I was reading chapter 75 of Srednicki's QFT book and I ran into this statement.

To determine the value of its integral, we make a Wick rotation to euclidean space, which yields a factor of i as usual; then we have \begin{equation} \int \frac{d^{4}l}{(2\pi)^4}\frac{\partial}{\partial l^{\beta}}f_{\alpha}(l)=i\lim_{l\rightarrow\infty}\int\frac{dS_{\beta}}{(2\pi)^{4}}f_{\alpha}(l) \end{equation} where $dS_{\beta}=l^{2}l_{\beta}d\Omega$ is a surface-area element, and $d\Omega$ is the differential solid angle in four dimensions.

I don't understand how the LHS of the equation can be written as the RHS. Especially, how did the derivative $\frac{\partial}{\partial l^{\beta}}$ disappear? Can someone give me a detailed explanation?

  • 1
    $\begingroup$ It seems OP is asking about the 4D version of the divergence theorem. $\endgroup$
    – Qmechanic
    Feb 16, 2019 at 12:46
  • $\begingroup$ Thanks for the comment. OK so he's using the divergence theorem, but how though? If the the form of LHS was \begin{equation} \frac{\partial}{\partial l^{\beta}}f_{\beta}(l) \end{equation} instead of \begin{equation} \frac{\partial}{\partial l^{\beta}}f_{\alpha}(l) \end{equation} then I can use divergence theorem I guess, but in this case can I? $\endgroup$
    – embreakin
    Feb 16, 2019 at 20:31
  • $\begingroup$ $f_\alpha(\ell)$ is not an arbitrary function here, but given by Eq (75.38). So this is not a general statement. Is it correct for the given function? $\endgroup$ Feb 17, 2019 at 12:51
  • $\begingroup$ Yes $f_{\alpha}(l)$ is not an arbitrary function. It is \begin{equation} f_{\alpha}(l)\equiv \frac{l_{\alpha}}{l^{2}(l+p+q)^{2}} \end{equation} to be specific. It must be correct at least for this specific function, otherwise it means that the book is incorrect. $\endgroup$
    – embreakin
    Feb 18, 2019 at 1:55

1 Answer 1


I believe Prof. Srednicki first performed the Wick rotation and then used a version of the 4-dimensional divergence theorem.

Given the comments below the question, people are unaware of the "component-wise" divergence theorem, so I'll give a derivation below starting from the typical expression of the divergence theorem.

We have $$\int dV\: \nabla \cdot \vec A = \int dS\: \hat n \cdot \vec A $$ from the typical divergence theorem in d-dimensions. Now consider an $\vec A = \vec c f$ where $\vec c $ is a constant vector. This yields, using the usual product rule for derivatives,

$$\vec c \cdot \int dV\: \nabla f = \vec c \cdot \int dS\: \hat n f $$

This is true for any $\vec c$, including the unit basis vectors, so we have

$$\int dV\: \nabla f = \int dS\: \hat n f. $$

Note that the above is a vector equation that holds componentwise:

$$\int dV\: \partial_\mu f = \int dS\: n_\mu f. $$

This is true for any $f$, so it is true for e.g. functions $g_0, g_1, ..., g_d$, which we can write as

$$\int dV\: \partial_\mu g_\nu = \int dS\: n_\mu g_\nu. $$

Please let me know if you need any clarifications or if that addressed the heart of your question. I beleive that should answer where the derivative goes on the right hand side of your equation. All the functions above are assumed to be well-enough behaved.

  • $\begingroup$ That explains it very clearly. Thanks a lot! $\endgroup$
    – embreakin
    Jun 8, 2019 at 10:10
  • $\begingroup$ @embreakin If you're satisfied with the answer, please consider hitting the green checkmark under the vote count to mark the answer as accepted. See physics.stackexchange.com/help/accepted-answer for some more details about what it means to accept answers. $\endgroup$
    – user196574
    Jun 20, 2019 at 6:13

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