With multipole expansions, we speak only of monopoles, dipoles, and $2^n$-poles. Why is there nothing like a tripole? So how would something like $rsin(3 \theta)$ be expressed with a multipole expansion? Is the reason for not having terms like this in a multipole expansion that it is redundant, or is there some more fundamental reason?
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1$\begingroup$ There is an infinite number of possible decompositions similar to multipoles. Multipoles arise as a natural (orthogonal) choice for the decomposition of solutions of wave equations in spherical coordinates, but there is no reason, why one couldn't chose a different decomposition, e.g. for a problem with a three-fold symmetry, in which a tripole term would make for a better base element. Those problems are simply not as common as the spherical problems, so we don't discuss them much in textbooks. A good place to look for 3,6,12- etc. poles may be the literature about electrical machines. $\endgroup$– CuriousOneCommented Aug 28, 2014 at 8:06
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$\begingroup$ Where, specifically? I'm quite confused, on how even to represent a tripole in a multipole expansion. $\endgroup$– user24082Commented Aug 28, 2014 at 8:11
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1$\begingroup$ A multipole expansion is similar to a Fourier transform, which maps functions to an infinite number of Fourier coefficients. One can write any function as a linear combination over the Fourier (multipole) base functions. The coefficients in these series can be calculated by integrating over the function multiplied with the base functions. In case of a function with three-fold symmetry, the multipole-series would have infinitely many non-zero terms, though. I believe the non-trivial implications of that are handled by a mathematical theory called "functional analysis". $\endgroup$– CuriousOneCommented Aug 28, 2014 at 8:22
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$\begingroup$ I see. It seems like the multipole basis only contains even functions, though... $\endgroup$– user24082Commented Aug 28, 2014 at 8:25
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$\begingroup$ That's why there will be infinitely many non-zero terms in the expansion of a function with three-fold symmetry over base functions that lack three-fold symmetry. This is part of the problem that you can also encounter in molecular physics and chemistry, where molecules can form three-fold and six-fold symmetries just fine, while the base functions that are solutions to a Schroedinger equation with spherical potential don't have that symmetry. As a result, approximating certain molecular orbitals by adding a few terms of spherical atomic orbitals has significant convergence problems. $\endgroup$– CuriousOneCommented Aug 28, 2014 at 8:34
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1 Answer
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The function $\sin 3\theta$ on the unit sphere is not an eigenfunction of the Laplacian on the sphere, i.e. the angular part of the Laplacian, i.e. of $L^2$, so it is not convenient a basis vector in problems whose Hamiltonian involves the Laplacian.
The function $\sin 3\theta$ may be written as a combination of spherical harmonics $Y_{lm}$ with many different values of $(l,m)$ so it is a "mixture" of multipoles of different "rank". For the more natural basis of functions on the sphere that may see as basis vectors, see
The table of spherical harmonics
https://en.wikipedia.org/wiki/Table_of_spherical_harmonics
For example, the spherical harmonics $Y_{3,\pm 3}$ are proportional to $$ Y_{3,\pm 3} \sim \exp(\pm 3i\phi) \sin^3 \theta $$ which is very similar to $\sin 3\theta$ but has the extra $\phi$-dependence. Similarly, one may look at the function $Y_{30}$ which is similar to $\sin 3\theta$ but prefers cosines and so on. Either $\sin^3 \theta$ or $\cos^3\theta$ (check it!) without any $\phi$-dependence is a combination of $Y_{30}$ and $Y_{10}$.
Once one realizes why the spherical harmonics are the preferred, more natural basis, we may carefully discuss the spherical harmonics' association with the multipole expansion. For example, we learn that $Y_{3,m}$ for any $m$, including the functions similar to yours above, are associated with octupoles, not "tripoles"!
More generally, $Y_{\ell m}$ is the angular part of the $2^\ell$-pole.
The powers of two are a natural way to describe the terms in the multipole expansions for reasons explained elsewhere, e.g. here:
In the multipole terminology, a "tripole" would correspond to a triplet (e.g. vertices of a triangle) of charges. If their total charge would be nonzero, there would be a leading "monopole" term. If the total charge cancelled, the system of 3 charges would still have a dipole moment. Unless the three (nonzero) charges would lie on the same line, the dipole moment couldn't be canceled.
answered Aug 28, 2014 at 8:33
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$\begingroup$ Does that mean that there is no potential that is of that form? $\endgroup$– user24082Commented Aug 28, 2014 at 8:38
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$\begingroup$ The particular potential you wrote down doesn't obey $\nabla^2 \phi = 0$ almost anywhere. It's easy to check. You can't fix this problem even by choosing a different dependence on $r$ exactly because the angular part isn't an eigenstate of the angular part of the Laplacian, i.e. the $L^2$ operator. $\endgroup$ Commented Aug 28, 2014 at 8:40
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$\begingroup$ Sorry, I was inaccurate. You chose the function so nicely that the laplacian vanishes almost everywhere - but it doesn't on the $z$-axis i.e. for $\theta=0$ or $\theta=\pi$. Or maybe it's an eigenstate of the Laplacian with a nonzero eigenvalue? You may calculate those things. $\endgroup$ Commented Aug 28, 2014 at 8:43
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$\begingroup$ So what would $sin(3\theta)$ be. in terms of multipoles? I feel like it's orthogonal to all the poles... $\endgroup$– user24082Commented Aug 28, 2014 at 8:44
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$\begingroup$ It is definitely not orthogonal to the other spherical harmonics, pretty much to none of them. The inner product with $Y_{l0}$ is nonzero for all odd values of $l$ - it's the integral over $\theta$ from zero to pi of the product of the functions times an extra $\sin\theta$. The spherical harmonics form a basis so they're complete - orthogonal to each other and you can't find any other that would be orthogonal to them. $\endgroup$ Commented Aug 28, 2014 at 8:46