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A Story of Charged Equal Spheres

07 February 2019

Problem

Imagine three conductive spheres A, B and C each having charges of respectively 3q, 4q , and 5q.

We know that if we connect all three spheres together, their charges will be the same value: which is the arithmetic mean of the system. So \frac{3 + 4 + 5}{3} = 4.

Now imagine we do not connect all three together, but only two at each time, alternating between the first two and the last two (i.e. AB, BC, AB, BC ...). Surprisingly, if you continue this process ad infinitum, the charge of each sphere will converge into the arithmetic mean - even though all three spheres never touch each other at any moment.

This seemed intuitive to me, but I have decided to check it numerically with a botched up script:

from random import randint

def mover(spheres, move):
    if move == 0:
        avg = (spheres[0] + spheres[1]) / 2
        spheres[0] = avg
        spheres[1] = avg
    else:
        avg = (spheres[1] + spheres[2]) / 2
        spheres[1] = avg
        spheres[2] = avg

move = 0
sph0 = randint(-10, +10)
sph1 = randint(-10, +10)
sph2 = randint(-10, +10)
spheres = [sph0, sph1, sph2]
print(spheres)
for i in range(0, 100):
    mover(spheres, move)
    move = int(not move)
    print(spheres)
print("Avg: " + str((sph0 + sph1 + sph2) / 3))

Here are some results to convince you:

[-7, -6, 0]
[-6.5, -6.5, 0]
[-6.5, -3.25, -3.25]
[-4.875, -4.875, -3.25]
[-4.875, -4.0625, -4.0625]
[-4.46875, -4.46875, -4.0625]
[-4.46875, -4.265625, -4.265625]
[-4.3671875, -4.3671875, -4.265625]
[-4.3671875, -4.31640625, -4.31640625]
[-4.341796875, -4.341796875, -4.31640625]
[-4.341796875, -4.3291015625, -4.3291015625]
[-4.33544921875, -4.33544921875, -4.3291015625]
[-4.33544921875, -4.332275390625, -4.332275390625]
[-4.3338623046875, -4.3338623046875, -4.332275390625]
[-4.3338623046875, -4.33306884765625, -4.33306884765625]
(...)
[-4.333333333333334, -4.333333333333334, -4.333333333333334]
Avg: -4.333333333333333```

[-4, 10, -1]
[3.0, 3.0, -1]
[3.0, 1.0, 1.0]
[2.0, 2.0, 1.0]
[2.0, 1.5, 1.5]
[1.75, 1.75, 1.5]
[1.75, 1.625, 1.625]
[1.6875, 1.6875, 1.625]
[1.6875, 1.65625, 1.65625]
[1.671875, 1.671875, 1.65625]
[1.671875, 1.6640625, 1.6640625]
[1.66796875, 1.66796875, 1.6640625]
[1.66796875, 1.666015625, 1.666015625]
[1.6669921875, 1.6669921875, 1.666015625]
[1.6669921875, 1.66650390625, 1.66650390625]
[1.666748046875, 1.666748046875, 1.66650390625]
[1.666748046875, 1.6666259765625, 1.6666259765625]
[1.66668701171875, 1.66668701171875, 1.6666259765625]
[1.66668701171875, 1.666656494140625, 1.666656494140625]
[1.6666717529296875, 1.6666717529296875, 1.666656494140625]
(...)
[1.6666666666666665, 1.6666666666666665, 1.6666666666666665]
Avg: 1.6666666666666667

After convincing myself that this is indeed the case, I have decided to come up with an analytical proof to show that the following relation holds:

\lim_{n\to\infty} {Q_A(n) = Q_B(n) = Q_C(n)} = \frac{Q_A + Q_B + Q_C}{3}

where n is the number of steps this procedure is repeated, Q(n) is the charge of any sphere at the step n and Q_{A, B, C} are the initial charges of the respective spheres. By running a similar python script but utilising symbolic math with sympy, we can see how the charge of the middle sphere B changes with n.

n = 1 --- a/2 + b/2
n = 2 --- a/4 + b/4 + c/2
n = 3 --- 3*a/8 + 3*b/8 + c/4
n = 4 --- 5*a/16 + 5*b/16 + 3*c/8
n = 5 --- 11*a/32 + 11*b/32 + 5*c/16
n = 6 --- 21*a/64 + 21*b/64 + 11*c/32
n = 7 --- 43*a/128 + 43*b/128 + 21*c/64
n = 8 --- 85*a/256 + 85*b/256 + 43*c/128
n = 9 --- 171*a/512 + 171*b/512 + 85*c/256

By some Googling, I have found that the sequence 1 - 1 - 3 - 5 - 11 - 21 - 43 - 85 - 171 ... is called the Jacobsthal sequence. The Wikipedia page sums it up as: “in simple terms, the sequence starts with 0 and 1, then each following number is found by adding the number before it to twice the number before that.”

It would seem by inspection that the charge of the middle sphere B can be written as:

Q_B(n) = \frac{J(n)}{2^n}a +\frac{J(n)}{2^n}b + \frac{J(n-1)}{2^{n-1}}c

At this point I ran out of my wits and my sympy skills. So using the formula $ J(n) = 2^{n-1} - J(n-1) J(n-1) = 2^{n-1} - J(n) $ and substituting J(n) with the closed form J(n) = \frac{2^n - (-1)^n}{3} in MATLAB:

syms J(n)
syms a b c

J(n) = (((2^n) - (-1)^n)) / 3
expr = (((J(n)) / (2^n)) * a) + (((J(n)) / (2^n)) * b) + ((2^(n-1) - J(n)) / (2^(n-1))) * c ;
simp_expr = simplify(expr, 'Steps', 10);

simp_expr

We get:

\frac{a}{3}+\frac{b}{3}+\frac{c}{3}-\frac{{\left(-1\right)}^n a}{3 \cdot 2^n}-\frac{{\left(-1\right)}^n b}{3 \cdot 2^n}+\frac{2{\left(-1\right)}^n c}{3 \cdot 2^n}

We finally have the \frac{a}{3}+\frac{b}{3}+\frac{c}{3} we are looking for, and the right hand side converges to 0, as \lim_{n\to\infty} (\frac{-1}{2})^n = 0.

Convergence Rate

This proof was about the special case N = 3 spheres, but I am guessing it can be extended to any finite number of spheres as long as no sphere is left unconnected. Also the convergence rate to the arithmetic mean probably is slower as the number of spheres increase. To get a sense of the convergence rate in the numerical results, I will calculate the standard deviation of the sphere charges at step N to see how quickly it decreases. The script I wrote checks when the standard deviation is 0 and reports the number of steps it took. Running it a few times, it seems that for the three sphere case, the standard deviation converges to 0 at about 55-60 steps. For the case of 10 spheres, the script ran for about 15 mins and the standard deviation failed to converge to 0 (of course checking equalities in floating-point arithmetic is a bit sloppy) after 8 million steps; so we can empirically conclude that the convergence rate indeed decreases for larger number of spheres.

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