Meaning of terms and interpretation in the electric multipole expansion

Question

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.

Answer 1

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.