Suppose you have an electric field in three dimensions created by some finite (but possibly arbitrarily high) number of point charges, each with charge equal to an integer multiple (positive or negative) of $e$. Now you pick a point $O$ in a three dimensional space, and a ray $\vec{\ell}$ originating from $O$ in any direction.

Is it true that for any such configuration of charges there always exist an rational number $\alpha$ such that $\displaystyle \lim_{r \rightarrow \infty} || E(P_r) || r^{\alpha}$ converges to some nonzero constant, where $P_r$ is a point in the ray $\ell$ such that distance from $O$ to $P_r$ is $r$ and $|| E(p_r) ||$ is the magnitude of the electic field at $P_r$ ?

  • $\begingroup$ I believe your question may be related to this one: is there a nontrivial finite charge configuration where all multipole moments are zero? $\endgroup$ – probably_someone May 9 '18 at 17:48
  • $\begingroup$ @probably_someone I have started reading electrostatics since only five days ago and I don't know what multi-pole moment is yet :( $\endgroup$ – katana_0 May 9 '18 at 17:50
  • $\begingroup$ Essentially, the multipole moments are the easiest ways to get the asymptotic behavior of an electromagnetic field. See the link in Kthaxt's answer. $\endgroup$ – probably_someone May 9 '18 at 17:52

I believe what you're saying is true, but maybe it could be asked in a simpler way. Would you agree another way of asking the question is, is the E field of a static assortment of charges always of the form: $$ E \propto r^{-\alpha}$$ Where alpha is > 0. If this is the case, then yes. I believe this is a result of the multipole expansion in Cartesian coordinates.

  • $\begingroup$ Thanks for pyping up this answer, but I don't know what multi-pole moment is yet (and I checked up the wikipedia link but although I know single variable calculus I don't know stuff like "spherical harmonic"). So I'll come up to this question later. But can $\alpha$ be an irrational number, e.g can you give a construction where $-\alpha = \frac{1+\sqrt{5}}{2}$ ? $\endgroup$ – katana_0 May 9 '18 at 17:54
  • $\begingroup$ Ah I see, I think I'm understanding the complexity of your question a little more. It seems to me no arrangement of charges could result in an irrational $\alpha$ for 3D space... but that's more from intuition. How can you get an irrational coefficient from superimposing rational ones? I once had a professor calculate for us the potential from a single charge in N dimensional space, where N was not necessarily an integer. Maybe if you pick a funkier space you could get irrational values of $\alpha$ $\endgroup$ – Kthaxt May 10 '18 at 18:58

My guess would be that what you are stating is true. You can even be sure that for $\alpha =2$ the limit will be finite.

Here's my reasoning. Pick your point O, and your ray. Then for $r$ sufficiently large, the sphere centered at $O$ and of radius $r$ contains all the point charges (since there is a finite number of them, they must lie in a finite region of space). If you want to be more explicit denote by $R$ the distance of the furthest point charge from $O$, then you can pick $r>R$.

Then, when $r\rightarrow \infty$, the details of the point particle distribution inside the sphere $R$ disappear. In other words, when looking from afar, your sphere $R$ just looks like a point particle with total charge the sum of the charges of the point particles. Hence, the electric field produced by your distribution, when $r\rightarrow \infty$, should be $E\propto \frac{\sum q}{r^2}$. Of course, if the charges cancel, we could have some faster decreasing field, but it is sure that it will decrease at least as fast as $r^{-2}$.

Let me be more specific. Consider the total electric field produced at our faraway point p : $$\vec{E(p)} \propto \sum_i \frac{q_i}{|\vec{r}-\vec{r}_i|^2}\vec{e}_{i}$$

Where $\vec{r}$ denotes the position of $p$, $\vec{r}_i$ the positions of the different charges measured from O, and $\vec{e}_i = \frac{\vec{r}-\vec{r}_i}{|r-r_i|}$.

Then, using the triangle inequality :

$$|E(p)|\leq \sum_i \frac{q_i}{|\vec{r}-\vec{r}_i|^2}$$

We now expand using the fact that $\vec{r}_i/r \rightarrow 0$, i.e. $r$ is really big :

$$|E(p)|\leq \sum_i \frac{q_i}{r^2}(1+2\frac{\vec{r}\cdot\vec{r}_i}{r^2}+O(\left(\frac{r_i}{r}\right)^2))$$

So, finally, $$lim_{r\rightarrow \infty}|E(p)|r^2 \leq \lim_{r\rightarrow \infty}\sum_i\frac{q_i}{r^2}r^{2} = \sum_i{q_i}<\infty$$

Concerning your comment, as you can see for two opposite charges, the limit is 0, meaning the norm of the magnetic fields decreases faster than $r^{-2}$ ($r^{-3}$, as you remarked).

Let's try and see if it is possible to determine the exponent of $r$ that we need in order to get a finite non-zero constant. First of all, we should exclude the case of no charges, since that clearly gives 0.

Let's start expanding the electric field without using the triangle inequality, since we want an exact limit. I will use $\vec{e}_r = \frac{\vec{r}}{r}$

$$\vec{E}(p) = \sum_i \frac{q_i}{|r-r_i|^3}(\vec{r}-\vec{r}_i) = \sum_i\frac{q_i}{r^2}(\vec{e}_r-\frac{\vec{r}_i}{r})(1+3\frac{\vec{e}_r\cdot\vec{r_i}}{r}+O(\frac{r_i^2}{r^2}))$$

Now, notice in the expansion above that no matter to which order we push the expansion, it is always rational exponents that we add. This is simply due to the fact that $\frac{1}{1-q} = \sum_i^\infty q^i$. The expansion at arbitrary order will thus be a function containing only monomials of the form $r^{-\alpha}$, where $\alpha$ is an integer. (I believe that's the multipole expansion people are referring to, it's been a long time for me so I'm not too sure).

Now, when $r\rightarrow \infty$, we only care about the lowest order term of the expansion. To illustrate, we first look at the first term, $\frac{|\sum_i q_i|}{r^2}$. If it is non-zero, then $\alpha = 2$ and we are done. If $|\sum_i q_i| = 0$ (example, the dipole) , then we go to next order, $r^{-3}$ and so on.

I believe this does prove that you can't have an non-integer exponent in $r$ for the expansion, let alone an irrational one.

Now there is however one last problem : what happens if the expansion is 0 for any order ? That would mean that the electric field is exactly 0 at infinity. Intuitively I would say this should be impossible to do with a finite number of charges, since we are essentially imposing an infinite number of constraint, but I did not find anyway to prove it rigorously yet.

In any case, it should not invalidate the claim that the exponents are integers.

  • $\begingroup$ Of course your claim that $\alpha = 2$ happens always is obviously wrong so I didn't read your answer fully: You place a point charge with charge $+e$ at $(0,0,0)$ and a point charge with charge $-e$ at $(1,0,0)$ and very far away in the x axis it would be proportional to $r^{-3}$, not $r^{-2}$ $\endgroup$ – katana_0 May 9 '18 at 17:20
  • $\begingroup$ Let me edit that in, I was a bit too hasty. Note however that it DOES work for $\alpha = 2$, since $\lim_{r\rightarrow \infty}r^{-3}r^{2} = 0$, which is finite. Maybe you wanted to ask about the asymptotic behavior, in which case you should also require the limit not to be 0. $\endgroup$ – Frotaur May 9 '18 at 17:22
  • $\begingroup$ Oh yeah sorry my wrong, I should clarify it that I'm asking for the exact exponent. BTW by induction you can give a construction for all positive integer $\alpha >1$. Also I don' think the mathematical justification is that easy (I'm not even sure if irrational exponents don't crop up if you take your configuration as limiting value of discrete configurations) $\endgroup$ – katana_0 May 9 '18 at 17:25
  • $\begingroup$ Re your edit just now: I mentioned you 15 minutes ago that I forgot to add that the limit to not be zero, but you seem to not address that. How you are sure that for any configuration, the exponent is a rational, not some irrational number like Euler's constant ? You seem to not address that issue too. $\endgroup$ – katana_0 May 9 '18 at 17:41

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.