Laplace problem-analytically

In which we show that the method of separation of variables gives an analytic solution to Laplace's equation.

Our prototypical problem is shown at right: We have an electric field which, at infinity is pointing uniformly in the $\uv{z}$ direction. (Perhaps we're between the flat plates of a large capacitor...) We insert a spherical conductor in the field. Now what's the electric field?

To solve this, we note that...

  • The whole (conducting) sphere is at the same potential, and
  • this problem has azimuthal symmetry--that is, the solution must be independent of the $\phi$ coordinate.
  • Laplace's equation holds outside of the sphere.
  • Inside the sphere, we know that the field is zero.

The Laplace equation (see your inside front cover) in spherical coordinates, assuming azimuthal symmetry, is...

$$\grad^2 V = \frac{1}{r^2}\frac{\del}{\del r}\left(r^2 \frac{\del V}{\del r}\right)+\frac{1}{r^2 \sin \theta}\frac{\del}{\del \theta}\left(\sin \theta \frac{\del V}{\del \theta}\right)=0$$.

Separation of variables

The method of separation of variables consists in guessing that the solutions to a partial differential equation consist of the products of functions, each of which is a function of just one coordinate.

Though it seems, erm, less than obvious that this will work, this approach to solving partial diff.eq.s actually has a pretty good track record!

Since we're using spherical polar coordinates, let's guess that the solution is the product of two functions: $$V(\theta,r) = \Theta(\theta)R(r)$$.

Substituting this in to Laplace's equation gives... $$\begineq\grad^2 V = 0 &=& \frac{1}{r^2}\frac{\del}{\del r}\(r^2 \frac{\del \Theta(\theta)R(r)}{\del r}\)+\frac{1}{r^2 \sin \theta}\frac{\del}{\del \theta}\(\sin \theta \frac{\del \Theta(\theta)R(r)}{\del \theta}\)\\ &=&\frac{\Theta(\theta)}{r^2}\frac{d}{d r}\(r^2 \frac{dR(r)}{d r}\)+\frac{R(r)}{r^2 \sin \theta}\frac{d}{d \theta}\(\sin \theta \frac{d \Theta(\theta)}{d \theta}\)\endeq$$

Multiplying through by $r^2/(\Theta(\theta)R(r))$:

$$ \frac{1}{R(r)}\frac{d}{d r}\(r^2 \frac{dR(r)}{d r}\)= -\frac{1}{\Theta(\theta) \sin \theta}\frac{d}{d \theta}\(\sin \theta \frac{d \Theta(\theta)}{d \theta}\).$$

The only way for this to hold, since $\theta$ and $r$ are independent variables, is if the left chunk (only a function of $r$) and the right chunk (only a function of $\theta$) are each equal to a constant--the same constant, called the separation constant $k$. So this one equation is really two equations.

$\Theta(\theta)$ equation

$$ -\frac{1}{\Theta(\theta) \sin \theta}\frac{d}{d \theta}\(\sin \theta \frac{d \Theta(\theta)}{d \theta}\)=k=l(l+1).$$

Taking the $\theta$ equation first... It turns out that solutions are only possible for certain values of the separation constant. These are $k=l(l+1)$ where $l>=0$ is an integer. Re-arranging our equation for $\Theta(\theta)$ we get: $$\frac{1}{\sin \theta}\frac{d}{d \theta}\(\sin \theta \frac{d \Theta(\theta)}{d \theta}\) + l(l+1)\Theta(\theta) = 0.$$

The solutions that work for physical situations are the Legendre polynomials $P_l(x)$:

$l$$P_l(\cos \theta)$
1$\cos \theta$
2$\frac{1}{2}(3 \cos^2 \theta -1)$
3$\frac{1}{2}(5\cos^3 \theta - 3\cos \theta)$

$R(r)$ equation

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

This is a 2nd order differential equation. $\Rightarrow$ There are two independent solutions: $$R_l(r) = r^l \ \hbox{and}\ R_l(r)=r^{-(l+1)}.$$

Now, if $V_1$ and $V_2$ are each separately solutions to Laplace's equation $\grad^2 V=0$, then $A_1V_1+B_2V_2$ is also a solution. So the most general solution to Laplace's equation in terms of these harmonic functions is...

Not $$V(r,\theta) = \sum_{l=0}^{\infty} \left[A_l r^l + B_l r^{-l+1}\right] P_l(\cos \theta).$$

Should be: $$V(r,\theta) = \sum_{l=0}^{\infty} \left[A_l r^l + B_l r^{-l-1}\right] P_l(\cos \theta).$$

Legendre polynomials as unit vectors

The Legendre polynomials have two key properties that sound like properties of the unit vectors we're more familiar with...


Let's define this integral as a "dot product" for functions $f(x)$ and $g(x)$: $$f(x) \cdot g(x) = \int_{-1}^{1} f(x)g(x) dx.$$

The Legendre polynomials are orthogonal because they behave like this... $$P_n(x) \cdot P_m(x) = \{ \begin{array}{c} 0 & \ \hbox{ if }\ n \neq m\\ \frac{2}{2n+1} & \ \hbox{ if }\ n=m \end{array}\right.$$

When the arguments of the polynomials are $\cos(\theta)$,

$$\begineq P_n(\cos\theta) \cdot P_m(\cos\theta) &=& \int_0^{\pi} P_n(\cos\theta)P_m(\cos\theta) \sin \theta \,d \theta\\ &=& \{\begin{array}{c} 0 & \ \hbox{ if } \ n \neq m\\ \frac{2}{2n+1} & \ \hbox{ if}\ n=m\end{array}\right.\endeq$$

Note that $d(\cos \theta) = \sin \theta d \theta$. So, for example... $$\begineq P_0(\cos\theta)\cdot P_2(\cos\theta) &=& \int_0^\pi1\cdot\frac{1}{2}(3\cos^2\theta-1) \sin \theta\, d \theta\\ &=&\int_{\cos 0}^{\cos\pi}\frac{1}{2}(3(\cos^2\theta-1)\, d (\cos\theta) \\ &=&\frac{1}{2}\left[3x^3\theta/3-x\right]_{\cos0}^{\cos\pi} \\ &=& \frac{1}{2}\left[(1-1)-(1-1)\right]=0\endeq$$


It is much harder to prove, but it can be shown that...

  • any bounded function $f(\theta)$ on the interval $0 <= \theta <= \pi$,
  • can be expressed as a linear combination of Legendre polynomials: $$f(\theta) = a_0P_0(\cos(\theta)) + a_1P_1(\cos(\theta)) + a_2P_2(\cos(\theta)) + ...$$

With these two properties in hand, it is not too much of a stretch to see the analogy with vectors in 3-d space: The Legendre polynomials act like a set of unit vectors $\uv{x}_l$ that "span" the space of bounded functions on the interval $[0,\pi]$. Any function can be expressed in terms of these unit vectors with a \sum like the one above. The only thing that remains is to figure out how to calculate the coefficients $a_l$--the "components" along each unit vector direction. The analogy with vectors in 3-d proves fruitful--the answer looks like the dot product $\myv v \cdot\uv{x}_l$: $$a_l = f\cdot P_l = \frac{1}{\pi}\int_0^{\pi} f(\theta) P_l(\cos \theta) d \theta.$$

This is exactly the same sort of thing we do in Analytical Mechanics with Fourier components of periodic functions.

Indeed, it turns out that if we tried to find solutions to Laplace's equations by separation of variables in Cartesian coordinates, like:

$$V(x,y,z) = X(x)Y(y)Z(z)$$,

The solutions $X$, $Y$, and $Z$ are Fourier functions (sines and cosines) of $x$, $y$, and $z$.

Back to our original problem

For a problem with azimuthal symmetry, we know the general solution in terms of the Legendre polynomials: $$V(r,\theta) = \sum_{l=0}^{\infty} \left[A_l r^l + B_l r^{-l-1}\right] P_l(\cos \theta),$$ where $P_0=1$, $P_1=\cos \theta$, $P_2=\frac{1}{2}(3 \cos^2 \theta -1)$, etc.

The particular solution should match the boundary conditions of the problem. There are two:

  1. The potential on the surface of the sphere is constant, let's take it to be zero, so that: $$V(r=R,\theta) = 0.$$
  2. The field was initially uniform in the $\uv{z}$ direction before the introduction of the sphere. Far, far away from the sphere, it should still be uniform in the $\uv z$ direction.

    One potential that does that... $$V(r\gg R,\theta)\to-E_0z + C$$

    Setting $V(z=0)=0 => C=0$. So, this now implies that, at large distances from the sphere: $$V(r \gg R,\theta) \to -E_0z = -E_0 r\cos \theta.$$

Apply these boundary conditions to find the precise solution.

Step 1: What terms in $V(r,\theta)$ can be neglected in the limit $r\to\infty$?

$\lim_{r\to\infty}B_lr^{-l-1}=0$, so $$\begineq V(r\gg R,\theta) &\to& \sum_{l=0}^{\infty} A_l r^l P_l(\cos \theta)\\ &=&A_0+A_1r\cos\theta+A_2r^2\frac 12(3 \cos^2\theta-1)+... \endeq $$

Step 2: Apply the boundary condition $V(r\gg R,\theta)\to -E_0 r \cos\theta$. Most of the $A_l$ have to be zero, except for one. Which one (which $l$) and what is it's value?

Ah, let $A_1=-E_0$ and $A_{l\neq 1}=0$. With these coefficients, the general solution now reads $$V(r,\theta)=-E_0r\cos\theta + \sum_{l=0}^\infty B_l r^{-l-1}P_l(\cos \theta).$$

Step 3: Now that you know something about the $A_l$'s, Write out $V(r=R)$ and apply the boundary condition $V(r=R)=0$ to decide which $B_l$'s are possible, and which are not.

$$\begineq 0&=&V(r=R,\theta)\\ &=&-E_0R\cos\theta + B_0 R^{-1} + B_1 R^{-2}\cos \theta+ B_2 R^{-3}\frac 12(3 \cos^2\theta+1)+... \endeq$$ The only way to get this expression to vanish for all $\theta$ is if $B_{l\neq 1}=0$ and $B_1=E_0 R^3$.

The solution which matches both boundary conditions is apparently $$\begineq V(r,\theta)&=&-E_0r\cos\theta+E_0\frac{R^3}{r^2}\cos\theta\\ &=&-E_0r\cos\theta\left(1-\frac{R^3}{r^3}\right). \endeq$$

Does the electric field behave as expected?

For $r\gg R$, we have $V(r\gg R,\theta)\to -E_0r\cos\theta=-E_0 z$, and therefore the electric field is $-\myv \grad V=\uv z \frac{\del V}{\del z}= E_0 \uv z$.

For $r=R$, the electric field should be perpendicular to the surface of the conducting sphere, that is, it should have a component in the $\uv r$ direction, but no $\theta$ or $\phi$ component.

We'll use the Spherical-Polar form of the gradient: $$\begineq \myv E(r,\theta, \phi)&=&-\myv\grad V\\ &=&-\frac{\del V}{\del r} \uv r-\frac 1r\frac{\del V}{\del \theta}\uv \theta -\frac{1}{r\sin \theta} \frac{\del V}{\del \phi}\uv \phi \\ \endeq $$

Taking the terms in turn:

$\phi$ component: $V$ does not depend on $\phi$, so $\frac{\del V}{\del \phi}=0$, and therefore $E_\phi=0$.

$\theta$ component: $$\begineq E_\theta &=&-\frac 1r\frac{\del V}{\del \theta}\\ &=&\frac 1r E_0 r(-\sin\theta)\left(1-\frac{R^3}{r^3}\right)\\ &=&-E_0\sin\theta\left(1-\frac{R^3}{r^3}\right)\\ \endeq $$ From this expression, we can tell that $E_\theta(r=R)=0$. So the field is perpendicular to the surface at $r=R$.

$r$ component: $$\begineq E_r&=&-\frac{\del V}{\del r}\\ &=&-\frac{\del }{\del r}\[-E_0\cos\theta\left(r-\frac{R^3}{r^2}\right)\] \\ &=&E_0\cos\theta\left(1+2\frac{R^3}{r^3}\right)\\ \endeq $$

Charge density, $\sigma$: On the surface of a conductor, the charge density $\sigma$ is related to the electric field just outside the conductor: $$\begineq \sigma&=&\epsilon_0 E=\epsilon_0 E_r(r=R)\\ &=&\epsilon_03E_0\cos\theta\\ \endeq $$