How often do quarks annihilate?

Quarks bond frequently with quarks, and antiquarks bond frequently with antiquarks. So why do quarks not annihilate antiquarks extremely quickly?

More to the point, how much mass is lost in a body to quark annihilation for a given duration of time?

  • $\begingroup$ I don't understand what you mean with "how often"? Compared to what? In what situation? And what do you mean with them being known to bond "very often"? $\endgroup$ – ACuriousMind Nov 20 '15 at 23:19
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    $\begingroup$ What you are asking for is essentially the decay rate of the quark antiquark pair. A $\pi^0$ particle is a combination of $u\bar{u}$ and $d\bar{d}$ and decays in about $10^{-16}$s. $\endgroup$ – octonion Nov 20 '15 at 23:28
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    $\begingroup$ @octonion It is worth noting, however, that the $\rho$ meson has the same valence quark content as the $\pi^0$ and has a much shorter lifetime (on order of $10^{-24}$ seconds). Which may be the reason that the question is receiving downvote, Mason, it is founded on an untrue idea (that there is a characteristic time for this this behavior). The last sentence "More to the point, how much mass is lost in a body to quark annihilation for a given duration of time?" has a worse version of the same problem: it's not even clear what you think is going on that would give that meaning. $\endgroup$ – dmckee Nov 20 '15 at 23:42
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    $\begingroup$ I also don't understand why a naive but fair question needs to be downvoted. $\endgroup$ – octonion Nov 20 '15 at 23:52
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    $\begingroup$ As I said, the notion that there is a time scale for quark annihilation is erroneous. In each case that time scale is driven by two things: (a) the coupling constant of the allowed reactions (why the $\rho$ and $\pi^0$ are so different) and the relative wave-functions of the participants (why the $\pi^0$ and the $\pi^\pm$ are so different). Worse is the notion that there is a mass loss to these processes in ambient matter; this suggests that you've taken some pop-sci description far too seriously and then spent some time thinking hard about it when it won't stand up to that. $\endgroup$ – dmckee Nov 21 '15 at 0:19

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