The Angular Component

Once again, it is reasonable to assume a certain dimensional independence which enables separation of variables in terms of the $\theta$ and $\phi$ components. Under this regime:

$$Y(\theta,\phi) = \Theta(\theta)\Phi(\phi) \Longrightarrow$$

$$\frac{1}{\Theta}\left(\sin(\theta)\frac{d}{d\theta}\left(\sin(\th... ...ta}\right) + \Delta \sin^2\theta\right) + \frac{1}{\Phi}\frac{d^2\Phi}{d\phi^2} = 0 \ni$$

$$\frac{1}{\Theta}\left(\sin(\theta)\frac{d}{d\theta}\left(\sin(\theta)\frac{d\Theta}{d\theta}\right) + \Delta \sin^2\theta\right) = m^2 ,$$

$$\frac{1}{\Phi}\frac{d^2\Phi}{d\phi^2}=-m^2 \Longrightarrow \Phi(\phi) = e^{im\phi}, m \in \mathbb{Z}$$ (3.1)

The $\Phi$ side of the equation was quickly solved. We can recast this equation in the following variations, where we let C represent $x=\cos(\theta)$ and S $\sqrt{1-x^2}=\sin(\theta)$ ; we introduce the notation $\blacktriangledown$ which represents an introduction of a new idea into the stream of derivation, and $\blacktriangleright$ is followed by the immediate consequence of this new idea; we then consider the $m=0$ case.

$$S\frac{d}{d\theta}\left(S\frac{d\Theta}{d\theta}\right) + (\Delta S^2\theta)\Theta =$$ 0

$$x=C(\theta) \rightarrow \frac{dx}{d\theta} = -S(\theta)\Longrightarrow S \frac{\partial}{\partial \theta}\left( S\frac{d\Theta}{dx}\frac{dx}{d\theta}\right)$$

$$=S\frac{d}{d\theta}\left(-S^2\frac{d\Theta}{dx}\right) = -2S^2C\frac{d\Theta}{dx} + \left(-S^3\frac{d}{dx} \frac{d\Theta}{d\theta} = S^4\frac{d^2\Theta}{dx^2}\right) \ni$$

$$\blacktriangledown \Theta = y \blacktriangleright S^2y''-2Cy'+\Delta y$$

$$= (1-x^2)y'' - 2xy' + \Delta y = \frac{d}{dx}\left((1-x^2)y'\right) + \Delta y =0$$

As we further press on this equation, we shall find it suggesting its own solution, even betraying the separation constant $\Delta$ .