Expanding around 0 for multipole expansion

[deuteron]

TL;DR Summary

why do we expand aound 0 for the multipole expansion?

For the multipole expansion of the electric potential, we expand $\frac 1 {|\vec r-\vec r'|}$ in the following way:
$$\frac 1 {|\vec r-\vec r'|} = \frac 1{\sqrt{(\vec r-\vec r')^2}} =\frac 1 {[ \vec r^2 +\vec r'^2 -2\vec r\cdot \vec r']^{-\frac 12}}=\frac 1{[\vec r^2 + \vec r'^2 - 2|\vec r||\vec r'| \cos\gamma]^{-\frac 12}}$$

Here, we define $r_<=\min\{ |\vec r|,|\vec r'|\}$ and $r_>=\max\{|\vec r|,|\vec r'|\}$, and using the symmetry of the equation, we get:

$$ = \frac 1{[r_<^2 + r_>^2 -2r_<r_> \cos\gamma]^{\frac 12}} = \frac 1 {r_>} \frac 1 {[ 1 + (\frac {r_<}{r_>})^2 -2(\frac {r_<}{r_>}) \cos\gamma]^{\frac 12}}= \frac 1 {r_>} [ 1+(\frac {r_<}{r_>})^2 -2(\frac {r_<}{r_>}) \cos\gamma] ^{-\frac 1 2}$$

Here, we expand $[ 1+(\frac {r_<}{r_>})^2 -2(\frac {r_<}{r_>}) \cos\gamma] ^{-\frac 1 2}$ with the expansion parameter $(\frac {r_<}{r_>})^2 -2(\frac {r_<}{r_>}) \cos\gamma$ around $0$ to get the expansion of $\frac 1 {|\vec r-\vec r'|}$ in terms of the Legendre polynomials.

What I don't understand here is, why do we expand around $0$? Doesn't that mean that we let $(\frac {r_<}{r_>})^2 -2(\frac {r_<}{r_>}) \cos\gamma=0$, which would mean, for $(\frac {r_<}{r_>})\approx 1$, $1-2\cos\gamma=0\ \Rightarrow\ 1=2\cos\gamma\ \Rightarrow\ \cos\gamma=\frac 12$. I don't understand what leads us to make the expansion around $\cos\gamma= \frac 12$.

 

[kuruman]

What I don't understand here is, why do we expand around $0$? Doesn't that mean that we let $(\frac {r_<}{r_>})^2 -2(\frac {r_<}{r_>}) \cos\gamma=0$, which would mean, for $(\frac {r_<}{r_>})\approx 1$, $1-2\cos\gamma=0\ \Rightarrow\ 1=2\cos\gamma\ \Rightarrow\ \cos\gamma=\frac 12$.

It doesn't mean that. It looks like you do not understand what expansions are all about. An expansion is an approximation of a function in the vicinity of certain fixed value of a variable. Expanding about $~\dfrac{r_<}{r_>}~$ does not mean "set $~\dfrac{r_<}{r_>}=0.$" It means find an approximate expression for $\dfrac{1}{|\mathbf{r}-\mathbf{r}'|}$ when $~\dfrac{r_<}{r_>}~$ is small. What you get is the multipole expansion which means a series of infinitely many terms. The more terms you keep, the more accurately the expansion matches the function.

Consider $f(x)=\dfrac{1}{1+x}$. For small values of $x$, the expansion is $$f(x)=\frac{1}{1+x}=1-x+x^2-x^3 +\dots$$Below is a plot of the $\dfrac{1}{1+x}$ (discrete points) and three solid lines representing increasingly more terms added to the expansion. As more terms are added, you can see how the expansion gets closer to the function at higher values of $x$.

The multipole expansion allows you to find an approximate expression to the potential when you have an amorphous blob of charge by finding its monopole, dipole, quadrupole, octupole, hexadecapole, etc. terms which are easier to handle mathematically than $\dfrac{1}{|\mathbf{r}-\mathbf{r}'|}$.

 

[deuteron]

It doesn't mean that. It looks like you do not understand what expansions are all about. An expansion is an approximation of a function in the vicinity of certain fixed value of a variable. Expanding about $~\dfrac{r_<}{r_>}~$ does not mean "set $~\dfrac{r_<}{r_>}=0.$" It means find an approximate expression for $\dfrac{1}{|\mathbf{r}-\mathbf{r}'|}$ when $~\dfrac{r_<}{r_>}~$ is small. What you get is the multipole expansion which means a series of infinitely many terms. The more terms you keep, the more accurately the expansion matches the function.

Consider $f(x)=\dfrac{1}{1+x}$. For small values of $x$, the expansion is $$f(x)=\frac{1}{1+x}=1-x+x^2-x^3 +\dots$$Below is a plot of the $\dfrac{1}{1+x}$ (discrete points) and three solid lines representing increasingly more terms added to the expansion. As more terms are added, you can see how the expansion gets closer to the function at higher values of $x$.

View attachment 341211
The multipole expansion allows you to find an approximate expression to the potential when you have an amorphous blob of charge by finding its monopole, dipole, quadrupole, octupole, hexadecapole, etc. terms which are easier to handle mathematically than $\dfrac{1}{|\mathbf{r}-\mathbf{r}'|}$.

Thank you! And I really like the quote below your post!

I understand from your post that the expansion of $f(x)$ with the expansion parameter $x$ around $0$ gives us an approximation of $f(x)$ for small $x$, where the approximation gets better with increasing $x$ as we add more and more terms

What confuses me is that the expansion parameter in the expansion of $\frac 1 {|\vec r-\vec r'|}$ is not $\frac {r_<}{r_>}$ but it is instead $(\frac {r_<}{r_>})^2-2(\frac {r_<}{r_>}) \cos\gamma$, which from what I'have understood would mean that, for the expansion around $0$, that the expansion parameter is small, that was my reasoning for setting $(\frac {r_<}{r_>})^2-2(\frac {r_<}{r_>}) \cos\gamma=0$ above.

 

[pasmith]

What confuses me is that the expansion parameter in the expansion of $\frac 1 {|\vec r-\vec r'|}$ is not $\frac {r_<}{r_>}$ but it is instead $(\frac {r_<}{r_>})^2-2(\frac {r_<}{r_>}) \cos\gamma$, which from what I'have understood would mean that, for the expansion around $0$, that the expansion parameter is small, that was my reasoning for setting $(\frac {r_<}{r_>})^2-2(\frac {r_<}{r_>}) \cos\gamma=0$ above.

Your expansion parameter is $f(r_{<}/r_{>})$, where $f(x) = x^2 - 2x\cos \gamma = (x - \cos \gamma)^2 - \cos^2 \gamma$ and by definition $0 < x < 1$. The binomial expansion $$(1 + f(x))^\alpha = 1 + \alpha f(x) + \frac{\alpha(\alpha - 1)}2 f(x)^2 + \dots$$ is valid when $$|f(x)| = |(x - \cos \gamma)^2 - \cos^2 \gamma|< 1$$, ie when $$0 < x < \cos \gamma + (1 + \cos^2 \gamma)^{1/2}.$$ When $\cos \gamma > 0$, this upp;er bound is actually greater than 1!

 

[TSny]

Your expansion parameter is $f(r_{<}/r_{>})$, where $f(x) = x^2 - 2x\cos \gamma = (x - \cos \gamma)^2 - \cos^2 \gamma$ and by definition $0 < x < 1$. The binomial expansion $$(1 + f(x))^\alpha = 1 + \alpha f(x) + \frac{\alpha(\alpha - 1)}2 f(x)^2 + \dots$$ is valid when $$|f(x)| = |(x - \cos \gamma)^2 - \cos^2 \gamma|< 1$$, ie when $$0 < x < \cos \gamma + (1 + \cos^2 \gamma)^{1/2}.$$ When $\cos \gamma > 0$, this upp;er bound is actually greater than 1!

This is interesting. So, start with the expression $$F(x)= \frac 1 {\sqrt{1 + x^2 - 2 x \cos \gamma }} = \left(1 + x^2 - 2 x \cos \gamma \right)^{-1/2}$$ Follow @pasmith and obtain an infinite series expansion of $F(x)$ in the following way. First, obtain the binomial expansion of $\left(1 + y \right)^{-1/2}$ : $$ \left(1 + y \right)^{-1/2} = 1 - \frac y 2 + \frac{3y^2}{8} - \frac {5y^3}{16} + . . . $$ Next, replace $y$ by $(x^2 - 2 x \cos \gamma)$ so we get $$ \left(1 + x^2 - 2 x \cos \gamma \right)^{-1/2} = 1 - \frac 1 2 (x^2 - 2 x \cos \gamma) + \frac 3 8 (x^2 - 2 x \cos \gamma)^2 - \frac {5}{16} (x^2 - 2 x \cos \gamma)^3 + . . . $$ I'll call this the "binomial series expansion" of $F(x)$. As shown by @pasmith, if $\cos \gamma > 0$, this infinite series has an upper bound of convergence greater than 1. For example, if $\cos \gamma = 1/2$, the series converges for values of $x$ up to $\frac 1 2 (1 + \sqrt{5}) \approx 1.618 \, .$

A different infinite series expansion of $F(x)$ is the Taylor Series expansion of $\left(1 + x^2 - 2 x \cos \gamma \right)^{-1/2}$:
$$ \left(1 + x^2 - 2 x \cos \gamma \right)^{-1/2} = 1 + (\cos \gamma ) x + \left( \frac 3 2 \cos^2 \gamma - \frac 1 2 \right) x^2 + \left( \frac 5 2 \cos^3 \gamma- \frac 3 2 \cos \gamma \right) x^3 + ...$$ This power series could have been obtained from the binomial expansion series by rearranging the terms in the binomial expansion series to collect together terms with the same power of $x$.

The power series corresponds to the "multipole expansion" of $F(x)$ where the coefficients of the powers of $x$ are the Legendre polynomials. But what I find interesting is that the power series has a smaller domain of convergence. The power series diverges for all $x > 1$. I guess this shouldn't be surprising since rearranging terms in an infinite series can change the convergence properties of the series. But I find it interesting.