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Understanding Jackson (again!)

  1. Nov 21, 2009 #1
    1. The problem statement, all variables and given/known data
    Jackson Section 5.10, the uniformly magnetized sphere, I'm trying to fill in the steps from his first equation to Equation 5.104. I get the same potential except I am lacking the cos[tex]\theta[/tex] term. My work shown below.

    2. Relevant equations



    3. The attempt at a solution
    [tex]$\begin{array}{l}
    \Phi _M = \frac{{M_0 a^2 }}{{4\pi }}\int {\frac{{\cos \theta '}}{{\left| {\user1{x - x'}} \right|}}d\Omega '} \\
    \Phi _M = \frac{{M_0 a^2 }}{2}\int {\frac{1}{{\left| {\user1{x - x'}} \right|}}\cos \theta 'd\left( {\cos \theta '} \right)} \\
    \Phi _M = \frac{{M_0 a^2 }}{2}\int {\left[ {\sum\limits_{l = 0}^\infty {\frac{{r_ < ^l }}{{r_ > ^{l + 1} }}P_l \left( {\cos \theta } \right)} } \right]P_1 \left( {\cos \theta '} \right)d\left( {\cos \theta '} \right)} \\
    \Phi _M = \frac{{M_0 a^2 }}{2}\int {\frac{{r_ < ^{} }}{{r_ > ^2 }}P_1 \left( {\cos \theta } \right)P_1 \left( {\cos \theta '} \right)d\left( {\cos \theta '} \right)} \\
    \Phi _M = \frac{{M_0 a^2 }}{2}\frac{{r_ < ^{} }}{{r_ > ^2 }}\int {P_1 \left( {\cos \theta } \right)P_1 \left( {\cos \theta '} \right)d\left( {\cos \theta '} \right)} \\
    \Phi _M = \frac{{M_0 a^2 }}{2}\frac{{r_ < ^{} }}{{r_ > ^2 }}\left( {\frac{2}{{2 + 1}}} \right) \\
    \Phi _M = \frac{{M_0 a^2 }}{3}\frac{{r_ < ^{} }}{{r_ > ^2 }} \\
    \end{array}$[/tex]
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 21, 2009 #2

    gabbagabbahey

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    Homework Helper
    Gold Member

    I see two things wrong with your solution:

    (1) You seem to be claiming that

    [tex]\frac{1}{|\textbf{x}-\textbf{x}'|}=\sum\limits_{l = 0}^\infty {\frac{{r_ < ^l }}{{r_ > ^{l + 1} }}P_l \left( {\cos \theta } \right)}[/tex]

    but this is only true if [itex]\theta[/itex]is defined to be the angle between [itex]\textbf{x}[/itex] and [itex]\textbf{x}'[/itex], not the polar angle of [itex]\textbf{x}[/itex]

    (2) You also seem to claim that [itex]P_l(\cos\theta)[/itex] is orthogonal (under the weight function [itex]\sin\theta'[/itex] ) to [itex]P_{l'}(\cos\theta')[/itex], but this is only true if [itex]\theta=\theta'[/itex]
     
  4. Nov 22, 2009 #3
    Thanks; I went through again but with the full spherical harmonics, and paid careful attention to the primed and unprimed values, and got it, but it raises a more general question I would appreciate some help on. I know that azimuthal symmetry allows me to automatically set m=0, and I understand why. I can also demonstrate that l must be 1 for this problem, but is there something (physically or mathmatically) that would allow me to know without further work that l can only =1?
     
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