# Homework Help: Stress energy tensor transformation

1. Sep 15, 2018

### BillKet

1. The problem statement, all variables and given/known data
Show that if you add a total derivative to the Lagrangian density $L \to L + \partial_\mu X^\mu$, the energy momentum tensor changes as $T^{\mu\nu} \to T^{\mu\nu}+\partial_\alpha B^{\alpha\mu\nu}$ with $B^{\alpha\mu\nu}=-B^{\mu\alpha\nu}$.

2. Relevant equations

3. The attempt at a solution
So we have $T_{\mu\nu}=\frac{\partial L}{\partial(\partial_\mu \phi)}\partial_\nu \phi-g_{\mu\nu}L$, where $\phi$ is the field that the Lagrangian depends on. If we do the given change on the Lagrangian, the change in $T_{\mu\nu}$ would be $\frac{\partial (\partial_\alpha X^\alpha)}{\partial(\partial_\mu \phi)}\partial_\nu \phi-g_{\mu\nu}\partial_\alpha X^\alpha =\partial_\alpha \frac{\partial X^\alpha}{\partial(\partial_\mu \phi)}\partial_\nu \phi-g_{\mu\nu}\partial_\alpha X^\alpha$. From here I thought of using this: $g_{\mu\nu}\partial_\alpha X^\alpha=g_{\mu\nu}\partial_\alpha \phi \frac{\partial X^\alpha}{\partial \phi}$ But I don't really know what to do from here. Mainly I don't know how to get rid of that $g_{\mu\nu}$. Can someone help me?

2. Sep 16, 2018

### Orodruin

Staff Emeritus
This is not correct. First of all, you are making the assumption that you can just insert $\partial_\mu X^\mu$ in the expression for $T_{\mu\nu}$ and obtain the correct $T_{\mu\nu}$. This is not so because the expression assumes a dependence only on derivatives up to first order existing in $L$. If your $X^\mu$ contains any derivatives of $\phi$ (as it would have to to make it Lorentz invariant), then $\partial_\mu X^\mu$ will contain derivatives of second order and higher. Second, you are assuming that
$$\frac{\partial (\partial_\alpha X^\alpha)}{\partial(\partial_\mu \phi)} = \partial_\alpha \frac{\partial X^\alpha}{\partial(\partial_\mu \phi)},$$
which is simply not true.

Instead, I suggest that you start from the expression for $\delta \mathscr S$ in terms of the Lagrangian for the transformation $\delta y^\mu = k^\mu$ and $\delta \phi = 0$, which is what gives the energy-momentum tensor as the current $J^\mu = T^{\mu}_\nu k^\nu$ and takes the form
$$\delta \mathscr S = \int [\{L(\phi(y-\delta y))-L(\phi(y))\} + \partial_\mu(L \delta y^\mu)] d^4x.$$

Edit: Also note that you have some errors in your index placements.
Edit 2: You should not have to make any assumptions on the actual field content.

Last edited: Sep 16, 2018
3. Sep 16, 2018

### BillKet

Thank you for your reply. So I tried this: $$\frac{\delta(L+\partial_\mu X^\mu)}{\delta y^\nu}=\Big(\frac{\partial L}{\partial \phi}+\frac{\partial(\partial_\mu X^\mu)}{\partial \phi}-\partial_\mu\frac{\partial L}{\partial (\partial_\mu \phi)}-\partial_\mu\frac{\partial (\partial_\mu X^\mu)}{\partial (\partial_\mu \phi)}\Big)\frac{\delta \phi}{\delta y^\nu}+\partial_\mu\Big(\frac{\partial L}{\partial(\partial_\mu\phi)}\frac{\delta \phi}{\delta y^\nu}+\frac{\partial (\partial_\mu X^\mu)}{\partial(\partial_\mu\phi)}\frac{\delta \phi}{\delta y^\nu}\Big)$$ So this is the normal derivation from my book, but with the replacement made to the Lagrangian. We also have $$\frac{\partial L}{\partial \phi}-\partial_\mu\frac{\partial L}{\partial (\partial_\mu \phi)}=0$$ as the field satisfies the equations of motion. Then we have $$\frac{\delta \phi}{\delta y^\nu}=\partial_\nu \phi$$ which is the infinitesimal transformation for a scalar field. As the Lagrangian, even wit the new term, is a scalar field we also have: $$\frac{\delta (L+\partial_\mu X^\mu)}{\delta y^\nu}=\partial_\nu L+\partial_\nu(\partial_\mu X^\mu)$$ Equating the 2 expressions we get:
$$\Big(\frac{\partial(\partial_\mu X^\mu)}{\partial \phi}-\partial_\mu\frac{\partial (\partial_\mu X^\mu)}{\partial (\partial_\mu \phi)}\Big)\partial_\nu\phi+\partial_\mu\Big(\frac{\partial L}{\partial(\partial_\mu\phi)}\partial_\nu\phi+\frac{\partial (\partial_\mu X^\mu)}{\partial(\partial_\mu\phi)}\partial_\nu\phi\Big)=\partial_\nu L+\partial_\nu(\partial_\mu X^\mu)$$ Now from the old stress energy tensor we have $$\frac{\partial L}{\partial(\partial_\mu\phi)}\partial_\nu\phi-g_{\mu\nu}L$$ so the part I need to write in the form $$\partial_\mu (M_{\mu\nu})$$ is what remains from there: $$\Big(\frac{\partial(\partial_\alpha X^\alpha)}{\partial \phi}-\partial_\mu\frac{\partial (\partial_\alpha X^\alpha)}{\partial (\partial_\mu \phi)}\Big)\partial_\nu\phi+\partial_\mu\Big(\frac{\partial (\partial_\alpha X^\alpha)}{\partial(\partial_\mu\phi)}\partial_\nu\phi\Big)=\partial_\nu(\partial_\alpha X^\alpha)$$ I replaced the index for the summation for X with $\alpha$. Based on what I was told my answer should look like this $$T_{\mu\nu} \to T_{\mu\nu} +\partial_\alpha\Big( \frac{\partial X^\alpha}{\partial(\partial_\mu \phi)}\partial\nu \phi - \frac{\partial X^\mu}{\partial(\partial_\alpha \phi)}\partial_\nu \phi \Big)$$. If we apply the product rule in the second term in our equation, we get rid of one more term and we end up with: $$\Big(\frac{\partial(\partial_\alpha X^\alpha)}{\partial \phi}-\partial_\mu\frac{\partial (\partial_\alpha X^\alpha)}{\partial (\partial_\mu \phi)}\Big)\partial_\nu\phi+\partial_\mu\Big(\frac{\partial (\partial_\alpha X^\alpha)}{\partial(\partial_\mu\phi)}\Big)\partial_\nu\phi+\frac{\partial (\partial_\alpha X^\alpha)}{\partial(\partial_\mu\phi)}\partial_\mu\partial_\nu\phi=\partial_\nu(\partial_\alpha X^\alpha)$$ $$\frac{\partial(\partial_\alpha X^\alpha)}{\partial \phi}\partial_\nu\phi+\frac{\partial (\partial_\alpha X^\alpha)}{\partial(\partial_\mu\phi)}\partial_\mu\partial_\nu\phi=\partial_\nu(\partial_\alpha X^\alpha)$$ Now I am not sure what to do. Is it ok so far? Any suggestions now? Thank you again for help!

4. Sep 16, 2018

### Orodruin

Staff Emeritus
You really need to stop thinking of the field $\phi$ altogether and start thinking more in terms of how the action changes when you do the $\delta y^\mu$ transformation and how that affects $\delta L$.

Also note that some of your terms have four $\mu$ indices. That is never a good sign.

5. Sep 16, 2018

### BillKet

I am a bit confused, don't I need the $\phi$, as what I am doing is like a chain rule?

6. Sep 16, 2018

### Orodruin

Staff Emeritus
No, it is irrelevant what is inside the $X^\mu$ as your transformation of the fields is $\delta \phi = 0$, all changes in the action will be expressible in terms of $X^\mu$ and its derivatives.

7. Sep 16, 2018

### BillKet

I am not sure why the change in $\phi$ is zero. When you derive E-L equations, you actually need $\delta \phi$ to be non-zero.

8. Sep 16, 2018

### Orodruin

Staff Emeritus
I am not talking about the equations of motion, I am talking about the 1-parameter symmetry related to the conserved current that is the energy-momentum tensor contracted with the spacetime translation vector $k^\nu$, i.e., $J^\mu = T^\mu_\nu k^\nu$.

9. Sep 16, 2018

### BillKet

Ok, so for the change in action we have: $$\frac{\delta S}{\delta y^\mu}=\int(\frac{\delta L}{\delta y^\mu}+\frac{\delta \partial_\alpha X^\alpha }{\delta y^\mu})$$ And the first term gives the old result. Is this ok so far?

10. Sep 17, 2018

### BillKet

Any further hint would be greatly appreciated

11. Sep 19, 2018

### Orodruin

Staff Emeritus
Did you get any constraints on the form of $X^\mu$? I suggest that you do the case where $X^\mu$ is a function of the fields only (and not derivatives of the fields). If $X^\mu$ depends on the field derivatives, then $\partial_\mu X^\mu$ will depend on the second derivatives of the fields.