Physics Help Forum Stress energy tensor transformation

 Nuclear and Particle Physics Nuclear and Particle Physics Help Forum

 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

Join Date: Apr 2008
Location: On the dance floor, baby!
Posts: 2,664
 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
__________________
Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup.

See the forum rules here.

 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: 2,664 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: 2,664 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)

 Thread Tools Display Modes Linear Mode

 Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post Razi Kinematics and Dynamics 7 May 17th 2017 12:44 PM laguna Special and General Relativity 0 Oct 1st 2016 01:10 PM thinhnghiem Light and Optics 1 Aug 30th 2016 09:51 AM jcocker Energy and Work 6 May 2nd 2016 09:55 AM botee Equilibrium and Elasticity 4 Feb 1st 2013 09:11 PM