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This comes from Griffiths' Introduction to Elementary Particles. After introducing the reader to four-momentum, he provides us with this example problem:

The Bevatron at Berkeley was built with the idea of producing antiprotons by the reaction $p + p \to p + p + p + \bar{p}$. That is, a high-energy proton strikes a proton at rest, creating (in addition to the original particles) a proton-antiproton pair. Question: What is the threshold energy for this reaction (i.e. the minimum energy of the incident proton)?

To solve this, he first starts by considering what happens before the collision from the view of the proton at rest. This gives us the total four-momentum vector of $$p_{\text{tot}}^\mu = (E + m_p, \mathbf{p})$$

I am using natural units. The $E$ corresponds to the total energy of the high-energy proton, $\mathbf{p}$ is its momentum, and $m_p$ is the mass of the resting proton.

Next, he considers the total four-momentum after the interaction from the centre of mass perspective and says that the four-momentum is
$$p_{\text{tot}'}^\mu = (4m_p, \mathbf{0})$$

Obviously, he sums over the 4 particle's four-vectors, but I'm not sure why he could conclude that every proton had energy $m_p$. Should it not have energy larger than $m_p$ since it is travelling?

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    $\begingroup$ The words “threshold” and “minimum” are very important. Think about what they imply. $\endgroup$
    – Ghoster
    Dec 11, 2022 at 3:55

1 Answer 1

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$$p_{\text{tot}'}^\mu = (4m_p, \mathbf{0})$$ ... Should it not have energy larger than $𝑚_𝑝$ since it is travelling?

ANS: Adding to the earlier comment,
these are the threshold-case components in the center of momentum frame, not the lab frame.


By the way,
to supplement Griffith's algebraic-approach, you may find it enlightening to draw an energy-momentum diagram of the process in the lab frame:

  • What is generally a polygon ( an irregular hexagon with Minkowski-congruent sides, two sides for the incoming particles and the remaining 4 for the outgoing particles)
    will be a Minkowski-isoceles triangle. (Think about why that would be.)
  • Use the Minkowski Law of [hyperbolic]-Cosines to find the hyperbolic-cosine (the time-dilation factor of the incident proton in the lab frame).

update: Here's a visualization of "proton-antiproton pair" production in energy-momentum space with Desmos ( https://www.desmos.com/calculator/kckm1cxuom )

  • The orange hyperbola is the mass-shell of the COM, centered at the origin.
  • The obscured violet hyperbola is the mass-shell of the COM-for-the-threshold case.
  • The green hyperbola is the unit-mass mass-shell of the incident proton (centered at the tip of the 4-momentum of the target-proton at rest in the LAB).
  • The threshold-case is located in energy-momentum space as the intersection of the green hyperbola and the violet hyperbola.

LAB frame
robphy-Desmos-energyMomentum-1

COM-threshold frame
(This reveals that this problem is, geometrically-speaking, essentially the the clock-effect [as seen in the twin paradox].)
robphy-Desmos-energyMomentum-2

The outgoing 4-momenta are chained by a sequence of unit-mass mass-shells (not shown by default)---a linkage. You can tune these 4-momenta to move off of the threshold condition (revealing the distinction between the orange and violet mass-shells).

beyond-threshold in the COM-threshold frame
robphy-Desmos-energyMomentum-3

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