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Is it possible theta to become negative in spherical harmonics? If so, whats the qualitative idea of when this might happen, and perhaps an example where it does?

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  • $\begingroup$ Is this like wondering where you get to when you stand at the North Pole and then walk north? $\endgroup$ – Lord Shark the Unknown Jul 25 '17 at 5:22
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It depends on which convention for spherical coordinates you are using. If you're using $\theta$ for the angle from the north pole, that is by definition between $0$ and $\pi$ radians ($180$ degrees). If you're using $\theta$ for the longitude, there are some who allow it to be negative, but usually it's required to be from $0$ to $2\pi$ radians ($360$ degrees).

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  • $\begingroup$ Thanks, Robert! $\endgroup$ – Adilah Jul 27 '17 at 1:11
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Supposing you use the convention that a vector ${\bf r}$ in polar coordinates is given by $$ {\bf r} = \left( \begin{array}{c} r \sin \theta \cos \phi \\ r \sin \theta \sin \phi \\ r \cos \theta \end{array} \right) $$ with $r \in [0,\infty), \theta \in [0,\pi],\phi \in [0,2 \pi]$ and the spherical harmonics $Y_l^m(\theta,\phi)$ then negative values of $\theta$ normally do not occur.

That does no mean however that negative values are forbidden, but that care should be taken. With the above restrictions each point in ${\mathbb R}^3$ has a unique $(r,\theta,\phi)$ description. Allowing negative values of $\theta$ would result in an additional different characterisation of points and is probably undesirable. This is usually solved by mapping the values back on the allowed domain, so if we consider a point $(r,\vartheta,\varphi)$ with $-\pi \leq \vartheta < 0$ we would obtain the point $$ {\bf r} = \left( \begin{array}{c} r \sin \vartheta \cos \varphi \\ r \sin \vartheta \sin \varphi \\ r \cos \vartheta \end{array} \right) = \left( \begin{array}{c} r \sin (-|\vartheta|) \cos \varphi \\ r \sin (-|\vartheta|) \sin \varphi \\ r \cos -|\vartheta| \end{array} \right) = \left( \begin{array}{c} - r \sin |\vartheta| \cos \varphi \\ - r \sin |\vartheta| \sin \varphi \\ r \cos |\vartheta| \end{array} \right) = \left( \begin{array}{c} r \sin |\vartheta| \cos (\varphi+\pi) \\ r \sin |\vartheta| \sin (\varphi+\pi) \\ r \cos |\vartheta| \end{array} \right) $$ So we map $(r, \vartheta,\varphi)$ for negative $\vartheta$ on $(r, |\vartheta|,\varphi+\pi)$. Strictly speaking we might have to use $\varphi-\pi$ to bring it back in to the proper domain.

The spherical harmonics themselves are not affected by this because $$Y_l^m(\theta,\phi)=Y_l^m(-\theta,\phi+\pi)$$ $$Y_l^m(\theta,\phi)=Y_l^m( \theta,\phi+2 \pi)$$ But other functions in polar coordinates might, for instance $\sin \theta$, which is not a spherical harmonic, would be an example where allowing negative angles would become an issue.

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  • $\begingroup$ Thanks a lot, Ronald! $\endgroup$ – Adilah Jul 27 '17 at 1:11

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