Jackson Electrodynamics problem 6.5b

[andrew1982]

Homework Statement

A localized electric charge distribution produces an electrostatic field,
$${\bf E}=-\nabla \phi$$
Into this field is placed a small localized time-independent current density J(x) which generates a magnetic field H.
a) show that the momentum of these electromagnetic fields, (6.117), can be transformed to
$${\bf P_{field}}=\frac{1}{c^2}\int \phi {\bf J} d^3x$$

b) Assuming that the current distribution is localized to a region small compared to the scale of variation of the electric field, expand the electrostatic potential in a Taylor series and show that
$${\bf P_{field}}=\frac{1}{c^2}{\bf E(0)\times m}$$
where E(0) is the electric field at the current distribution and m is the magnetic moment (5.54), caused by the current.

Homework Equations

(6.117):
$${\bf P_{field}}=\mu_0 \epsilon_0 \int {\bf E \times H} d^3x$$
(5.54):
$${\bf m}=\frac{1}{2} \int {\bf x' \times J(x')} d^3x'$$

The Attempt at a Solution

Part a) was straight forward: subsituting E=- grad phi and integrating by parts gives the answer plus a surface integral that goes to 0 if phi*H goes to 0 faster than 1/r^2.

Part b): This is where I get stuck. I tried to put
$$\phi=\phi(0)+\nabla \phi(0)\cdot{\bf x}$$
which replaced in the integral for P_field from a) gives
$${\bf P_{field}}=-\frac{1}{c^2} \int {\bf (E(0)\cdot x) J)} d^3x$$
if I choose the potential to zero at the origin. Further, using
$${\bf a\times (b\times c)=(a\cdot c) b-(a\cdot b)c}$$
on the integrand I get
$${\bf P_{field}}=\frac{1}{c^2} (\int {\bf E(0)\times (x\times J) }d^3x-\int{\bf (E(0)\cdot J)x}\,d^3x)$$
The first integral is as far I can see
$$\frac{2}{c^2} {\bf E(0)\times m}$$
that is, twice the answer. The second integral gets me stuck. I guess I should show that it is equal to minus half of the answer (if I did everything correctly so far), but I don't see how to do this.

I would appreciate if anyone could give me a hint on how to continue or if I'm on the right track at all. Thanks in advance!

 

[Meir Achuz]

Somewhere else in Jackson (and other texts) it is shown the second integral
equals EXm. It's in Sec. 5.6 of teh 2nd Edition, if you can follow it.

 

[dhris]

If you believe that the second integral is indeed equal to Exm, then why not simply write them both out in components to prove it? Granted it's not really a proper derivation, but you were only asked to show that the formula is true.

 

[andrew1982]

Thanks for your replies, it was very helpful! Using section 5.6 of Jackson as you said I saw that I could show directly that the first integral
$${\bf P_{field}}=-\frac{1}{c^2} \int {\bf (E(0)\cdot x) J)} d^3x$$
is equal to the sought answer without using the abc vector rule. Looking below eq. 5.52 (in the 3rd edition) and substituting x by E(0) the whole derivation is there.

 

[shuangfengde]

How about the part C