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

In summary: I think I misunderstood your question. You said you don't know how to approach the problem, could you please clarify?
  • #1
JD_PM
1,131
158
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.
 
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  • #2
JD_PM said:
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?)
 
  • #3
nrqed said:
You really mean ##\delta \mathcal{L}## here.

My bad.

nrqed said:
What is the Lagrangian?

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

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

nrqed said:
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.
 
  • #4
JD_PM said:
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.
JD_PM said:
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.
 
  • #5
nrqed said:
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).

nrqed said:
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 😬
 

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

1. What is the stress-energy tensor formula in general relativity?

The stress-energy tensor formula in general relativity is a mathematical expression that describes the distribution of energy and momentum in a given space-time. It is a 4x4 matrix that represents the energy density, momentum density, and stress (pressure) in each direction of space and time.

2. How is the stress-energy tensor formula derived?

The stress-energy tensor formula is derived from the Einstein field equations, which relate the curvature of space-time to the distribution of matter and energy. It is also derived from the conservation laws of energy and momentum.

3. What are the components of the stress-energy tensor formula?

The components of the stress-energy tensor formula include the energy density, momentum density, and stress (pressure) in each direction of space and time. These components are represented by the 4x4 matrix in the formula.

4. Why is the stress-energy tensor formula well-established in general relativity?

The stress-energy tensor formula is well-established in general relativity because it has been extensively tested and verified through various experiments and observations. It is also a fundamental part of the Einstein field equations, which have been proven to accurately describe the behavior of gravity.

5. How is the stress-energy tensor formula used in general relativity?

The stress-energy tensor formula is used in general relativity to calculate the curvature of space-time and the resulting gravitational effects. It also helps to predict the behavior of matter and energy in the presence of strong gravitational fields, such as those around black holes.

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