billket Posted September 18, 2018 Report Share Posted September 18, 2018 Hello! This is actually from a homework problem: Show that if you add a total derivative to the Lagrangian density \( L \to L + \partial_\mu X^\mu \), the energy momentum tensor changes as \( T^{\mu\nu} \to T^{\mu\nu}+\partial_\alpha B^{\alpha\mu\nu}\) with \( B^{\alpha\mu\nu}=-B^{\mu\alpha\nu}\). The Lagrangian can depend on higher order derivatives of the field. Attempted Solution: So we have \( T_{\mu\nu}=\frac{\partial L}{\partial(\partial_\mu \phi)}\partial_\nu \phi-g_{\mu\nu}L\), where \( \phi\) is the field that the Lagrangian depends on. If we do the given change on the Lagrangian, the change in \( T_{\mu\nu}\) would be \( \frac{\partial (\partial_\alpha X^\alpha)}{\partial(\partial_\mu \phi)}\partial_\nu \phi-g_{\mu\nu}\partial_\alpha X^\alpha =\partial_\alpha \frac{\partial X^\alpha}{\partial(\partial_\mu \phi)}\partial_\nu \phi-g_{\mu\nu}\partial_\alpha X^\alpha\). From here I thought of using this: \( g_{\mu\nu}\partial_\alpha X^\alpha=g_{\mu\nu}\partial_\alpha \phi \frac{\partial X^\alpha}{\partial \phi}\) But I don't really know what to do from here. Mainly I don't know how to get rid of that \(g_{\mu\nu}\). Also I am not sure if what I did so far is correct. Can someone help me? Quote Link to comment Share on other sites More sharing options...
Vmedvil2 Posted September 18, 2018 Report Share Posted September 18, 2018 Just take the derivative until guv disappears to 1. Quote Link to comment Share on other sites More sharing options...
billket Posted September 18, 2018 Author Report Share Posted September 18, 2018 Just take the derivative until guv disappears to 1.I am not sure I understand what you mean. Quote Link to comment Share on other sites More sharing options...
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