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.

  • $\begingroup$ 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. $\endgroup$ – nijankowski Jul 7 '13 at 1:39
  • $\begingroup$ 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. $\endgroup$ – nijankowski Jul 7 '13 at 1:47
  • $\begingroup$ 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? $\endgroup$ – 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.

  • $\begingroup$ 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! $\endgroup$ – Astrum Jul 7 '13 at 4:06
  • $\begingroup$ 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. $\endgroup$ – user24999 Jul 7 '13 at 4:27
  • 1
    $\begingroup$ 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. $\endgroup$ – Astrum Jul 7 '13 at 6:18
  • $\begingroup$ 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. $\endgroup$ – user24999 Jul 7 '13 at 13:59
  • $\begingroup$ Many thanks to you sir. Very helpful responses. I finally understand! $\endgroup$ – Astrum Jul 7 '13 at 19:55

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