Register to reply

Jackson Electrodynamics problem 6.5b

by andrew1982
Tags: electrodynamics, jackson
Share this thread:
andrew1982
#1
Mar19-07, 08:20 AM
P: 8
1. The problem statement, all variables and given/known data
A localized electric charge distribution produces an electrostatic field,
[tex]
{\bf E}=-\nabla \phi
[/tex]
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
[tex]
{\bf P_{field}}=\frac{1}{c^2}\int \phi {\bf J} d^3x
[/tex]

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
[tex]
{\bf P_{field}}=\frac{1}{c^2}{\bf E(0)\times m}
[/tex]
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):
[tex]
{\bf P_{field}}=\mu_0 \epsilon_0 \int {\bf E \times H} d^3x
[/tex]
(5.54):
[tex]
{\bf m}=\frac{1}{2} \int {\bf x' \times J(x')} d^3x'
[/tex]

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
[tex]
\phi=\phi(0)+\nabla \phi(0)\cdot{\bf x}
[/tex]
which replaced in the integral for P_field from a) gives
[tex]
{\bf P_{field}}=-\frac{1}{c^2} \int {\bf (E(0)\cdot x) J)} d^3x
[/tex]
if I choose the potential to zero at the origin. Further, using
[tex]
{\bf a\times (b\times c)=(a\cdot c) b-(a\cdot b)c}
[/tex]
on the integrand I get
[tex]
{\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)
[/tex]
The first integral is as far I can see
[tex]
\frac{2}{c^2} {\bf E(0)\times m}
[/tex]
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!
Phys.Org News Partner Science news on Phys.org
Experts defend operational earthquake forecasting, counter critiques
EU urged to convert TV frequencies to mobile broadband
Sierra Nevada freshwater runoff could drop 26 percent by 2100
Meir Achuz
#2
Mar19-07, 09:00 AM
Sci Advisor
HW Helper
PF Gold
P: 2,015
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
#3
Mar19-07, 09:52 AM
P: 80
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
#4
Mar19-07, 10:46 AM
P: 8
Jackson Electrodynamics problem 6.5b

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
[tex]
{\bf P_{field}}=-\frac{1}{c^2} \int {\bf (E(0)\cdot x) J)} d^3x
[/tex]
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
#5
Apr27-11, 12:20 AM
P: 1
How about the part C


Register to reply

Related Discussions
Jackson Electrodynamics problem 9.8a Advanced Physics Homework 2
Jackson Electrodynamics Problem 2.3 Advanced Physics Homework 4
Jackson Electrodynamics Academic Guidance 10
? is to QM as Jackson is to electrodynamics. General Physics 3
Need help with EM/Jackson problem Classical Physics 2