Monday Math 52

Last week, we looked at solving Laplace’s equation, $\nabla^2f=0$ in three dimensions, specifically Cartesian and cylindrical coordinates. Now, let us consider spherical coordinates (ρ,φ,θ) (we use the physicist’s convention of polar angle θ and azimuthal angle φ).

In spherical coordinates, Laplace’s equation is
$\nabla^2f=\frac{1}{\rho^2}\frac{\partial}{\partial{\rho}}\left(\rho^2\frac{\partial{f}}{\partial{\rho}}\right)+\frac{1}{\rho^2\sin\theta}\frac{\partial}{\partial{\theta}}\left(\sin\theta\frac{\partial{f}}{\partial{\theta}}\right)+\frac{1}{\rho^2\sin^2\theta}\frac{\partial^2f}{\partial\phi^2}=0$
Trying a solution of the form f(ρ,φ,θ)=R(ρ)Φ(φ)Θ(θ), we have:
$\begin{eqnarray}\nabla^2R(\rho)\Phi(\phi)\Theta(\theta)&=&{0}\\R''\Phi\Theta+\frac{2}{\rho}R'\Phi\Theta+\frac{1}{r^2}R\Phi\Theta''+\frac{\cos\theta}{\rho^2\sin\theta}R\Phi\Theta'+\frac{1}{\rho^2\sin^2\theta}R\Phi''\Theta&=&{0}\\\left[\frac{R''}{R}+\frac{2}{\rho}\frac{R'}{R}\right]+\frac{1}{r^2}\left[\frac{\Theta''}{\Theta}+\frac{\cos\theta}{\sin\theta}\frac{\Theta'}{\Theta}+\frac{1}{\sin^2\theta}\frac{\Phi''}{\Phi}\right]&=&{0}\\\left[\rho^2\frac{R''}{R}+2\rho\frac{R'}{R}\right]+\left[\frac{\Theta''}{\Theta}+\frac{\cos\theta}{\sin\theta}\frac{\Theta'}{\Theta}+\frac{1}{\sin^2\theta}\frac{\Phi''}{\Phi}\right]&=&{0}\end{eqnarray}$
The first bracket depends only on ρ, while the second depends only on the angular coordinates. We choose a constant to which the first bracket equals:
$\rho^2\frac{R''}{R}+2\rho\frac{R'}{R}=k$
$\rho^2R''+2\rho{R'}-kR={0}$
The angular portion is then:
$\begin{eqnarray}k+\frac{\Theta''}{\Theta}+\frac{\cos\theta}{\sin\theta}\frac{\Theta'}{\Theta}+\frac{1}{\sin^2\theta}\frac{\Phi''}{\Phi}&=&{0}\\\sin^2\theta\frac{\Theta''}{\Theta}+\sin\theta\cos\theta\frac{\Theta'}{\Theta}+k\sin^2\theta+\frac{\Phi''}{\Phi}&=&{0}\end{eqnarray}$
and we have separated φ and θ terms. Expecting solutions that are continuous and periodic in 0≤φ<2π, we set
$\frac{\Phi''}{\Phi}=-m^2$
$\Phi''+m^2\Phi={0}$
where m is an integer.
This gives us the remaining differential equation
$\sin^2\theta\frac{\Theta''}{\Theta}+\sin\theta\cos\theta\frac{\Theta'}{\Theta}+k\sin^2\theta-m^2=0$
$(\sin^2\theta)\Theta''+(\sin\theta\cos\theta)\Theta'+[k\sin^2\theta-m^2]\Theta=0$ .

To examine the solutions to this last equation, we first consider the m=0 case. Here, we have:
$(\sin^2\theta)\Theta''+(\sin\theta\cos\theta)\Theta'+(k\sin^2\theta)\Theta=0$ .
Let us make the change of variables $x=\cos\theta$ . Then we have
$(1-x^2)\frac{d^2\Theta}{dx^2}-2x\frac{d\Theta}{dx}+k\Theta=0$
$\frac{d}{dx}\left[(1-x^2)\frac{d\Theta}{dx}\right]+k\Theta=0$
This is amenable to solution by the method of Frobenius. One finds that the resulting power series converges for all -1<x<1 (0<θ<π), but that it diverges at x=±1 (blows up at the polar axis) unless the series terminates. As the recursive formula for the coefficients of the power series $P(x)=\sum_{n=0}^{\infty}a_nx^n$ is $a_{n+2}=\frac{n(n+1)-k}{(n+1)(n+2)}a_n$ , the polynomial terminates only if k=l(l+1), where l is a non-negative integer. Then the solution is a polynomial $P_l(x)$ with degree l, which is an even polynomial for even l and an odd polynomial for odd l. The initial coefficient, a₀ or a₁, is arbitrary, and is usually chosen so that $P_l(1)=1$ . These polynomials are known as Legendre polynomials.
They can be generated via Rodrigues’ formula:
$P_n(x)=\frac{1}{2^nn!}\frac{d^n}{dx^n}\left[(x^2-1)^n\right]$ .
The first few polynomials are:
$P_0(x)=1$
$P_1(x)=x$
$P_2(x)=\frac{1}{2}(3x^2-1)$
$P_3(x)=\frac{1}{2}(5x^3-3x)$
$P_4(x)=\frac{1}{8}(35x^4-30x+3)$ .

Now, for m≠0, we have, using k=l(l+1) (here l is not necessarily an integer),
$(\sin^2\theta)\Theta''+(\sin\theta\cos\theta)\Theta'+[l(l+1)\sin^2\theta-m^2]\Theta=0$
$\frac{d}{dx}\left[(1-x^2)\frac{d\Theta}{dx}\right]+[l(l+1)-\frac{m^2}{1-x^2}]\Theta=0$ ,
called the “general Legendre equation.” It has solutions called the associated Legendre functions, written $P_l^{(m)}(x)$ , where l is called the degree and m the order of the function. We already have that
$P_l^{(0)}(x)=P_l(x)$ , the legendre polynomial of degree l, when l is a non-negative integer.
It can be shown that the associated Legendre functions are well behaved on -1≤x≤1 for l and m integers with |m|≤l, l≥0, with the solutions for non-negative m m in terms of Legendre polynomials being
$P_l^{(m)}(x)=(-1)^m(1-x^2)^{m/2}\frac{d^m}{dx^m}\left[P_l(x)\right]$
$P_l^{(m)}(x)=\frac{(-1)^m}{2^ll!}(1-x^2)^{m/2}\frac{d^{l+m}}{dx^{l+m}}\left[(x^2-1)^l\right]$ ,
and the negative m solutions given by
$P_l^{(-m)}(x)=(-1)^m\frac{(l-m)!}{(l+m)!}P_l^{(m)}(x)$ .
(I have used here the Condon-Shortley phase.)
So our solutions will generally have a polar angle component
$\Theta{\theta}=P_l^{(m)}(\cos{\theta})$ .
If we combine these with the azimuthal angular component written in terms of complex exponentials, we then have:
$Y_l^m(\theta,\phi)=A_{l,m}P_l^{(m)}(\cos{\theta})e^{\imath{m}\phi}$
where l and m are integers with l≥0 and |m|≤l. Choosing our constant to produce the othonormal relation:
$\int_{0}^{2\pi}\int_{0}^{\pi}Y_l^m(\theta,\phi){Y_{l'}^{m'}}*(\theta,\phi)\sin\theta\,d\theta\,d\phi=\delta_{l\,l'}\delta_{m\,m'}$
(where the integration is over all solid angles, ^* indicates the complex conjugate, and $\delta_{m\/n}$ is the Kronecker Delta),
we get
$Y_l^m(\theta,\phi)=\sqrt{\frac{2l+1}{4\pi}\frac{(l-m)!}{(l+m)!}}P_l^{(m)}(\cos{\theta})e^{\imath{m}\phi}$ .
These are called spherical harmonics.
The first few are:
$Y_0^0(\theta,\phi)=\frac{1}{2\sqrt{\pi}}$
$Y_1^0(\theta,\phi)=\frac{1}{2}\sqrt{\frac{3}{\pi}}\cos\theta$
$Y_1^{\pm1}(\theta,\phi)=\mp\frac{1}{2}\sqrt{\frac{3}{2\pi}}\sin\theta{e^{\pm\imath\phi}}$
$Y_2^{0}(\theta,\phi)=\frac{1}{4}\sqrt{\frac{5}{\pi}}(3cos^2\theta-1)$
$Y_2^{\pm1}(\theta,\phi)=\mp\frac{1}{2}\sqrt{\frac{15}{2\pi}}\sin\theta\cos\theta{e^{\pm\imath\phi}}$
$Y_2^{\pm2}(\theta,\phi)=\frac{1}{4}\sqrt{\frac{15}{2\pi}}\sin^2\theta{e^{\pm2\imath\phi}}$

Now, for our radial component, we have $\rho^2R''+2\rho{R'}-kR={0}$ ; with k=l(l+1), we get
$\rho^2R''+2\rho{R'}-l(l+1)R={0}$
Trying a solution of the form $\rho^n$ , we find
$\rho^2\cdot{n}(n-1)\rho^{n-2}+2\rho\cdot{n}\rho^{n-1}-l(l+1)\rho^n={0}$
$[n(n-1)+2n-l(l+1)]\rho^n={0}$
$n^2+n-l(l+1)={0}$
$n=l$ or $n=-(l+1)$
So our radial component is of the form $R(\rho)=A_lr^l+B_lr^{-(l+1)}$ , and our solutions to Laplace’s equation in spherical coordinates will be of the form
$f(\rho,\phi,\theta)=\sum_{l=0}^{\infty}\sum_{m=-l}^{l}[A_{l,m}r^l+B_{l,m}r^{-(l+1)}]Y_l^m(\theta,\phi)$ .

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Tags: Laplace's Equation, Math, Monday Math, Separation of Variables, Spherical Coordinates, Spherical Harmonics

This entry was posted on December 29, 2008 at 5:10 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

4 Responses to “Monday Math 52”

Monday Math 53 « Twisted One 151’s Weblog Says:
January 5, 2009 at 12:05 am | Reply
[...] Math 53 By twistedone151 Previously, we introduced the spherical harmonics. We note that as written, the spherical harmonic is real [...]
GreenTeaTropical Says:
January 21, 2009 at 6:46 am | Reply
thanks for the info !
Physics Friday 59 « Twisted One 151’s Weblog Says:
February 13, 2009 at 12:15 am | Reply
[...] separation of variables, the angular components will be given by the spherical harmonics (see here and here). We will then have . With this in place, our Schrödinger equation becomes, for the [...]
Physics Friday 60 « Twisted One 151’s Weblog Says:
February 20, 2009 at 2:24 am | Reply
[...] variables in spherical coordinates, with the angular components being the spherical harmonics (see here and here). As in here, when we perform the spherical coordinate separation , we obtain radial [...]

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Monday Math 52

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