$$=F_{\mu\nu} F_{\lambda\kappa} \left[ \delta^{\mu}_{\alpha} \delta^{\lambda}_{\beta} g^{\nu\kappa} + g^{\mu\lambda} \delta^{\nu}_{\alpha} \delta^{\kappa}_{\beta} \right] = 2 g^{\mu\nu} F_{\alpha\mu}F_{\beta\nu}$$
But we also have:
$$t_{\alpha\beta} = g_{\alpha\mu} g_{\beta\nu} t^{\mu\nu} = -2 g^{\mu\nu} F_{\alpha\mu}F_{\beta\nu} \neq 2 g^{\mu\nu} F_{\alpha\mu}F_{\beta\nu}$$
Which is clearly a contradiction. I don't know if I missed or misunderstood something in his explanation, but as I see it, that is the consequence of what he explained.

Also, from thread (1):
lonelyphysicist asks about terms like $V^{\alpha} V_{\alpha}$ in the lagrangian:
Physics Monkey promptly replies him, but what if instead of terms like $V^{\alpha} V_{\alpha}$ we had something like $A^{\alpha} B_{\alpha}$, where A and B are arbitrary vector fields.
This question is related to my other thread (2) and to Barbarabados post in thread (1) (last post).​

The point of all of this being:
I'm not sure what are the technical procedures for deriving the stress-energy tensor for an arbitrary lagrangian using the usual definition given in GR books. It seems to me that the usual method for differentiating with respect to the the metric lacks consistency sometimes. Although, I think it's much more probable that I'm just not getting something very simple.

2. Dec 21, 2013

### Bill_K

The energy-momentum tensor TÎ¼Î½ is defined as the functional derivative of the matter action IM with respect to gÎ¼Î½. That is,

Î´IM = Â½ âˆ«d4x âˆšg TÎ¼Î½ Î´gÎ¼Î½

This always gives a conserved symmetric tensor.

As part of the definition, one must state which field variables are independent of gÎ¼Î½ and therefore held fixed in this variation. It's not an ambiguity, but simply part of the definition - you must specify the interaction of the fields with gravity, and you'll get a different theory and a different TÎ¼Î½ depending on your answer. In the case of electromagnetism, AÎ¼ is the independent variable, not AÎ¼.

3. Dec 21, 2013

### maykot

I found out what was wrong with the above derivation that $t_{\alpha\beta} \neq g_{\alpha\mu} g_{\beta\nu} t^{\mu\nu}$, I implicitly, and erroneously, assumed that $g^{\alpha\mu} g^{\beta\nu} \frac{\partial}{\partial g^{\alpha\beta}} = \frac{\partial}{\partial g_{\mu\nu}}$.

But I'm still uncomfortable with the fact that the expression $\frac{\partial g^{\alpha\beta}}{\partial g^{\mu\nu}} = \delta ^\alpha _\mu \delta ^\beta _\nu$ is not symmetric under $\alpha \to \beta$ and separately under $\mu \to \nu$.
To see where this could go wrong lets take for example a lagrangian of the form $\mathcal{L} = A^\alpha B_\alpha$. Lets see what happens in two separate cases: (i) $A_\alpha$ and $B_\alpha$ are both defined to be metric invariant; (ii) $A^\alpha$ is metric invariant and $B_\alpha$ is metric invariant.
(i):
$$T_{\mu\nu} = 2 \frac{\partial \mathcal{L}}{\partial g^{\mu\nu}} - g_{\mu\nu} \mathcal{L} = 2 \frac{\partial}{\partial g^{\mu\nu}} \left( A^\alpha B_\alpha \right) - g_{\mu\nu} A^{\alpha} B_{\alpha} = 2 \frac{\partial}{\partial g^{\mu\nu}} \left( g^{\alpha\beta} A_\beta B_\alpha \right) - g_{\mu\nu} A^{\alpha} B_{\alpha} = 2 A_\nu B_\mu - g_{\mu\nu} A^{\alpha} B_{\alpha}$$

(ii):
$$T_{\mu\nu} = 2 \frac{\partial}{\partial g^{\mu\nu}} \left( A^\alpha B_\alpha \right) - g_{\mu\nu} A^{\alpha} B_{\alpha} = - g_{\mu\nu} A^{\alpha} B_{\alpha}.$$

It's evident that expression (i) is not symmetric (given that $A \neq B$).
To analyze better these expressions lets take A to be the negative of the electromagnetic current and B the electromagnetic potential. The lagrangian above will then represent the source part of the electromagnetic lagrangian. If we compare them to the actual SET for the electromagnetic field with sources:
$$T^{\mu\nu} = \frac{1}{4} g^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta} + g_{\alpha\beta} F^{\mu\alpha} F^{\beta\nu} + g^{\mu\nu}j_{\alpha} A^{\alpha} - j^{\mu} A^{\nu}$$
*Extracted directly from Greiner's Field Quantization (equation 6.40).
We can see that neither (i) nor (ii) matches with this SET.

Greiner obtained this SET through the Belinfante procedure, so the expressions should differ only by a surface term, but it doesn't matter how much I try I cannot find any surface term that makes them equivalent.

4. Dec 21, 2013

### Bill_K

5. Dec 21, 2013

### maykot

Kinda...
I don't see why I have to vary $A_{\mu}$ as well as $g_{\mu\nu}$. You yourself said that $A_{\mu}$ is metric independent, so it shouldn't add anything to the SET.
However, Ron Maimon's last paragraph sounds reasonable to me. If you fix $\sqrt{g} j^{\alpha}$ while varying the metric (due to conservation issues), and since the covariant potential is already fixed as discussed, then the whole interaction term gives no contribution to the SET.

Do you know how can I show that $\sqrt{g} j^{\alpha}$ is metric independent? Or can you point me to any books/articles that treat this in a more rigorous way?

And thank you again for your time :)

PS: If the interaction term doesn't contribute to the SET does that mean Greiner's expression is wrong?

6. Dec 21, 2013

### Bill_K

It's certainly true that AÎ¼ and JÎ¼ are the components which are metric independent, and therefore the interaction term AÎ¼JÎ¼ makes zero contribution to the (correctly defined ) stress-energy tensor.

I'm not familiar with Greiner's book, but he may be talking about one of the other forms, the canonical tensor or Belinfante's symmetrized tensor.

7. Dec 22, 2013

### maykot

Actually, according to the link you posted, I don't think that it's $j^{\mu} A_{\mu}$ that is independent, but $\sqrt{g} j^{\mu} A_{\mu}$. If it was the former, then the SET would correspond to case (ii) mentioned earlier, instead of not contributing to the SET at all.

And yes, Greiner derives Belinfante's SET. But aren't Belinfante's and Hilbert's SET supposed to be equal? Or at least equal up to a surface term?
Given that they have the following form:
$$T_{(Greiner)}^{\mu\nu} = \frac{1}{4} g^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta} + g_{\alpha\beta} F^{\mu\alpha} F^{\beta\nu} + g^{\mu\nu}j^{\alpha} A_{\alpha} - j^{\mu} A^{\nu}$$
$$T_{(Hilbert)}^{\mu\nu} = \frac{1}{4} g^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta} + g_{\alpha\beta} F^{\mu\alpha} F^{\beta\nu}$$
If we say they differ only by a surface term, this means that:
$$T_{(Greiner)}^{\mu\nu} - T_{(Hilbert)}^{\mu\nu} = g^{\mu\nu}j^{\alpha} A_{\alpha} - j^{\mu} A^{\nu} = \nabla _\alpha\psi ^{\alpha\mu\nu}$$
I could not find any such $\psi$ and am very inclined to say that there isn't one.