Where is the Anti-matter Hiding?

Around the time of the Big Bang there was, according to one theory, unequal amounts of matter and anti-matter. When matter met anti-matter, they annihilated each other, albeit a little bit of matter was left over--that matter is the matter of our universe. Below is a Feynman diagram illustrating the process of matter-anti-matter creation and destruction.

According the above diagram, a particle (+A) and an anti-particle (-A) destroy each other and produce a boson (B). The boson then goes on to produce another +A/-A pair. This is all done in time (+t) and minus time (-t). (Anti-particles allegedly have anti-time.)

To my knowledge, the above theory has yet to be experimentally verified. It would be great if some scientist could show that a photon, for example, could produce an electron without the positron. That would surely establish why our universe is predominately matter.

The next best thing is to demonstrate that an anti-particle always comes with a particle, but in spite this, a universe could still end up with mostly matter or anti-matter. That is what we are going to demonstrate below.

(Note: What we call matter could very well be anti-matter. However, we are biased and like to think positive--so we label what we have "matter" and its opposite "anti-matter.")

Let's kick things off with a diagram of two electrically charged plates. Let's pretend the positively charged plate is matter and the negatively charged plate is anti-matter. The arrows represent the field lines. Let's also assume the plates are identical in every way except for the charge--this will be analogous to equal amounts of matter and anti-matter.

The arrows originating from the positive plate point away from the source. One arrow goes up and another goes down and through the negative plate:

The arrows originating from the negative plate behave in the opposite fashion:

Now let's add the arrows. Arrows pointing in opposite directions shall cancel each other. Arrows pointing in the same direction shall enhance each other.

Well would you look at that! The arrows didn't cancel each other out completely. We end up with two left over. So it is possible to start with equal amounts of opposite charge, put them together and not end up with zero. However, what if we had a second pair of plates that are reversed?

As you can see the arrows left over are pointing in the opposite direction. If we add those arrows to our original left-over arrows, we get zero.

So whether we get zero or left-over arrows depends on whether we have an odd or even number of plate pairs. An odd number will always give us left-over arrows. Even-numbered pairs will sometimes give us left-over arrows and sometimes not.

We could ask, what is the probability we will get an odd or even number of pairs? I'd say .50 is a reasonable estimate. (P1(a) stands for probability of getting an odd or even number of pairs. The "a" stands for annihilation.)

Let's assume our luck is bad and there are two pairs of plates. What is the probability (P2(a)) that the left-over arrows will be opposite and cancel each other?

How about .50? Finally, what is the total probability (P(a)) that we will end up with zero arrows? What is the probability (P(!a)) we will have something left over when matter and anti-matter annihilate each other?

It should be obvious our universe had at least a .75 (or 75%) chance of having some matter (anti-matter) left over. The odds improve when bigger numbers are crunched. For example, four pairs of plates have a .375 probability of cancelling each other out and yielding zero. That raises the chance of left-over matter to .78 (or 78%) (1- (.5 * .375)= .81; [.81+.75]/2=.78).