How significant are energy losses in subatomic inelastic particle collisions?

[buffordboy23]

I think of all of the energy losses associated with macroscopic collisions and wonder how significant this effect is in the subatomic realm of inelastic particle collisions. For example, consider the inelastic pion photoproduction reaction: $$\gamma + p \rightarrow p + \pi^{0}$$. Obviously, there is a minimum threshold energy needed for the reaction to proceed. How do the energy losses by such a reaction compare to the threshold energy? Are they significant?

EDIT: I realize that we are trying to compare two different worlds, the classical vs. the quantum, but intuition tells me that there must be similar effects although they may not be precisely clear.

 

[humanino]

Various sorts of energy losses mostly depend on your experimental apparatus, and how well you understand it, model it, and can reproduce its fine details in a simulation. For the photon and the neutral pion, they are absent. Also there will obviously not be any for the initial proton if it sits at rest in a target (or at least, its motion is supposed to be negligible for sufficiently cold liquid hydrogen). Only energy losses in this specific case will be for the scattered proton. They could be anywhere between fully negligible to overwhelming the kinetic energy of your proton to zero, in the entire spectrum from a clean and well designed detector, to having forgotten a lead brick between your target and your drift chamber.

 

[buffordboy23]

Okay, but you are mainly discussing efficiencies in design. Assume that our design is 100% efficient. Can the inherent energy losses associated with the collision be significant? Are there any losses at all?

 

[humanino]

What kind of loss ? You have 2 particles entering and two particles exiting. Nothing else here to carry energy around, right ? So where will energy go if it had been lost ? Do you question energy conservation ? If for instance the proton keeps some energy for itself, elevating its internal energy to an excited state, then we don't call it a proton (in the final state) anymore. Same for the neutral pion.

 

[buffordboy23]

No no, I don't want to question energy conservation. That is a pretty solid law. I guess I was wondering if there was some mechanism for energy loss while maintaining the integrity of the two exiting particles.

The particles interact on some background, which contains the quantum vacuum. The proton is a composite of three quarks and some number of gluons. However, the proton's total mass is not well approximated by the mass of the three individual quarks, but requires the explanation of a sea of virtual quarks from the vacuum and gluon field interactions. I was wondering if it was plausible that some small, but finite energy which we cannot measure with our current technology, could be dissipated to the background and thus affects the outgoing momenta and energy of the outgoing particles. So this effect would likely be negligible for our current and practical considerations when designing experiments.

 

[humanino]

It's actually pretty simple : if the proton absorbs some energy in its virtual gluon or quark structure, it becomes something else, like a $\Delta$ for instance. In the reaction you mentioned, this does not happen. So there is strictly no absorption of energy in its structure, which you have fixed to begin with.

 

[hamster143]

Energy does not just dissipate to the background. All energy that goes in must be accounted in the form of some particle leaving the scene of the crime. Most particles have non-negligible masses, and there are no free gluons. The only option is that there may be additional photons in the final state, which (if their energy goes towards zero) could be called negligible. In addition to $$\gamma + p \rightarrow p + \pi^{0}$$ you'd have $$\gamma + p \rightarrow p + \pi^{0} + \gamma'$$. In practice, we'd set an infrared cutoff and talk about interactions that don't result in anything above the cutoff except the pion and the proton.

 

[Bob S]

In pi zero photoproduction, the reaction cannot occur by itself, because the reaction cannot simultaneously conserve both momentum and energy. The proton recoils, carrying off mainly momentum. The proton energy probably creates a short visible ionization track in a bubble chamber.