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Take a field $\phi(\bf{x})$ created from a charge distribution contained within a radius $R$. The multipole expansion in spherical harmonics $Y_{\ell,m}$ outside of $R$ is approximated by:

$$ \phi({\bf x}) \approx \frac{1}{4\pi \epsilon_0} \sum_{\ell=0}^{\ell_{MAX}} \sum_{m=-\ell}^{\ell} \frac{4\pi}{2\ell +1} \alpha_{\ell, m} \frac{Y_{\ell,m}(\theta, \phi)}{r^{\ell +1}} $$

Given only the finite set of the multipole moments $\alpha_{\ell,m}$, is it possible to find a charge distribution of $N$ discrete charges $(q_i, r_i, \theta_i, \phi_i)$ that "best-fit" this potential? Right now I'm using a simple minima finder over the $3^N$ variables (see note 1), but I'm wondering if there is any prior work/observations on inverting the multipole moments.

Note 1

If the positions of the charges are fixed, the charge magnitudes $q_i$ are simply linear combinations:

$$ \alpha_{\ell, m} = \sum_i^N q_i Y^*_{\ell,m}(\theta_i, \phi_i) r_i^\ell $$

thus simple linear algebra gives the best-fit charge magnitudes.

Note 2

It's easy to see that this charge distribution need not be unique, indeed if $\ell_{MAX}=0$ any combination such that $\sum_i^N q_i \propto \alpha_{0,0}$ should work. This is ok, I'm more concerned with finding good approximations than the absolute best fit.

Note 3

If the charge magnitudes are real (which is a given for a physical problem), the number of unique moments are reduced by roughly a factor of 2 since:

$$ \alpha_{\ell,-m} = (-1)^m \alpha_{\ell,m}^* $$

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It's a nice problem but the solution is indeed very non-unique - for example, you may scale the distances from the origin $r$ and the charges in various inverse ways, and rotate the four $+-+-$ charges defining a quadrupole around the axis, and do many other things. And because the solution isn't unique, it seems unlikely that there is a "canonical formula" of any sort. –  Luboš Motl Nov 2 '12 at 20:37
    
@LubošMotl I agree, but the problem is motivated by a practical concern, making (pseudo) accurate "toy models" of electrostatics for protein solutions. As such any thoughts on the matter, canonical or not, would suffice as solutions in this case. I would expect that the symmetry of the problem should suggest something better than brute force optimization. –  Hooked Nov 2 '12 at 21:55
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2 Answers

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This problem has a unique solution if you only allow charges of one sign; it is known as the moment problem and is one of the central problems of measure theory. The wikipedia article on it should provide a good starting point for reading on it.

However, as Luboš points out, for a signed measure the moment problem is usually indeterminate. One way to phrase this is that there is a big set of charge distributions for which all moments vanish. (This includes, for instance, all bounded charge distributions with a conducting shell around them.) I don't know of results for finding solutions in the indeterminate case but phrasing the problem in these terms might help.

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For a given $\alpha_{\ell,m}$ (i.e. for a given term in your potential), it is straightforward to find a set of $2^\ell$ charges whose potential will be exactly the same as that term to leading order. In fact, there are systematic ways to do it - you can create a "basis" so that each $\alpha_{\ell,m}$ has a particular set of charges associated with it. The problem is that if the charges and their separations are finite, their potential will also include higher-order terms that will "mess up" the agreement between their potential and the one you're trying to match. The $\ell$th order will agree, but higher orders won't.

However, if you take the $2^{\ell}$ charges and squeeze them together while letting their magnitudes get bigger in such a way that $\alpha_{\ell,m}$ does not change, i.e. the leading order of their potential stays constant, then the higher order terms get smaller quickly. In the ideal limit that means replacing your $2^\ell$ charges with a $\ell$th order point multipole. Depending on your analytical or numerical needs, you could stop short of the infinitesimal limit and keep the charges and their separations finite, but you will lose accuracy in the higher-order terms of the potential.

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The observations are nice and indeed useful when $N=2^\ell$, but my original idea was to approximate the potential with some value of $N$ such that $N < 2^\ell$. In this case I'd like to approximate the basis vectors as "best" as possible using only a limited subset. –  Hooked May 28 '13 at 21:03
    
In fact, you need not specify each charge separately. For a given $\ell$, there are only $2\ell+1$ charge configurations needed. Once you've specified how those configurations are laid out, you just need one number/magnitude for each. (And if you know more about the physical system, e.g. if the charge distribution on the proteins is linear, you can do much better than that.) –  pwf May 28 '13 at 21:25
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