Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question
    
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. –  CuriousOne Aug 28 at 8:06
    
Where, specifically? I'm quite confused, on how even to represent a tripole in a multipole expansion. –  Anthony Aug 28 at 8:11
    
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". –  CuriousOne Aug 28 at 8:22
    
I see. It seems like the multipole basis only contains even functions, though... –  Anthony Aug 28 at 8:25
    
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. –  CuriousOne Aug 28 at 8:34

1 Answer 1

up vote 9 down vote accepted

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:

http://physics.stackexchange.com/a/127496/1236

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.

share|improve this answer
    
Does that mean that there is no potential that is of that form? –  Anthony Aug 28 at 8:38
    
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. –  Luboš Motl Aug 28 at 8:40
    
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. –  Luboš Motl Aug 28 at 8:43
    
So what would $sin(3\theta)$ be. in terms of multipoles? I feel like it's orthogonal to all the poles... –  Anthony Aug 28 at 8:44
    
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. –  Luboš Motl Aug 28 at 8:46

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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