Problem 3.3 b) of Schwartz's Quantum Field Theory

Question

In problem 3.3 b) of Schwartz's Quantum field theory you are asked to prove that $ Q = \int T_{00} d^3x $ is invariant under changing the Lagrangian $ \mathcal{L} \rightarrow \mathcal{L} + \partial_\mu X^\mu $ by a total derivative. In 3.3 a) I already wrote down the transformation of the stress-energy tensor $$ T_{\mu\nu} \rightarrow T_{\mu\nu} + \frac{\partial \left( \partial_\lambda X^\lambda \right)}{\partial \left( \partial_\mu \phi \right)} \partial_\nu \phi - g_{\mu\nu} \partial_\lambda X^\lambda $$

Plugging this into the formula for $Q$ you get $$ \delta Q = \int \frac{\partial \left(\partial_\lambda X^\lambda \right)}{\partial \dot{\phi}} \dot{\phi} - \partial_\lambda X^\lambda \, d^3x $$

From here I don't know how to proceed to show that $\delta Q = 0$. You could use the product rule with repsect to $\partial_t$ to then utilize the Euler-Lagrange equations to get $$ \delta Q= \int d^3x \, \partial_t \left(\frac{\partial \left(\partial_\lambda X^\lambda\right)}{\partial \dot{\phi}} \phi\right) + \phi \left(\frac{\partial \left(\partial_\lambda X^\lambda\right)}{\partial \phi}- \partial_i \frac{\partial \left(\partial_\lambda X^\lambda\right)}{\partial \left(\partial_i \phi\right)}\right) -\partial_\lambda X^\lambda,$$ but I don't see how this has helped me.

The solutions available online by To Chin Yu reads $$ \delta Q = \int \frac{\partial \left(\partial_\lambda X^\lambda \right)}{\partial \dot{\phi}} \dot{\phi} - \partial_\lambda X^\lambda \, d^3x \\ = -2 \int \partial_\lambda X^\lambda d^3x \\ = - 2 \partial_0 \int X^0d^3x \\ =0. $$ I understand the step from the second to the third equation where we have used $\int \partial_i X^i dx_i = X^i|_{\partial M} =0$ (no Einstein summation implied). But I don't see why the first line should be equal to the second line and why the third line should be zero.

If you can help me I would be most grateful :)

Answer 1

To get from the first line to the second, you integrate by parts underneath the integral sign. This is a classic technique used in many classical field theory problems. You would have had to do it already in (3.1) and (3.2), as I see it. You can always drop out the boundary term gained from the IBP, allowing you to change

\begin{equation} A \partial_{\mu} B = - (\partial_{\mu} A) B \end{equation} inside Lagrangians. You do the same here. Identify

\begin{equation} B =\partial_{\lambda} X^{\lambda}, \, \, A = \dot{\phi}, \end{equation} and make the switch (remembering that you are differentiating w.r.t $\dot{\phi}$), giving

\begin{align} &- \frac{\partial \dot{\phi}}{\partial \dot{\phi}} \partial_{\lambda} X^{\lambda} \\ &= - \partial_{\lambda} X^{\lambda} \end{align} This obviously combines with the other term, giving the factor of $-2$.