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 ($c=1$):


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)


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.

  • 1
    $\begingroup$ It's \bar{q} :) $\endgroup$ – 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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.