# Understanding Jackson (again!)

1. Nov 21, 2009

### Old Guy

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$$\theta$$ term. My work shown below.

2. Relevant equations

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

2. Relevant equations

3. The attempt at a solution

2. Nov 21, 2009

### gabbagabbahey

I see two things wrong with your solution:

(1) You seem to be claiming that

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

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

(2) You also seem to claim that $P_l(\cos\theta)$ is orthogonal (under the weight function $\sin\theta'$ ) to $P_{l'}(\cos\theta')$, but this is only true if $\theta=\theta'$

3. Nov 22, 2009

### Old Guy

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?