Laplace equation in spherical polar co-ordinate

Solution of Laplace equation in spherical polar co-ordinate:

We know the laplace equation in spherical polar co-ordinate is,

$$\nabla^2 \phi= \frac{1}{r^2} \frac{\delta}{\delta r}(r^2 \frac{\delta \psi}{\delta r})+\frac{1}{r^2 sin\theta}\frac{\delta}{\delta \theta}(sin\theta \frac{\delta \psi}{\delta \theta})+\frac{1}{r^2 sin^2 \theta}\frac{\delta^2 \phi}{\delta \phi^2}=0$$

Since, \(\psi(r,\theta,\phi)= R(r) \Theta \theta \Phi (\phi)\)

Using variable separation method, we get,

$$\frac{\Theta \Phi}{r^2}\frac{\delta}{\delta r}(r^2 \frac{\delta R}{\delta r})+ R\frac{\Phi}{r^2 sin\theta} \frac{\delta}{\delta \theta}(sin\theta\cdot \frac{\delta \Theta}{\delta \theta})+\frac{R\Theta}{r^2 sin^2 \theta}\frac{\delta^2 \Phi}{\delta \Phi^2}=0$$

Multiplying both sides by \(\frac{r^2 sin^2\theta}{R\Theta \Phi} \) we get,

$$\frac{sin^2\theta}{R}\frac{\delta}{\delta r}\biggl(r^2\frac{\delta R}{\delta r}\biggr)+\frac{sin\theta}{\Theta}\frac{\delta}{\delta \theta}\biggl(sin\theta \frac{\delta\Theta}{\delta \theta}\biggr)+\frac{1}{\Phi}\frac{\delta^2\phi}{\delta \phi^2}=0$$

$$\frac{1}{\phi}\frac{d^2\phi}{d\phi^2}=-m^2$$

$$\Rightarrow \frac{d^2\Phi}{d\phi^2}+m^2\Phi=0$$

$$\Rightarrow \Phi(\phi)= Ae^{im\phi}\dotsm(2)$$

$$or,\;\; \frac{sin^2\theta}{R}\frac{\delta}{\delta r}(r^2 \frac{\delta R}{\delta r})+\frac{sin\theta}{\Theta}\frac{\delta}{\delta \theta}(sin\theta\cdot\frac{\delta \Theta}{\delta \theta})-m^2=0$$

Dividing both sides by \(sin^2\theta\) we get,

$$\frac{1}{R}\cdot \frac{\delta}{\delta r}\biggl(r^2 \frac{\delta R}{\delta r}\biggr)+ \frac{1}{sin\theta\Theta}\frac{\delta}{\delta\theta}(sin\theta\frac{\delta\Theta}{\delta\phi})-\frac{m^2}{sin^2\theta}=0$$

$$\frac{1}{R}\cdot \frac{\delta}{\delta r}\biggl(r^2 \frac{\delta R}{\delta r}\biggr)= \frac{1}{sin\theta\Theta}\frac{\delta}{\delta\theta}(sin\theta\frac{\delta\Theta}{\delta\phi})-\frac{m^2}{sin^2\theta}=\lambda^2$$

Taking second term and third term only, we get

$$\frac{1}{sin\theta}\frac{\delta}{\delta \theta}(sin\theta\frac{\delta\Theta}{\delta \theta})+\lambda^2- \frac{m^2}{sin^2\theta}=0$$

Multiplying both sides by \(\Theta\) we get,

$$\frac{1}{sin\theta}\frac{\delta}{\delta \theta}(sin\theta \frac{\delta \Theta}{\delta \theta})+(\lambda^2-\frac{m^2}{sin^2\theta}\biggr)\Theta=0$$

Put,

$$x=cos\theta$$

$$\delta x=-sin\theta d\theta$$

$$\frac{\delta}{\delta \theta}=-sin\theta \frac{\delta}{\delta x}$$

$$or, \frac{1}{sin\theta}\biggl[-sin\theta(-sin\theta)\frac{\delta \Theta}{\delta x})\biggr]+(\lambda^2-\frac{m^2}{1-cos\theta})\Theta=0$$

$$\frac{\delta}{\delta x}\biggl[(1-cos^2\theta)\frac{\delta \Theta}{\delta x}\biggr]+\biggl[\lambda^2-\frac{m^2}{1-x^2}\biggr]\Theta=0$$

$$\frac{\delta}{\delta x}\biggl[(1-x^2)\frac{\delta\Theta}{\delta x}\biggr]+\biggl(\lambda^2-\frac{m^2}{1-x^2}\biggr) \Theta=0$$

$$(1-x^2)\frac{\delta^2\Theta}{\delta x^2}-2x \frac{\delta \Theta}{\delta x}+(\lambda^2-\frac{m^2}{1-x^2})\Theta=0$$

The above equation reduces to associated Legendre if \(\lambda^2=n(n+1)\)

$$\therefore\;\; (1-x^2)\frac{d^2\Theta}{dx^2}-2x\frac{\delta \Theta}{\delta x}+\biggl[n(n+1)-\frac{m^2}{(1-n^2)}b\biggr]\Theta=0$$

The solution of above equation is

$$\Theta(\theta)=P_n^m(x)$$

$$\therefore\;\; \Theta(\theta)=BP_n^m (cos\theta)$$

Now,

$$\frac{1}{R} \frac{\delta}{\delta r}\biggl(r^2\frac{\delta R}{\delta r}\biggr)=\lambda^2$$

$$\frac1R \frac{\delta}{\delta r}\biggl(r^2\frac{\delta R}{\delta r}\biggr)=n(n+1)$$

$$\frac{\delta}{\delta r}\biggl(r^2 \frac{\delta R}{\delta r}\biggr)-n(n+1)R=0$$

Put $$r=e^{\rho}$$

$$\delta r= e^{\rho}\cdot \frac{\delta}{\delta \rho}$$

$$\Rightarrow \frac{\delta}{\delta r}=\frac{1}{e^{\rho}}\cdot \frac{\delta}{\delta \rho}$$

Then,

$$\frac{1}{e^{\rho}} \cdot \frac{\delta}{\delta \rho}\biggl(e^{2\rho }\cdot \frac{1}{e^{\rho}} \cdot \frac{\delta R}{\delta \rho}\biggr)-n(n+1)R=0$$

$$or,\;\; \frac{1}{e^{\rho}}\cdot \frac{\delta}{\delta \rho}\biggl(e^{\rho} \frac{\delta R}{\delta \rho}\biggr)-n(n+1)R=0$$

$$or,\;\; \frac{1}{e^{\rho}}\cdot e^{\rho} \frac{\delta^2 R}{\delta \rho^2}+\frac{1}{e^{\rho}} \cdot e^{\rho}\cdot \frac{\delta R}{\delta \rho}-n(n+1)R=0$$

$$or,\;\; \frac{\delta^2 R}{\delta \rho^2}+\frac{\delta R}{\delta \rho}-n(n+1)R=0$$

Let,\(\frac{\delta}{\delta\rho}=E\)

$$[(D^2+D-n(n+1)]R=0$$

$$or,\;\; [(D^2-n^2)+(D-n)]R=0$$

$$or,\;\;[(D-n)(D+n)+(D-n)]R=0$$

$$or,\;\;[(D-n)(D+n+1)]R=0$$

Either,

$$(D-n)R_1=0$$

$$\frac{\delta R_1}{\delta \rho}=nR_1$$

$$\frac{\delta R_1}{R_1}=n\delta \rho$$

On integrating ,

$$l_n R_1= n\rho+l_n c$$

$$\Rightarrow R_1= Ce^n \rho$$

Now, $$(D+n+1)R_2=0$$

$$\frac{\delta R_2}{\delta \rho}+(n+1)R_2=0$$

$$\frac{\delta R_2}{\delta \rho}=-(n+1)R_2$$

$$\frac{\delta R_2}{R_2}=-(n+1)\delta \rho$$

On integrating we get,

$$l_n R_2=-(n+1)\rho + l_n D$$

$$l_n R_2-l_nD=-(n+1)D$$

$$l_n(\frac{R_2}{D})=-(n+1)\rho$$

$$R_2= De^{-n(n+1)\rho}$$

Now, $$\psi (r,\theta,\phi)= R[(r) \Theta \theta \Phi (\theta)]$$

$$=\sum_{n=1}^\infty\biggl[C_n e^{n\rho}+D_n e^{-[(n+1)\rho][Be^{im\theta}][Ae^{im\theta}]}\biggr]$$

Reference:

1. Adhikari, Pitri Bhakta, and Dya Nidhi Chhatkuli.A Textbook of Physics. Third Revised Edition ed. Vol. III. Kathmandu: SUKUNDA PUSTAK BHAWAN, 2072.
2. Vaughn, Michael T.Introduction to Mathematical Physics. Weinheim: Wiley-VCH, 2007
3. Butkov, Eugene.Mathematical Physics. Reading, MA: Addison-Wesley Pub., 1968.
4. Carroll, Robert W.Mathematical Physics. Amsterdam: North-Holland, 1988.
5. Dass, H. K.Mathematical Physics. New Delhi: S. Chand, 2005.