Inelastic scattering problem [closed]

Question

Regarding this example, I am unable to get the last step. The confusion I have is: I think that the threshold energy remains the same for both CM frame and lab frame(i.e $4m_p$). Hence, to get 4 protons(/anti-proptons) in the output (at rest), with 2 incoming proton (with one at rest) implies $2m_p$=Kinetic energy of the colliding proton. Thus $2m_p=(\gamma -1)m_p \implies \gamma=3\implies v_{p,lab frame}=\sqrt(8)/3.$ Thus momentum of colliding proton in lab frame is $\gamma \times m_p \times v_{p,lab frame} =2.65$. I can't understand where I am going wrong.

Answer 1

Something appears to be off in this example. Given the mass $m$ and the momentum $p$, the energy is determined by:

$$ \frac{E}{c} = \sqrt{m^2c^2 + |\mathbf{p}|^2} $$

Putting in the momentum from the last step the energy must be:

$$ E = 6.56737 \; \text{GeV} $$

Which is twice as much as the threshold energy in the example, while in fact it should be $E^* - mc^2$ because one particle is at rest.