Transition of Electric Charge In Collision Between Proton And Antiproton

Question

I know that

$$p+\bar{p}\to 4\pi^++4\pi^-+(\gamma)$$

Before the collision, the sum of absolute electric charge value is $2$.

$$\left | +1 \right |+\left | -1 \right |=2$$

After the collision, the $p$ collapses to $4\pi^+$, and $\bar{p}$ collapses to $4\pi^-$.

After that, $4\pi^+$ collapse to $4\mu^++(\nu)$, and $4\pi^-$ collapse to $4\mu^-+(\bar{\nu})$.

$$\pi^+\to\mu^++\nu$$ $$\pi^-\to\mu^-+\bar{\nu}$$

And then, $4\mu^+$ collapse to $4e^++(\nu+\bar\nu)$, and $4\mu^-$ collapse to $4e^-+(\nu+\bar\nu)$

$$\mu^+\to e^++(\nu+\bar\nu)$$ $$\mu^-\to e^-+(\nu+\bar\nu)$$

The result is

$$p+\bar{p}\to 4\pi^++4\pi^-+(\gamma)\to 4\mu^++4\mu^-+(\gamma) \to 4e^++4e^-+(\gamma)$$

In this situation, the sum of absolute electric charge value is $8$.

$$\left | +4 \right |+\left | -4 \right |=8$$

Question. How could it possible that the sum of absolute electric charge value has increased?

Added. 'So, what's the significance?'

Electric force bwtween $p$ and $e^-$

$$F=k\frac{e^2}{r^2}$$

Electric force bwtween $4e^+$ and $e^-$

$$\sum_{n=1}^{4} k\frac{e^2}{(r_{n})^2}$$

$p$'s electric charge is +1, but after this collision, $p$ collapses to $4e^+$, it means the total electric charge has increased $+1$ to $+4$.

Answer 1

It's certainly possible to have an absolute electric charge value that increases, as long as total charge still sums up to 0. After all - this is why virtual particles (which consist of a pair of positive/negative charges) can be created

See http://en.wikipedia.org/wiki/Virtual_particle

Is there any physical significance to the absolute electric charge value? Also - I'm just curious - where did you get the reaction equation from?