- #1
andrew1982
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Homework Statement
A localized electric charge distribution produces an electrostatic field,
<br /> {\bf E}=-\nabla \phi <br />
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
<br /> {\bf P_{field}}=\frac{1}{c^2}\int \phi {\bf J} d^3x <br />
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
<br /> {\bf P_{field}}=\frac{1}{c^2}{\bf E(0)\times m}<br />
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):
<br /> {\bf P_{field}}=\mu_0 \epsilon_0 \int {\bf E \times H} d^3x<br />
(5.54):
<br /> {\bf m}=\frac{1}{2} \int {\bf x' \times J(x')} d^3x'<br />
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
<br /> \phi=\phi(0)+\nabla \phi(0)\cdot{\bf x}<br />
which replaced in the integral for P_field from a) gives
<br /> {\bf P_{field}}=-\frac{1}{c^2} \int {\bf (E(0)\cdot x) J)} d^3x<br />
if I choose the potential to zero at the origin. Further, using
<br /> {\bf a\times (b\times c)=(a\cdot c) b-(a\cdot b)c}<br />
on the integrand I get
<br /> {\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)<br />
The first integral is as far I can see
<br /> \frac{2}{c^2} {\bf E(0)\times m}<br />
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!
