# boundary condition at rotation axis of a spherically-symmetric system

The quantity I am interested in is not the rotation rate $$\Omega$$, but I will use this quantity nonetheless to make the problem clearer. I am interested in a spherically-symmetric system and in the boundary condition of the rotation rate $$\Omega=u/r\sin\theta$$. Now, at $$\theta=0$$, we have an undetermined form $$0/0$$ and we must use de L'Hopital rule: $$\Omega (r,0)=\text{lim}_{\theta\rightarrow 0}\frac{du/d\theta}{r\cos\theta}\neq 0$$

which is not zero as expected. Using central differences (I assume this is still valid):

$$\frac{du}{d\theta}|_{\theta=0}=\frac{u_1-u-1}{2\Delta\theta},$$

where the index for radius was ignored (the minus 1 should be as an index but it somehow doesn't work within the fraction). Now, $$u_{-1}=-u_1$$ and:

$$\frac{du}{d\theta}|_{\theta=0}=\frac{u_1}{\Delta\theta}=\frac{\Omega_1 r\sin\theta_1}{\Delta\theta}\simeq\Omega_1 r\quad\quad\text{[for \theta_1=\Delta\theta small]}.$$

Finally: $$\Omega_0=\Omega_1.$$

Now, if I decide to discretize the equations regularly in $$\mu=\cos\theta$$, so that:

$$\sin\theta=\sqrt{1-\mu^2}\quad\quad\frac{1}{d\theta}=-\frac{\sin\theta}{d\cos\theta}=-\frac{\sqrt{1-\mu^2}}{d\mu},$$

then:

$$\Omega(r,1)=\text{lim}_{\mu\rightarrow 1}\frac{du/d\mu}{rd\sqrt{1-\mu^2}/d\mu},$$

and

$$\Omega(r,1)=\text{lim}_{\mu\rightarrow 1}-\frac{\sqrt{1-\mu^2}}{r\mu}\frac{du}{d\mu}=0$$

What am I missing here? This should definitely not be zero at the pole (or anywhere).

• What is $u$ in your example? Feb 25, 2023 at 20:03
• It's the velocity in the $\phi$-direction $u=r\sin\theta\Omega$ Feb 25, 2023 at 20:28