Question about mathematical equality

[Mr. Rho]

Hi there, I am reading Chapter 9 of Jackson Classic Electrodynamics 3rd edition, and I don't see why this equality is true, it says "integrating by parts", but I still don't know... any help?

http://imageshack.com/a/img673/9201/4WYcXs.png

 

[ShayanJ]

Write $\mathbf J$ as the sum of its Cartesian components and then write the integral as the sum of three integrals of the three Cartesian components. Then integrate by parts each integral, but the $J_x$ w.r.t. to x and so on! If you consider the integrals to be from $-\infty$ to $\infty$ and take $\mathbf J$ to be zero at infinities, you'll get what you want.

 

[Mr. Rho]

Thank you Shyan, that's it, I just found this awesome article that explains in great detail https://www.phy.duke.edu/~rgb/Class/phy319/phy319/node14.html

 

[Meir Achuz]

That equality depends on div J=0 (or d\rho/dt=0).

 

[vanhees71]

Just my derivation. For an arbitrary constant vector $\vec{n}$ we have
$$\vec{\nabla} \cdot [(\vec{n} \cdot \vec{x}) \vec{j}]=(\vec{n} \cdot \vec{x}) \vec{\nabla} \cdot \vec{j} + \vec{n} \cdot \vec{j}.$$
Now integrate this over the whole space and assume that $\vec{j}$ goes to 0 quickly enough at infinity (or most realistically that it has compact support). Then the left-hand side vanishes, because it's a divergence, and thus can be transformed to a surface integral, which vanishes at infinity, if $\vec{j}$ vanishes quickly enough. This implies
$$\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \; \vec{n} \cdot \vec{j} = -\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \; (\vec{n} \cdot \vec{x} ) \vec{\nabla} \cdot \vec{j}.$$
Now use for $\vec{n}$ the three basis vectors of a cartesian reference frame, and you see that the vector equation stated by Jackson holds.

It's independent of whether $\vec{j}$ is a solenoidal field or not. Of course in the case of stationary currents it must be one, but the integral identity is generally valid.

 

[Meir Achuz]

Sorry about that. I was thinking cases where j appears with other variables.