# Meaning of terms and interpretation in the electric multipole expansion

In section 3.4.1 of Griffiths' Introduction to Electrodynamics, he discusses electric multipole expansion.

He derives the formula or the electric potential of a dipole, which I follow, but right after, he begins talking about the electric potential at a large distance, which is as follows:

$$V( \vec{r}) = \frac{1}{4 \pi \epsilon _{0}} \int \frac{\rho (\vec{r}')}{ℛ} dV$$

$ℛ$ is from some point inside the charge distribution to the point P. What exactly does $\vec{r}'$ denote? Is it the distance from the center of the charge distribution?

Next, he uses law of cosines to find the expression for $ℛ^{2} = r^{2}+(r')^{2}-2rr'\cos\theta'$. This gives the same question as above, what does the symbol $r'$ mean? Then he defines $ℛ = r(1+ \epsilon)^{1/2}$, where $\epsilon \equiv \left(\frac{r'}{r}\right)^{2}\left(\frac{r'}{r}-2\cos\theta^\prime\right)$. The proceeding part is where I really get lost:

For points well outside the charge distribution, $\epsilon$ is much less than 1, and this invites a binomial expansion. $$\frac{1}{ℛ} = \frac{1}{r}\left[1- (1/2) \epsilon+ (3/8) \epsilon ^{2} - (5/16) \epsilon ^{3}+\ldots\right]$$

What is going on in this last step?

And finally, we eventually derive the formula $$V(\vec{r}) = \frac{1}{4 \pi \epsilon _{0}} \sum ^{\infty}_{n=0}\frac{1}{r^{n+1}} \int(r')^n\,P_{n}(\cos \theta)\,\rho( \vec{r}')dV$$

Why did we go for all this trouble? If this is supposed to be the electric potential at a large distance, couldn't we have just used $V( \vec{r}) = \frac{1}{4 \pi \epsilon _{0}} \int \frac{\rho (\vec{r}')}{ℛ} dV$? I don't understand what this equation actually means in physical terms.

And in regards to the actual equation, what is $P_{n}$?

Any help is much appreciated.

• Figure 3.28 from the book should clear the meaning of $\vec{r}'$, you can look at it as an integration variable that helps you account for the whole charge distribution that "creates" an electric potential at point $P$. $P_n$ are the Legendre polynomials, look it up on wiki. He could have skipped all the intermediate steps and just used the expansion of $\frac{1}{|\vec{r}-\vec{r}'|}$ using the condition $r>r'$, but I think he just wanted to be more pedagogical in his approach. Not just plant the final result after one expansion. – nijankowski Jul 7 '13 at 1:39
• And you don't understand from where he got that binomial expansion? What is the series for $\frac{1}{1+\epsilon}$ or $\frac{1}{(1+\epsilon)^{\alpha}}$ ? Its just simple mathematics. – nijankowski Jul 7 '13 at 1:47
• You misunderstood my question with the binomial expansion. Why do we need to expand this out? I don't understand the need to do this. And the meaning of $\vec{r}'$ still confuses me. Is it the distance from a reference point inside the charge distribution? – Astrum Jul 7 '13 at 2:36

First, $\vec{r}^\prime$ is a vector that goes from the origin to the source of charge. If the source is a volumetric distribution, one must sum all contributions of charge, that's why one integrates over all the volume, say $\mathcal{V}$; the (correct) expression for the potential should be $$V( \vec{r}) = \frac{1}{4 \pi \epsilon _{0}} \int_\mathcal{V} \frac{\rho (\vec{r}^\prime)}{ℛ} d\mathcal{V}^\prime$$ so that all dependence of $V$ remains on $\vec{r}$. Then, $r^\prime$ is just the magnitude $|\vec{r}^\prime|$, being the distance from the origin to the source of charge.
Second, usually, the series expansion of a function $f(x)$ about some point $x_0$ is useful because if you want to know the value of $f$ near $x_0$, you may just take some few terms of the expansion; it is as seeing the plot of $f$ with a magnifying glass. You should remember this from your first calculus courses, it is done a lot in physics. Here the expansion about $\epsilon=0$ will be useful since $\epsilon\to0$ implies $r\to\infty$ (just really big, if you will). The (correct) expression $$V(\vec{r}) = \frac{1}{4 \pi \epsilon _{0}} \sum ^{\infty}_{n=0}\frac{1}{r^{n+1}} \int(r')^n\,P_{n}(\cos \theta^\prime)\,\rho( \vec{r}')\,d\mathcal{V}'$$ is just another way of writing the series expansion in terms of $r$, $r^\prime$ and $\theta^\prime$, where $P_n$ are the Legendre polynomials (Griffiths defines them there, ain't he?). This expression is useful, as it means, explicitly, that $$V(\vec{r})=\frac{1}{4\pi\epsilon_0}\left[\frac{1}{r}\int\rho(\vec{r}')\,d\mathcal{V}'+\frac{1}{r^2}\int{r'}\cos\theta'\,\rho(\vec{r}')\,d\mathcal{V}'+\frac{1}{r^3}\left(\cdots\right)+\ldots\right]$$ so that if you want to evaluate the potential for points far from the source (big $r$), then you may just neglect higher order terms in $r$ and just take the $1/r$ (monopole) term; and so on if you're considering a better approximation, you may take the $1/r^2$ (dipole) term, etc... That's the real usefulness of the series expansion; in a lot of situations evaluating $V( \vec{r}) = \frac{1}{4 \pi \epsilon _{0}} \int \frac{\rho (\vec{r}')}{ℛ} d\mathcal{V}'$ will get really ugly, and then, mostly, is when the multipole approximation will be useful.
• Thanks, so really, this is equal to $V( \vec{r}) = \frac{1}{4 \pi \epsilon _{0}} \int \frac{\rho (\vec{r}')}{ℛ} d\mathcal{V}'$? I did example 3.26, and it turned out to be really ugly as well, egad! – Astrum Jul 7 '13 at 4:06
• I'm glad I could help. The full expansion (the term with the sum and the Legendre polynomials) yes. When one cuts the series, it's just an approximation, but a good one if you consider large $r$. In the analogy I made about the magnifying glass, here it would be exactly the opposite, just as seeing $V$ so far away that you don't need to worry about how contrived the source is, taking it just as a monopole or dipole or cuadrupole, and so on, depending how much detail you want or how good the approximation. – user24999 Jul 7 '13 at 4:27
• Yes, you've been really helpful, just one things that I still don't get. When doing the problem in the book, problem 3.26, where we are asked to find the electric potential far from the sphere with a charge density of $\rho ( \theta , r)= \frac{kR}{r^{2}}(R-2r)(sin \theta)$ , the monopole and dipole terms disappeared, why is that? We're left with only a quadrupole term. – Astrum Jul 7 '13 at 6:18
• In particular cases multipole terms may vanish because of charge distribution symmetries. For the monopole term, $\frac{1}{r}\int\rho(r')\,dV'=\frac{q}{r}$, so that if it vanishes, then the total charge is zero, $q=0$ (the integral vanishes when you sum all contributions in $r'$). Conversely, if the source is a point charge, $\rho(\vec{r}')\sim{q}\,\delta(\vec{r}')$, all multipole terms vanish except the monopole one. In your case for the dipole term, $\int_0^\pi\sin^2\theta'\cos\theta'd\theta'=0$ so that the dipole moment of this distribution is symmetric in the polar direction. – user24999 Jul 7 '13 at 13:59