Randomly distributed charges seem to have an oriented dipole moment

Question

Yesterday, I uploaded a post in which I asked about the total dipole moment of a system composed of $N$ arbitrarily placed identical charges. Nonetheless, as it was pointed out by Ján Lalinský, among others, I forgot that the dipole moment of the system was not really representative, since it was dependent on the origin (because the charge was evidently nonzero). Therefore, I now propose a minor modification to my system, in which half of the charges are positive ($q$) and the other half are negative ($-q$). The total dipole moment of my system would be:

$$\vec{p}_T=\displaystyle\sum_{i=1}^N \vec{p}_i=\displaystyle\sum_{i=1}^{N/2}q\vec{r}_i-\displaystyle\sum_{i=N/2+1}^N q\vec{r}_i$$

Now, treating my distributions of + and - charges separately, I can define a center of mass for each of them as follows:

$$\vec{r}_+=\frac{1}{M_T/2}\displaystyle\sum_{i=1}^Nm\vec{r}_i$$

and an analogous one for the negative charge distribution. Therefore, I can now write the net dipole moment of the system as such:

$$\vec{p}_T=q\left[\frac{2}{M_T}\displaystyle\sum_{i=1}^{N/2}m\vec{r}_i - \frac{2}{M_T}\displaystyle\sum_{i=N/2+1}^N m\vec{r}_i\right]\frac{M_T}{2m}$$

Recalling $M_T=Nm$, this expression is simplified as follows:

$$\vec{p}_T=Nq(\vec{r}_+-\vec{r}_-)$$

Which, as stated by Ján, does not depend on where we set the origin of the coordinate system, since the relative distance between the two centers of mass will always be the same. If I choose to set the origin on the center of mass of the negative distribution, then I get: $\vec{r}_-=0$, so the final expression is:

$$\vec{p}_T=Nq\vec{r}_+=N\vec{p}_+$$

Therefore, although my system of charges is randomly distributed and therefore the dipole moments are randomly oriented, the system's total dipole moment seems to always be pointing in the same direction, as if the system was indeed polarised in a specific direction. How can this be? An interesting conclusion I draw from my result is that, if the positive charge distribution is symmetric with respect to the center of mass of the negative distribution, then the systems net dipole moment is zero (and the other way around also holds).

Answer 1

Nowhere in your derivation did you treat randomness. You started with an arbitrary neutral distribution of identical charged particles, and you've found that if the center of positive charge is separated from the center of negative charge, the system has a dipole moment. This is not a surprise. Or, in short, you can't stop me from plugging in $N=2,\vec r_1=(1,0,0),\vec r_2=(-1,0,0),$ which is a configuration that genuinely has a dipole. And if I placed $500$ positive charges near $\vec r_1$ and $500$ negative ones near $\vec r_2,$ then this is a configuration with large $N$ but that genuinely has a large dipole moment.

In order to treat randomness, the fundamental method of analysis is to average the quantity you're interested in over some space of possibilities for the random variables. This is what allows you to use the fact that arranging all the negative and positive charges on opposite sides of the origin is "unlikely" to show that the dipole moment should be small. You have not done this. Here's what it should look like:

Let the $\vec r_i$ be random variables, each with expected value (average value) the origin. Already, for all even $N,$ the expected value of the dipole moment is always $\langle\vec p_T\rangle=\vec0.$ However, this does not mean the magnitude is likely to be near zero, since it could be the system tends to have a large moment but the direction is uncertain. Indeed, for all finite $N,$ the expected dipole moment magnitude is nonzero: $\langle\|\vec p_T\|\rangle>0$ (but it could be small).

In order to learn more about the magnitude of the dipole moment, assume the $\vec r_i$ are identically and independently distributed, and that their distribution has finite variance. Then the central limit theorem ensures, as $N\to\infty,$ $\langle\|\vec p_T\|\rangle\to0.$ This statement confirms that a large neutral system of independently and identically distributed charges tends to have no dipole moment. In fact, I believe you are actually guaranteed $\sqrt{N}\langle\|\vec p_T\|\rangle\to0$ as $N\to\infty.$ This means, for large $N,$ quadrupling the number of particles tends to at least halve the strength of the dipole moment.

Answer 2

If your "system of charges is randomly distributed", $\vec{r}_+=\vec{r}_-$, so $\vec{p}_T=0.$