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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 $+4$.

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2  
I didn't downvote, but I'll offer my opinion. I think that even with the "Added" part, the question's very unclear. You're asking why the total absolute value of charge fails to be conserved, but I don't really know why you'd expect it to be conserved in the first place! –  Ted Bunn Jul 21 '11 at 17:39
    
Like Ted Bunn, I don't feel that the "Added" part clarified anything. The electrical force would actually be a vector sum, not a sum of magnitudes like $|F_1|+|F_2|+\ldots$, and in any case, it doesn't seem to connect to your statement that "the total electric charge has increased +1 to +4." There is no family tree of relationships between the initial particles and the final particles. It's not as though the p is the father of the four e+'s. –  Ben Crowell Jul 21 '11 at 22:35
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@ks0830: I assume you're talking about me, but what would make you think that I downvoted your question? (For what it's worth, I don't think it's a good question for the same reason Ted mentioned. You haven't explained why you would expect the absolute value of charge to be conserved.) –  David Z Jul 22 '11 at 5:36
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up vote 5 down vote accepted

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?

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2  
What book though? –  InquilineKea Jul 21 '11 at 4:12
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Hm, what's the chapter/page number? I have a hard time believing that the book would ask a paticle physics question like this (but I haven't looked at the book too closely yet) –  InquilineKea Jul 21 '11 at 4:47
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@kso830 sit tight, because this is actually the right answer; also, absolute charge does not have any physical meaning, and more importantly, it is experimentally known to not be conserved, precisely by the mechanism that InquilineKea tried to explain to you –  lurscher Jul 21 '11 at 16:10
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