Current density and theorem of curl of curl

[mingshey]

Jackson("Classical Electrodynamics", Ch.6)
uses the theorem of curl of curl to separate current density into transverse and parallel,
\vec J = \vec{J_p}+\vec{J_t} to say,

\begin{align*}\vec{J}(\vec{x}) &amp;= \int\vec{J}(\vec{x&#039;})\delta(\vec{x}-\vec{x&#039;})d^{3}x&#039;\\<br /> &amp;= -{1\over{4\pi}}\int\vec{J}(\vec{x&#039;})\nabla^2 \left({1\over|\vec{x}-\vec{x&#039;}|}\right)d^{3}x&#039;<br /> \end{align*}
Since the del is about x and independent of the integral variable,
\begin{align*}{}&amp;=-{1\over{4\pi}}\nabla^2\int{\vec{J}(\vec{x&#039;})<br /> \over|\vec{x}-\vec{x&#039;}|}d^{3}x&#039;<br /> \end{align*}
And using the theorem
\nabla\times(\nabla\times\vec{A})=\nabla(\nabla\cdot\vec{A})-\nabla^2\vec{A}

\begin{align*}\vec{J}(\vec{x}) &amp;= <br /> {1\over{4\pi}}\nabla\times\nabla\times\int{\vec{J}(\vec{x&#039;})<br /> \over|\vec{x}-\vec{x&#039;}|}d^{3}x&#039;-{1\over{4\pi}}\nabla\left(\nabla\cdot\int{\vec{J}(\vec{x&#039;})<br /> \over|\vec{x}-\vec{x&#039;}|}d^{3}x&#039;\right)\end{align*}
But here Jackson take some hidden procedure to get from the second term of ther right side
-{1\over{4\pi}}\nabla\left(\int{\nabla&#039;\cdot\vec{J}(\vec{x&#039;})<br /> \over|\vec{x}-\vec{x&#039;}|}d^{3}x&#039;\right)={1\over{4\pi}}\nabla\left(\int{\partial\rho(\vec{x&#039;})/\partial t<br /> \over|\vec{x}-\vec{x&#039;}|}d^{3}x&#039;\right)
to use the continuity theorem to get the term about a time derivative of charge density at x&#039;.

And I cannot see how is the differential about x changed into a differential about x&#039; and got inside the integral, and is only applied to the current density, but not the denominator.

Can somebody explain it for me? Thank you.

 

[MalleusScientiarum]

He used the conservation of charge theorem thingy:

\frac{\partial \rho}{\partial t} = - \nabla \cdot \vec{J}

 

[George Jones]

I'll give in some hints in words. If the magic behind Jackson's sleight of hand still remains elusive, I'll give some of the mathematical details.

1) Move the unprimed divergence into the integral.

2) Explicitly take the unprimed divergence of the integrand, so that no derivative symbols remain under the integral.

3) Note the symmetry between x and x', and also note that the product rule for primed coordinates gives that the integrand is the required final expression plus/minus a total primed divergence.

4) Use the divergence theorem (for primed coordinates) to turn this total divergence into a surface integral, and argue that the surface integral vanishes at infinity, and thus can be neglected.

Regards,
George

 

[mingshey]

Thanks

Thanks, George.
I couldn't think of number 4).

Thanks to you Malleus, too, for your hand of help. It was the easy part but I diffused the point there. Sorry. ;)