Why neutrino energy threshold in inverse muon decay is so high

Question

In the interaction:

$$\rm \nu_\mu + e\to\nu_e+\mu$$

the $\nu_\mu$ energy threshold to produce the process is $m_\mu^2c^2/2m_e$, which gives a $\sim10~\rm GeV$ threshold.

For the same muon neutrino scattering, but implying a nucleon, e.g., $$\rm\nu_\mu + n \to\mu+p$$ the threshold is of the order of the muon mass $\sim100~\rm MeV$.

I would like to know what is the physical reason that makes the first interaction have a much larger energy threshold.

Answer 1

Quick note: I'm suppressing factors of $c$ here. Anywhere you see a $p$ imagine it's really $pc$ and anywhere you see an $m$ it should be $mc^2$.

It's just kinematics. The limitation in both cases is that, in the center of mass frame, there is enough energy to produce both products. (Since, in this frame, if you are right on threshold, both particles are created at rest). The center of mass energy squared for two particles is given by: $$s=(E_1+E_2)^2-(\vec p_1+\vec p_2)^2$$

where $E$ is the energy and $\vec p$ is the momentum of a given particle.

For the case where one particle is massless and the other is at rest, this simplifies to:

$$ s=(E_1+m_2)^2-p_1^2=2m_2E_1+m_2^2$$

And this needs to be greater than or equal to the sum of the masses of the products squared:

$$ 2m_2E_1+m_2^2 \ge (m^\prime_1+m^\prime_2)^2 $$

For the first reaction, $m^2_e$ is negligible compared to $m_\mu^2$, and so the threshold is:

$$ E_\nu\ge\frac{m_\mu^2-m_e^2}{2m_e}\approx\frac{m_\mu^2}{2m_e} $$

On the other hand, $m_p^2\approx m_n^2$ and $m_n\gg m_\mu$, so we get a different simplification:

$$ E_\nu\ge\frac{(m_n+m_\mu)^2-m_p^2}{2m_p}\approx m_\mu $$

The general idea is that you need enough energy in the center of mass frame to create particles. Since the first reaction occurs far from the center of mass frame (i.e. $\vec p_1+\vec p_2\not\approx 0$) most of the energy is wasted in giving both the created particles momentum. The second process is nearly in the center of mass frame (since $m_p$ is large), and so most of the particle's energy can go to particle creation.

Answer 2

I am not sure what "physical reason" is meant to help you intuit the different relativistic kinematics involved.

You are getting the threshold by calculating the invariant mass in the lab frame on the left hand side and in the cm frame on the right hand side of the reaction, assuming the product are produced at rest, and setting the negligible neutrino masses to zero only in the end--after all you have a massive neutrino at rest in the first reaction, $$ (E+m_e)^2-E^2= m_\mu^2 \qquad \Longrightarrow \qquad 2Em_e\approx m_\mu^2 \qquad \Longrightarrow \qquad E\approx m_\mu^2/2m_e~, $$ where we also dropped the electron mass in the subdominant term; versus $$ (\tilde E+m_n)^2-\tilde E^2= (m_p+m_\mu)^2 \qquad \Longrightarrow \qquad 2\tilde Em_n\approx m_\mu^2 +2m_\mu m_p \approx 2m_\mu m_p \\ \qquad \Longrightarrow \qquad \tilde E \approx m_\mu ~. $$

The dominant balance of the two cases differs both on the l.h.side, where the small target mass enhances the threshold energy--you want more kinetic energy to achieve a comparable effect, and the r.h.side, where the bulk of the kinetic energy thrust on a heavy target goes to the cross term with the nucleon, instead of the square of the muon mass, as in the first case.

Are you sure a "physical" picture could clarify things further? Like: a marble bounces off an apple converting it to a watermelon; versus: a marble bounces off a truck turning itself into a watermelon? Stop that cartoon with extreme prejudice!