Stress-Energy Tensor from Lagrangian: Technical Question II

[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,

This obviously begs the question, "How do I know which of the raised or lowered components are the ones independent of the metric?" The answer has to come from context. Suppose I define some vector as tangent to a curve. Well this object naturally comes with a raised index and isn't defined using the metric, so the raised components must be metric independent. However, when you use the metric to find the equivalent 1 form, the 1 form you obtain obviously depends to the metric you use. Thus the lowered components do depend on the metric. This subtlety means that you have to be very careful about how you define various tensors. For instance, is the electromagnetic potential most naturally a 1 form or a vector? Elementary treatments often start with it as a vector, but it gauge theory it appears as a connection 1 form. Consistency must be you guide.

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}} = <br /> \frac{\partial}{g_{\alpha\beta}} \left( g^{\mu\lambda} g^{\nu\kappa} F_{\mu\nu} F_{\lambda\kappa} \right) = <br /> 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] =<br />
<br /> =-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] =<br /> -2 g_{\mu\nu} F^{\alpha\mu}F^{\beta\nu}<br />
And:
<br /> t_{\alpha\beta} \equiv \frac{\partial \left( F_{\mu\nu}F^{\mu\nu} \right)}{\partial g^{\alpha\beta}} =<br /> \frac{\partial}{g^{\alpha\beta}} \left( g^{\mu\lambda} g^{\nu\kappa} F_{\mu\nu} F_{\lambda\kappa} \right) =<br /> 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] =<br />
<br /> =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] =<br /> 2 g^{\mu\nu} F_{\alpha\mu}F_{\beta\nu}<br />
But we also have:
<br /> 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}<br />
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:

what is
\frac{\partial}{\partial g_{\mu \nu}} \left( g_{\alpha \beta} V^{\alpha} V^{\beta} \right)
Is it
\frac{\partial}{\partial g_{\mu \nu}} \left( g_{\alpha \beta} V^{\alpha} V^{\beta} \right) = \delta^{\mu}_{\phantom{\mu}\alpha} \delta^{\nu}_{\phantom{\mu}\beta} V^{\alpha} V^{\beta} = V^{\mu} V^{\nu}
or is it
\frac{\partial}{\partial g_{\mu \nu}} \left( g^{\alpha \beta} V_{\alpha} V_{\beta} \right) = - g^{\mu \alpha} g^{\nu \beta} V_{\alpha} V_{\beta} = -V^{\mu} V^{\nu}
?

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.

 

[Bill_K]

The energy-momentum tensor T^μν is defined as the functional derivative of the matter action I_M with respect to g_μν. That is,

δI_M = ½ ∫d⁴x √g T^μν δg_μν

This always gives a conserved symmetric tensor.

As part of the definition, one must state which field variables are independent of g_μν 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 T^μν depending on your answer. In the case of electromagnetism, A_μ is the independent variable, not A^μ.

 

[maykot]

Thank you for your reply Bill_K.
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 let's take for example a lagrangian of the form \mathcal{L} = A^\alpha B_\alpha. Let's 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):
<br /> T_{\mu\nu} = 2 \frac{\partial \mathcal{L}}{\partial g^{\mu\nu}} - g_{\mu\nu} \mathcal{L} =<br /> 2 \frac{\partial}{\partial g^{\mu\nu}} \left( A^\alpha B_\alpha \right) - g_{\mu\nu} A^{\alpha} B_{\alpha} =<br /> 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}<br />

​

(ii):
<br /> 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}.<br /> <br />
​

It's evident that expression (i) is not symmetric (given that A \neq B).
To analyze better these expressions let's 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:
<br /> 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}<br />
*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.

 

[Bill_K]

See if this discussion is any help.

 

[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?

 

[Bill_K]

It's certainly true that A_μ and J^μ are the components which are metric independent, and therefore the interaction term A_μJ^μ 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.

 

[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:
<br /> 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}<br />
<br /> 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} <br />
If we say they differ only by a surface term, this means that:
<br /> 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}<br />
I could not find any such \psi and am very inclined to say that there isn't one.