# Where can I find the equations for “quasi” elastic collisions?

Yes, you all talk about neutrinos and spins, but I came out with this basic s**t :D

All of us learnt the basic equations of collisions, elastic (everything bounces and energy remains the same), or non-elastic (the classic example where objects remain stick together). What if objects don't remain together and some energy is lost (a fraction of the kinetic energy of course)? I tried to solve the maths and since I'm just a Java programmer, I realized that after 6 years away from University I can only do 2+2 and little more. Apart from joking, the equations come out as a system with an elliptic curve (for energys) and a line (for momentums). Anyway I'd like to get a solution that is good enough to do some analisys over initial speeds, energys and so on.

Please don't flame me, probably my english for physics is not exact, I studied physics in Italian and my everyday bread is IT.

Thanks

## 1 Answer

"Quasi-elastic" has two meaning that I am aware of.

• In neutrino physics we talk of $$\nu + p \to l + n$$ and similar reaction as being "quasi-elastic". It is not elastic because the products do not have the same individual masses as the inputs, but we can analyze it with exactly the same math as a elastic reaction if $$m_n - m_p \ll \{ |Q|, E_{\nu} \}$$ $$m_l - m_{\nu} \approx m_l \ll \{ |Q|, E_{\nu} \}$$ (where $Q$ is the four-momentum transfer of the reaction) because the corrections due to the mass changes are small compared to all the energies and momenta involved.

• In nuclear physics we sometimes talk about reactions like $$e + A \to e + p + B$$ where B represents a remnant nucleus that is intact and mostly unexcited. That is, we've interacted (elastically) with a constituent nucleon more-or-less without disturbing the rest of the nucleus. There are some corrections due to initial- and final-state interactions, but these tend to be of the scale of nuclear binding energies, so we can use elastic kinematics if the electron beam energy and the four momentum transfer are much larger than a few MeV.

In either case, the reaction isn't exactly elastic, but it is pretty close and we can use elastic kinematics. Thus "quasi-elastic".

Let me say the important part again...if the above conditions are met you can analyze these reactions with elastic kinematics.

The fun case is, of course, $$\nu + A \to l + n + B$$ in which we get both meanings at once.

Thankfully this is a small fraction of events in beam-neutrino experiments