Physics Help Forum Stress energy tensor transformation

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 Sep 17th 2018, 01:39 PM #1 Junior Member   Join Date: Sep 2018 Posts: 5 Stress energy tensor transformation 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 by BillKet; Sep 17th 2018 at 03:13 PM.
Sep 17th 2018, 03:29 PM   #2

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 Originally Posted by BillKet 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
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 Sep 17th 2018, 06:48 PM #3 Junior Member   Join Date: Sep 2018 Posts: 5 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.
 Sep 17th 2018, 07:33 PM #4 Forum Admin     Join Date: Apr 2008 Location: On the dance floor, baby! Posts: 2,554 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 __________________ Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup. See the forum rules here.
 Sep 18th 2018, 08:06 AM #5 Junior Member   Join Date: Sep 2018 Posts: 5 Thank you so much! I would really appreciate if you can help me with this.
 Sep 18th 2018, 07:54 PM #6 Junior Member   Join Date: Sep 2018 Posts: 5 Hey! Any update on this?
 Sep 18th 2018, 08:02 PM #7 Forum Admin     Join Date: Apr 2008 Location: On the dance floor, baby! Posts: 2,554 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 __________________ Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup. See the forum rules here.
 Sep 18th 2018, 09:17 PM #8 Junior Member   Join Date: Sep 2018 Posts: 5 I found something related to the Belinfante tensor, but I am not sure how to get to that (if that is the solution)

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