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.
Best Answer
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.