Additional Theorem for Spherical Harmonics

One can use the static multipole solution in order to infer the behavior of the spherical harmonics under rotation. We consider an arbitrary orthogonal coordinate rotation

For any fixed point ${\cal P}$ the effect of this change is given

$$(x({\cal P}),y({\cal P}),z({\cal P})) \sim\!\!\to (x'({\cal P}),y' ({\cal P}),z'({\cal P}))$$

$$(\theta ,\varphi ) \sim\!\!\to (\theta ',\varphi ')$$

but the radius and the Laplacian remain fixed;

$$\sqrt{x^2+y^2+z^2} = r =\sqrt{x^{\prime 2} +y^{\prime 2}+z^{\prime 2}}$$

$$\frac{\partial^2}{\partial x^2} +\frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} = \frac{\partial^2}{\partial x^{\prime 2}} +\frac{\partial^2}{\partial y^{\prime 2}}+\frac{\partial^2}{\partial z^{\prime 2}}\,.$$

Thus for any fixed point ${\cal P}$ the potential $\psi$ determined relative to the rotated coordinate system is the same as that relative to the unrotated one,

$$\psi (r({\cal P}),\theta ({\cal P}),\varphi ({\cal P})) = \psi (r'({\cal P}), \theta '({\cal P}),\varphi '({\cal P}))\,.$$

In other words, relabelling the coordinates of a fixed point has no effect on the (measured) potential at this point. This equality holds for all radii $r$ .

It follows that the coresponding $2^\ell$ -pole fields are equal for each integral. Thus

$$\sum^\ell_{m=-\ell} Y^m_\ell (\theta ,\varphi )\langle Y^m_\ell ,... ...\sum^\ell_{m=-\ell} Y^m_\ell (\theta ',\varphi ')\langle Y^m_\ell ,g\rangle\,.$$

This equality also holds for all boundary value functions $g$ . Consequently,

$$\sum^\ell_{m=-\ell} Y^m_\ell (\theta ,\varphi )\overline{Y}^m_\el... ...ell_{m=-\ell} Y^m_\ell (\theta ',\varphi ')Y^m_\ell (\theta_0',\varphi_0')\,,$$

$\ell =0,1,2,\dots$ . This is a remarkable statement: it says that this particular sum of products is unchanged under a rotation. It is a scalar, even though each individual factor does get altered.

Question: What is this rotationally invariant sum equal to?

To find out, orient the $z'$ -axis so that it passes through the source point ${\cal P}_0$ , which now becomes the new North pole. Relative to the new $(x',y',z')$ frame the spherical coordinates of the source point ${\cal P}_0$ and the observation point ${\cal P}$ are given by

$$\textrm{Source~point}~{\cal P}_0~\textrm{(\lq\lq new North Pole'')} \colon \left\{\begin{array}{l} \theta_0' = 0\\ \varphi_0'=\textrm{indeterminate}\end{array}\right.$$

$$\textrm{Observation~point}~{\cal P} \colon \left\{\begin{array}{l} \theta '=\Theta\\ \varphi '=\Phi\end{array}\right.$$

Suppose we reexpress the spherical harmonics in terms of the associated Legendre polynomials $P^m_\ell (\cos\theta )$ . Then

$$Y^m_\ell (\theta ,\varphi ) = \sqrt{\frac{2\ell +1}{2}}~\sqrt{\fr... ... {\ell +m!}} (-1)^m P^m_\ell (\cos\theta )\frac{e^{im\varphi}}{\sqrt{2\pi}}\,.$$

These polynomials satisfy

$$P^m_\ell (\cos 0)=0~~\qquad~~m\not= 0$$

and

$$P^0_\ell (\cos 0)=1\,.$$

Consequently,

$$\overline{Y}^m_\ell (0,\varphi_0') = \delta_{m0}\sqrt{\frac{2\ell +1} {4\pi}}$$

$$Y^0_\ell (\Theta ,\Phi ) = \sqrt{\frac{2\ell +1}{4\pi}} P_\ell (\cos\Theta )\,.$$

The rotationally invariant sum simplifies into

$$P_\ell (\cos\Theta ) = \sum^\ell_{m=-\ell}\frac{(\ell -m)!}{(\ell... ...} P^m_\ell (\cos\theta )P^m_\ell (\cos\theta_0) e^{im(\varphi -\varphi_0)}\,.$$

We conclude that $P_\ell (\cos\Theta )$ , which is an amplitude pattern azimuthally symmetric around the new North pole $(\theta_0,\varphi_0)$ , is a finite linear combination of amplitude patterns centered around the old North pole.