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#+title: Solving the laplace equation in Spherical coordinates
#+author: Thomas Albers Raviola
#+date: 2022-12-15
#+setupfile: ../../math_options.org
* Disclaimer
This site as of now just a technology demonstration and its claims
should not be taken as true (even though I myself am pretty confident
they are)
* Laplace equation
{{{beg-eqn}}}
\Delta \phi = 0
{{{end-eqn}}}
Solutions to this equation are called harmonics
* Laplace equation in spherical coordinates
Using the definition for the laplace operator in spherical coordinates, it
follows:
{{{beg-align}}}
\frac{1}{r^2}\pderiv{}{r}\left(r^2 \pderiv{\phi}{r}\right) + \frac{1}{r^2\sin\theta} \pderiv{}{\theta}\left(\sin\theta\pderiv{\phi}{\theta}\right) + \frac{1}{r^2\sin^2\theta}\frac{\partial^2\phi}{\partial\varphi^2} &= 0\\
\pderiv{}{r}\left(r^2 \pderiv{\phi}{r}\right) + \frac{1}{\sin\theta} \pderiv{}{\theta}\left(\sin\theta\pderiv{\phi}{\theta}\right) + \frac{1}{\sin^2\theta}\frac{\partial^2\phi}{\partial\varphi^2} &= 0
{{{end-align}}}
Using sepration of variables $\phi$ becomes
{{{beg-eqn}}}
\phi(r, \theta, \varphi) = R(r)Y(\theta, \varphi)
{{{end-eqn}}}
Applying this to the original equation produces
{{{beg-align}}}
\pderiv{}{r}\left(r^2 \pderiv{R}{r}\right)Y(\theta, \phi) + \left(\frac{1}{\sin\theta} \pderiv{}{\theta}\left(\sin\theta\pderiv{\phi}{\theta}\right) + \frac{1}{\sin^2\theta}\frac{\partial^2\phi}{\partial\varphi^2}\right)R(r) &= 0\\
\frac{1}{R(r)}\pderiv{}{r}\left(r^2 \pderiv{R}{r}\right) + \frac{1}{Y(\theta, \phi)}\left(\frac{1}{\sin\theta} \pderiv{}{\theta}\left(\sin\theta\pderiv{\phi}{\theta}\right) + \frac{1}{\sin^2\theta}\frac{\partial^2\phi}{\partial\varphi^2}\right) &= 0
{{{end-align}}}
Because the terms depend on different independant variables, the only way the
equation holds is if both terms are constant.
{{{beg-align}}}
&\frac{1}{R(r)}\pderiv{}{r}\left(r^2 \pderiv{R}{r}\right) = \lambda\\
&\frac{1}{Y(\theta, \phi)}\left(\frac{1}{\sin\theta} \pderiv{}{\theta}\left(\sin\theta\pderiv{Y}{\theta}\right)+ \frac{1}{\sin^2\theta}\frac{\partial^2Y}{\partial\varphi^2}\right) = -\lambda
{{{end-align}}}
* Angle dependant term
We again use separation of variables to solve the partial differential equation
of the angle dependant term.
{{{beg-eqn}}}
Y(\theta, \varphi) = \Theta(\theta)\Phi(\varphi)
{{{end-eqn}}}
Replacing into the equation
{{{beg-eqn}}}
\lambda\sin^2\theta + \frac{\sin\theta}{\Theta(\theta)}\pderiv{}{\theta}\left(\sin\theta\pderiv{\Theta}{\theta}\right)
+ \frac{1}{\Phi(\varphi)}\frac{\partial^2\Phi}{\partial\varphi^2} = 0
{{{end-eqn}}}
Based on a similar argument it follows that both terms must be constant, with
this we may now solve for $\Phi$
{{{beg-eqn}}}
\frac{1}{\Phi(\varphi)}\frac{\partial^2\Phi}{\partial\varphi^2} = -m^2
{{{end-eqn}}}
{{{beg-eqn}}}
\Phi(\varphi) = e^{-m\varphi{}i}
{{{end-eqn}}}
{{{beg-eqn}}}
\lambda\sin^2\theta + \frac{\sin\theta}{\Theta(\theta)}\pderiv{}{\theta}\left(\sin\theta\pderiv{\Theta}{\theta}\right) = m^2
{{{end-eqn}}}
{{{beg-eqn}}}
l(l+1)\sin^2(\theta)\Theta(\theta) + \pderiv{}{\theta}\left(\sin\theta\pderiv{\Theta}{\theta}\right) = m^2\Theta(\theta)
{{{end-eqn}}}
{{{beg-eqn}}}
l(l+1)\sin^2(\theta)\Theta(\theta) + \sin(\theta)\left(\cos(\theta) \Theta'(\theta) + \sin(\theta) \Theta''(\theta)\right) = m^2\Theta(\theta)
{{{end-eqn}}}
{{{beg-eqn}}}
\sin^2(\theta) \Theta''(\theta) + \sin(\theta)\cos(\theta) \Theta'(\theta) + (l(l+1)\sin^2(\theta) - m^2)\Theta(\theta) = 0
{{{end-eqn}}}
{{{beg-eqn}}}
\Theta''(\theta) + \cot(\theta)\Theta'(\theta) + \left(l(l+1) - \frac{m^2}{\sin^2\theta}\right)\Theta(\theta) = 0
{{{end-eqn}}}
{{{beg-eqn}}}
\Theta(\theta) = P(\cos \theta)
{{{end-eqn}}}
{{{beg-eqn}}}
\cot(\theta)\Theta'(\theta) = \cot(\theta) \deriv{}{\theta}P(\cos \theta)=-\cos(\theta) P'(\cos\theta)
{{{end-eqn}}}
{{{beg-eqn}}}
\frac{\text{d}^2}{\text{d}\theta^2}(P(\cos\theta)) = \sin^2(\theta) P''(\cos\theta) - \cos(\theta) P'(\cos\theta)
{{{end-eqn}}}
{{{beg-eqn}}}
\sin^2(\theta) P''(\cos\theta) - 2\cos(\theta) P'(\cos\theta) + \left(l(l+1)-\frac{m^2}{\sin^2\theta}\right)P(\cos\theta) = 0
{{{end-eqn}}}
{{{beg-eqn}}}
x = \cos\theta
{{{end-eqn}}}
{{{beg-eqn}}}
(1 - x^2)P''(x) - 2xP'(x) + \left(l(l+1)-\frac{m^2}{1-x^2}\right)P(x) = 0
{{{end-eqn}}}
Solve using Frobenius Method
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