The sphere has radius $R$ and is missing its "pole" - meaning that in the region $\theta\leq\alpha$ there is nothing. The object has a homogenous charge density $\sigma=\frac{Q}{4\pi R^2}$

I'm trying to derive what the field inside and outside is. It should be a Legendre polynomials excercise. I know this is an axially symetrical problem, so the general solution of the Laplace equation should be: $$\phi(r,\theta)=\sum_{l=0}^{+\infty}\left(A_l r^l+B_l r^{-(l+1)}\right)P_l(\cos\theta)$$ Where $P_l$ are the Legendre polynomials. In my case, I shoud separate the results in two areas: $$\phi(r<R,\theta)=\sum_{l=0}^{+\infty}A_l r^l P_l(\cos\theta)$$ $$\phi(r>R,\theta)=\sum_{l=0}^{+\infty}B_l r^{-(l+1)}P_l(\cos\theta)$$ in order for the potential not to diverge in $r=0$ or $r\rightarrow+\infty$ There's a hint in the book to use the fact, that potential should be continuous and that the difference of derivatives (i.e. the difference in electrical intensity: $[\vec{E}]$) in the direction of the normal is the charge density.

The potential continuity is clear. It gives the following condition: $$\sum_{l=0}^{+\infty}\left(A_l R^l-B_l R^{-(l+1)}\right)P_l(\cos\theta)=0\ \ \ \ \forall\theta\in[0,\pi]$$ which implies $$\frac{A_l}{B_l}=\frac{1}{R^{2l+1}}$$ I'm not sure how to use the electrical intensity condition, because $$\frac{\partial \phi(r\rightarrow R_-,\theta\leq\alpha)}{\partial r}-\frac{\partial \phi(r\rightarrow R_+,\theta\leq\alpha)}{\partial r}=0$$ $$\frac{\partial \phi(r\rightarrow R_-,\theta>\alpha)}{\partial r}-\frac{\partial \phi(r\rightarrow R_+,\theta>\alpha)}{\partial r}=\sigma$$ I'm not sure how to deal with the fact, that the condition is differenct for $\theta\leq, >\alpha$.

The result should be: $$\phi(r<R,\theta)=\frac{Q}{8\pi\epsilon_0}\sum_{l=0}^{+\infty}\frac{1}{2l+1}\left[P_{l+1}(\cos\alpha)-P_{l-1}(\cos\alpha)\right]\frac{r^l}{R^{l+1}}P_l(\cos\theta)$$ $$\phi(r>R,\theta)=\frac{Q}{8\pi\epsilon_0}\sum_{l=0}^{+\infty}\frac{1}{2l+1}\left[P_{l+1}(\cos\alpha)-P_{l-1}(\cos\alpha)\right]\frac{R^l}{r^{l+1}}P_l(\cos\theta)$$ Can you tell me how to deal with the other condition?



2 Answers 2


It is easier to write the charge condition as $$\frac{\partial \phi(r\rightarrow R_-,\theta)}{\partial r}-\frac{\partial \phi(r\rightarrow R_+,\theta)}{\partial r}=\sigma(\theta)$$ and start by calculating the operator on the left. This is straightforward: $$ \frac{\partial \phi(r\rightarrow R_-,\theta)}{\partial r} = \frac{\partial }{\partial r}\sum_{l=0}^{+\infty}A_l r^{l} P_l(\cos\theta) = \sum_{l=0}^{+\infty}l\,A_l R^{l-1} P_l(\cos\theta) $$ and $$ \frac{\partial \phi(r\rightarrow R_+,\theta)}{\partial r} = \frac{\partial }{\partial r}\sum_{l=0}^{+\infty}B_l r^{-(l+1)} P_l(\cos\theta) = \sum_{l=0}^{+\infty}(-l-1)\,B_l R^{-l-2} P_l(\cos\theta). $$ Their difference is therefore \begin{align} \frac{\partial \phi(r\rightarrow R_-,\theta)}{\partial r}-\frac{\partial \phi(r\rightarrow R_+,\theta)}{\partial r} &= \sum_{l=0}^{+\infty}\left[l R^{l-1}\,A_l +(l+1) R^{-l-2}B_l\right]P_l(\cos\theta) \\&= \sum_{l=0}^{+\infty}(2l +1)R^{l-1}A_lP_l(\cos\theta), \end{align} using the condition $A_l/B_l=R^{-(2l+1)}$ which you already have. What you're trying to do, then, is obtaining the coefficients in the Legendre expansion of the charge function $\sigma(\theta)$, or in other words inverting the equation $$ \sum_{l=0}^{+\infty}(2l +1)R^{l-1}A_lP_l(\cos\theta)=\sigma(\theta). $$

This is done, of course, by appealing to the polynomials' orthogonality condition, $$\int_0^\pi P_l(\cos\theta) P_n(\cos\theta)\sin\theta\mathrm d \theta=2\delta_{ln}/(2l+1).$$ Integrating both sides against an arbitrary $P_l$, you get

$$ 2R^{l-1}A_l = \int_0^\pi P_l(\cos\theta) \sigma(\theta)\sin\theta\mathrm d \theta = \sigma_0\int_\alpha^\pi P_l(\cos\theta) \sin\theta\mathrm d \theta, $$ so all you need to do now is integrate. This looks like a chore, but it's a useful thing to know that, as a general principle, the single derivatives of orthogonal polynomials are usually expressible as small linear combinations of adjacent polynomials. For the Legendre polynomials, the relevant identity is $$ (2l+1)P_l(u)=\frac{d}{du}\left[P_{l+1}(u)-P_{l+1}(u)\right]. $$ Implementing this, you get \begin{align} A_l &= \frac{\sigma_0}{2R^{l-1}} \int_{-1}^{\cos(\alpha)} P_l(u) \mathrm du \\ &= \frac{1}{2l+1}\frac{\sigma_0}{2R^{l-1}} \int_{-1}^{\cos(\alpha)}\frac{d}{du}\left[P_{l+1}(u)-P_{l+1}(u)\right]\mathrm du \\ &= \frac{1}{2l+1}\frac{\sigma_0}{2R^{l-1}} \left[P_{l+1}(u)-P_{l+1}(u)\right]_{-1}^{\cos(\alpha)} \\ &= \frac{1}{2l+1}\frac{\sigma_0}{2R^{l-1}} \left[P_{l+1}(\cos(\alpha))-P_{l+1}(-1)+P_{l-1}(-1)-P_{l+1}(\cos(\alpha))\right] \\ &= \frac{1}{2l+1}\frac{\sigma_0}{2R^{l-1}} \left[P_{l+1}(\cos(\alpha))-P_{l+1}(\cos(\alpha))\right]. \end{align} This then leads to the result you want, modulo constants you should be careful with.

  • $\begingroup$ Is Legendre polynomial of negative order defined? The first term in the series includes $P_{-1}(\cos\alpha)$ (there's a typo in the answer). $\endgroup$ Commented Dec 3, 2023 at 10:05

Outside of charge distributions, the general (non-$\phi$-dependent) solution of $\Delta\Phi=0$ takes the form that you described. Now, supposing that the solutions for $r \lt R$ and $r \gt R$ take the form that you mention, then the electrical field $\overrightarrow{E} = -\nabla\Phi$ will have the following values in spherical coordinates:

$$r \lt R: \overrightarrow{E}_{r \theta \phi}=-\sum_{l=0}^\infty \left( \begin{array}{c}A_ll r^{l-1} P_l(\cos\theta)\\A_lr^{l-1}P_l^{'}(\cos\theta) \cdot -sin\theta\\0\end{array} \right)$$ $$r \gt R: \overrightarrow{E}_{r \theta \phi}=-\sum_{l=0}^\infty \left( \begin{array}{c}-B_l(l+1) r^{-(l+2)} P_l(\cos\theta)\\B_lr^{-(l+2)}P_l^{'}(\cos\theta) \cdot -sin\theta\\0\end{array} \right)$$

Thus, making use of $B_l=R^{2l+1}A_l$, the magnitude of the normal part $E_r$ inside and outside the sphere for r=R, has the following values:

$$r \lt R: \overrightarrow{E}_{r=R_-}=-\sum_{l=0}^\infty A_ll R^{l-1} P_l(\cos\theta) * \overrightarrow{e_r}$$ $$r \gt R: \overrightarrow{E}_{r=R_+}=\sum_{l=0}^\infty A_l(l+1)R^{l-1}P_l(\cos\theta) * \overrightarrow{e_r}$$

Now look at the surface just enclosing the charge distribution, and apply Gauss' law $\oint_{S}\overrightarrow{E}\overrightarrow{dS}=\int_V div\overrightarrow{E}d^3x$

Here $\overrightarrow{dS}$ is pointing outward, thus towards the origin for the inside, and away from the origin for the outside of the sphere. Maxwell equations tell us that $div\overrightarrow{E}=\frac{\rho}{\epsilon_0}$, so we find that

$$\int div\overrightarrow{E}d^3x=\int_0^{2\pi}d\phi\int_\alpha^\pi \frac{\sigma}{\epsilon_0} R^2sin\theta d\theta=2\pi\int_\alpha^\pi \frac{Q}{\pi\epsilon_0}sin\theta d\theta=2\frac{Q}{\epsilon_0}(1+cos\alpha)$$

The surface integrals evaluate to

$$\int\!\!\!\int_{S_{inside}}\overrightarrow{E}\overrightarrow{dS}=\int_0^{2\pi}d\phi\int_\alpha^\pi \sum_{l=0}^\infty A_ll R^{l-1} P_l(\cos\theta) Rsin\theta d\theta=\\2\pi \sum_{l=0}^\infty A_ll R^l \int_\alpha^\pi P_l(\cos\theta) sin\theta d\theta$$ and $$\int\!\!\!\int_{S_{outside}}\overrightarrow{E}\overrightarrow{dS}=\int_0^{2\pi}d\phi\int_\alpha^\pi \sum_{l=0}^\infty A_l(l+1) R^{l-1} P_l(\cos\theta) Rsin\theta d\theta=\\2\pi \sum_{l=0}^\infty A_l(l+1) R^l \int_\alpha^\pi P_l(\cos\theta) sin\theta d\theta$$ So in total $$\oint_{S}\overrightarrow{E}\overrightarrow{dS}=2\pi \sum_{l=0}^\infty A_l(2l+1) R^l \int_\alpha^\pi P_l(\cos\theta) sin\theta d\theta=\\2\pi \sum_{l=0}^\infty A_l(2l+1) R^l \cdot -\int_{cos\alpha}^{-1} P_l(x)dx$$ Making use of the identity $(2l+1)P_l = P_{l+1}^{'} - P_{l-1}^{'}$, we get $$\oint_{S}\overrightarrow{E}\overrightarrow{dS}=2\pi \sum_{l=0}^\infty A_l R^l \int_{-1}^{cos\alpha} [P_{l+1}^{'}(x) - P_{l-1}^{'}(x)]dx=\\2\pi \sum_{l=0}^\infty A_l R^l [P_{l+1}(cos\alpha) - P_{l-1}(cos\alpha)]$$

This gives us a restriction on the values of $A_l$, namely $$ \sum_{l=0}^\infty A_l R^l [P_{l+1}(cos\alpha) - P_{l-1}(cos\alpha)] = \frac{Q}{\pi\epsilon_0}(1+cos\alpha)$$ So the following will work: $$A_l = \frac{Q}{\pi\epsilon_0 R^l}\cdot\frac{1}{2^{l+1}}\cdot\frac{1+cos\alpha}{P_{l+1}(cos\alpha) - P_{l-1}(cos\alpha)}$$ but I can't see how to arrive at the value given by you, $$A_l = \frac{QR^{-(l+1)}}{8\pi\epsilon_0(2l+1)}[P_{l+1}(cos\alpha) - P_{l-1}(cos\alpha)]$$

By the way, a similar argument as the one above, now for the surface enclosing the non-charged pole of the sphere, shows us that $$\oint_{S^{'}}\overrightarrow{E}\overrightarrow{dS}=2\pi \sum_{l=0}^\infty A_l R^l \cdot -[P_{l+1}(cos\alpha) - P_{l-1}(cos\alpha)] = \int div\overrightarrow{E}d^3x = 0$$

Clearly this forms a contradiction together with the other formula - so the solutions inside and outside the sphere must be different than the ones assumed. Not illogical, as the open pole allows fields from one side of the sphere to penetrate the other side, but it is unclear what this amounts to.

Obviously close to the origin all $B_l$ must be zero, and in infinity all $A_l$ likewise must be zero, but where do these basic solutions merge? Where's their boundaries?


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