Computing an Energy-Momentum tensor given a Lagrangian

[JD_PM]

Homework Statement

a) Given the Lagrangian density

$$\mathcal{L} = -\frac 1 2 (\partial_{\mu} A_{\nu})( \partial^{\mu} A^{\nu})$$

Show that the Energy-Momentum tensor is

$$T_{\mu \nu} = -\partial_{\mu} A_{\rho} \partial_{\nu}A^{\rho} + \frac 1 2 \eta_{\mu \nu} \partial_{\rho} A_{\sigma} \partial^{\rho} A^{\sigma}$$

b) Show that $\partial_{\nu} T^{\mu \nu} = 0$

Relevant Equations

The given Lagrangian

$$\mathcal{L} = -\frac 1 2 \partial_{\mu} A_{\nu} \partial^{\mu} A^{\nu}$$

The Energy-momentum tensor to show

$$T^{\mu \nu} = -\partial_{\mu} A_{\rho} \partial_{\nu}A^{\rho} + \frac 1 2 \eta_{\mu \nu} \partial_{\rho} A_{\sigma} \partial^{\rho} A^{\sigma}$$

We also have to show that $\partial_{\nu} T^{\mu \nu} =0$

REMARK: First of all I have to say that this Lagrangian reminds me of the Lagrangian from which we can derive Maxwell's equations, which is (reference: Tong QFT lecture notes, equation 1.18; I have attached the PDF).

$$\mathcal{L} = -\frac 1 2 (\partial_{\mu} A_{\nu} )(\partial^{\mu} A^{\nu}) + \frac 1 2(\partial_{\mu} A^{\mu})^2$$

I guess here we are dealing with a simplified version of it:$$\mathcal{L} = -\frac 1 2 (\partial_{\mu} A_{\nu})( \partial^{\mu} A^{\nu})$$a) At first glace we see that the given Lagrangian is invariant under Lorentz transformations because the Lorentz indices $\mu$ and $\nu$ are contracted with the metric (which is a Lorentz invariant object); let's show this explicitely:

$$\mathcal{L} = -\frac 1 2 (\eta_{\mu \lambda}\partial^{\lambda} A_{\nu})( \eta^{\mu \rho} \partial_{\rho} A^{\nu})$$

Beautiful! OK that's good because we already know that, by Noether's theorem, there has to be a conserved current associated to the transformation we decide to apply.

I have studied that the Energy-momentum tensor is defined to be the conserved current obtained from applying the following transformation: translations. Thus the logic step now is to apply translation to the given Lagrangian to get the current and then compute our specific case.

The general formula for the current $j^{\mu}$ when $\delta \mathcal{L} = 0$ is given by

$$j^{\mu} = \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi_a)}\partial \phi_a$$

I wanted to be careful here and I thought of working with the (I think) equivalent Lagrangian

$$\mathcal{L} = -\frac 1 2 (\partial_{\mu} \phi)( \partial^{\mu} \phi)$$

I did the following: first I focused on computing $\frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi)}$

$$\frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi)} = -\frac 1 2 \eta^{\rho \sigma} \frac{\partial (\partial_{\rho} \phi \partial_{\sigma} \phi)}{\partial(\partial_{\mu} \phi)} = -\frac 1 2 \eta^{\rho \sigma} (\delta_{\rho}^{\mu} \partial_{\sigma} \phi + \delta_{\sigma}^{\mu} \partial_{\rho} \phi)$$

Where, note, I carefully swapped $\rho$ by $\mu$ in the original Lagrangian so we can avoid any index confusions

OK, so noticing that we can contract the metric like $\eta^{\rho \sigma} \delta_{\rho}^{\mu}$ I get

$$\frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi)} = -\frac 1 2 (\eta^{\mu \sigma} \partial_{\sigma} \phi + \eta^{\mu \rho} \partial_{\rho} \phi )$$

We're just left multiplying by $\partial \phi$ to get the current

$$j^{\mu} = \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi)}\partial \phi = -\frac 1 2 (\eta^{\mu \sigma} \partial_{\sigma} \phi + \eta^{\mu \rho} \partial_{\rho} \phi )\partial \phi$$

Thus the Energy-Momentum tensor is

$$T^{\mu}_{ \ \nu} = j^{\mu}_{ \ \nu} = \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi)}\partial_{\nu} \phi = -\frac 1 2 (\eta^{\mu \sigma} \partial_{\sigma} \phi + \eta^{\mu \rho} \partial_{\rho} \phi )\partial_{\nu} \phi$$

Now the idea would be to generalize this result and be able to get the desired tensor. I do not see how to do so...

How can I proceed from here?

Notice that I am not sure of the equivalence of the following Lagrangians

$$\mathcal{L} = -\frac 1 2 (\partial_{\mu} \phi)( \partial^{\mu} \phi)$$

$$\mathcal{L} = -\frac 1 2 (\partial_{\mu} A_{\nu})( \partial^{\mu} A^{\nu})$$

I am sure there is a straightforward way to solve a) anyways.
b) To be honest I do not know how to actually prove that $\partial_{\nu} T^{\mu \nu} = 0$. In the notes I am studying (Tong, page 15 example 1.3.2) he asserts that it does satisfy it but he doesn't prove it explicitly.

Could you please give me a hint on this one?

Thank you.

 

[nrqed]

...
Thus the Energy-Momentum tensor is

$$T^{\mu}_{ \ \nu} = j^{\mu}_{ \ \nu} = \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi)}\partial_{\nu} \phi = -\frac 1 2 (\eta^{\mu \sigma} \partial_{\sigma} \phi + \eta^{\mu \rho} \partial_{\rho} \phi )\partial_{\nu} \phi$$

So far so good. This can be written in a more compact form, right? You can contract the $\eta's$ to get a more compact form.

Now, you can simply follow the same steps for your problem. I don't see what difficulty you are encountering. It's basically the same approach.

 

[JD_PM]

So far so good. This can be written in a more compact form, right? You can contract the $\eta's$ to get a more compact form.

Do you mean we can compact $\eta's$ with partial derivatives? How?

 

[nrqed]

Do you mean we can compact $\eta's$ with partial derivatives? How?

You can contract derivatives with $\eta's$ like you contract anything. For example,
$\eta^{\mu \sigma} \partial_\sigma \phi = \partial^\mu \phi$.

 

[JD_PM]

Let's explicitly show where I'm stuck

We have to compute the following current (I am not sure if I've written it correctly):

$$j^{\mu}_{ \ \nu} = \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} A_{\nu} )}\partial_{w} A^w$$

Let's first compute $\frac{\partial \mathcal{L}}{\partial (\partial_{\mu} A_{\nu} )}$

$$\frac{\partial \mathcal{L}}{\partial (\partial_{\mu} A_{\nu} )} = -\frac 1 2 \eta^{\rho \sigma} \frac{\partial (\partial_{\rho} A_{\nu} \partial_{\sigma} A^{\nu})}{\partial(\partial_{\mu} A_{\nu})} = -\frac 1 2 \eta^{\rho \sigma} (\delta_{\rho}^{\mu} \partial_{\sigma} A^{\nu} + \delta_{\sigma}^{\mu} \partial_{\rho} A_{\nu}) = -\frac 1 2 (\eta^{\mu \sigma} \partial_{\sigma} A^{\nu} + \eta^{\mu \rho} \partial_{\rho} A_{\nu}) = -\frac 1 2 (\partial^{\mu} A^{\nu} + \partial^{\mu} A_{\nu})$$

Mmm I am close but this is not correct. The correct solution is

$$\frac{\partial \mathcal{L}}{\partial (\partial_{\mu} A_{\nu} )} = -\partial^{\mu} A^{\nu} + (\partial_{\rho} A^{\rho}) \eta^{\mu \nu}$$

It seems I am missing something ovvio but I do not see it..

 

[JD_PM]

OK I will be even more explicit this time. I did a slight modification wrt #5

$$\frac{\partial \mathcal{L}}{\partial (\partial_{\mu} A_{\nu} )} = \frac{\partial}{\partial (\partial_{\mu} A_{\nu})} \Big(-\frac 1 2 \eta^{\rho \sigma} \partial_{\rho} A_{\nu} \partial_{\sigma} A^{\nu} \Big) = -\frac 1 2 \eta^{\rho \sigma} \partial_{\sigma} A^{\nu} \frac{\partial(\partial_{\rho} A_{\nu})}{\partial (\partial_{\mu} A_{\nu})} - \frac 1 2 \eta^{\rho \sigma} \partial_{\rho} A_{\nu} \frac{\partial(\partial_{\sigma} A^{\nu})}{\partial (\partial_{\mu} A_{\nu})}$$

I think my mistake has to be in how I am computing the term $\frac{\partial(\partial_{\sigma} A^{\nu})}{\partial (\partial_{\mu} A_{\nu})}$. I say that it is as follows

$$\eta^{\epsilon \nu} \frac{\partial(\partial_{\sigma} A_{\epsilon})}{\partial (\partial_{\mu} A_{\nu})} = \eta^{\epsilon \nu} \delta_{\sigma}^{\mu}$$

Thus I end up getting

$$\frac{\partial \mathcal{L}}{\partial (\partial_{\mu} A_{\nu} )} = -\frac 1 2 \partial^{\mu} A^{\nu} - \frac 1 2 \partial^{\mu} A_{\nu} \eta^{\epsilon \nu}$$

This is slightly better that what I got at #5 but still not correct.

$$\frac{\partial \mathcal{L}}{\partial (\partial_{\mu} A_{\nu} )} = -\frac 1 2 \partial^{\mu} A^{\nu} - \frac 1 2 (\partial^{\mu} A_{\nu}) \eta^{\epsilon \nu} \neq -\partial^{\mu} A^{\nu} + (\partial_{\rho} A^{\rho}) \eta^{\mu \nu} $$

Any help is appreciated.

Thanks.

 

[nrqed]

OK I will be even more explicit this time. I did a slight modification wrt #5

$$\frac{\partial \mathcal{L}}{\partial (\partial_{\mu} A_{\nu} )} = \frac{\partial}{\partial (\partial_{\mu} A_{\nu})} \Big(-\frac 1 2 \eta^{\rho \sigma} \partial_{\rho} A_{\nu} \partial_{\sigma} A^{\nu} \Big) ...$$Thanks.

Watch out, you should be careful about not using the same index three times in an expression. If you take a derivative with respect to $\partial_{\mu} A_{\nu}$, you should change the index $\nu$ on the $A$ in the Lagrangian, say to $\delta$.

 

[JD_PM]

Thank you nrqed.

Alright, let's swap $\nu$ by $k$. Then we get:

$$\frac{\partial \mathcal{L}}{\partial (\partial_{\mu} A_{\nu} )} = \frac{\partial}{\partial (\partial_{\mu} A_{\nu})} \Big(-\frac 1 2 \eta^{\rho \sigma} \partial_{\rho} A_{k} \partial_{\sigma} A^{k} \Big) = -\frac 1 2 \eta^{\rho \sigma} \partial_{\sigma} A^{k} \frac{\partial(\partial_{\rho} A_{k})}{\partial (\partial_{\mu} A_{\nu})} - \frac 1 2 \eta^{\rho \sigma} \partial_{\rho} A_{k} \frac{\partial(\partial_{\sigma} A^{k})}{\partial (\partial_{\mu} A_{\nu})}$$

Let's work out the second term of the RHS of the above equation

$$\eta^{k a} \frac{\partial(\partial_{\sigma} A_{a})}{\partial (\partial_{\mu} A_{\nu})} = \eta^{k a} \delta_{\sigma}^{\mu} \delta_{a}^{\nu} = \delta_{\sigma}^{\mu} \eta^{k \nu}$$

Thus we get

$$\frac{\partial \mathcal{L}}{\partial (\partial_{\mu} A_{\nu} )} = -\frac 1 2 \partial^{\rho} A^k \delta_{\rho}^{\mu} \delta_{k}^{\nu} -\frac 1 2 \eta^{\rho \sigma} \partial_{\rho} A_k \delta_{\sigma}^{\mu} \eta^{k \nu} = -\frac 1 2 \partial^{\mu} A^{\nu} -\frac 1 2 \partial^{\mu} A^{\nu} = - \partial^{\mu} A^{\nu}$$

Do you agree at this point? Or should it be

$$\frac{\partial \mathcal{L}}{\partial (\partial_{\mu} A_{\nu} )} = -\partial^{\mu} A^{\nu} + (\partial_{\rho} A^{\rho}) \eta^{\mu \nu}$$

?

Thank you.

 

[nrqed]

Thank you nrqed.

Alright, let's swap $\nu$ by $k$. Then we get:

$$\frac{\partial \mathcal{L}}{\partial (\partial_{\mu} A_{\nu} )} = \frac{\partial}{\partial (\partial_{\mu} A_{\nu})} \Big(-\frac 1 2 \eta^{\rho \sigma} \partial_{\rho} A_{k} \partial_{\sigma} A^{k} \Big) = -\frac 1 2 \eta^{\rho \sigma} \partial_{\sigma} A^{k} \frac{\partial(\partial_{\rho} A_{k})}{\partial (\partial_{\mu} A_{\nu})} - \frac 1 2 \eta^{\rho \sigma} \partial_{\rho} A_{k} \frac{\partial(\partial_{\sigma} A^{k})}{\partial (\partial_{\mu} A_{\nu})}$$

Let's work out the second term of the RHS of the above equation

$$\eta^{k a} \frac{\partial(\partial_{\sigma} A_{a})}{\partial (\partial_{\mu} A_{\nu})} = \eta^{k a} \delta_{\sigma}^{\mu} \delta_{a}^{\nu} = \delta_{\sigma}^{\mu} \eta^{k \nu}$$

Thus we get

$$\frac{\partial \mathcal{L}}{\partial (\partial_{\mu} A_{\nu} )} = -\frac 1 2 \partial^{\rho} A^k \delta_{\rho}^{\mu} \delta_{k}^{\nu} -\frac 1 2 \eta^{\rho \sigma} \partial_{\rho} A_k \delta_{\sigma}^{\mu} \eta^{k \nu} = -\frac 1 2 \partial^{\mu} A^{\nu} -\frac 1 2 \partial^{\mu} A^{\nu} = - \partial^{\mu} A^{\nu}$$

Do you agree at this point?

Hi. Sorry for taking time to reply. I do agree with that result.
Now the question is: what formula do you use to find $T_{\mu \nu}$? I think the problem comes from that.

 

[JD_PM]

Now the question is: what formula do you use to find $T_{\mu \nu}$?

$$T^{\mu \nu} \equiv j^{\mu \nu} = \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} A_{\nu} )}\partial_{w} A^w = -\partial^{\mu} A^{\nu} \partial_{w} A^{w}$$

Thus I get

$$T_{\mu \nu} = -\eta^{\mu \rho} \partial_{\rho} \eta^{\nu \sigma} A_{\sigma} \partial_{w} A^{w}$$

But the above is incorrect; the correct answer is:

$$T_{\mu \nu} = -\partial_{\mu} A_{\rho} \partial_{\nu}A^{\rho} + \frac 1 2 \eta_{\mu \nu} \partial_{\rho} A_{\sigma} \partial^{\rho} A^{\sigma}$$

Mmm what am I missing?

 

[nrqed]

$$T^{\mu \nu} \equiv j^{\mu \nu} = \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} A_{\nu} )}\partial_{w} A^w = -\partial^{\mu} A^{\nu} \partial_{w} A^{w}$$

Thus I get

$$T_{\mu \nu} = -\eta^{\mu \rho} \partial_{\rho} \eta^{\nu \sigma} A_{\sigma} \partial_{w} A^{w}$$

But the above is incorrect; the correct answer is:

$$T_{\mu \nu} = -\partial_{\mu} A_{\rho} \partial_{\nu}A^{\rho} + \frac 1 2 \eta_{\mu \nu} \partial_{\rho} A_{\sigma} \partial^{\rho} A^{\sigma}$$

Mmm what am I missing?

I think that they are using the formula for the canonical stress-energy momentum tensor, which is (I have seen the form below but sometimes minus that expression)
$$ \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} A_{\nu} )}\partial_{w} A^w - \eta_{\mu \nu} {\cal L} $$

 

[JD_PM]

I think that they are using the formula for the canonical stress-energy momentum tensor, which is (I have seen the form below but sometimes minus that expression)
$$ \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} A_{\nu} )}\partial_{w} A^w - \eta_{\mu \nu} {\cal L} $$

Ahhh so apparently they did not assume $\delta L = 0$; they assumed $\delta L = \partial_{w} A^{w}$

I will go step by step now.

We swap $\mu \rightarrow \rho$ and $\nu \rightarrow \sigma$ in the original Lagrangian; doing so we avoid mixing up indices.

OK so assuming the expression you proposed at #11 we get

$$T_{\mu \nu} = -\partial^{\mu} A^{\nu} \partial_{w} A^w + \frac 1 2 \eta_{\mu \nu} \partial_{\rho} A_{\sigma} \partial^{\rho} A^{\sigma}$$

The second term on the RHS is OK but we need to fix $-\partial^{\mu} A^{\nu} \partial_{w} A^w$, because $\mu$ and $\nu$ must be at the bottom. Thus

$$-\partial^{\mu} A^{\nu} \partial_{w} A^w = - \eta^{\mu z} \partial_z \eta^{\nu \rho} A_{\rho} \partial_{w} A^w$$

At this point I assumed that the metric tensors $\eta^{\mu z}$ and $\eta^{\nu \rho}$ are $4 \times 4$ identity matrices. Then I set $\mu = z$, $\nu = \rho$ and got:

$$-\eta^{\mu z} \partial_z \eta^{\nu \rho} A_{\rho} \partial_w A^w = -\partial_{\mu} A_{\rho} \partial_w A^w$$

OK and here comes the point where I feel like I am cheating: as $w$ is a dummy index I can set it to whatever value I want; to match the answer I used $w = \nu$ and the fact that $\nu = \rho$ to get the desired answer

$$T_{\mu \nu} = -\partial_{\mu} A_{\rho} \partial_{\nu} A^{\rho} + \frac 1 2 \eta_{\mu \nu} \partial_{\rho} A_{\sigma} \partial^{\rho} A^{\sigma}$$

 

[JD_PM]

I'll be back soon to (try to) show that $\partial_{\nu} T^{\mu \nu} = 0$

 

[nrqed]

Ahhh so apparently they did not assume $\delta L = 0$; they assumed $\delta L = \partial_{w} A^{w}$

I will go step by step now.

We swap $\mu \rightarrow \rho$ and $\nu \rightarrow \sigma$ in the original Lagrangian; doing so we avoid mixing up indices.

OK so assuming the expression you proposed at #11 we get

$$T_{\mu \nu} = -\partial^{\mu} A^{\nu} \partial_{w} A^w + \frac 1 2 \eta_{\mu \nu} \partial_{\rho} A_{\sigma} \partial^{\rho} A^{\sigma}$$
...

I a sorry, I should have written
$$ T_{\mu \nu} = \frac{\partial \mathcal{L}}{\partial (\partial^{\mu} A^{\nu} )}\partial_{w} A^w - \eta_{\mu \nu} {\cal L} $$ or
$$ T^{\mu \nu} = \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} A_{\nu} )}\partial_{w} A^w - \eta^{\mu \nu} {\cal L} $$
Sorry about that.

I am not sure what you mean by $\delta {\cal{ L}}$ zero or not. I am not talking about the Euler-Lagrange equations here, just the canonical stress-energy momentum tensor. In any case, if you agree with the expression I gave, one gets the energy momentum tensor of the answer.

 

[JD_PM]

Hi nrqed no problem! Now I understand why I was getting the wrong answer.

I will post what I get for b).

 

[JD_PM]

Alright, let's show that $\partial^{\mu} T_{\mu \nu} = 0$

$$\partial^{\mu} \Big[ -\partial_{\mu} A_{\rho} \partial_{\nu}A^{\rho} + \frac 1 2 \eta_{\mu \nu} \partial_{\rho} A_{\sigma} \partial^{\rho} A^{\sigma} \Big] = -(\partial^{\mu} \partial_{\mu} A_{\rho}) \partial_{\nu}A^{\rho} - \partial_{\mu} A_{\rho} (\partial^{\mu} \partial_{\nu} A^{\rho}) + \frac 1 2(\partial_{\nu} \partial_{\rho} A_{\sigma})(\partial^{\rho} A^{\sigma}) + \frac 1 2(\partial_{\rho} A_{\sigma})(\partial_{\nu} \partial^{\rho} A^{\sigma})$$

We can compact the third and forth terms, as $(\partial_{\rho} A_{\sigma})(\partial^{\rho} A^{\sigma}) = (\partial^{\rho} A^{\sigma})(\partial_{\rho} A_{\sigma})$

$$-(\partial^{\mu} \partial_{\mu} A_{\rho}) \partial_{\nu}A^{\rho} - \partial_{\mu} A_{\rho} (\partial^{\mu} \partial_{\nu} A^{\rho}) + \frac 1 2(\partial_{\nu} \partial_{\rho} A_{\sigma})(\partial^{\rho} A^{\sigma}) + \frac 1 2(\partial_{\rho} A_{\sigma})(\partial_{\nu} \partial^{\rho} A^{\sigma}) = -(\partial^{\mu} \partial_{\mu} A_{\rho}) \partial_{\nu}A^{\rho} - \partial_{\mu} A_{\rho} (\partial^{\mu} \partial_{\nu} A^{\rho}) + (\partial^{\rho} A^{\sigma})(\partial_{\nu} \partial^{\rho} A^{\sigma})$$

By swapping $\mu \rightarrow \rho$ and $\rho \rightarrow \sigma$ in the second term of the RHS of the above equation we see that the second and third term cancel each other out; thus we end up with

$$\partial^{\mu} T_{\mu \nu} = -(\partial^{\mu} \partial_{\mu} A_{\rho}) \partial_{\nu}A^{\rho}$$
Mmm but I do not see why $-(\partial^{\mu} \partial_{\mu} A_{\rho}) \partial_{\nu}A^{\rho}=0$ ...

Thanks.

 

[nrqed]

Alright, let's show that $\partial^{\mu} T_{\mu \nu} = 0$

$$\partial^{\mu} \Big[ -\partial_{\mu} A_{\rho} \partial_{\nu}A^{\rho} + \frac 1 2 \eta_{\mu \nu} \partial_{\rho} A_{\sigma} \partial^{\rho} A^{\sigma} \Big] = -(\partial^{\mu} \partial_{\mu} A_{\rho}) \partial_{\nu}A^{\rho} - \partial_{\mu} A_{\rho} (\partial^{\mu} \partial_{\nu} A^{\rho}) + \frac 1 2(\partial_{\nu} \partial_{\rho} A_{\sigma})(\partial^{\rho} A^{\sigma}) + \frac 1 2(\partial_{\rho} A_{\sigma})(\partial_{\nu} \partial^{\rho} A^{\sigma})$$

We can compact the third and forth terms, as $(\partial_{\rho} A_{\sigma})(\partial^{\rho} A^{\sigma}) = (\partial^{\rho} A^{\sigma})(\partial_{\rho} A_{\sigma})$

$$-(\partial^{\mu} \partial_{\mu} A_{\rho}) \partial_{\nu}A^{\rho} - \partial_{\mu} A_{\rho} (\partial^{\mu} \partial_{\nu} A^{\rho}) + \frac 1 2(\partial_{\nu} \partial_{\rho} A_{\sigma})(\partial^{\rho} A^{\sigma}) + \frac 1 2(\partial_{\rho} A_{\sigma})(\partial_{\nu} \partial^{\rho} A^{\sigma}) = -(\partial^{\mu} \partial_{\mu} A_{\rho}) \partial_{\nu}A^{\rho} - \partial_{\mu} A_{\rho} (\partial^{\mu} \partial_{\nu} A^{\rho}) + (\partial^{\rho} A^{\sigma})(\partial_{\nu} \partial^{\rho} A^{\sigma})$$

By swapping $\mu \rightarrow \rho$ and $\rho \rightarrow \sigma$ in the second term of the RHS of the above equation we see that the second and third term cancel each other out; thus we end up with

$$\partial^{\mu} T_{\mu \nu} = -(\partial^{\mu} \partial_{\mu} A_{\rho}) \partial_{\nu}A^{\rho}$$
Mmm but I do not see why $-(\partial^{\mu} \partial_{\mu} A_{\rho}) \partial_{\nu}A^{\rho}=0$ ...

Thanks.

Indeed, $\partial^{\mu} T_{\mu \nu}$ is not zero here. But what made you think that it should be? I am a bit confused because in your initial post you refer to page 15 of Tong's notes but that example does not involve a vector field at all, it is about a scalar field. So I am not sure how you concluded that the derivative of your stress energy tensor should be zero. Is this a question from an assignment ?

 

[JD_PM]

I am a bit confused because in your initial post you refer to page 15 of Tong's notes but that example does not involve a vector field at all, it is about a scalar field.

I just wanted to show one of the sources I was using to solve the problem; Tong uses a scalar field in his 1.3.2 example on translation and energy momentum tensor indeed. My problem is about a vector field, but the idea is the same. This is a homework question.

Indeed, $\partial^{\mu} T_{\mu \nu}$ is not zero here. But what made you think that it should be?

Please note that Section b) of my problem asks for showing that $\partial_{\nu} T^{\mu \nu} = 0$, so I must be missing something...

 

[nrqed]

I just wanted to show one of the sources I was using to solve the problem; Tong uses a scalar field in his 1.3.2 example on translation and energy momentum tensor indeed. My problem is about a vector field, but the idea is the same. This is a homework question.
Please note that Section b) of my problem asks for showing that $\partial_{\nu} T^{\mu \nu} = 0$, so I must be missing something...

Ok, thanks for the clarification.
My gut reaction is that this is classical physics, not quantum field theory. In that case, did you work out the Euler-Lagrange equation? What condition does that impose on the field $A_\mu$?

 

[JD_PM]

My gut reaction is that this is classical physics, not quantum field theory.

Well, to be precise, it is coming from a QFT course, from the chapter 'Classical field theory'.

But just to be sure: don't you see anything wrong at #16?

 

[nrqed]

Well, to be precise, it is coming from a QFT course, from the chapter 'Classical field theory'.

But just to be sure: don't you see anything wrong at #16?

I do agree with your post 16.
Ok, since you are doing classical field theory for now, then you can use the equations of motion.

 

[JD_PM]

OK so do you suggest writing the equations of motion of the given Lagrangian then?

What is the aim of it?

 

[nrqed]

OK so do you suggest writing the equations of motion of the given Lagrangian then?

What is the aim of it?

If the field must satisfy $\partial^\mu \partial_\mu A_\nu = 0$, then ...

 

[JD_PM]

If the field must satisfy $\partial^\mu \partial_\mu A_\nu = 0$, then ...

Mmm I’m afraid I don’t get what you mean...

 

[nrqed]

Mmm I’m afraid I don’t get what you mean...

You found that the derivative of the stress-energy momentum tensor contains a factor $\partial_\mu \partial^\mu A_\nu$, right? So if this is zero, the derivative of the stress energy tensor is zero.

 

[JD_PM]

Alright but my point now is that I set $\partial_{\mu}\partial^{\mu} A_{\nu} =0$ because I was given that the derivative of the stress energy tensor is zero (i.e. I did it to match the answer). Imagine we are not given that fact. How can we justify that $\partial_{\mu}\partial^{\mu} A_{\nu} =0$ then?

 

[nrqed]

Alright but my point now is that I set $\partial_{\mu}\partial^{\mu} A_{\nu} =0$ because I was given that the derivative of the stress energy tensor is zero (i.e. I did it to match the answer). Imagine we are not given that fact. How can we justify that $\partial_{\mu}\partial^{\mu} A_{\nu} =0$ then?

This was my point about finding the E-L equation. A classical field must obey that equation. So one must find the E-L equation and consider only the field configurations that obey them.

 

[JD_PM]

Ahhh so the point is that, by finding the equations of motion, I should find $\partial_{\mu}\partial^{\mu} A_{\nu} =0$ and then we'd be done.

I'll work it out and post what I get.

 

[JD_PM]

The Euler-Lagrange equation is

$$\partial_{\mu} \Big( \frac{\partial \mathcal{L}}{\partial(\partial_{\mu} A_{\nu})} \Big) - \frac{\partial \mathcal{L}}{\partial A_{\nu}} = 0$$

At #8 I showed that

$$\frac{\partial \mathcal{L}}{\partial(\partial_{\mu} A_{\nu})} = -\partial^{\mu} A^{\nu}$$

The given Lagrangian doesn't depend on $A_{\nu}$ explicitly. Thus:

$$\frac{\partial \mathcal{L}}{\partial A_{\nu}} = 0$$

Thus our equation of motion is

$$-\partial_{\mu} \partial^{\mu} A^{\nu} = -\eta^{\nu \rho} \partial_{\mu} \partial^{\mu} A_{\rho} = 0$$

$\eta^{\nu \rho}$ is a non-zero element, so we can always multiply both sides of the above equation by $\frac{1}{\eta^{\nu \rho}}$ to get

$$\partial_{\mu} \partial^{\mu} A_{\rho} = 0$$

Which is what we needed to finish b)

Thank you nrqed!

 

[nrqed]

The Euler-Lagrange equation is

$$\partial_{\mu} \Big( \frac{\partial \mathcal{L}}{\partial(\partial_{\mu} A_{\nu})} \Big) - \frac{\partial \mathcal{L}}{\partial A_{\nu}} = 0$$

At #8 I showed that

$$\frac{\partial \mathcal{L}}{\partial(\partial_{\mu} A_{\nu})} = -\partial^{\mu} A^{\nu}$$

The given Lagrangian doesn't depend on $A_{\nu}$ explicitly. Thus:

$$\frac{\partial \mathcal{L}}{\partial A_{\nu}} = 0$$

Thus our equation of motion is

$$-\partial_{\mu} \partial^{\mu} A^{\nu} = -\eta^{\nu \rho} \partial_{\mu} \partial^{\mu} A_{\rho} = 0$$

$\eta^{\nu \rho}$ is a non-zero element, so we can always multiply both sides of the above equation by $\frac{1}{\eta^{\nu \rho}}$ to get

$$\partial_{\mu} \partial^{\mu} A_{\rho} = 0$$

Which is what we needed to finish b)

Thank you nrqed!

Good job!

Although you cannot divide by $\eta^{\nu \rho}$. Note that $\rho$ is summed over so it does not make sense to divide by that. Even worse, some of the components are zero. It is better to notice that $\partial_{\mu} \partial^{\mu} A^{\nu}=0$ implies that

$$\eta_{\rho \nu} \partial_{\mu} \partial^{\mu} A^{\nu}= 0 $$ and now we can move the $\eta$ inside the derivative, yielding what you wanted.

 

[JD_PM]

Even worse, some of the components are zero.

Ahhh you're right.

This equation

$$\partial_{\mu} \partial^{\mu} A^{\nu} = \eta^{\nu \rho} \partial_{\mu} \partial^{\mu} A_{\rho} = 0$$

It is equivalent to

$$\partial_{\mu} \partial^{\mu} A_{\nu} = \eta_{\nu \rho} \partial_{\mu} \partial^{\mu} A^{\rho} = \partial_{\mu} \partial^{\mu} A_{\nu} = 0$$

Which is what we want.