• March 27th 2012, 03:44 AM
paiggey
Hey Guys

Literally do not know where to start on this one. I've only learnt Multiplicative separation but I think I might be over thinking it.

Assuming that \displaystyle \begin{align*} E \end{align*} is a constant, you have \displaystyle \begin{align*} \left(\frac{\partial S}{\partial r}\right)^2 + \frac{1}{r^2} \left(\frac{\partial S}{\partial \theta}\right)^2 + \frac{a\cos{\theta}}{r^2} = E \end{align*} and if we let \displaystyle \begin{align*} S(r, \theta) = R(r) + \Theta(\theta) \end{align*}, since the first term is only a function of \displaystyle \begin{align*} r \end{align*} and the second is only a function of \displaystyle \begin{align*} \theta \end{align*}, then evaluating derivatives gives \displaystyle \begin{align*} \frac{\partial S}{\partial r} = \frac{dR}{dr} \end{align*} and \displaystyle \begin{align*} \frac{\partial S}{\partial \theta} = \frac{d\Theta}{d\theta} \end{align*}, and substituting gives
\displaystyle \begin{align*} \left(\frac{dR}{dr}\right)^2 + \frac{1}{r^2}\left(\frac{d\Theta}{d\theta}\right)^ 2 + \frac{a\cos{\theta}}{r^2} &= E \\ \frac{1}{r^2}\left(\frac{d\Theta}{d\theta}\right)^ 2 + \frac{a\cos{\theta}}{r^2} &= E - \left(\frac{dR}{dr}\right)^2 \\ \left(\frac{d\Theta}{d\theta}\right)^2 + a\cos{\theta} &= r^2E - r^2\left(\frac{dR}{dr}\right)^2 \end{align*}
Since the left hand side is only a function of \displaystyle \begin{align*} r \end{align*} and the right hand side is only a function of \displaystyle \begin{align*} \theta \end{align*}, the only way they can be equal is if they are both equal to some constant value \displaystyle \begin{align*} \lambda \end{align*}. This gives
\displaystyle \begin{align*} r^2E - r^2\left(\frac{dR}{dr}\right)^2 &= \lambda \\ \left(\frac{d\Theta}{d\theta}\right)^2 + a\cos{\theta} &= \lambda \end{align*}