Potencial generado por una esfera hueca con un agujero

Es más fácil escribir la condición de carga como

$$\frac{\partial \phi(r\rightarrow R_-,\theta)}{\partial r}-\frac{\partial \phi(r\rightarrow R_+,\theta)}{\partial r}=\sigma(\theta)$$ y comienza calculando el operador de la izquierda. Esto es sencillo: $$\frac{\partial \phi(r\rightarrow R_-,\theta)}{\partial r} = \frac{\partial }{\partial r}\sum_{l=0}^{+\infty}A_l r^{l} P_l(\cos\theta) = \sum_{l=0}^{+\infty}l\,A_l R^{l-1} P_l(\cos\theta)$$ y $$\frac{\partial \phi(r\rightarrow R_+,\theta)}{\partial r} = \frac{\partial }{\partial r}\sum_{l=0}^{+\infty}B_l r^{-(l+1)} P_l(\cos\theta) = \sum_{l=0}^{+\infty}(-l-1)\,B_l R^{-l-2} P_l(\cos\theta).$$ Their difference is therefore $$\begin{align} \frac{\partial \phi(r\rightarrow R_-,\theta)}{\partial r}-\frac{\partial \phi(r\rightarrow R_+,\theta)}{\partial r} &= \sum_{l=0}^{+\infty}\left[l R^{l-1}\,A_l +(l+1) R^{-l-2}B_l\right]P_l(\cos\theta) \\&= \sum_{l=0}^{+\infty}(2l +1)R^{l-1}A_lP_l(\cos\theta), \end{align}$$ usando la condición $A_l/B_l=R^{-(2l+1)}$ 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

$$2R^{l-1}A_l = \int_0^\pi P_l(\cos\theta) \sigma(\theta)\sin\theta\mathrm d \theta = \sigma_0\int_\alpha^\pi P_l(\cos\theta) \sin\theta\mathrm d \theta,$$ así que todo lo que necesita hacer ahora es integrar. Esto parece una tarea, pero es útil saber que, como principio general, las derivadas simples de polinomios ortogonales generalmente se expresan como pequeñas combinaciones lineales de polinomios adyacentes. Para los polinomios de Legendre, la identidad relevante es $$(2l+1)P_l(u)=\frac{d}{du}\left[P_{l+1}(u)-P_{l+1}(u)\right].$$ Implementing this, you get $$\begin{align} A_l &= \frac{\sigma_0}{2R^{l-1}} \int_{-1}^{\cos(\alpha)} P_l(u) \mathrm du \\ &= \frac{1}{2l+1}\frac{\sigma_0}{2R^{l-1}} \int_{-1}^{\cos(\alpha)}\frac{d}{du}\left[P_{l+1}(u)-P_{l+1}(u)\right]\mathrm du \\ &= \frac{1}{2l+1}\frac{\sigma_0}{2R^{l-1}} \left[P_{l+1}(u)-P_{l+1}(u)\right]_{-1}^{\cos(\alpha)} \\ &= \frac{1}{2l+1}\frac{\sigma_0}{2R^{l-1}} \left[P_{l+1}(\cos(\alpha))-P_{l+1}(-1)+P_{l-1}(-1)-P_{l+1}(\cos(\alpha))\right] \\ &= \frac{1}{2l+1}\frac{\sigma_0}{2R^{l-1}} \left[P_{l+1}(\cos(\alpha))-P_{l+1}(\cos(\alpha))\right]. \end{align}$$ Esto lleva al resultado que desea, constantes de módulo con las que debe tener cuidado.

Fuera de las distribuciones de carga, el general (no$\phi$-dependiente) solución de$\Delta\Phi=0$ 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

$$\oint_{S^{'}}\overrightarrow{E}\overrightarrow{dS}=2\pi \sum_{l=0}^\infty A_l R^l \cdot -[P_{l+1}(cos\alpha) - P_{l-1}(cos\alpha)] = \int div\overrightarrow{E}d^3x = 0$$

Claramente, esto forma una contradicción junto con la otra fórmula, por lo que las soluciones dentro y fuera de la esfera deben ser diferentes a las asumidas. No es ilógico, ya que el polo abierto permite que los campos de un lado de la esfera penetren en el otro lado, pero no está claro a qué equivale esto.

Obviamente cerca del origen todo$B_l$debe ser cero, y en el infinito todos$A_l$igualmente debe ser cero, pero ¿dónde se fusionan estas soluciones básicas? ¿Dónde están sus límites?