I Struggling with vector calculus identity used in E&M derivation

[mhirschb]

TL;DR

I don't recognize this vector calculus identity

This one comes from the Landau & Lifschitz series, where they love to make you figure out what they did :).

This is from Chapter 2, section 6 of the Electrodynamics in Continuous Media volume. In showing that the polarization vector is the dipole moment per unit volume of a dielectric, we get this equation:

$$ -\int\vec{r}(\nabla\cdot\vec{P}) dV = - \oint \vec{r} (d\vec{f} \cdot \vec{P}) + \int (\vec{P} \cdot \nabla) \vec{r} dV $$

Off the top of my head I think there's some integration by parts happening and the surface integral comes from an application of Gauss' Divergence theorem. Otherwise it's some vector calculus identity I don't recognize.

I appreciate any help!

 

[ahmadphy]

I think the last term is equal to the integral of the polarization over the volume am I right?

 

[julian]

Consider $\vec{\nabla} \cdot (x \vec{P})$. Suppose $\Omega$ is a volume with a boundary $\Gamma = \partial \Omega$. Integrating over $\Omega$ with respect to the volume $dV$, and applying the divergence theorem, gives:

\begin{align*}
\int_\Gamma x \vec{P} \cdot \vec{n} d S & = \int_\Omega \vec{\nabla} \cdot (x \vec{P}) d V
\nonumber \\
& = \int_\Omega x \vec{\nabla} \cdot \vec{P} d V + \int_\Omega \vec{P} \cdot \vec{\nabla} x d V
\end{align*}

There are similar expressions for $\vec{\nabla} \cdot (y \vec{P})$ and $\vec{\nabla} \cdot (z \vec{P})$.

 

[logos_spermatikos]

Yes, it is is vector integration by parts. Here is resource about it which might be helpful to you: https://webhome.phy.duke.edu/~rgb/Class/Electrodynamics/Electrodynamics/node51.html?

And making use of the fact that $\vec{P} = \vec{r} ( \vec{P} \cdot \nabla )$