Sunday, 5 July 2015

symmetry - Derivation of correction to canonical stress energy tensor due to addition of total divergence to Lagrangian


It is mentioned in almost every text book that equations of motions are not modified if we add a total divergence of some vector $$\partial_\mu \ X^{\mu}$$ to Lagrangian but canonical stress energy tensor:


$$T^{\mu}_{\ \ \ \ \nu} = \frac{\delta {\cal L}}{\delta (\partial_\mu \phi_s)} \partial_\nu \phi_s - \eta^{\mu}_{\ \ \ \nu} \ {\cal L}$$


is modified. However, I was unable to find any text book where a derivation of such correction to canonical stress energy tensor is thoroughly performed. Can anyone, please, either provide a reference where such derivation is made or illustrate it.


The following thread: Symmetrizing the Canonical Energy-Momentum Tensor but no derivation was provided.


One of the comments provided a reference to Bandyopadhyay PhD thesis (http://research.physics.illinois.edu/Publications/theses/copies/Bandyopadhyay/Chapter_3.pdf, end of page 18, paragraph after Eq. 3.9). For simplicity I repeat the statement from there. If we perform a gauge transformation of a Lagrangian


$${\cal{L}}(x) \rightarrow {\cal{L}}(x) + \partial_{\mu} Z^{\mu}(x)$$



where $Z^{\mu}$ is zero at the boundary then treating $Z^{\mu}$ as independent vector field in the Lagrangian results in the following transformation of the canonical stress energy tensor: $${\Theta^{\mu}}_{\nu} \rightarrow {\Theta^{\mu}}_{\nu}+\left( \partial_{\nu}Z^{\mu}-{\delta^{\mu}}_{\nu}\partial_{\alpha}Z^{\alpha}\right)$$


Fine. As an exercise (but with not much physics), let's take a standard EM Lagrangian:


$${\cal L}=-\frac{1}{4}F^{\mu \nu}F_{\mu \nu}$$


and add to it a total divergence constructed from vector potential $A_{\mu}$. We can construct at least 3 such terms with the proper "dimension": $${\cal L}_1=\partial_{\mu} \left(A^{\mu} \partial^{\nu}A_{\nu} \right)$$ $${\cal L}_2=\partial_{\mu} \left(A^{\nu} \partial^{\mu}A_{\nu} \right)$$ $${\cal L}_3=\partial_{\mu} \left(A^{\nu} \partial_{\nu}A^{\mu} \right)$$


and add (any of) them to Lagrangian with some constant multipliers $c_1,c_2,c_3$: $${\cal L} \rightarrow {\cal L} + c_1 {\cal L}_1 + c_2 {\cal L}_2 + c_3 {\cal L}_3$$


As we added total divergence we can integrate it out to infinity using Gauss theorem and provided that the field is zero at the infinity (which is always assumed?) the equation of motions should not change? But we can "forget" about Gauss theorem and just apply Euler Lagrange equations to new Lagrangian. As all of the ${\cal L}_i$ contains the second derivatives of $A_{\mu}$ and unless we use some appropriate values of $c_i$ so that to have the second derivatives cancelled the equation of motions and canonical stress energy tensor will be updated to reflect second derivatives of the field (for example, http://relativity.livingreviews.org/Articles/lrr-2009-4/, paragraph 2.1.1, pp. 11, 12). The additional terms are clearly field dependent and thus above-mentioned gauge transformation approach to find a transformation of stress energy tensor cannot be applied. Or am I missing something?


Back to the original question. The procedure to correct canonical EM stress energy tensor is well-known. Does that mean that we could add a divergence of some field to standard EM Lagrangian so that to obtain corrected (symmetric canonical EM stress energy tensor)? If yes, then what shall we add?




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