electricity - Sum total distance of electrons on a spherical surface

What is the sum total distance between every possible pair of point charges when there are n point charges on a spherical surface? All point charges can only and are located on the infinitesimal spherical surface.

Basically, we're going to have a bunch of points staying as far apart as possible on the surface of a sphere. Is there a general equation for this?

And whats the sum total distance between every pair of points? (Both the distances by travelling the shortest distance on the spherical surface and the shortest distance travelling through the interior volume of the sphere. I'm confident that the relationship between those two values require a relatively simple equation.

Is there a similar mathematical model/problem which can be used to solve this problem?

Answer

To begin with, I'll have to assume you mean surplus electron charges on a sphere. Obviously, the integral of the inverse of distances between charges would be comparatively easy to relate to physical quantities since that's basically potential. So let's talk about this sphere, radius $R$, with a charge $Q$ on it, made up of $Q/e$ electrons. The self capacitance of a conducting sphere is $4 \pi \epsilon_0 R = R/k$. Then we can establish that the total potential energy is $\frac{1}{2}C V^2 = \frac{1}{2} \frac{Q^2}{C}$.

$$E=\frac{Q^2}{2 C} = k e^2 \frac{N^2 }{2 R} = k e^2 \sum_{i=1}^N \sum_{j=1}^{i-1} \frac{1}{|r_i-r_j|} $$

For the sum of distances (not the inverse), you can get a formula like this.

$$\sum_{i=1}^N \sum_{j=1}^{i-1} |r_i-r_j| = \frac{2}{3} N^2 R$$

This falls out from the calculus of the summation, just like the capacitance value. I didn't do that, instead I just wrote a code that discovered the relationship to my satisfaction.

program sphere
  implicit none

  double precision :: mu, theta
  double precision :: mu2, theta2
  double precision, dimension(3) :: r1, r2, rand
  integer :: i, j, N
  double precision :: thesum,thesum2, ind
  double precision, parameter :: pi = 3.14159265

  double precision :: d, rad

  N = 5000
  rad = 2.

  ind = 0
  thesum = 0.
  thesum2 = 0.
  do i = 1,N
    r1 = random_points(rad)

    do j = 1,i-1
      ind = ind+1
      r2 = random_points(rad)
      d = sqrt(sum((r1-r2)**2))
      thesum = thesum + 1./d
      thesum2 = thesum2 + d
    end do
  end do

  write(*,*) ' N= ',N,'  number= ',ind

  write(*,*) ' pot/N^2 ',thesum/N**2
  write(*,*) ' len/N^2 ',thesum2/N**2

  contains

    function random_points(r)
      implicit none
      double precision, dimension(3) :: random_points
      double precision, intent(in) ::  r
      double precision :: theta, mu

      double precision, dimension(3) :: rand
      call random_number(rand(1))
      call random_number(rand(2))
      theta = 2.*pi*rand(1)
      mu = rand(2)*2.-1.
      random_points(1) = cos(theta)*sqrt(1.-mu**2)*rad
      random_points(2) = sin(theta)*sqrt(1.-mu**2)*rad
      random_points(3) = mu*rad
    end function random_points



end program sphere

I don't know if this value has any physical utility. To begin with, we can put it in terms of more familiar physical values.

$$\frac{2 N^2 R}{3} = \frac{2 Q^2 R}{3 e^2}$$

The problem with doing a distance integral is that I can't think of any physical thing for which this would matter. Field and potential are $1/r^2$ and $1/r$ and if you integrate again, you get $ln(r)$. I suppose some forces do grow with distance, and maybe they're proportional to distance.

I'm confident that the relationship between those two values require a relatively simple equation.

As long as the geometry is sufficiently simple, this will be true for many similar questions. That's in the domain of math.

EDIT:

I think that the revise problem is to constrain:

$$i=1 .. N$$

$$ |\vec{r}_i| < R$$

Then show that 1 being true implies 2 below

1. the sum of $1/r$ integral of all the charges over the entire volume is constant


2. the summed distance between all points is at a maximum

I think this could be what is being asked. It's a little lofty, but I'm sure it's entirely doable. My suspicion is that it would apply for any arbitrary region.

THE CALCULUS:

I'll present the way to get these numbers by integrating, partly because I think it would be helpful for an incoming freshman to see. The problem my above code solves is to integrate the values of $1/r$ and $r$ over all pairs of points over a sphere. When I do the actual integral I'll multiply it by 1/2 because the integral would otherwise double-count the pairs. Integrating is basically a way of using math to describe a problem with infinite points.

Let's start out. The surface area of the sphere is:

$$SA = 4 \pi R^2$$

The charge density is the number divided by the surface area. This is the number of charges per unit area on the surface of the sphere.

$$\sigma = \frac{N}{SA}$$

You can integrate over any variable you'd like, I'll chose the angle between the x-axis and the vector. I'm going to denote this with a prime to indicate that it is the "second" piont in the pair, and the "first" point in the pair will simply be fixed.

$$ \vec{r}' = = < R \cos{\theta}, 0, R \sin{\theta} >$$

$$ \vec{r} = < 1, 0, 0 >$$

Define the distance between them. This is a scalar.

$$ d = | \vec{r} - \vec{r}' | $$

I'm going to do a surface integral to cover all the $\vec{r}'$ and then do another surface integral (times 1/2) over all the $\vec{r}$. That's the jist of it, but there are lots of symmetries involved. These symmetries reduce the dimensionality of the $\vec{r}'$ integral by 1 and the $\vec{r}$ integral by 2. That means the latter isn't even an integral. The proposition behind these is that

1. You can rotate the $\vec{r}'$ point around the x-axis and it doesn't change the distance


2. You can move the $\vec{r}$ point all around the sphere and it doesn't change the distance

Now I'm at a point where I can write the integral. First for the total electrostatic energy. Note that $e \sigma$ is the charge density since I used $\sigma$ as the number density. The first expression of charge density time surface area is the multiplier that I use in lieu of an outer integral. The $2 \pi y'$ is needed to correctly use the x-axis symmetry, it equates to multiplying by the perimeter of a washer centered about the x-axis.

$$ N = (e \sigma SA) \frac{1}{2} \int_0^{\pi} 2 \pi y' \frac{k e \sigma}{d} d \theta = k e^2 \frac{ N^2}{2} $$

This is the result I wanted. The same manner is used almost identically to reproduce the number for the sum of distances between points. I'll leave out the charges because there's no clear physical interpretation.

$$ sum = ( \sigma SA) \frac{1}{2} \int_0^{\pi} 2 \pi y' \sigma d d \theta = \frac{2 N^2}{3} $$

And that's the integral. Using a computational algebra package is helpful, but everything should be sufficiently defined here.