Minimum $E$ of $p\bar{p}$-collision for $q\bar{q}$ pair with mass $m_q$

I am currently working out the energy required to create a particle anti-particle pair from a collision of a proton travelling along the x-direction with an anti-proton which is at rest. The particle has a mass $m_q$.

Conservation of 4-momentum in the rest frame of the target anti-proton ($c=1$):

$p_1=(m_\text{p},0,0,0)$

The moving proton has this minimum energy, the quantity we are trying to find:

$p_2=\left(E, \sqrt{E^2-m_p^2 },0,0\right)$

Then what I find confusing is the terminology, it says that $p_{q\bar{q}}=\left(\sqrt{p^2+4 m_q^2 },p,0,0\right)$

Where has this $p$ come from and what is it? Is it the momentum of the center of mass of the $q\bar{q}$ system? Also, we don't seem to have taken much care about reference frames here, or rather I havn't thought about them really which is worrying.

Finally, it says that the minimum energy, E, the $q\bar{q}$-pair will be at rest in the CMF frame. What is this frame referring to? The center of mass of the particle/anti-particle pair? I understand that the threshold energy to produce these will be a combination of the energy of the proton and their rest masses but I don't know how we have the equation (above)

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It seems you are correct the $p$ is the momentum of the center of mass (COM) of your $qq$ (sorry don't know how to add bars) system in the lab frame. Due to conversation of momentum the $qq$ system may only have momentum in the x direction since that is your initial conditions.

With regards to the 2nd part of your question in the CMF (center of mass frame) of the $qq$ system the $qq$ pair will have equal and opposite momentums; however, in the case where the E of the initial condition is just sufficient to produce the $qq$ pair they can have no additional kinetic energy, thus they will be at rest in their CMF. Again due to conservation of momentum the $qq$ COM must still be moving in the x direction in the lab frame.

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It's \bar{q} :) –  Michael Brown Mar 28 '13 at 14:51

I think $p$ is indeed the momentum of the center of mass of the $q\bar{q}$ system. As the $q\bar{q}$-pair is created with the minimum energy required there is no relative momentum and all the momentum is in the center of mass momentum $p$. The energy of the $q\bar{q}$-pair is then given by $E = \sqrt{ p^2 + (2m_{q})^2 } = \sqrt{ p^2 + 4m_{q}^2 }$. As the mass of the $q\bar{q}$ pair is just $2m_{q}$.

So $p_{q\bar{q}}$ is the four vector of the $q\bar{q}$-pair with center of mass momentum $p$ along the $x$-axis.

In respect to your answer about the frame of reference. All four vectors you mentioned are in the lab-frame (and not in the center of mass frame).

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