# The electrical potential of charge-neutral masses

1. Aug 19, 2011

### kmarinas86

I have noticed something quite peculiar.

If you have equal numbers of e+ and e- charge, there will be more "force pairs" between opposite charges than there are between like charges.

Generally speaking, if X is the number of e+ charges and Y is the number of e- charges, then number of "force pairs" between opposite charges is X*Y, and the number of "force pairs" between like charges is X*(X-1)/2 + Y*(Y-1)/2

When X=Y, as X and Y both get larger, the % difference between both figures approaches 0. A difference of 0 can be obtained also from certain combinations of X and Y. The simplest case is where you have 3 e+ charges and 1 e- charge [i.e. (X,Y)=(3,1)], or conversely, 3 e- charges and 1 e- charge [i.e. (X,Y)=(1,3)].

If
X*Y > X*(X-1)/2 + Y*(Y-1)/2 [i.e. opposite charge "force pairs" > like charge "force pairs"]
2*X*Y > X*(X-1) + Y*(Y-1)
0 > X*(X-1)+ Y*(Y-1) - 2*X*Y
0 > X^2 - X + Y^2 - Y - 2*X*Y
X + Y > X^2 + Y^2 - 2*X*Y
(X+Y) > (X-Y)^2

If
X*Y = X*(X-1)/2 + Y*(Y-1)/2 [i.e. opposite charge "force pairs" = like charge "force pairs"]
(X+Y) = (X-Y)^2
Y=(2*X+1±sqrt(8*X+1))/2 [solution for solving Y in terms of X - quickmath.com]
All integer solutions for X and Y in this case are pairs of triangular numbers which are next to each other on the sequence "1, 3, 6, 10, 15, 21, 28, 36, 45, 55, .....", the sum of each results in a square, beginning with 1+3=4.

If
X*Y < X*(X-1)/2 + Y*(Y-1)/2 [i.e. opposite charge "force pairs" < like charge "force pairs"]
(X+Y) < (X-Y)^2

Knowing that opposite charges attract and like charges repel, it would appear that any electrical neutral system (i.e. consisting of equal positive and negative charge) may be "self-attracting" on the basis of electric forces, provided that the distances between charges is randomized. If an electrically-neutral system suddenly stops collapsing, then it would appear that the forces between like charges must have caught up with the forces between opposite charges. What causes this is probably a growing magnetic field component of the Lorentz force that begins to overtake the electrical component, where the like charges co-gyrate around an axis (corresponding to the magnetic attraction seen between two parallel e- currents) while opposite charges counter-gyrate.

Not only does the above indicate that it is quite possible from electrical forces alone to have some residual net attraction for electrically-neutral bodies, you could have situations where like charges could attract electrically so as long as they actually consisted of a certain number of both e- and e+ of charges satisfying one of the conditions provided above, which is:

(X+Y) > (X-Y)^2

Again, where:
X is the number of e+ charges
Y is the number of e- charges

With the added condition that X does not equal Y.

If we fix "X-Y" to a constant (i.e. constant=X-Y), by increasing X and Y by equal amounts, X-Y as a percentage of X+Y decreases, and the result is that the difference between X+Y and X-Y reaches a proportionality with non-signed sum of charges X+Y.

Of course, the formula requires sufficiently random charge distributions to have any direct connection with whether a given mass will either expand, contract, or remain at steady-state. Otherwise, one must obviously use more complicated calculations.

Whatever you might think of the above math, the fact that for equal numbers of opposite charges, more "force pairs" exist between opposite charges than between like charges, which suggests that supposedly charge-neutral objects will exert unusual "attractive" forces. Or perhaps this effect is so usual, that we call it by another name: Gravity?

Question: Has this apparent asymmetry ever been recognized in peer-reviewed physics research? Or am I just barking up a new tree?

2. Feb 14, 2012

### kmarinas86

Conclusion: No way! They don't even realize this.

3. Feb 14, 2012

### DennisN

Hi, interesting thinking (combinatorics). I am afraid it has no physical relevance, as you are only counting the number of vector components, and not the individual vector magnitudes. Further, the vector components for each charge can be added together, yielding only one vector for every single charge. Then there will be no X and Y, only, let's say, Z = number of vectors (forces) = number of charges. (see Vector space).

4. Feb 14, 2012

### kmarinas86

In the case where the distribution of charges is random, it would diminish the effect that spread of magnitudes it may have though, could it not? So for a randomly distributed neutral mass, we may have a net attractive force on-the-average.

In reality, of course, a net attractive force would cause some stability problems, but that could be counteracted by centrifugal spin.

Also, the electrons and protons preside in distinguishable regions in atoms. That should help the increase repulsive forces to be on par with the attractive forces.

The most interesting result out of all of this is that, if we consider only electrical forces, a homogenous, neutral object with no spin is destined to collapse. Conversely, a neutral object that is heterogeneous and with significant centrifugal spin is destined to expand. An object oscillating between these two states would produce a kind of spherical radiation of sorts. Any object not emitting much radiation corresponding to its "physical wavelength" would be a kind of compromise (e.g. a heterogeneous object with relatively slow spin (planetary scales), or conversely, a spinning object with relative uniformity (sources of static curl fields)).

Last edited: Feb 15, 2012
5. Feb 15, 2012

### DennisN

(First, by spin I presume you mean classical rotation, not the spin used in particle physics? Because they are very different.)

Yes, I think I see what you are implying, and I have thought in somewhat similar ways many years ago. Yes, you can get a net attractive force with particular configurations. The easiest visualization I can think of is to arrange four charges in a squarelike fashion (using two dimensions, not three);

O *

* O

(O = +, * = -)
Using Coulomb's law in this configuation, every charge will have a net force pointing towards the center of the square.

But I disagree with your conclusion that this neutral configuration will implode. Using what we know of physics, every charge, i.e. charged particle, has a mass. Disregarding anything else, what will happen is that the configuration first will implode and then explode (as a sidenote I actually have simulated this once with a computer program). If O=protons and *=electrons, they would end up flying away from eachother. If O=positrons (antielectrons) and *=electrons, they would either do the same (or annihilate eachother, resulting in photons flying away at the speed of light).

Furthermore, there is a problem with applying this reasoning concerning mass and gravity, if that is what you're thinking of? An example is the electron, which we know both have mass me and charge -e. Today, there is no indication of any substructure of the electron (it's the same story with the quarks), down to a certain spatial resolution (10-18 m, if I remember correctly).

So, how do you model the mass of the electron in this fashion? If you assume the electron is made up of a lot of tiny fractionary charges (probably with masses), the sum of them must be -e. And to keep the electron from exploding (it might implode first), you'll probably need to come up with new physics to hold it together (gluonlike or something). Tricky thing, lots of unknowns. But interesting to think about . So, basically, I agree with some of your reasoning concerning net forces, but perhaps not your conclusions.

Also, on the atomic/molecular level, the Van der Waals force is the result of the sum of the attractive or repulsive forces.

Last edited: Feb 15, 2012
6. Feb 16, 2012

### DennisN

Hi again, I came to think of another thing which may enlighten you further. You said:
That is correct. But it's even more fundamental than this; no collection of charges can be stable based solely on electrostatic interaction. Earnshaw's theorem: "Earnshaw's theorem states that a collection of point charges cannot be maintained in a stable stationary equilibrium configuration solely by the electrostatic interaction of the charges." See also Earnshaw's theorem - Impact on physics.

So, whatever we do, we have to consider more things than just electrostatics to get stable configurations.