# Electric quadrupole moment and electric octupole moment

From a Chinese electrodynamics textbook, first we take a $$1/r$$ Taylor expansion into the electric potential equation and simplify it, then we get the final equation, and we call each term different name. But why are these terms named electric dipole moment, electric quadrupole moment and electric octupole moment? Could I name them, say, electric tripole moment or electric pentapole moment, or electric hexapole moments? (I know they are wrong.)

I am curious about deeper physical significance of this naming convention, and the relation between it and electric polarizability Χ's order (electric polarization).

• It might be clearer if you included a (translated?) copy of the section of your text that you are looking at. For use of the word "moment", you might want to see the Wikipedia entry Moment (physics). – Kyle Kanos Feb 22 at 11:19
• "di" means 2, and a dipole moment is equivalent to two charges with a separation. "quad" means 4, and a quadrapole moment is equivalent to 4 charges with appropriate separations. Any arrangement of charges can be represented by an expansion of these (and higher) moments, with all moments being powers of 2. If you prefer to represent it with other expansions (spherical harmonics, say), go ahead... – Jon Custer Feb 22 at 15:17
• @JonCuster your comment looks like an answer – GiorgioP Feb 23 at 8:16

This is just a historic notation, and by now the only real reason for why we use it is plain inertia.

Originally, the reasons terms like dipole and quadrupole were chosen is that, as pointed out in the existing answers, you can indeed make dipole-like and quadrupole-like fields, and even octupole-like fields, by having two, or four, or eight point charges ("poles") of opposite signs and arranging them in such a way that all of the previous orders vanish.

However, this has several problems.

• For one, from the quadrupole onwards, this does not represent the only way to get fields at that order of multipolarity. As a simple example, you can get a quadrupole field with three point charges, by having a $$-2q$$ charge at the origin and $$+q$$ charges at $$(0,0,\pm d)$$. But when you get to octupoles, depending on the configuration, you can make do with as few as six or even four point charges.

• Moreover, if you want to get past the octupole level, you're completely stuck: you cannot fit a hypercube in three dimensions, which is what you would need to do to keep replicating the mantra "a $$2^{\ell+1}$$-pole is just two opposite $$2^\ell$$-poles side by side".

From the formal perspective, these problems are all solved - there is a very clear understanding of what we mean by a multipole, and the respective behaviours are very firmly understood.

The only real hitch is the nomenclature, which is kind of stuck at keeping up with the names given to these objects when we only understood a few particular cases of this hierarchy. But as far as nomenclature goes, there is a bigger problem in that there isn't really any clear alternative that would be 'better' in any objective sense, so we just keep the existing names, in the understanding that they only work well for the base cases, and that after that they are only imperfect renditions of the mathematical concept they embody.

As for this, though,

Could I name them, say, electric tripole moment or electric pentapole moment, or electric hexapole moments?

you can always make up your own private notation, particularly if your goal is to confuse your readers. If you're trying to do anything else, though, then I would strongly advise against it.

(Though, that said, the notation differs somewhat depending on the dimensionality. If you're working in 2D, then the mathematical structures change, and it becomes much more meaningful to use the name "hexapole" (or even "sextupole") for the $$\ell=3$$ sector, and this does get used in practice (and also e.g. here). But unless you know exactly what you're doing, stick to the $$2^\ell$$-pole notation from the 3D formalism.)

The way I think about it is this:

• we can take two opposite monopoles and put them together to make a dipole, it has no monopole moment but there is still an asymptotic field, we call this the dipole field,
• we can then take two opposite dipoles and put them together to make a quadrupole, it had no monopole or dipole moment, but there is still an asymptotic field, we call this the quadrupole field,
• we can then take two opposite quadrupoles, put them together to make an octupole, it has no monopole or dipole or quadrupole moment, but there is still an asymptotic field, we call this the octupole field...

You can see that in the most obvious way to define such things the number of charges that have to be brought together doubles with each iteration, justifying the name.

• While not specifically wrong, this answer is limited in its applicability. For one, from quadrupole onwards, not all multipoles are obtained that way. And for another, as was expressed clearly in a previous thread, you cannot fit a hyper-cube in 3 dimensions, i.e. your post implies that you can keep going as is, when in fact the recipe stops short at the octupole. – Emilio Pisanty Feb 22 at 14:30
• That's definitely fair, I cannot extend this argument to the -- what does one even call it? Hexadecupole? – CR Drost Feb 22 at 20:18
• I normally see it spelled as 'hexadecapole'. Frankly, I don't really see any reasonable use of the term in a real-world sentence other than "from the hexadecapole onwards, it makes very little sense to use explicit names instead of systematic mathematical notation". But then again Google Scholar comes up with 9k hits, so I guess it does have some uses. (And, of note, "hexadecupole" does produce some 900 hits, so I guess that's also a thing.) – Emilio Pisanty Feb 22 at 22:31

Let's begin with two opposite charges, $$+q$$ and $$-q$$ placed at a short distance form each other. I assume you know that in the potential they generate the term $$1/r$$ (the "monopole") is lacking whereas the "dipole" $$1/r^2$$ term is present. If positions of charges are $$+q:\quad (d/2, 0, 0) \qquad -q:\quad (-d/2, 0, 0)$$ then the dipole moment is vector $$(qd, 0, 0).$$ To get a pure dipole potential you must take the limit $$q\to\infty$$, $$d\to0$$, with $$q\,d$$ constant.

Now consider 4 charges, $$\#1$$ positive $$+q$$, $$\#2$$ negative $$-q$$, $$\#3$$ still $$-q$$, $$\#4$$ equal $$+q$$ again. Let's place them as follows: \eqalign{ \#1:& +q \quad (d/2, d/2, 0) \cr \#2:& -q \quad (d/2, -d/2, 0) \cr \#3:& -q \quad (-d/2, d/2, 0) \cr \#4:& +q \quad (-d/2, -d/2, 0).\cr} You can see that these are two dipoles, of opposite moments, one near the other. Now potential lacks the $$1/r^2$$ term too; its series begins with $$1/r^3$$. Needless to say, this is a quadrupole. In order to get a pure quadrupole a limit is to be taken again: $$q\to\infty$$, $$d\to0$$, $$q\,d^2$$ constant.

In general to jump from a multipole to the following one we must pair a multipole with another exactly alike, but for charge signs. So the number of charges gets doubled.

But quadrupole moment is a more complex entity: not a vector but a tensor. Actually a quadrupole may be built in other ways. Just an example. \eqalign{ \#1:& +q \quad (0, 0, d) \cr \#2:& -2q \quad (0, 0, 0) \cr \#3:& +q \quad (0, 0, -d).\cr} Its moment is a different tensor and can't be transformed into the former even if you subject it to a rotation.

The instance above is only an apparent exception to the "doubling" rule, as $$+q, -2q, +q$$ is just two opposite dipoles aligned.

If you've got the game, I can save to show how an octupole is built.

• This has good information. I think you should give your answer the final push towards answering the OP's question and explicitly say why we don't use the terms "tripole" or "hexapole" etc. – Aaron Stevens Feb 22 at 13:51