# Forcing quadrupole moments to vanish for a neutral system

For a system of electric charges $q_i$, at positions $\mathbf{r}_i$, with a nonzero net charge $Q=\sum_i q_i$, one can define a "centre of charge" in the obvious way as $$\mathbf{r}_c=\frac{1}{Q}\sum_i q_i\mathbf{r}_i.$$ This concept is definitely not as useful as one might naively hope, but it does have a physical significance as the position of the origin that sets the system's dipole moment (and therefore the dipolar term in a multipolar expansion) to zero. That means, then, that the monopole approximation is far better in the far field than normally: the electrostatic potential goes as $$\Phi(\mathbf{r})=\frac{Q}{|\mathbf{r}-\mathbf{r}_c|}+O\left(\frac{1}{|\mathbf{r}-\mathbf{r}_c|^{3}}\right)$$ where the subleading term is of order $1/r^3$, instead of $1/r^2$ as usual.

For a neutral system with $Q=0$, however, the concept of centre of charge is meaningless and there is no monopole term in the expansion. The relevant concept is then the dipole moment, $$\mathbf{d}=\sum_i q_i(\mathbf{r}_i-\mathbf{r}_0),$$ which is independent of the position $\mathbf{r}_0$ of the origin. However, this means that the relative importance of the subleading term in the multipole expansion is higher than above: $$\Phi(\mathbf{r})=\frac{\mathbf{d}\cdot\mathbf{r}}{|\mathbf{r}-\mathbf{r}_0|^2}+O\left(\frac{1}{|\mathbf{r}-\mathbf{r}_0|^{3}}\right)$$ My question is: for a neutral system, is it possible to find a suitable position $\mathbf{r}_0$ for the origin that will set the subleading, quadrupole term to zero? Since this entails a system of five equations (linear in $\mathbf{r}_0$ when $Q=0$), I suspect this is impossible in general geometries. If this is the case, which geometries allow for vanishing quadrupole moments and which ones don't? In the cases where one can do this, does this position have a special name? More generally, if all multipole moments up to some $l\geq0$ are zero, (when) can the subleading term be made to vanish?

-
Choosing the origin so as to make the dipole moment vanish is like transforming $x\rightarrow x+c$ to make the average value of $x$ be zero. Trying to make the quadrupole moment equal zero is like trying to make the variance of a distribution zero. You can't do that by adding a constant to the variable. – Ben Crowell May 7 '13 at 14:55
@BenCrowell No, that can't be quite right, since there definitely are situations where you can eliminate the quadrupole moments, such as a displaced point dipole. The difference with a variance is that multipole moments are signed polynomials whereas variances are always positive. – Emilio Pisanty May 7 '13 at 18:04
The question is then: where do you draw the line? Are there systems for which this is provably not possible? If so, how does one decide whether it will? – Emilio Pisanty May 7 '13 at 18:06

I think it's easier to see in Cartesian coordinates. Define the "primitive" moments

$q = \int \rho\, \text{d}^3\mathbf{r}$

$p_i = \int r_i \rho\, \text{d}^3 \mathbf{r}$

$Q_{ij} = \int r_i r_j \rho\, \text{d}^3 \mathbf{r}$

Assume $q = 0$, $p_i$ not all 0.

If you displace the origin by $\mathbf{d}$ and call the new quadrupole moments $Q'_{ij}$, then

$$Q'_{ij} = \int (r_i - d_i)(r_j - d_j) \rho \text{d}^3 \mathbf{r}\\ = Q_{ij} - p_i d_j - p_j d_i + q d_i d_j.$$

If q = 0, finding a translation that makes $Q'_{ij} = 0$ is now a (overspecified) linear problem of six equations in three unknowns. To make it even simpler, assume the dipole is aligned along the z-axis ($p_1 = p_2 = 0$). Then it is easy to see that no translation will change $Q_{11}, Q_{12}, \text{ or } Q_{22}$, so if any of them are nonzero then translating the distribution will not make them zero. If they are all zero, then the quadrupole moment can be entirely zeroed out by setting $d_1 = Q_{13}/p_3$, $d_2 = Q_{23}/p_3$, and $d_3 = Q_{33}/p_3$.

-

Thank you to Ben Cromwell for jogging my mind in the right direction: here's a (rather) partial answer.

Consider the charge density $\rho=\left[d_z z + Q_{zz}(x^2+y^2-2z^2)\right]e^{-r^2/2\sigma^2}$, which is a superposition of dipole and quadrupole gaussians. The system is neutral with a nonvanishing dipole moment, so the leading term will remain, but no displacement of the origin can change the quadrupole moment: $$Q_{zz}=\frac{1}{3\pi^{3/2}}\int\mathrm{d}\mathbf{r}(x^2+y^2-2z^2)\rho(\mathbf{r}-\mathbf{r}_0)$$ is independent of $\mathbf{r}_0$.

The answer is therefore a partial no: there exist systems, with zero net charge and nonzero dipole moment, where no displacement of the origin can eliminate the subleading term.

Edit:

That is not correct. The integral in question comes out to \begin{align} Q_{zz} &=\frac{1}{3\pi^{3/2}}\int\mathrm{d}\mathbf{r}(x^2+y^2-2z^2)\rho(\mathbf{r}-\mathbf{r}_0) \\&=\sqrt{32\pi^3}\sigma^5(-d_z z_0+3Q_{zz}\sigma^2) \end{align} so in this particular case choosing $z_0=Q_zz\sigma^2/d_z$ does cause a vanishing of the quadrupole moment.

On the other hand, some systems do have such behaviour: displaced pure-dipole systems such as the charge density $\rho=d_z(z-z_0)e^{-(\mathbf{r}-\mathbf{r}_0)^2/2\sigma^2}$ do have nonzero quadrupole moments which vanish for appropriate choices of the origin. I do not know, however, where the dividing line is between the two is, or what general criterion might be used to decide if this elimination will be possible or not.

-
a zero quadrupole moment indicates a spherically symmetric charge distribution. The displaced pure-dipole system is fundamentally spherically symmetric, just that is centre could be off. Your conclusion that a translation of the origin is not sufficient to make $Q_{ij}=0$ is correct. Only a transformation that turns an ellipsoid into a sphere will make $Q_{ij}$ vanish. – Amey Joshi May 23 '13 at 7:11
A zero quadrupole moment does not necessarily indicate spherical symmetry. Consider, for example, a cube with alternating charges at the corners, or similarly, a spherical shell with alternating octants having alternating charge. In both cases, the leading moment is octupolar and the quadrupole moment is zero, but they are clearly not spherically symmetrical. Conversely, consider a uniform ball of charge of nonzero radius. This has a nonzero quadrupole moment (regardless of its location relative to the origin), but is clearly spherically symmetrical. – pwf Aug 16 '14 at 0:05