Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,047 questions , 2,200 unanswered
5,345 answers , 22,709 comments
1,470 users with positive rep
816 active unimported users
More ...

  Symmetrizing the Canonical Energy-Momentum Tensor

+ 6 like - 0 dislike
1645 views

The Canonical energy momentum tensor is given by $$T_{\mu\nu} = \frac{\delta {\cal L}}{\delta (\partial^\mu \phi_s)} \partial_\nu \phi_s - g_{\mu\nu} {\cal L} $$ A priori, there is no reason to believe that the EM tensor above is symmetric. To symmetrize it we do the following trick.

To any EM tensor we can add the following term without changing its divergence and the conserved charges. $${\tilde T}_{\mu\nu} = T_{\mu\nu} + \partial^\beta \chi_{\beta\mu\nu} $$ where $\chi_{\beta\mu\nu} = - \chi_{\mu\beta\nu}$. The antisymmetry of $\chi$ in its $\mu\beta$ indices implies that ${\tilde T}_{\mu\nu}$ conserved. Also, all the conserved charges stay the same.

Now even though $T_{\mu\nu}$ is not a symmetric tensor, it is possible to choose $\chi_{\beta\mu\nu}$ in such a way so as to make ${\tilde T}_{\mu\nu}$ symmetric. It can be shown that choosing

$$\chi_{\lambda\mu\nu} = - \frac{i}{2}\left[ \frac{\delta {\cal L}}{\delta (\partial^\mu \phi_r) } (I_{\nu\lambda})_{rs} \phi_s + \frac{\delta {\cal L}}{\delta (\partial^\lambda \phi_r) } (I_{\mu\nu})_{rs} \phi_s + \frac{\delta {\cal L} }{\delta (\partial^\nu \phi_r) } (I_{\mu\lambda})_{rs} \phi_s \right]$$ makes the new EM tensor symmetric. Here $(I_{\mu\nu})_{rs}$ is the representation of the Lorentz Algebra under which the fields $\phi_s$ transform.

Here's my question - Is it possible to obtain the symmetric EM tensor directly from variational principles by adding a total derivative term to the Lagrangian. In other words, by shifting ${\cal L} \to {\cal L} + \partial_\mu X^\mu$, and choosing $X^\mu$ appropriately, can be exactly get the shift in the EM tensor required, in order to make the EM tensor symmetric?

What I've done so far - It is possible to show that under a shift in the Lagrangian by a total derivative, one shifts the EM tensor by $T_{\mu\nu} \to T_{\mu\nu} + \partial^\lambda \chi_{\lambda\mu\nu}$ where

$$\chi_{\lambda\mu\nu} = \frac{1}{2} \frac{\delta X_\lambda}{\delta (\partial^\mu \phi_r)} \partial_\nu \phi_r - \frac{1}{2} \frac{ \delta X_\mu }{\delta (\partial_\lambda \phi_r)} \partial_\nu \phi_r + X_\mu g_{\lambda\nu} - X_\lambda g_{\mu\nu} $$

What I wish to do next - I now have a differential equation that I wish to solve.

\begin{align} &\frac{1}{2} \frac{\delta X_\lambda}{\delta (\partial^\mu \phi_r)} \partial_\nu \phi_r - \frac{1}{2} \frac{ \delta X_\mu }{\delta (\partial_\lambda \phi_r)} \partial_\nu \phi_r + X_\mu g_{\lambda\nu} - X_\lambda g_{\mu\nu} \\ &~~~~~~= - \frac{i}{2}\left[ \frac{\delta {\cal L}}{\delta (\partial^\mu \phi_r) } (I_{\nu\lambda})_{rs} \phi_s + \frac{\delta {\cal L}}{\delta (\partial^\lambda \phi_r) } (I_{\mu\nu})_{rs} \phi_s + \frac{\delta {\cal L} }{\delta (\partial^\nu \phi_r) } (I_{\mu\lambda})_{rs} \phi_s \right] \end{align}

Any ideas on how to solve this?


This post imported from StackExchange Physics at 2014-12-01 21:54 (UTC), posted by SE-user Prahar

asked Sep 19, 2012 in Theoretical Physics by prahar21 (545 points) [ revision history ]
edited Dec 1, 2014 by Dilaton
Related: physics.stackexchange.com/q/27048/2451 and links therein.

This post imported from StackExchange Physics at 2014-12-01 21:54 (UTC), posted by SE-user Qmechanic
Symmetrizability is equivalent with Lorentz invariance. Thus you need to assume that in your arguments.

This post imported from StackExchange Physics at 2014-12-01 21:54 (UTC), posted by SE-user Arnold Neumaier
Can you explain that a bit more? I don't undestand what you're trying to say. Thanks!

This post imported from StackExchange Physics at 2014-12-01 21:54 (UTC), posted by SE-user Prahar
Without the assumption of Lorentz invariance of the action, there is no symmetric e/m tensor, and the standard recipe fails. Lorentz invariance gives you additional properties that you must exploit in your derivation; otherwise you will not be able to arrive at the conclusion (since it might not hold). - If you reply to a comment, you should mention the name, like in @Prahar, so that the original commenter is informed. I noticed your comment only by chance (and hence very late).

This post imported from StackExchange Physics at 2014-12-01 21:54 (UTC), posted by SE-user Arnold Neumaier

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
$\varnothing\hbar$ysicsOverflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...