Hi there, im 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 [Broken]

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ShayanJ
Gold Member
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.

Meir Achuz
Homework Helper
Gold Member
That equality depends on div J=0 (or d\rho/dt=0).

vanhees71
Gold Member
2019 Award
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