I am trying to answer problem 3.3 part b in Schwartz's QFT book:
(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 $\mathcal{L}+\partial_{\mu}X^{\mu}$.
$$\begin{align} \mathcal{T'}^{\mu \nu}&=\mathcal{T}^{\mu \nu}+ \sum_{i} \frac{\partial \left(\partial_{\rho}X^{\rho}\right)}{\partial (\partial_{\mu} \phi_i)}\partial^{\nu}\phi_i - \eta^{\mu\nu}\left(\partial_{\rho}X^{\rho}\right)\\ &=\mathcal{T}^{\mu \nu}+ \sum_{i} \partial_{\rho}\left(\frac{\partial X^{\rho}}{\partial (\partial_{\mu} \phi_i)}\right)\partial^{\nu}\phi_i - \eta^{\mu\nu}\partial_{\rho}X^{\rho}\\ \end{align}$$
The new energy $Q'$ is given by:
$$\begin{align} Q'&=\int d^3 x \mathcal{T}'^{00}\\ &=Q+\int d^3 x \, \left[ \sum_{i} \partial_{\rho}\left(\frac{\partial X^{\rho}}{\partial \dot{\phi}_i}\right)\dot{\phi}_i -\partial_{\rho} X^{\rho}\right] \end{align}$$
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?
