Separation of variables : spherical coordinate system

In spherical coordinate system, the Laplacian is

$$\nabla^2 = \frac{1}{r} \frac{\partial^2}{\partial r^2} r + \frac{... ...artial \theta} + \frac{1}{r^2 \sin^2 \theta} \frac{\partial^2}{\partial \phi^2}$$ (73)

Writing the solution of Laplace's equation $\nabla \phi(\vec r) =0$ in the form $\phi(\vec r) = R(r)\Theta(\theta)\Phi(\phi)$, we have

$$\frac{\partial^2}{\partial \phi^2} \Phi(\phi) = m^2 \Phi(\phi).$$ (74)

As $\Phi(\phi+2 \pi)=\Phi(\phi)$, the solution is $\Phi(\phi) = e^{im \phi}$ with integral value of $m$.

Using the variable $x=\cos(\theta)$ instead of $\theta$, the equation for for $\Theta(x)$ becomes

$$\left [ (1 - x^2) \frac{d^2}{dx^2} - 2x \frac{d}{dx} + \left ( C - \frac{m^2} {1 - x^2} \right ) \right ] \Theta(x) = 0$$ (75)

where $m$ above is same as $m$ appearing in the solution of $\Phi(\phi)$. We need a regular solution for $-1 \le x \le 1$. This is the generalized Legendre equation. Without proof we state that a regular solution is possible if $C = l(l+1)$ with integral $l$ and $\vert m\vert \le l$. The solutions are associated Legendre polynomials $P_l^m(x)$. The properties of associated Legendre polynomials are given in Appendix(). The associated Legendre polynomials form an orthogonal set of functions in interval $\vert x\vert \le 1$ with the orthogonality relation

$$\int_{-1}^1 dx P_l^m(x) P_{l'}^m(x) = \frac{2}{2l+1}\frac{(l+m)!} {(l-m)!} \delta_{l,l'}.$$

Combining the solutions $\Phi$ and $\Theta$, we define orthonormal set of functions

$$Y_{l,m}(\theta, \phi) = \sqrt{\frac{2l+1}{4 \pi}\frac{(l-m)!} {(l+m)!}} P_l^m(\cos(\theta) e^{im \phi}$$

which satisfy the orthonormality condition

$$\int d\phi d(\cos \theta) Y^*_{l',m'}(\theta,\phi) Y_{l,m} (\theta,\phi) = \delta_{l,l'} \delta_{m,m'}.$$

With this, the differential equation for the radial function $R(r)$ is

$$\frac{1}{r} \frac{d^2}{dr^2} r R(r) - \frac{l(l+1)}{r^2} R(r... ...\frac{2}{r}\frac{d}{dr} - \frac{l(l+1)}{r^2} \right ) R(r) = 0.$$

The general solution is $R(r) = A r^l + B / r^{l+1}$.

Collecting these results, the general solution of Poisson's equation in spherical coordinate system can be written as

$$\phi(\vec r) = \sum_{l,m} \left ( A_{l,m} r^l + B_{l,m} r^{-(l+1)} \right ) Y_{l,m}(\theta, \phi).$$ (76)

If the problem has cylindrical symmetry, the solution does not depend on $\phi$ if the z axis is along the symmetry axis. Then we have¹⁰

$$\phi(\vec r) = \sum_l \left ( A_l r^l + B_l r^{-(l+1)} \right ) P_l(\cos \theta).$$

As an example consider the problem in which the potential on two hemispheres of a conducting sphere of radius a is maintained at $V_0$ and $-V_0$. We choose z-axis so that the northern hemisphere has potential $V_0$. Then the potential does not depend on $\phi$. Further, the potential must vanish at infinity. So the potential can be written as

$$\phi(\vec r) = \sum_l B_l r^{-(l+1)} P_l(\cos \theta)$$

for $r>a$. For $r < a$, the coefficients $B_l$'s are zero as the potential should not be singular at $r = 0$. So

$$\phi(\vec r) = \sum_l A_l r^l P_l( \cos \theta ).$$

The coefficients $A_l$'s and $B_l$'s are obtained by using the boundary condition at $r=a$. The coefficients for even $l$'s vanish and for odd $l$'s $A_l = \frac{2l+1} {a^l} V_0 \int_0^1 dx P_l(x)$ and $B_l = (2l+1) a^{(l+1)} V_0 \int_0^1 dx P_l(x)$. We therefore have

$$\phi(\vec r) = V_0 \left [ \frac{3}{2} \left ( \frac{a}{r} \right... ... P_3(x) + \frac{11}{16} \left ( \frac{a}{r} \right )^6 P_5(x) + \cdots \right ]$$ (77)

for $r>a$ and

$$\phi(\vec r) = V_0 \left [ \frac{3}{2} \left ( \frac{r}{a} \right... ... P_3(x) + \frac{11}{16} \left ( \frac{r}{a} \right )^5 P_5(x) + \cdots \right ]$$ (78)

for $r < a$.

As another example of the application of spherical coordinate system consider a conical conducting surface held at zero potential. Let $\beta$ be the opening angle. For $\beta < \pi /2$, we are interested in the computation of potential inside a conical hole inside a conductor. For $\beta > \pi /2$, on the other hand, we have a conical conducting surface projecting out ( see Fig (2.6.3)). We choose z axis along the direction of the axis of the cone so that, in spherical coordinate system $0 \le \phi \le 2 \pi$ and $0 \le \theta \le \beta$. The potential is independent of $\phi$ and therefore $\Phi(\phi) = 1$ or $m = 0$. The differential equation for $\Theta(\theta)$ is

$$\left [ (1-x^2)\frac{d^2}{dx^2} - 2 x\frac{d}{dx} + \nu(\nu+1) \right ] \Theta_\nu(x) = 0$$

where $x=\cos(\theta)$ and $\cos \beta \le x \le 1$. Corresponding radial differential equation is

$$\left ( \frac{d^2}{dr^2} + \frac{2}{r}\frac{d}{dr} - \nu(\nu+1) \right ) R(r) = 0$$

and the solution is a linear combination of $r^\nu$ and $r^{-(\nu+1)}$.

Figure 11: A conical conducting surface. For $\beta < \pi /2$ ( a ) it represents a conicalhole in a metalic surface and for $\beta > \pi /2$ ( b ) it represents a conical surface projecting out of a conducting plane.

For integral values of $\nu$ the functions $\Theta_\nu(x)$ are the Legendre polynomials having $\nu$ zeros between $1$ and $-1$. For nonintegral values of $\nu$, $\Theta_\nu(1) = 1$. As $\nu$ increases, the value of x at which $\Theta_\nu$ vanishes increases continuously. In any case, $\nu > 0$ to have at least one zero between $-1$ and $1$. For example, as $\nu$ increases from 0 to 1, the zero of $\Theta_\nu(x)$ moves from $x=-1$ to $x=0$. Our problem requires that the potential vanishes at $x'=\cos\beta$. This can happen for a number of values of $\nu$, say $\nu_m$ and then the solution for the potential can be written as

$$\phi(\vec r) = \sum_m \left ( A_m r{^\nu_m} + B_m r^{-(\nu_m+1)} \right ) \Theta_{\nu_m}(x)$$ (79)

satisfying the required boundary condition. Corresponding electric field components are

$$E_r( \vec r) = - \frac{\partial}{\partial r} \phi(\vec r)$$

$$= - \sum_m \left ( \nu_m A_m r^{\nu_m-1} - ( \nu_m + 1) B_m r^{-(\nu_m+2)} \right ) \Theta_{\nu_m}(x)$$ (80)

$${\mathrm {and}}$$

$$E_\theta (\vec r) = - \frac{1}{r} \frac{\partial}{\partial \theta} \phi( \vec r)$$

$$= \sum_m \left ( A_m r^ {\nu_m-1} + B_m r^{-(\nu_m+2)} \right ) \frac{d}{dx} \Theta_{\nu_m}(x)$$ (81)

From the electric field in $\theta$ direction at the surface of the cone we can compute the surface charge density induced on the cone by using Gauss theorem. We have $\sigma(\vec r) = \frac{1}{4 \pi} E_\theta(\vec r)$.

If we are interested in the solution at small value of $r$ ( i.e. the tip of the cone ), the coefficients $B_m$'s should be zero as the potential should be nonsingular. Further, for the same reason, the smallest power of $r$ corresponding to $A_1$ would dominate and the behaviour of the potential and field can be understood in terms of that term. Thus, we shall choose

$$\phi(\vec r) \sim A_1 r^{\nu_1} \Theta_{\nu_1}(\cos \theta).$$

Let us first consider $\beta$ increasing from 0 to $\pi/2$, so that $x'$ decreases from 1 to 0. The value of $\nu$ at which $\Theta_\nu(\theta)$ vanishes decreases from infinity to 1. Thus, for $\beta$ between $0$ and $\pi/2$ ( corresponding to a conducting cone drilled in a metal ), the leading term in the potential is $\propto r^{\nu_1}$ with $\nu_1$ between $\infty$ and 1. The electric field is then proportional to $r^{\nu_1-1}$. This means, the potential and field valishes as a power law near the tip of the cone. Further, the Gauss theorem implies that the induced charge density at the tip of the cone also vanishes as a power law. When $\beta$ is $\pi/2$, the potential is linear in r and the field is constant. This is the same result as that for the conducting plane.

For $\beta$ increasing from $\pi/2$ to $\pi$, the geometry is like a conducting wireprojecting into space. For this case, $\nu$ decreases from 1 to zero and the electric field behaves as $r^{\nu_1-1}$ with $\nu_1 - 1 < 0$ for small r. Thus, the electric field becomes singular near the tip of the cone. This also means that the induced charge density becomes singular at the tip.

This result explains the functioning of lightening arrestor. The lightening arrestor has a pointed conducting material. The presence of charges in the atmosphere induces charge density on the arrestor. This charge density is very large at the tip of the arrestor and that helps in inducing electrical breakdown resulting in a discharge between the atmospheric charge and the lightening conductor. Since the conductor is grounded, the discharge is safely conducted to the earth.