Prove the well-established (in GR) stress-energy tensor formula

[JD_PM]

Homework Statement

Prove that the stress-energy tensor is given by the functional derivative of the action with respect to ##\delta g^{\mu \nu}##



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

Relevant Equations

$$ \mathcal{L} = (\partial_{\mu} \alpha) h^{\mu} (\phi) $$

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

Prove that the stress-energy tensor is given by the functional derivative of the action with respect to ##\delta g^{\mu \nu}##

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



Tong suggests (around 25:30) we could get the desired tensor by performing a transformation ##\delta \phi = \alpha \phi##, where ##\alpha## is not a constant; ##\alpha := \alpha(x)##.

Thus the action is not invariant and we get

$$ \mathcal{L} = (\partial_{\mu} \alpha) h^{\mu} (\phi) $$

Then the change of the action is

$$\delta S = \int d^4 x \delta L = -\int d^4 x \alpha (x) \partial_{\mu} h^{\mu}$$

But I do not really know how to proceed from here. Could you please give me a hint?Any help is appreciated.

Thank you.

 

[nrqed]

Homework Statement:: Prove that the stress-energy tensor is given by the functional derivative of the action with respect to ##\delta g^{\mu \nu}##
$$T_{\mu \nu} = \frac{-2}{\sqrt{-g}} \frac{\delta S}{\delta g^{\mu \nu}}$$
Relevant Equations:: $$ \mathcal{L} = (\partial_{\mu} \alpha) h^{\mu} (\phi) $$

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

Prove that the stress-energy tensor is given by the functional derivative of the action with respect to ##\delta g^{\mu \nu}##

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



Tong suggests (around 25:30) we could get the desired tensor by performing a transformation ##\delta \phi = \alpha \phi##, where ##\alpha## is not a constant; ##\alpha := \alpha(x)##.

Thus the action is not invariant and we get

$$ \mathcal{L} = (\partial_{\mu} \alpha) h^{\mu} (\phi) $$

You really mean ##\delta \mathcal{L}## here.

What is the Lagrangian? And how does ##h^\mu (\phi)## transforms? (or if you prefer, how does ##\phi## transform?)

 

[JD_PM]

You really mean ##\delta \mathcal{L}## here.

My bad.

What is the Lagrangian?

Tong doesn't specify it but let's pick the Lagrangian from Einstein-Maxwell's theory

$$\mathcal{L}=-\frac 1 4 F_{\mu \nu} F^{\mu \nu}, \ \ \ \ F_{\mu \nu}:=\partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}$$

how does ##\phi## transform

Tong doesn't specify it but I am sure that the transformation has to yield an internal symmetry ##\delta \phi = \alpha \phi##. Let's also choose the transformation we are going to work with.

The only transformation I know related to such a Lagrangian is

$$A^{\mu} \rightarrow \Lambda^{\mu}_{ \ \nu} A^{\nu} (\Lambda^{-1} x)$$

I think we can work with this one.

My issue with this problem is that I do not really know how to approach it.

 

[nrqed]

My bad.
Tong doesn't specify it but let's pick the Lagrangian from Einstein-Maxwell's theory

$$\mathcal{L}=-\frac 1 4 F_{\mu \nu} F^{\mu \nu}, \ \ \ \ F_{\mu \nu}:=\partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}$$
Tong doesn't specify it but I am sure that the transformation has to yield an internal symmetry ##\delta \phi = \alpha \phi##. Let's also choose the transformation we are going to work with.

The only transformation I know related to such a Lagrangian is

$$A^{\mu} \rightarrow \Lambda^{\mu}_{ \ \nu} A^{\nu} (\Lambda^{-1} x)$$

I think we can work with this one.

My issue with this problem is that I do not really know how to approach it.

My bad.
Tong doesn't specify it but let's pick the Lagrangian from Einstein-Maxwell's theory

$$\mathcal{L}=-\frac 1 4 F_{\mu \nu} F^{\mu \nu}, \ \ \ \ F_{\mu \nu}:=\partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}$$
Tong doesn't specify it but I am sure that the transformation has to yield an internal symmetry ##\delta \phi = \alpha \phi##. Let's also choose the transformation we are going to work with.

The only transformation I know related to such a Lagrangian is

$$A^{\mu} \rightarrow \Lambda^{\mu}_{ \ \nu} A^{\nu} (\Lambda^{-1} x)$$

I think we can work with this one.

My issue with this problem is that I do not really know how to approach it.

They are talking about the variation with respect to the metric, so you need to couple the theory to gravity.

More context is needed for your question: you want to prove that the stress tensor is given by the variation with respect to the metric and then you mention the trick to obtain the current associated to a symmetry transformation using a space-time dependent parameter. These are two distinct concepts in general, in that the space-time dependent parameter trick can be used for gauge transformation, for example, and that has nothing to do with the stress energy momentum tensor. If you want to obtain the S-E-M tensor, you need a theory coupled to gravity. Did you take the question from a specific source? The context around the question would help understand what they want to do.

 

[JD_PM]

Did you take the question from a specific source?

I took it from Tong's lecture, which I shared at #1 (around min 25:30).

The context around the question would help understand what they want to do.

Tong suggests we use the internal symmetry trick, but I am getting nothing out of it at the moment. I found another way of deriving the formula, which is in Carroll's book (page 164, EQs 4.77 and 4.78). However, it does not really look intuitive to me.

I am going to ask in the Relativity Forum to see if they are acquainted with Carroll's way.

EDIT: Sorry for my late reply