The Canonical energy momentum tensor is given by Tμν=δLδ(∂μϕs)∂νϕs−gμνL
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. ˜Tμν=Tμν+∂βχβμν
where
χβμν=−χμβν. The antisymmetry of
χ in its
μβ indices implies that
˜Tμν conserved. Also, all the conserved charges stay the same.
Now even though Tμν is not a symmetric tensor, it is possible to choose χβμν in such a way so as to make ˜Tμν symmetric. It can be shown that choosing
χλμν=−i2[δLδ(∂μϕr)(Iνλ)rsϕs+δLδ(∂λϕr)(Iμν)rsϕs+δLδ(∂νϕr)(Iμλ)rsϕs]
makes the new EM tensor symmetric. Here
(Iμν)rs is the representation of the Lorentz Algebra under which the fields
ϕs 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 L→L+∂μXμ, and choosing Xμ 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 Tμν→Tμν+∂λχλμν where
χλμν=12δXλδ(∂μϕr)∂νϕr−12δXμδ(∂λϕr)∂νϕr+Xμgλν−Xλgμν
What I wish to do next - I now have a differential equation that I wish to solve.
12δXλδ(∂μϕr)∂νϕr−12δXμδ(∂λϕr)∂νϕr+Xμgλν−Xλgμν =−i2[δLδ(∂μϕr)(Iνλ)rsϕs+δLδ(∂λϕr)(Iμν)rsϕs+δLδ(∂νϕr)(Iμλ)rsϕs]
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