I am trying to better understand what the multipole expansion means from a phyiscal point of view. Although mathematically, one may say it is just another form of a series expansion, in this case, the terms in the expansion are themselves physically meaningful, namely: monopoles, dipoles, quadrupoles and so on.
Naively, as an instance, if one has a good understanding of these terms, then purely from physical considerations one would be able to say: "in this case, due to this and that aspect, we cannot have monopoles, so the leading term must be a dipole…". A bit similar to how physicists exploit symmetry reasonings in order to reduce or simplify their calculations.
Example
To make the question more concrete, let us take an example of a charged line segment along an axis, with the segment having the length $l.$ So if our total charge is $q,$ the charge density is simply $\rho=q/l.$ We are interested in describing the electric field generated by this charged line segment, and to do that we are told to do a multipole expansion and study leading terms.
My take on multipole expansion
Now how I understand multipole expansion: it's a useful method for approximating the electric field at a far distance from the source, so it has a special domain of applicability (though I don't know what limits it to far-field study). The expansion itself, entails Taylor expanding the free-field Green's function $G_{\text{free}}$ with respect to $d$ which is roughly the radius of the spherical region containing the charge source. To clarify,
$$
\phi=G_f(\vec{r}-\vec{d}) := \frac{1}{4\pi\epsilon_0}\frac{1}{|\vec{r}-\vec{d}|}, \text{The potential for a charge distribution equal to a delta peak.}
$$
Moreover, having $G_f,$ the idea is to superpose the Green's function solution in order to find solutions for any other charge distribution $\rho$ in order to solve for the potential $\phi.$ That is, roughly, $\phi \propto \int G_f \rho \mathrm{d}r \tag{1}$ Now the final idea is to, Taylor expand $G_f$ around $\vec{d}$ and re-insert back into $(1),$ which can be rewritten as:
$$
\phi(\vec{r})= \frac{1}{4\pi\epsilon_0} \sum_{l=0}^{\infty}\frac{1}{l!*2^{2l+1}} Q_{i_1\dots i_l}^l x_{i_1}\dots x_{i_l} \tag{2}
$$
The $Q^l$ terms are then said to be the different multipole moments.
Returning to the example, let us do the expansion: For the monopole moment, we have $Q^0=\int \rho(r') \mathrm{d}V' = q, \text{where }\rho=q/l,$ so the corresponding field is $\phi^0 \propto q/r.$ The dipole moment is $0$ as $\frac{q}{l} \int_{-l/2}^{l/2} x' \mathrm{d}x' = 0.$ Here we have assumed the segment to be lying on the x-axis. To find the next leading term, we have to go one order higher, so to calculate the quadrupole moment: $Q^2_{zz}=Q^2_{yy}\propto -q l^2$ and $Q^2_{xx}\propto \frac{q}{3}l^2.$ Inserting back these moments into Eq.$(2),$ we can approximate the electric field far from the source as
$$
\phi (\vec{r}) \approx \frac{q}{r}+ql^2 \left(\frac{3x^2}{r^5}-\frac{1}{r^3}\right) \tag{3}
$$
Questions:
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We saw in the solution, Eq. $(3),$ that far from the charged line segment, the electric field generated can be approximated upto two leading terms by a monopole plus a quadrupole. Could we have physically expected to find these two to be the leading terms in the expansion? How do we interpret again physically the composition of these different pole orders, e.g., a monopole plus a quadrupole? I don't even understand why this composition is allowed, because a monopole is only a scalar, and a quadrupole a higher order tensor.
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Is it ever possible that a far field of a certain charge distribution $\rho$ does not even exhibit a monopole for instance, in the expansion?
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Why is the multipole expansion primarily an approximation for the description of the far-field only?
Again my aim is to find a better grip over these concepts from a general physical point of view, because I know these are recurring concepts throughout physics (for instance that gravitational waves are quadrupolar…). But for myself, I have no way of translating such statements into simpler terms.
Best Answer
$$ +\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - $$ $$ + $$ $$ -\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + $$ 2. Yes it is possible - if you consider Gauss' law \begin{equation} \oint_{S} \vec{E}\cdot d\vec{S} = \frac{Q_{total}}{\epsilon_0} \end{equation} and take large enough sphere $S$ centered around $0$ with radius $r$ which envelopes all the charges. Then you can write the integral as $$ \oint_{\partial V} \vec{E}\cdot d\vec{S} = \left < E_{\perp} \right >_S \cdot 4\pi r^2 $$ where $\left < E_{\perp} \right >$ is the average of the perpendicular component of $E$ to the sphere. Since there is no monopole moment, field dies faster than $1/r^2$. Therefore the integral dies at least as $1/r$ (or faster if the dipole moment is 0) which means that it is $0$ (as the right side of Gauss' law is constant) which means that $Q_{total}= 0$. You can arrive at the same answer by deriving multipole expansion from scratch and seeing that $$Q_0 = \int_{\mathbb{R}^3} \rho({\vec{r}})\, d^3 r$$ Multipole expansion is not an approximation. It is exactly valid (in this form) if you have a charge distribution confined in some finite volume $V$ which can be enveloped by a sphere $S_V$, for points outside the sphere. This doesn't mean that the distances need to be much larger than some characteristic distance of the system. Since it is basically a Taylor expansion around $|\vec{r}| = \infty$ you get fast convergence (of the series) for large $|\vec{r}|$ which means that the higher contributions can be neglected with small error in far field regime.