# Threshold energy for pair production from proton-electron collision

In this undergrad nuclear physics problem I am asked to find the kinetic energy threshold for an electron colliding with a still proton to create an electron-positron pair. So in short: $$e^- + p^+ \rightarrow e^- + e^+$$

First I am not sure whether this reaction can even happen in real life. Knowledge expected from me at this point is to apply 4-momentum conservation laws.

So I approached it with a standard procedure: I wrote the 4-momentum conservation law, and then squared it. $$P^2$$ is the same in any reference frame, so I chose the lab reference frame for the left side ($$e^- + P^+$$) and the center-of-mass reference frame for the $$e^--e^+$$ pair. And to get the threshold energy for incoming electron, I just chose the kinetic energy of the resulting pair in their center of mass frame to be zero.

The general result for kinetic energy threshold for reaction type $$a+b \rightarrow c+d$$ is: $$$$\label{gen} T_a^{threshold} = \frac{(m_c+m_d)^2c^2-(m_a+m_b)^2c^2}{2m_b}$$$$

The result I got was in line with the previous one: $$T_e^{threshold} = \frac{3m_e^2-m_p^2-2m_em_p}{2m_p}$$

Note that the result $$T_e^{threshold} < 0$$

My questions are:

• How do I interpret the result?
• Is the general formula for $$T_a^{threshold}$$ valid for any value of masses $$m_a,m_b,m_c,m_d$$, ie. even if it yields $$T_a^{threshold} < 0$$?
• Is this reaction even possible, or it is merely a dummy reaction with a sole purpose of illustrating the calculation?
• It violates baryon and lepton number conservation. Commented Jul 4, 2022 at 14:14
• the negative energy threshold arises because you made a 938 GeV proton turn into a 511 keV positron.
– JEB
Commented Jul 4, 2022 at 15:53
• @JEB So calculation is sound, were it not for the fact that the reaction is altogether impossible, as anna v explained. But is there such a case where the threshold energy is negative for a real possible reaction? Commented Jul 4, 2022 at 16:14
• Surely your assignment is $\rm e^- + p^+ \to e^- + p^+ + e^-e^+$?
– rob
Commented Jul 4, 2022 at 19:04
• @rob Yeah, now I'm pretty sure it is. But it still might be possible that: $$T_a^{threshold} = \frac{(m_c+m_d)^2c^2-(m_a+m_b)^2c^2}{2m_b}<0$$ in a reaction? On the other hand there might be a law I'm currently unaware of that forbids this... Commented Jul 4, 2022 at 19:14

$$\rm e^- p^+ \not\rightarrow e^-e^+$$

violates the conservation of baryon number and lepton number, and is therefore forbidden. Your homework assignment is almost certainly about

$$\rm e^- p^+ \to e^-p^+e^-e^+$$

which has a positive threshold energy.

However, you clarify that your real question is about whether it's possible to have a reaction whose threshold energy is negative, and how you would interpret such a system. The answer is yes, it's possible, and the meaning is that the reaction occurs at zero interaction energy.

For example, consider the capture of a free neutron on some other nucleus $$^A Z$$ with proton number $$Z$$ and mass number $$A$$. For nuclei which are not on the neutron drip line, the bound nucleus $$^{A+1}Z$$ has less total mass than the unbound system $$^AZ+\rm n$$. (Proof: if the unbound system were less massive, the nucleus could decay by neutron emission.) Consider a capture which results in a photon emission,

$$\rm n + {}^1H \to {}^2H + \gamma + 2.2\,MeV$$

or in nucleon or cluster emission,

\begin{align} \rm n + {}^{14}N &\to \rm {}^{14}C + p \\ \rm n + {}^6Li &\to \rm{}^3H + {}^4He \end{align}

For each of these, you can verify that the total mass on the right-hand side is less than the total mass on the left-hand side. The observable effect is that these reactions can take place when the kinetic energy on the left-hand side is zero. All of these reactions take place with thermal neutrons, with milli-eV kinetic energies, even though the energy released in the reaction is mega-eV.

In these negative-threshold reactions, we talk about the threshold of the reverse reaction. For instance, some of the literature on the low-energy mass-two nuclear system is based on neutron capture on hydrogen, but other literature talks about "threshold photodissociation of deuterium," which means

$$\rm \gamma + {}^2H \to {}^1H + n$$

where the photon energy is close to the 2.2 MeV minimum.