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I have a question, raised form kinetic theory (pure mathematical). Imagine, that $\Psi (\overrightarrow{r},\overrightarrow{p},t)$ - sufficiently smooth function, where $\overrightarrow{r}$ - radius vector, $\overrightarrow{p}$ - unit vector of orientation, $t$ - time.

Can someone explain me, how to proove this identities (I think, Stoke's theorem must be used here):

$$ \int_{S} \overrightarrow{p} \left(\nabla_p \cdot \left( \frac {d\overrightarrow{p}} {dt} \Psi\right) \right)d\overrightarrow{p}=-\int_{S} \frac {d\overrightarrow{p}} {dt} \Psi d\overrightarrow{p} $$

$$ \int_{S} \nabla_p \cdot \left( \frac {d\overrightarrow{p}} {dt} \Psi\right) d\overrightarrow{p}=0 $$

Here $S$ is the surface of the unit sphere, so we integrate over a unit sphere.

This formula are used in physical article. I am sure, that you don't need more information for this identities.

Edit: If this is simple Integration by parts, I am interested, what boundary term is, and why it is zero

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    $\begingroup$ hint: integration by parts $\endgroup$ – tired Sep 3 '15 at 19:39
  • $\begingroup$ Thanks, I figured that. Can you explain me, why boundary term disappears? I don't know, what boundary is here $\endgroup$ – Mikhail Genkin Sep 3 '15 at 19:44
  • $\begingroup$ I have an explanation that avoids differential geometry in the appendix of this article: faculty.missouri.edu/~stephen/preprints/pde-sphere.html $\endgroup$ – Stephen Montgomery-Smith Sep 3 '15 at 19:48
  • $\begingroup$ But if you want to use Stoke's Theorem, the clever thing is that on the surface of the sphere, the boundary is the empty set. $\endgroup$ – Stephen Montgomery-Smith Sep 3 '15 at 19:49
  • $\begingroup$ You said that $\vec p$ is a unit vector. Is that correct? And, what is the relevance or $\vec r$ then? $\endgroup$ – Mark Viola Sep 3 '15 at 19:51

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