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I try to rephrase here a my question (https://math.stackexchange.com/q/4661784/), explaining more specifically the case.

Given a lagrangian $L=L(\theta_{\mu\nu},\phi)$ , the conserved Noether current is \begin{equation}\tag{1} j_{\beta}^{\alpha} \doteq - \dfrac{\partial L^{(2)}}{\partial (\partial_{\alpha}\theta_{\mu\nu})}\partial_{\beta}\theta_{\mu\nu} - \dfrac{\partial L^{(2)}}{\partial (\partial_{\alpha}\phi)}\partial_{\beta}\phi + \delta_{\beta}^{\alpha} L^{(2)} \end{equation}

where $L^{(2)}$ is $L$ at second order in the fields (but this is not important for my question). For a scalar-tensor theory of gravity, one finds something like

\begin{equation}\tag{2} j_{\beta}^{\alpha} = \dfrac{\phi_{0}}{32 \pi G} \left[ - 2 \partial^{\nu} \theta^{\mu \alpha} \partial_{\beta}\theta_{\mu\nu} + \partial^{\alpha} \theta^{\mu\nu}\partial_{\beta}\theta_{\mu\nu} + ... + \delta_{\beta}^{\alpha}\partial_{\nu} \theta_{\mu\gamma} \partial^{\gamma} \theta_{\mu\nu} + ... \right] \end{equation} I'm pretty sure about this result but, again, this is not important to my question. What is important is that the authors of the paper say that they integrate by parts some terms in (2), in order to exploit the gauge conditions and the field equations for $\phi$ and $\theta_{\mu\nu}$, and so to cancel some terms. The first term, for example, becomes $2 \theta^{\mu\alpha}\partial_{\beta}\partial^{\nu} \theta_{\mu\nu}$ and so on. Ok, if I do this, I actually get the same result as them. My question is why this is possible: I can't integrate (by parts) any tensor (or pieces of it) and neglect border terms; it would be like saying that $ln(x)=-1$ (I have integrated by parts the rhs) !

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  • $\begingroup$ Why don't you heed the advice on mathematics.SE to give a concrete example so that we can see your partial integration in action? $\endgroup$
    – kricheli
    Mar 19 at 20:04
  • $\begingroup$ Without further context I would be wary about partial integration in (2), but it may be possible. You can perform partial integration in the action functional, or, equivalently, use so-called null-Lagrangians, cf. my answer here: physics.stackexchange.com/questions/719309/… $\endgroup$
    – kricheli
    Mar 19 at 20:05
  • $\begingroup$ As I said in your previous question, you need to provide further context to make the question meaningful. For example, maybe they were being sloppy and they meant to say they integrate by parts in the action, which produces a different current. Or perhaps they mean they're adding something like Belinfante improvement terms which do affect the Noether currents but do not affect the Noether charges. $\endgroup$
    – knzhou
    Mar 19 at 20:11
  • $\begingroup$ @kricheli I know that one can integrate by parts the lagrangian terms, but $j_{\beta}^{\alpha}$ is not a lagrangian, is just a tensor defined as in (1). $\endgroup$
    – gravitone123
    Mar 19 at 20:20
  • $\begingroup$ @knzhou The context is just the computation of the pseudotensor of scalar-tensor theory of gravity. The paper is arxiv.org/pdf/1710.08863v1.pdf (eq.81). The authors simply start from the action of Brans-Dicke theory, compute the second order lagrangin $L^{(2)}$ and then use eq. (1) above. Nothing else. They need $j_{\beta}^{\alpha}$ to obtain the gravitational pseudotensor $\endgroup$
    – gravitone123
    Mar 19 at 20:24

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