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I have a few questions on multipole expansions and I have read about the topic in many places but could not find an answer to my questions, so please be patient with me.

The electrostatic potential due to an arbitrary charge distribution $\rho(\mathbf{r}')$ at a given point $\mathbf{r}$ is given (up to a factor of $1/4\pi\epsilon_0$) by

$$ V(\mathbf{r})=\int_{V'}\frac{\rho(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|} dV'$$

In case where $r\gg r'$, $V(\mathbf{r})$ can be multipole expanded to give

$$V(\mathbf{r})=V(\mathbf{r})_\text{mon}+V(\mathbf{r})_\text{dip}+V(\mathbf{r})_\text{quad}+\cdots$$

where

\begin{align} V(\mathbf{r})_\text{mon}& =\frac{1}{r}\int_{V~`}\rho(\mathbf{r}') dV',\\ V(\mathbf{r})_\text{dip}&=\frac{1}{r^2}\int_{V~`}\rho(\mathbf{r}') ~\hat{\mathbf{r}}\cdot\mathbf{r}'dV', \\ V(\mathbf{r})_\text{quad}&=\frac{1}{r^3}\int_{V~`}\rho(\mathbf{r}') ~\left(3(\hat{\mathbf{r}}\cdot\mathbf{r}')^2-r'^2\right)dV', \end{align} and so on.

Now here are my questions:

  1. Is there an intuitive meaning of every one of these terms? For example, I can make sense of the monopole term in the following way: to the 1st approximation the charge distribution will look like a point charge sitting at the origin, which mathematically corresponds to what is called a monopole term, which is nothing but $Q/r$. Is this correct?

  2. Now what is the meaning of the dipole term? I know that the word dipole comes from having 2 opposite charges, and the potential due to that configuration, if the charges are aligned along the $z$ axis symmetrically say, goes like $\frac{\cos\theta}{r^2}$. But from the multipole expansion there is a nonzero dipole term even, say, in the case of a single charge sitting at some distance from the origin. Why is it called a dipole term then? Is there a way to make sense of this term in the same way I made sense of the monopole term?

  3. What is the intuitive meaning of the quadrupole term?

  4. Is the multipole expansion an expansion in powers of $1/r$ only? or of $\cos\theta$ too?

  5. Maybe this is not an independent question but I am wondering if there is something like a geometrical/pictorial meaning of every term in the multipole expansion.

Emilio Pisanty
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Revo
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3 Answers3

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For question 2: ("Why does a single charge away from the origin have a dipole term?")

Let's say you have a charge of +3 at point (5,6,7). Using the superposition principle, you can imagine that this is the superposition of two charge distributions

  • Charge distribution A: A charge of +3 at point (0,0,0)

  • Charge distribution B: A charge of -3 at point (0,0,0) and a charge of +3 at (5,6,7).

Obviously, when you add these together, you get the real charge distribution:

$$ (\text{real charge distribution}) = (\text{charge distribution A}) + (\text{charge distribution B}). $$

By the superposition principle: $$ (\text{Real }\mathbf E\text{ field}) = (\mathbf E\text{ field of charge distribution A}) + (\mathbf E\text{ field of charge distribution B}). $$

And, since the multipole expansion also obeys the superposition principle:

\begin{align} (\text{real monopole term}) & = (\text{monopole term of distribution A}) + (\text{monopole term of distribution B}),\\ (\text{real dipole term}) & = (\text{dipole term of distribution A}) + (\text{dipole term of distribution B}),\\ (\text{real quadrupole term}) & = (\text{quadrupole term of distribution A}) + (\text{quadrupole term of distribution B}), \end{align} and so on.

The field of charge distribution A is a pure monopole field, while the field of charge distribution B has no monopole term, only dipole, quadrupole, etc. Therefore, \begin{align} (\text{real monopole term}) & = (\text{monopole term of distribution A}), \\ (\text{real dipole term}) & = (\text{dipole term of distribution B}),\\ (\text{real quadrupole term}) & = (\text{quadrupole term of distribution B}), \end{align} and so on.

Even though it's unintuitive that the real charge distribution has a dipole component, it is not at all surprising that charge distribution B has a dipole component: It is two equal and opposite separated charges! And charge distribution B is exactly what you get after subtracting off the monopole component to look at the subleading terms of the expansion.

Emilio Pisanty
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Steve Byrnes
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  • But this is physically different from having just one charge. You altered the charge distribution. Your new charge distribution will not affect the monopole term, but it will have a physical effect on all the other terms. – Revo Nov 16 '11 at 16:56
  • @Revo: I did not alter the charge distribution, I split it into two pieces. I edited, hopefully it's clearer now how this works. – Steve Byrnes Nov 16 '11 at 17:32
  • No, Revo, @SteveB didn't change the charge distribution. He took your very distribution - a single charged particle away from the center - and wrote it as a combination of a charge in the center and the "true dipole" consisting of a pair of oppositely charged particles. The latter clear has a dipole moment: it's the canonical representation of a "pure dipole". Well, except that this dipole isn't center at zero, so it will also carry a quadrupole moment (and maybe higher ones, too), but the latter is subleading in this expansion. – Luboš Motl Nov 16 '11 at 17:34
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    SteveB's explanation is of course perfectly valid but it doesn't directly "attack" the core of Revo's misconception. Revo correctly knows that the "canonical dipole" is a pair of oppositely charged particles. However, that doesn't mean that every object that doesn't have this form carries a zero dipole moment. Quite on the contrary, generic objects (charge distributions) carry a nonzero amount of each multipole moment. One needs some "special" or "very special" distributions for some of these moments to be zero. – Luboš Motl Nov 16 '11 at 17:36
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    In particular, it's not true that everything that isn't of the form of "the exact naive children's dipole" carries a vanishing dipole moment - which seems to be the (incorrect) assumption in Revo's question. @Revo, try this analogy: this assumption is analogous to saying that one kilogram is the mass of the platinum prototype in France, or whatever it is. Now, your reasoning is that the mass of a person has to be zero because the person isn't even made of platinum: she can't be the platinum stick. This sounds like a joke but your logic applied to the dipole moment is exactly the same. – Luboš Motl Nov 16 '11 at 17:38
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    Just to be sure, the dipole moment is called the dipole because the simplest way to realize a charge distribution with a nonzero dipole moment - but vanishing all other moments - is a pair (therefore "di") of nearby charges. Similarly for quadrupoles, octupoles etc. (powers of two). However, that doesn't mean that the number of charges always has to be a power of two (and they have to be pointlike): almost any charge distribution carries almost all the multipole moments. Just trust the formulae instead of words, and don't misinterpret the words. – Luboš Motl Nov 16 '11 at 17:43
  • @SteveB Okay, I did not notice that you put the 2 new opposite charges at the center, my bad. – Revo Nov 16 '11 at 17:51
  • @LubošMotl Yeah thanks a lot Lubos, you attacked exactly the core of my question. I think my misconception came from the fact that in books they start talking about dipoles by considering the 2 opposite charges example, they never said explicitly that this is a special case of what a dipole term look like. They should have said that whatever we call a dipole term is characterized by $1/r^2$ regardless of the charge distribution. I guess it is another misnomer in physics that must be taken carefully, trust formulae instead of words is a perfect advice. – Revo Nov 16 '11 at 17:56
  • @LubošMotl I was thinking that this way of thinking about the existence of dipole term of a single charge, in terms of superposition of 2 charge distributions will not be valid in case if the multipole expansion were for masses instead of charges. Because mass is always positive, so how to interpret the dipole term in this case for a single mass? – Revo Nov 18 '11 at 11:16
  • @SteveB could you please provide the name of the reference book or link from where you found out this explanation. The explanation is really beautiful. – user103515 May 18 '16 at 07:52
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To understand the meaning of multipole expansion,firstly we need to ask ourselves about the evaluation of potential of a very random charge distribution.Recall that, if you forget about the multipole expansion you have no any simple device to know the potential of a random charge distribution, either you want potential near the distribution or very far away.

In our basic electrostatic courses we are usually taught about how to evaluate the potential of some symmetric charge distribution, for instance, potential for distributions which are spherically and cylindrically symmetric,and in most toughest cases the distribution would be of some parameter of polar angles or radial distances.

Now, the second thing which one always ignores (and that's why got stuck with the confusion about intuition) during the study of multipole expansions is that the dipole has nothing to do with a pair of positive and negative charge, or an octupole has nothing to do with a group of 4 positive and 4 negative charge. There charge distribution only take care of the variation of potential,

Emilio Pisanty
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Kushagra Yadav
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  1. Is there an intuitive meaning of every one of these terms? For example, I can make sense of the monopole term in the following way: to the 1st approximation, the charge distribution will look like a point charge sitting at the origin, which mathematically corresponds to what is called a monopole term, which is nothing but $Q/r$. Is this correct?

First off, the nomenclature is rather unfortunate (goes as $2^n$, i.e $2^2=\text{quadrupole}$) and can be misleading. Mathematically, the monopole term is the zeroth order Legendre Polynomial ($L_0(x)=1$, $L_1(x)=x$, $L_2(x)=(3x^2-1)/2$) and so on (up to a normalization).

Physically speaking, the first term tells us something about the symmetry of the system depending on the location of the observer. Suppose the potential at a point A of the system of charges only has a monopole structure, this means that the charge distribution has full spatial invariance. More importantly, the first term tells you that if we want the total energy of the system, we need to only worry about the Potential at A because the monopole only couples to the electric Potential.

  1. Now what is the meaning of the dipole term? I know that the word dipole comes from having 2 opposite charges, and the potential due to that configuration, if the charges are aligned along the z axis symmetrically say, goes like $\cos\theta/r^2$. But from the multipole expansion there is a nonzero dipole term even in the case of a single charge sitting at some distance from the origin say. Why is it called a dipole term then? Is there a way to make sense of this term in the same way I made sense of the monopole term?

The dipole term is the 1st order Legendre polynomial ($2^1=\text{dipole}$). It is possible to have higher order terms even when the net charge is zero. This means, the energy of the system depends on the interaction of the dipole moment with the Electric Field of the test charge. $\vec{d} \cdot\vec{E}$ couplings are studied in light-matter interactions. Another interesting point to note is that there is some kind of spatial symmetry breaking that emerges because dipole interactions can setup a preferred spatial axis (ex along the line joining the charges).

  1. What is the intuitive meaning of the quadrupole term?

  2. Is the multipole expansion an expansion in powers of $1/r$ only? or of $\cos\theta$ too?

  3. Maybe this is not an independent question, but I am wondering if there is something like a geometrical/pictorial meaning of every term in the multipole expansion.

The quadrupole term is so named because it is the second order Legendre polynomial $2^2=4$. The quadrupole moment couples with the gradient of the electric field.

Anyway, this is my personal take on the subject. Mathematically, it is very interesting to ask why do we get Legendre polynomials out of this? It turns out that the Legendre polynomials can be generated by ortho-normalizing monomials (basis for this series expansion).

Emilio Pisanty
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Antillar Maximus
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