Billket Posted September 18, 2018 Share Posted September 18, 2018 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? Link to comment Share on other sites More sharing options...
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