Stress energy tensor transformation

Sep 2018
6
0
Show that if you add a total derivative to the Lagrangian density \(\displaystyle L \to L + \partial_\mu X^\mu\), the energy momentum tensor changes as \(\displaystyle T^{\mu\nu} \to T^{\mu\nu}+\partial_\alpha B^{\alpha\mu\nu}\) with\(\displaystyle B^{\alpha\mu\nu}=-B^{\mu\alpha\nu}\). (Note: the Lagrangian can depend on higher order derivatives of the field)

The attempt at a solution:

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

topsquark

Forum Staff
Apr 2008
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On the dance floor, baby!
Show that if you add a total derivative to the Lagrangian density \(\displaystyle L \to L + \partial_\mu X^\mu\), the energy momentum tensor changes as \(\displaystyle T^{\mu\nu} \to T^{\mu\nu}+\partial_\alpha B^{\alpha\mu\nu}\) with\(\displaystyle B^{\alpha\mu\nu}=-B^{\mu\alpha\nu}\). (Note: the Lagrangian can depend on higher order derivatives of the field)

The attempt at a solution:

So we have \(\displaystyle T_{\mu\nu}=\frac{\partial L}{\partial(\partial_\mu \phi)}\partial_\nu \phi-g_{\mu\nu}L\), where \(\displaystyle \phi\) is the field that the Lagrangian depends on. If we do the given change on the Lagrangian, the change in \(\displaystyle T_{\mu\nu}\) would be \(\displaystyle \frac{\partial (\partial_\alpha X^\alpha)}{\partial(\partial_\mu \phi)}\partial_\nu \phi-g_{\mu\nu}\partial_\alpha X^\alpha =\partial_\alpha \frac{\partial X^\alpha}{\partial(\partial_\mu \phi)}\partial_\nu \phi-g_{\mu\nu}\partial_\alpha X^\alpha\). From here I thought of using this: \(\displaystyle g_{\mu\nu}\partial_\alpha X^\alpha=g_{\mu\nu}\partial_\alpha \phi \frac{\partial X^\alpha}{\partial \phi}\) But I don't really know what to do from here. Mainly I don't know how to get rid of that \(\displaystyle g_{\mu\nu}\). Can someone help me?
Since you are posting this in Particle Physics I presume we can take \(\displaystyle g_{\mu \nu}\) to be the Minkowski metric.

For the last term on the RHS I don't see why you can't just use
\(\displaystyle g_{\mu \nu} \partial _{\alpha } X^{\alpha } = \partial _{\alpha } \left ( g_{ \mu \nu } X^{\alpha } \right )\)

and factor out the \(\displaystyle \partial _{\alpha }\) to get
\(\displaystyle \partial _{\alpha } \frac{ \partial X^{\alpha } }{\partial ( \partial _{\mu } \phi ) } \partial _{\nu } \phi - g_{\mu \nu } \partial _{\alpha }X^{\alpha } = \partial _{\alpha } \left ( \frac{ \partial X^{\alpha } }{\partial ( \partial _{\mu } \phi ) } \partial _{\nu }\phi - g_{\mu \nu } X^{\alpha } \right )\)

so we get
\(\displaystyle B^{\alpha }_{~~ \mu \nu } = \frac{ \partial X^{\alpha } }{\partial ( \partial _{\mu } \phi ) } \partial _{\nu } \phi - g_{\mu \nu } X^{\alpha }\)

-Dan
 
Sep 2018
6
0
Since you are posting this in Particle Physics I presume we can take \(\displaystyle g_{\mu \nu}\) to be the Minkowski metric.

For the last term on the RHS I don't see why you can't just use
\(\displaystyle g_{\mu \nu} \partial _{\alpha } X^{\alpha } = \partial _{\alpha } \left ( g_{ \mu \nu } X^{\alpha } \right )\)

and factor out the \(\displaystyle \partial _{\alpha }\) to get
\(\displaystyle \partial _{\alpha } \frac{ \partial X^{\alpha } }{\partial ( \partial _{\mu } \phi ) } \partial _{\nu } \phi - g_{\mu \nu } \partial _{\alpha }X^{\alpha } = \partial _{\alpha } \left ( \frac{ \partial X^{\alpha } }{\partial ( \partial _{\mu } \phi ) } \partial _{\nu }\phi - g_{\mu \nu } X^{\alpha } \right )\)

so we get
\(\displaystyle B^{\alpha }_{~~ \mu \nu } = \frac{ \partial X^{\alpha } }{\partial ( \partial _{\mu } \phi ) } \partial _{\nu } \phi - g_{\mu \nu } X^{\alpha }\)

-Dan
Thank you for your reply! I am not sure if what I did is actually correct (I think I reached your answer, too), but that doesn't satisfy the required condition \(\displaystyle B_{\alpha\mu\nu}=-B_{\mu\alpha\nu}\) I found this question here too, but the solution involved some assumptions which I am explicitly told not to make. However in the question they put the desired answer for my tensor which should look like this: \(\displaystyle B_{\alpha\mu\nu} = \frac{\partial X^\alpha}{\partial(\partial_\mu\phi)}\partial_\nu \phi-\frac{\partial X^\mu}{\partial(\partial_\alpha\phi)}\partial_\nu \phi\) and which indeed satisfies the condition. But I am not sure how to reach that.
 
Sep 2018
6
0
Thank you for your reply. I am actually not sure that what I did so far is right, but I think I reached your result at a point. However that from doesn't satisfy the required condition \(\displaystyle B_{\alpha\mu\nu}=-B_{\mu\alpha\nu}\). I found a similar question here, but the solution involves an assumption that I am explicitly told not to make. However in the question they have the expected form for \(\displaystyle B_{\alpha\mu\nu}\). I just don't know how to get there.
 

topsquark

Forum Staff
Apr 2008
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651
On the dance floor, baby!
I know I've seen this derivation in one of my references. The "baby" form of it is non-relativistic and deals with a time derivative. But I just can't seem to find the relativistic version. (It has something to do with Noether's theorem.)

I'm giving up for the night but I'll get back to it tomorrow.

-Dan
 
Sep 2018
6
0
Thank you so much! I would really appreciate if you can help me with this.
 

topsquark

Forum Staff
Apr 2008
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651
On the dance floor, baby!
Sorry, I'm having one of my Hell days.

I have decided to approach the problem essentially from scratch. The stress-energy tensor is derived form Noether's theorem aka conservation of momentum. I plan to start the whole thing over with the added total derivative and derive the stress-energy tensor with the added term. My other thought (which is unlikely, but maybe worth looking into) is that we are actually dealing with the the Belinfante tensor. As I said, it's unlikely because you didn't define it to be that, but it's easy enough to double check.

Sorry for the delay.

-Dan
 
Sep 2018
6
0
I found something related to the Belinfante tensor, but I am not sure how to get to that (if that is the solution)
 

topsquark

Forum Staff
Apr 2008
3,055
651
On the dance floor, baby!
I recalled what I was trying to come with earlier. First I was mistaken about the total derivative: The action is invariant to a change in total derivative, not the Lagrangian. Second: Your proposed solution looks like two stress energy tensors that have been subtracted. Since \(\displaystyle O^{\mu \nu}\) depends on \(\displaystyle T^{\mu \nu} - T^{\nu \mu}\) I thought they might be related. However upon working a couple of steps out it looks like they have little to do with each other.

I went through the whole derivation hoping that I could find some sort of extra condition on X but I found nothing that I could use. (Since the action is invariant I thought I could perhaps come up with another equation X would require and use that to work out a new form for \(\displaystyle T^{\mu \nu}\) but the idea fell through.)

I am left with the following questions for you:
1) I don't understand the physical meaning behind the \(\displaystyle \mu \leftrightarrow \alpha\). It just doesn't make any sense to me. Two indices are space-time indices on the stress energy tensor but the other (\(\displaystyle \alpha\)) is an internal index and has nothing to do with these.

2) Your proposed answer leaves me thinking that we might need to introduce yet another total derivative to get the answer to that form. Which makes no sense because we could just absorb that into the original total derivative. But a second total derivative may give us a condition to remove that pesky metric term.

3) We really need to fix the indices. We've both been a little sloppy. The original problem statement was that the change in the stress energy tensor is denoted \(\displaystyle B^{ \alpha \mu \nu}\) so the equation should read:
\(\displaystyle B^{\alpha\mu\nu} = \frac{\partial X^\alpha}{\partial(\partial_\mu\phi)}\partial ^\nu \phi-\frac{\partial X^\mu}{\partial(\partial_\alpha\phi)}\partial^\nu \phi\)

And finally
4) Are there any conditions on X you haven't mentioned? The outcome looks suspiciously like a Lorentz symmetry. I know it isn't but I'm thinking there has to be some kind of condition on it to put it in the form of your proposed solution. Like, for example, that X can be considered to be dependent on a field with a specific construction.

Sorry, I was hoping to help out more.

-Dan