andrew1982
Mar19-07, 08:20 AM
1. The problem statement, all variables and given/known data
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.
2. Relevant 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'
3. 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!
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.
2. Relevant 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'
3. 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!