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In a previous question, I noted that if you have a charge distribution with nonzero charge, then it is possible to choose an origin (at the centre of charge) which makes its dipole moment vanish, and which therefore takes out the subleading order in the multipole expansion of the resulting electrostatic potential $$ \Phi(\mathbf{r})=\frac{Q}{|\mathbf{r}-\mathbf{r}_c|}+O\left(\frac{1}{|\mathbf{r}-\mathbf{r}_c|^{3}}\right). $$ Following on that standard observation, I asked for a result for the next layer up on the multipole expansion: given a neutral system, with a nonzero dipole moment, is it possible to choose an origin which will completely kill the coefficients on the subleading multipole, i.e. which will make all the quadrupole moments vanish.

The answer to that, as it turns out, is much more restrictive than the scalar case: this is possible if and only if the quadrupole is cylindrically symmetric (i.e. if and only if it has two equal eigenvalues) and the dipole moment lies along that axis of symmetry.


I now want to ratchet up the position on the multipole ladder, and ask about the next nontrivial step:

  • Given a charge distribution with zero total charge and dipole moment, is it possible to choose an origin such that the octupole moments will vanish completely? What restrictions need to be imposed on the quadrupole and octupole moments for this to be possible?

There is also the obvious question of what happens for some arbitrary multipole $\ell$ at some finite point up the ladder, but given how hard the quadrupole case was, I'd like to keep it simple: kind of like in this other question along similar lines, I feel that there isn't yet a useful pattern that can be generalized to arbitrary multipoles, particularly since the passage from quadrupoles to higher-rank tensor takes out the eigenvectors as a tool of analysis, and I don't really know of good equivalent tools for higher-rank tensors. So I'd like to keep things focused for now and keep the general case for later.

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This may not fully answer your question, but there is a general theorem that the lowest order (ell) non-vanishing moment is independent of origin.

Let's see how this helps. If the monopole moment does not vanish, then its value is independent of origin. This means that the dipole moment is not independent of origin. If we change the origin through $x^i\to x^i+a^i$, then the dipole changes as \begin{equation} d^i \to d^i +a^i M \end{equation} where M is the non vanishing monopole moment. Since the dipole moment has 3 components, it is possible to choose the origin to force the dipole moment to vanish. This same choice of origin will not necessarily cause any other multipole moment to vanish.

If the monopole moment vanishes, then the value of the dipole moment is independent of origin. This means that changing the origin will affect the quadrupole moment. It has 5 degrees of freedom and so, in general, a change of origin cannot make the quadrupole moment vanish. You would need two additional condition make its vanishing even possible. The quadrupole moment changes as \begin{equation} Q^{ij} \to Q^{ij} + 2\left(a^{(i}d^{j)}-\frac13\delta^{ij}a^k d_k\right). \end{equation} This shows the exact form that the quadrupole moment must have in order for there to be a chance for it to vanish.

You can continue this argument to the quadrupole moment being the first non-vanishing multipole and see what form the octupole moment must have for it to be possible for a coordinate change to make that moment vanish. It will be even more restrictive since you only have 3 degrees of freedom to make 7 degrees of freedom vanish.

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  • $\begingroup$ No, this does not address the question at all. I'm well aware of the theorem you mention -- the question is about what happens to the multipole moment immediately after the first nonvanishing one. To the extent that you address it here, it is not close to the level of detail that would begin to count as an answer -- see the previous thread linked in the question above for the level of detail at the dipole-to-quadrupole layer. (Otherwise, though, welcome to the site!) $\endgroup$ – Emilio Pisanty Sep 18 '19 at 23:37

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