1
$\begingroup$

In general we can say that if the potential is specified in the surface of a sphere with azimuthal symmetry $(m=0)$ has a solution:

$$ \Phi(r,\theta)= \sum_{l=0}^\infty \left[ A_lr^l + B_lr^{-(l+1)}\right]P_l\left(cos\theta \right) $$

and the potential for a sphere without azimuthal symmetry:

$$ \Phi(r,\theta,\phi)= \sum_{l=0}^\infty \sum_{m=-l}^l \left[ A_{l,m}r^l + B_{l,m}r^{-(l+1)}\right]Y_{l,m}\left( \theta,\phi \right)$$

Lets think in the case (for azimuthal symmetry) where the sphere has potentials $\Phi=\Phi_0$ for the northern hemisphere and $\Phi=-\Phi_0$ for the southern.

That will give me certain solution.

If I rotate that sphere $\frac{\pi}{2}$ only in the $z-axis$ (left hemisphere $\Phi=-\Phi_o$ and right hemisphere $\Phi=\Phi_0$), that will give me another solution using the second ecuation.

The question is how can I get from one solution to the other using a rotation matrix?

$\endgroup$
1
$\begingroup$

This question is very similar to this one

Is there a relation between the Legendre generator function and Spherical Harmonics for a Potential?

See Luc's answer and the paper referenced.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.