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I know that two protons colliding at high energies can produce a pion $$p+p \rightarrow p+p+\pi^0$$

But is it possible for two protons to produce a pion pair? $$p+p \rightarrow p+p+\pi^{-}+\pi^{+}$$

and if yes how would one be able to calculate the necessary kinetic energy for this to happen?

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  • $\begingroup$ Think about it in the center-of-momentum frame. $\endgroup$ – G. Smith Jun 25 at 21:44
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Yes. As long as you obey the symmetries of the Standard model (for example, in this case you have already made sure to conserve elecrtic charge) and have enough energy, any production is possible. To see the kinetic energy you need, simply open up your Particle data group booklet and look up the mass of the charged pion and the proton. You see that it is 140 MeV and 938 MeV respectively. Now you need to conserve both energy and momentum. Following the same calculations as in

Pion production in proton-proton collision

but changing the $\pi^0$ to two $\pi^\pm$ we then have

$E_p = m_p + 4m_\pi + \dfrac{2m_\pi^2}{m_p} = 1540 \text{ MeV}$

$E_{p, \text{kin}} = E_p - m_p = 602 \text{ MeV}$

Here we used the fact that $\pi^+$ and $\pi^-$ have the same mass. It is however unlikely that you will create them at this energy, the likelyhood will increase with higher energy as the pions have more possible final states (more possible different values of their momenta). Also, it is much more likely to create neutral pions than charged pions in general, nature likes to keep things neutrally charged if possible.

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  • $\begingroup$ I am not really sure if what you are saying is correct. I have found following post that contradicts your statement physics.stackexchange.com/a/106809/235334 $\endgroup$ – Alessio Popovic Jun 26 at 7:05
  • $\begingroup$ Yea I forgot about momentum xD it should be fine now. $\endgroup$ – Hotlab Jun 26 at 8:22
  • $\begingroup$ Great answer now everything is clear, but didn't you forget multiplying by $c^2$? $E_p=m_p+4m_{\pi}c^2$... $\endgroup$ – Alessio Popovic Jun 26 at 10:01
  • $\begingroup$ Yes there should be a $c^2$ on all the terms (also on $m_p$). Often people will skip writing it out, like I did here, because it is undrestood that it should be there. $\endgroup$ – Hotlab Jun 26 at 12:44

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