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Jackson Electrodynamics problem 6.5b 
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#1
Mar1907, 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 timeindependent 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! 


#2
Mar1907, 09:00 AM

Sci Advisor
HW Helper
PF Gold
P: 2,026

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. 


#3
Mar1907, 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.



#4
Mar1907, 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. 


#5
Apr2711, 12:20 AM

P: 1

How about the part C



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