Laplace Equation in Spherical Coordinates
Laplace Equation in spherical coordinates \((r, \theta, \phi)\) becomes
\[\begin{aligned} \nabla^2 V = \frac{1}{r^2}\frac{\partial}{\partial r}\Big( r^2 \frac{\partial V}{\partial r}\Big) + \frac{1}{r^2 \sin\theta}\frac{\partial}{\partial \theta} \Big( \sin\theta \frac{\partial V}{\partial \theta}\Big) + \frac{1}{r^2 \sin^2 \theta}\frac{\partial^2 V}{\partial \phi^2}=0\end{aligned}\]
\[\begin{aligned} \nabla^2 V = \frac{1}{r}\frac{\partial^2}{\partial r^2}\Big(r V\Big) + \frac{1}{r^2 \sin\theta}\frac{\partial}{\partial \theta}\Big(\sin\theta \frac{\partial V}{\partial \theta}\Big) + \frac{1}{r^2 \sin^2\theta}\frac{\partial V}{\partial \phi^2}=0\end{aligned}\]
Step 1: Separation of variables assumption
Assume a product form for the potential \(V\).
\[\begin{aligned} \boxed{V = \frac{R(r)}{r}\Theta(\theta)\Phi(\theta)}\end{aligned}\]
Here, we put \(\frac{R(r)}{r}\) for easy calculation. Then the result equation becomes
\[\begin{aligned} \frac{\Phi \Theta}{r}\frac{d^2}{d r^2}R(r)+\frac{R\Phi}{r^3 \sin\theta}\frac{d}{d \theta}\Big(\sin\theta \frac{d \Theta(\theta)}{d \theta}\Big)+\frac{R\Theta}{r^3\sin^2\theta}\frac{d \Phi(\phi)}{d \phi^2}=0 \label{eq:1} \end{aligned}\]
Step 2: Find the separated equations
If we multiply \(\frac{1}{R \Theta\Phi}\) to \(Eq.\eqref{eq:1}\).
\[\begin{aligned} \frac{1}{rR}\frac{d^2}{dr^2}R(r)+\frac{1}{r^3\sin{\theta}}\frac{1}{\Theta }\frac{d}{d\theta}\Big(\sin\theta \frac{d\Theta(\theta)}{d\theta}\Big)+\frac{1}{r^3\sin^2\theta}\frac{1}{\Phi}\frac{d\Phi(\phi)}{d\phi^2} =0\end{aligned}\]
Nothing separates until we multiply \(r^3\sin^2\theta\).
\[\begin{aligned} \frac{r^2 \sin^2\theta}{R}\frac{d^2}{dr^2}R(r) + \frac{\sin\theta}{\Theta}\frac{d}{d\theta}\Big(\sin\theta \frac{d\Theta(\theta)}{d\theta}\Big) + \frac{1}{\Phi}\frac{d\Phi(\phi)}{d\phi^2} =0\end{aligned}\]
Now, \(\Phi\) term has completely separated. If we take partial derivative with respect to \(\phi\)
\[\begin{aligned} \frac{\partial}{\partial \phi}\Big(\frac{1}{\Phi}\frac{d^2 \Phi}{d\phi^2}\Big) =0\end{aligned}\]
since the first two terms depend only on \(r\) and \(\theta\). Therefor
\[\begin{aligned} \frac{1}{\Phi}\frac{d^2 \Phi}{d\phi^2} = -m^2\end{aligned}\]
The equation for \(\Phi\) is then,
\[\begin{aligned} \boxed{\frac{d^2\Phi}{d\phi^2}+m^2 \Phi=0} \label{eq:sv1}\end{aligned}\]
The remainder of the Laplace equation is
\[\begin{aligned} \frac{r^2 R^{\prime\prime}}{R}+\frac{1}{\Theta \sin\theta}\frac{d}{d\theta}\Big(\sin{\theta}\frac{d\Theta}{d\theta}\Big) -\frac{m^2}{\sin^2\theta}=0\end{aligned}\]
Now, we can separate the functions between \(r\) and \(\theta\). Let us define each term as \(\alpha\).
\[\begin{aligned} {\frac{r^2 R^{\prime\prime}}{R}=\alpha} \label{eq:sv_2}\\ {\frac{1}{\Theta \sin\theta}\frac{d}{d\theta}\Big(\sin{\theta}\frac{d\Theta}{d\theta}\Big)-\frac{m^2}{\sin^2\theta}=-\alpha } \label{eq:sv_3}\end{aligned}\]
Step 3:Solving the equations for each variables
Eq.[eq:sv1] for \(\Phi\)
\[\begin{aligned} \sin^2\theta \{\frac{r^2 R\prime\prime}{R} + \frac{1}{\Theta \sin\theta}\frac{d}{d\theta}(\sin\theta\frac{d\Theta}{d\theta}) \}=-\frac{\Phi\prime\prime}{\Phi}= m^2\end{aligned}\]
We let \(m^2\) to be positive because we want periodic answer for \(Q\). Hence
\[\begin{aligned} \Phi \propto e^{\pm im\phi}\end{aligned}\]
Eq.[eq:sv_2] for R
\[\begin{aligned} r^2 R^{\prime\prime}-\alpha R = 0\end{aligned}\]
Notice that re-scaling \(r\) leaves this equation unchanged. This is a clue that powers of \(r\) may work. Substituting a power-law solution, \(\frac{R(r)}{r}=r^{l}\); (notice that in the beginning, we divide extra \(r\) for convenience).
\[\begin{aligned} r^2 (r^{(l+1) \prime\prime})-\alpha r^{l+1}&=0 \\ (l(l+1)-\alpha)r^{l+1} &= 0\\ \alpha &= l(l+1)\end{aligned}\]
with \(a=l(l+1)\), we have a solution for every number \(l\).Since it is a quadratic equation, there are two values of \(l\).
\[\begin{aligned} l^2 + l - \alpha =0\end{aligned}\]
Let \(l\) have some value \(x\), so that \(\alpha = x(x+1)\). Then, the value \(l=-(x+1)\) gives the same \(\alpha\).
\[\begin{aligned} \alpha = (-(x+1))(-(x+1)+1)=x(x+1)\end{aligned}\]
with \(\alpha = l(l+1)\), we now have to solve for \(\Theta\).
\[\begin{aligned} \frac{1}{\sin\theta}\frac{d}{d\theta}\Big( \sin\theta \frac{d\Theta}{d\theta}\Big) + \Big( l(l+1) - \frac{m^2}{\sin^2\theta}\Big)\Theta = 0 \label{eq:theta_1}\end{aligned}\]
Legendre polynomials: The solution to the \(\theta\) equation with \(m=0\).
associated Legendre polynomials: Solutions with general \(m\).
Eq.[eq:sv_3] for \(\Theta\)
Legendre polynomials, \(m=0\)
Now \(\sin\theta d\theta = d\cos\theta\). Hence, define \(x=\cos\theta\). Then,
\[\begin{aligned} \frac{d}{d\theta}&=\frac{dx}{d\theta}\frac{d}{dx}\\ &=-\sin\theta \frac{d}{dx}\\ \frac{d}{dx}&=-\frac{1}{\sin\theta}\frac{d}{d\theta}\end{aligned}\]
When we substitute this to Eq.[eq:theta_1] with \(m=0\),
\[\begin{aligned} \frac{d}{dx}\Big((1-x^2)\frac{d\Theta}{dx}\Big) + l(l+1)\Theta = 0\end{aligned}\]
Notice that \(\sin^2\theta=1-\cos^2\theta=1-x^2\). The solutions are polynomials, \(P_l(x)=P_l(\cos\theta)\)
Legendre Polynomial
Let us consider azimuthally symmetric case, meaning \(m=0\).
\[\begin{aligned} \frac{d}{dx}[(1-x^2)\frac{dP}{dx}]+[l(l+1)]P=0\end{aligned}\]
This equation is called Legendre Equation. The range of \(x\) is \(-1 \leq x \leq 1\) because the range of \(\cos\theta\) is \(0 \leq \cos\theta \leq \pi\)
\[\begin{aligned} \sum_{j=0}^{\infty} & \{[(a+j)(a+j-1)]a_j x^{\alpha+j-2} \nonumber \\ & -[(a+j)(a+j+1)+2(a+j)-l(l+1)]a_jx^{\alpha+j}\}\end{aligned}\]
\[\begin{aligned} l=0 P_0(x)&=1 \nonumber \\ l=1 P_1(x)&=x \nonumber \\ l=2 P_2(x)&=\frac{1}{2}(3x^2-1) \nonumber \\ l=3 P_3(x)&=\frac{1}{2}(5x^3-3x) \nonumber \\ l=4 P_4(x)&=\frac{1}{8}(35x^4-30x^2+3) \end{aligned}\]
\(\rightarrow\) \(P_1\) is an even function for even \(l\), and odd for odd \(l\).
\(\rightarrow\) Legendre polynomials can be represented by Rodrigues’ formula
\[\begin{aligned} P_l(x)=\frac{1}{2^{l}l!}\frac{d^l}{dx^l}(x^2-1)^l\end{aligned}\]
Example
\[\begin{aligned} P_2(x)&=\frac{1}{2^2 2!}\frac{d^2}{dx^2}(x^2-1)^2 \nonumber \\ &=\frac{1}{8} \frac{d^2}{dx^2}(x^4-2x^2+1) \nonumber \\ &=\frac{1}{8}(12x^2-4)=\frac{3x^2-1}{2} \nonumber\end{aligned}\]
\(\rightarrow\) \(P_l\) is orthogonal
\[\begin{aligned} \int_{-1}^{1} P_l(x) P_{l\prime}(x)dx =\frac{2}{2l+1}\delta_{ll\prime}\end{aligned}\]
This means that any function defined for \([-1,1]\) can be represented by expansion by \(P_i\)
\[\begin{aligned} f(x) = \sum_{l=0}^{\infty}A_l P_l(x), \quad \quad A_l=\frac{2l+1}{2}\int_{-1}^{1}f(x)P_l(x)dx\end{aligned}\]
Step 4:Putting all together.
For general \(m\), the full solution is
\[\boxed{V(r,\theta,\phi) = \sum_{l=0}^{\infty} \sum_{m=-l}^{l} \Big( A_{lm}r^l + \frac{B_{lm}}{r^{l+1}} \Big) P_{l}^{m}( \cos \theta) e^{i m \phi}}\]
