I am trying to answer problem 3.3 part b in :     (b) Show that the total energy $Q=\int \mathcal{T}^{00} d^3 x$ is invariant under the addition of a total derivative to the Lagrangian, $\mathcal{L}\rightarrow \mathcal{L}+\partial_{\mu} X^{\mu}$.   When I try to show this, I find that it is in general not invariant, meaning I am making an error somewhere. Here is my attempt:  The (canonical) energy-momentum tensor $\mathcal{T}^{\mu \nu}$ is defined by:  $$\mathcal{T}^{\mu \nu}\equiv \sum_{i} \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \phi_i)}\partial^{\nu}\phi_i - \eta^{\mu\nu}\mathcal{L}$$  Since a total derivative $\partial_{\mu}X^{\mu}$ doesn't change the action, and $\mathcal{T}^{\mu \nu}$ is defined through the action, the new energy-momentum tensor will simply be the old one with $\mathcal{L}$ replaced with $\mathcal{L}+\partial_{\mu}X^{\mu}$.  $$\begin{align} \mathcal{T'}^{\mu \nu}&amp;=\mathcal{T}^{\mu \nu}+ \sum_{i} \frac{\partial \left(\partial_{\rho}X^{\rho}\right)}{\partial (\partial_{\mu} \phi_i)}\partial^{\nu}\phi_i - \eta^{\mu\nu}\left(\partial_{\rho}X^{\rho}\right)\\ &amp;=\mathcal{T}^{\mu \nu}+  \sum_{i} \partial_{\rho}\left(\frac{\partial X^{\rho}}{\partial (\partial_{\mu} \phi_i)}\right)\partial^{\nu}\phi_i - \eta^{\mu\nu}\partial_{\rho}X^{\rho}\\ \end{align}$$  The new energy $Q'$ is given by:  $$\begin{align} Q'&amp;=\int d^3 x \mathcal{T}'^{00}\\ &amp;=Q+\int d^3 x \, \left[ \sum_{i} \partial_{\rho}\left(\frac{\partial X^{\rho}}{\partial \dot{\phi}_i}\right)\dot{\phi}_i -\partial_{\rho} X^{\rho}\right] \end{align}$$  So I need to show that the second term vanishes, but I don't know how. That's a 4-derivative sitting inside the integral, so if it was $d^4 x$ I'd just use the divergence theorem and say "oh well assuming $X^{\rho}$ vanishes at infinity we get our desired result". But it's an integral over $d^3 x$.   What do I do?
There is a problem 3.3 in :     Ambiguities in the energy-momentum tensor:      (a) If you add a total derivative to Lagrangian ${\cal{L}} \rightarrow {\cal{L}} + \partial_\mu X^\mu$, how does the energy-momentum tensor change?      (b) Show that the total energy $Q = \int d^3x \; {\cal{T^{00}}}$ is invariant under such changes   Given the definition of energy-momentum tensor:  $$ {\cal{T}}^\mu{}_\nu = \sum_n \frac{\partial{\cal{L}}}{\partial \left( \partial_\mu \phi_n \right)} \partial_\nu \phi_n - g_{\mu\nu} {\cal{L}} $$  It changes as:  $$ \delta {\cal{T}}^\mu{}_\nu =  \sum_n \left( \partial_\rho \frac{\partial{X^\rho}}{\partial \left( \partial_\mu \phi_n \right)} \right) \partial_\nu \phi_n - g_{\mu\nu} {\partial_\rho X^\rho} $$  $T^{00}$ changes as:  $$ \delta {\cal{T}}^{00}= \sum_n \left( \partial_\rho \frac{\partial{X^\rho}}{\partial \dot{\phi_n}} \right) \dot{\phi_n} - \partial_\rho X^\rho $$   From this point I’m stuck. The first term doesn’t seem to disappear at all after integration over space, and the second disappears only partially:  $$ \int d^3x \; \dot{X^0} $$  So even if I’d have assumed that $X$ depends on position, and not on the fields - the total energy doesn’t seem to be invariant under it.  What is the best way to proceed forward with this?
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