Proton-Antiproton colliding to produce top-antitop pair

[Je m'appelle]

Homework Statement

Suppose a proton and an antiproton collide producing a pair of top-antitop quarks.

What would be the minimum required momenta of both proton and antiproton in order for this pair creation to occur?

p + \bar{p} \rightarrow t + \bar{t}

(Answer: 173 \frac{GeV}{c^2}, 59.9 \frac{TeV}{c^2})

Homework Equations

I. Energy-momentum relation
E^2 = (pc)^2 + (m_0 c^2)^2

II. Relativistic kinetic energy equation
E_{ki} = m_i c^2 \left(\frac{1}{\sqrt{1-\frac{v_i^2}{c^2}}} -1 \right)

III. Relativistic momentum equation
p_i = \frac{m_i v_i}{\sqrt{1-\frac{v_i^2}{c^2}}}

IV. Rest masses of proton and top quark

m_p = 938 \frac{MeV}{c^2}, m_t = 173 \frac{GeV}{c^2}

The Attempt at a Solution

1. Conservation of energy:
E_i = E_f \Rightarrow E_{p} + E_{\bar{p}} = E_{t} + E_{\bar{t}}

\sqrt{ (p_{p}c)^2 + (m_p c^2)^2 } + \sqrt{ (p_{\bar{p}}c)^2 + (m_{\bar{p}} c^2)^2 } = \sqrt{ (p_{t}c)^2 + (m_t c^2)^2 } + \sqrt{ (p_{\bar{t}}c)^2 + (m_{\bar{t}} c^2)^2 }

2. Conservation of momentum:
p_i = p_f \Rightarrow p_{p} + p_{\bar{p}} = p_{t} + p_{\bar{t}}

I'm really stuck here because I feel like there isn't enough information and I'm missing something, one thing that occurred to me is that since I'm trying to find the minimum momenta of the proton-antiproton, I could argue that the momenta of the quarks are zero, i.e. they're created at rest, but then on a second thought, I'm not certain about this because it would provide an equal momentum in magnitude for the proton-antiproton, which isn't the case.

Can anyone point me in the right direction here, please?

 

[BvU]

Hello MyNameIs,

Your book answer surprises me: why does it consist of two numbers ?

Are there two questions, perhaps? Like: one in a colliding beam machine and one in a stationary target machine (in which case one of the momenta is 0) ?