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Solving the Integral present in Dirac's quantization of charge problem

  1. Oct 16, 2011 #1
    1. The problem statement, all variables and given/known data
    This problem involves an electric monopole placed at the origin and a magnetic monopole placed a distance D away, say arbitrarily, along the z-axis.

    I need to compute the angular momentum of the EM fieds:
    [itex]\vec{J_{field}} = \frac{1}{4\pi c}\int{d\tau \vec{r} \times (\vec{E}\times \vec{B})}[/itex]

    The integral is over all space. After trudging along, I am left with the following integral:

    [itex]\int_0^\infty{\frac{r dr}{[r^2 +D^2 -2Dru]^{3/2}}}[/itex]

    If I can solve this last integral, I am home free.

    2. Relevant equations

    All terms in the integral, other than r, are constants with respect to r.

    3. The attempt at a solution
    I tried using integration by parts but was not able to get anywhere. "u" substitution taking the argument of the cube root in the denominator as "u" doesn't seem productive, either.

    Unless my above two attempts were done wrong, I must say I am out of ideas.

    Can anyone suggest anything?
    Last edited: Oct 16, 2011
  2. jcsd
  3. Oct 16, 2011 #2


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    I would try writing the denominator as

    [tex]r^2-2rDu+D^2u^2 -D^2u^2 +D^2=(r-Du)^2 + D^2(1-u^2)[/tex]

    then try letting v = r - Du, dv = dr and see what that gets you.
  4. Oct 16, 2011 #3
    Hi LCKurtz,

    I can't seem to get much further with that approach as I'm left with a [tex](v+Du)dv[/tex] in the numerator and can't see where to go from there.
  5. Oct 17, 2011 #4


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    And so the integrand is

    [tex]\frac{v+Du}{(v^2 + D^2(1-u^2))^\frac 3 2}[/tex]

    right? Certainly you can break that into two fractions and do the v part of the numerator with a substitution. Doing the other part it probably makes a difference if 1-u2 is positive or negative. If it is positive that part has the form

    [tex]\int \frac 1 {(v^2+k^2)^{\frac 3 2}}[/tex]

    and I think a trig substitution would work. And if 1-u2 is negative a different trig or maybe a hyperbolic trig substitution looks like it should work. Did you try those?
  6. Oct 17, 2011 #5

    That does it.

    More specifically, [itex] 1-u^2 \geq 0 [/itex], since, for my problem, [itex] u \equiv cos(\theta) [/itex].

    Thanks, LCKurtz.
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