1
$\begingroup$

The multiple expansion of a potential V has contributing terms proportional to $\frac{1}{r^{n+1}}$ where $n=0,1,2...$.

First, why are we interested only in integer powers of r?

Second, why are we interested in symmetry pole arrangements? In other words, the $1/r$ term is a monopole, the $1/r^2$ term is a dipole, the $1/r^3$ term a quadrupole -- all of these are symmetric configurations. Why do we not consider an asymmetric 'tripole' formed in an isosceles triangle?

The reason I pose these questions is that my understanding of this formalism is that it represents the potential far away from an arbitrarily complicated charge configuration. But if the charge configuration is arbitrarily complicated, then why do we assume such perfectly symmetric poles?

Improve this question
$\endgroup$
  • $\begingroup$ As the Wikipedia article tells you, we are doing a Laurent expansion around the origin. That is by definition in integer powers, and has nothing to do with our "interest" - it's simply what comes out when using that approximation. $\endgroup$ – ACuriousMind♦ Apr 11 '15 at 3:38
  • $\begingroup$ @ACuriousMind Is this really a Laurent expansion? A Laurent series is an expansion of a function of one complex variable, while this is an expansion of a function of three real variables. The number of real degrees of freedom don't match up. $\endgroup$ – tparker Jan 10 '17 at 14:16
  • $\begingroup$ @tparker I think what I meant is that you're writing the function as a function of spherical coordinates and then Laurent expand the radial coordinate. You're right that it's not really a Laurent expansion because the function is real but the result is the same as if you analytically continued the real-valued function, Laurent expanded it around the origin, and then restricted it to the reals again. $\endgroup$ – ACuriousMind♦ Jan 10 '17 at 14:27
  • $\begingroup$ @ACuriousMind Not sure if I buy that. A Laurent expansion is parameterized by a set of coefficients $c_n$ that naturally depends on one integer. A multipole expansion is parameterized by a set of coefficients $C_lm$ that naturally depends on one whole number $l$ and one integer $m$ between $-l$ and $l$. How exactly do you match these two sequences up? $\endgroup$ – tparker Jan 10 '17 at 15:22
  • $\begingroup$ @tparker Oh, we're talking about two different things in the expansion. The $C_{lm}$ are the coefficients of the expansion of the function into spherical harmonics. They are themselves functions of $r$ which you may Laurent expand. It is this $1/r^n$ expansion OP is talking about, not the spherical expansion. $\endgroup$ – ACuriousMind♦ Jan 10 '17 at 15:38
0
$\begingroup$

First, why are we interested only in integer powers of r?

Because when you expand $$ \frac{1}{|\vec r - \vec r'|} $$ about $r'=0$ you get only integer valued terms in the power series expansion.

So, why does this matter? It's because the potential at $\vec r$ due to a localized charge distribution $\rho$ is $$ V(\vec r)=\int \frac{\rho(\vec r')}{|\vec r - \vec r'|}d^3r' =\frac{1}{r}\int \rho (\vec r')d^3r' + \frac{\hat r}{r^2}\cdot \int \rho(\vec r')\vec r'+\mathcal{O}\left(\frac{1}{r^3}\right) $$

Improve this answer
$\endgroup$
1
$\begingroup$

As ACuriousMind pointed out, the integer powers of $1/r$ come from the fact that the multipole expansion is a Laurent expansion, and therefore by definition is an expansion in integer powers of $1/r$.

Regarding your question about symmetry, the point is that each term in the multipole expansion must be "independent" of the others in the sense that it cannot be expanded in terms of them. For example, any configuration of charges that do not sum to zero will have a monopole contribution, while any configuration is which the positive and negative charges are not symmetric under an appropriate reflection will have a dipole contribution, etc. Now, your "tripole" configuration can actually be expanded in terms of a monopole term (since any configuration of three equal magnitude charges cannot be electrically neutral) and a dipole term (since the charge distribution of the triangle isn't symmetric under reflections), plus (I believe) other terms. So in this sense it's not independent of the other multipole moments.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy