To put it simply, solving Laplace’s equation in spherical coordinates is a mathematical method used to figure out how a property like heat or electric potential is distributed in a three-dimensional, spherical object.
Imagine you have a ball and you want to know how the temperature is spread out on and inside it. Instead of using a standard x, y, z grid, which would be complicated for a sphere, we use spherical coordinates:
- r for the distance from the centre.
- θ for the angle up or down from the top.
- φ for the angle around the middle.
By using this special coordinate system, the problem becomes much easier to solve. The solution breaks down the complex problem into three simpler parts, one for each coordinate, and then combines them to give you the complete picture of how the property is distributed throughout the sphere.
The Laplace equation in spherical coordinates can be written as:
$$\frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \frac{\partial V}{\partial r} \right) + \frac{1}{r^2 \sin\theta} \frac{\partial}{\partial \theta} \left( \sin\theta \frac{\partial V}{\partial \theta} \right) + \frac{1}{r^2\sin^2\theta} \frac {\partial^2 V}{\partial\phi^2} = 0 \tag{1}$$
Since potential \(V\) is a function of \(r\), \(\theta\), and \(\phi\), i.e.,
$$V = R(r) \Theta(\theta) \Phi(\phi) = R \Theta \Phi $$
Therefore:
$$\frac{\partial V}{\partial r} = \Theta \Phi \frac{\partial R}{\partial r} $$
$$\frac{\partial V}{\partial \theta} = R \Phi \frac{\partial \Theta}{\partial \theta} $$
$$\frac{\partial V}{\partial \phi} = R \Theta \frac{\partial \Phi}{\partial \phi} $$
$$\frac{\partial^2 V}{\partial \phi^2} = R \Theta \frac{\partial^2 \Phi}{\partial \phi^2}$$
Substituting the above in eq. (1), we get:
$$\frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \Theta \Phi \frac{\partial R}{\partial r} \right) + \frac{1}{r^2 \sin\theta} \frac{\partial}{\partial \theta} \left( \sin\theta R \Phi \frac{\partial \Theta}{\partial \theta} \right) + \frac{1}{r^2 \sin^2\theta} \left( R \Theta \frac{\partial^2 \Phi}{\partial \phi^2} \right) = 0 \tag{2}$$
Multiplying eq. (2) by \(\frac{r^2 \sin^2\theta}{R \Theta \Phi}\), we have:
$$\frac{\sin^2\theta}{R} \frac{\partial}{\partial r} \left( r^2 \frac{\partial R}{\partial r} \right) + \frac{\sin\theta}{\Theta} \frac{\partial}{\partial \theta} \left( \sin\theta \frac{\partial \Theta}{\partial \theta} \right) + \frac{1}{\Phi} \frac{\partial^2 \Phi}{\partial \phi^2} = 0 \tag{3} $$
Let:
\begin{equation}
\frac{1}{\Phi} \frac{\partial^2 \Phi}{\partial \phi^2} = -m^2\tag {4}
\end{equation}
or
\begin{equation}
\frac{\partial^2 \Phi}{\partial \phi^2} + m^2 \Phi = 0,\end{equation}
whose general solution is:
\begin{equation}
\Phi = C_1 e^{\pm i m \phi}
\end{equation}
Now, using eq. (4) in eq. (3), we have:
\begin{equation}
\frac{\sin^2\theta}{R} \frac{\partial}{\partial r} \left( r^2 \frac{\partial R}{\partial r} \right) + \frac{\sin\theta}{\Theta} \frac{\partial}{\partial \theta} \left( \sin\theta \frac{\partial \Theta}{\partial \theta} \right) + m^2 =0 \tag {5}\end{equation}
Let:
\begin{equation}
\frac{1}{R} \frac{\partial}{\partial r} \left( r^2 \frac{\partial R}{\partial r} \right) = n(n+1),
\tag{6}
\end{equation}
Then eq. (5) becomes:
\begin{equation}
\frac{\sin\theta}{\Theta} \frac{\partial}{\partial \theta} \left( \sin\theta \frac{\partial \Theta}{\partial \theta} \right) + \sin^2\theta \, n(n+1) – m^2 = 0.
\tag{7}
\end{equation}
Or:
\begin{equation}
\frac{1}{\sin\theta} \frac{\partial}{\partial \theta} \left( \sin\theta \frac{\partial \Theta}{\partial \theta} \right) + \left[ n(n+1) – \frac{m^2}{\sin^2\theta} \right] \Theta = 0 \end{equation}
Let \(x = \cos\theta\), then:
$$\frac{\partial}{\partial \theta} = -\sin\theta \frac{\partial}{\partial x},
\quad \sin\theta \frac{\partial}{\partial \theta} = – (1 – x^2)^{1/2} \frac{\partial}{\partial x} $$
Substituting, we get:
\begin{equation}
(1 – x^2) \frac{d^2 \Theta}{dx^2} – 2x \frac{d\Theta}{dx} + \left[ n(n+1) – \frac{m^2}{1 – x^2} \right] \Theta = 0.
\tag{8}
\end{equation}
This is the associated Legendre equation, whose solution is given by:
\begin{equation}
\Theta = (1 – x^2)^{m/2} \frac{d^m}{dx^m} P_n(x)
\end{equation}
or
\begin{equation}
\Theta = \sum_{n=0}^{\infty}(\sin\theta)^m \frac{d^m}{d(\cos\theta)^m} P_n(\cos\theta).
\end{equation}
In case of azimuthal symmetry, \(m = 0\), then:
\begin{equation}
\Theta = \sum_{n=0}^{\infty} P_n(\cos\theta).
\end{equation}
Now consider eq. (6), which is:
\begin{equation}
r^2 \frac{d^2 R}{dr^2} + 2r \frac{dR}{dr} – n(n+1) R = 0.
\tag{9}
\end{equation}
Let \(r=e^z\), then\(\frac{dR}{dr}\) = \(e^{-z}\) \(\frac{dR}{dz}\) and \(r^2 \frac{d^2 R}{dr^2} = e^{-2z} \frac{d^2 R}{dz^2} – e^{-2z} \frac{dR}{dz}\)
Substituting the above in eq. (9), we get:
\begin{equation}
\frac{d^2 R}{dz^2} + \frac{dR}{dz} – n(n+1) R = 0.
\tag{10}
\end{equation}
The solution of eq. (10) is:
\begin{equation}
R = A_n e^{nz} + B_n e^{-(n+1)z} \quad \Rightarrow \quad R = A_n r^n + B_n r^{-(n+1)}
\end{equation}
Therefore, the complete solution is:
\begin{align}
V(r,\theta,\phi) &= R \Theta \Phi \nonumber\
&= \left( C_1 e^{i m \phi} \right)
\frac{\sin^m\theta \, d^m P_n(\cos\theta)}{d(\cos\theta)^m}
\left[ A_n r^n + B_n r^{-(n+1)} \right]
\end{align}
Or in summation form:
\begin{equation}
V(r,\theta,\phi) = \sum_{n=0}^\infty \left[ C_1 e^{i m \phi} \left( A_n r^n + B_n r^{-(n+1)} \right)
\frac{\sin^m\theta \, d^m P_n(\cos\theta)}{d(\cos\theta)^m} \right]
\tag{11}
\end{equation}