Forced Nuclear capture?

TWest

Has anyone launched a high energy positron into a nucleus with no electrons that had a neutron such as Deuterium ---> Tritium, Can you force positron capture? has it happened before and if so what happened?


Because I was thinking couldn't you do this with a highly vacuumed out supercollider taking the positrons from Na^22 dacay like the guys at massachetts university?

Is this possible or has it been done?

mfb

Electrons or photons colliding with matter provide a better source of collimated positrons.

You want to see positron capture like ##T+e^+ \to {}^3He + \nu##?
Hmm, interesting, should be possible but I did not see it yet.

TWest

I would like to see what the neutron turns into if you pump a positron full of Energy like CERN does to protons then have it captured, because if it enters the nucleus you should have a charge exchange or something.

hydrogen Isotope 2(Deutrium) + e+(High Energy) -----> ?????/God Proton + P+ + Ve/ P+ + P+ + Vx(Super massive Neutrino)/P+2 + N + Ve


I dunno I am just guessing but how would the decay force react to that?

mfb

If the energy is very high, you produce a lot of new particles (like the LHC and other accelerators do), but you don't get positron capture.

D + e⁺ -> many hadrons, one or more leptons

I don't see the point in using deuterium and positrons, however. Electron/positron is cleaner, proton/proton or proton/antiproton allows higher energies and collision rates.

TWest

So, even though we are launching a anti lepton into a atom that is about to positron capture anyway and not a hadron into a hadron collision would be the result would still be the same as the collision?

The Point is that I want all of the positron (energy/Mass) to be absorbed naturally through a decay reaction instead of just bashing them together. e+[112(TEV)] + N -----> ????? +Ve instead of P+[112(Tev)] + P+[112(Tev)] -------> Quark Stew

I think that a different effect may happen, but I guess not.

Could we do a effect like that with electron capture or would it just beta minus decay?

mfb

If you mean "many particles are produced" with "same effect": Yes.
If you want the collision to be sensitive to specific particles, you need lower energies.

I have no idea what you are talking about here.

TWest

Okay gotcha, I guess making a supermassive proton is more difficult than I believed. We cannot toss a high energy positron at a neutron and force it to capture it and turn into a Tauon level proton. e+ + (D,D,U) ----> (T,T,B) and sorry I was still editing when you replied.

mfb

A proton has a fixed mass, you cannot change this. You can create other baryons, which are heavier.
How is that related to tau particles?

e⁺ + ddu -> duu + ν is possible. You won't get top-quarks in hadrons (they decay before they form hadrons) and if the collision is so high-energetic that you create bottom quarks, they can go to other hadrons as well.

TWest

I realize this but a Top and Bottom are just heavier forms of Up and Down quarks can we get them all to transform by giving enough energy at once in a decay reaction, can we use the Color Field to make these quarks stable enough to not instantly decay? making the quarks of a stable neutron turn into a "God proton"(not actually a proton but u get my point a higher level baryon) that is containing Top and Bottom Quarks. The reason why I ask this is because neutrons decay after 20 min outside a nucleus and protons last forever but neutrons in a nucleus last forever, maybe higher generation quarks are the same way.

Normal

e+(Low E) + (U,D,D) -------> (U,U,D) + Ve [While Still Gluoned to another particle]

my Idea

e+(High E= to difference between U,D,D to T,T,B) + (U,D,D) ------> (T,T,B) + Vt(tauon Neutrino) [While gluoned to another particle]

My point is what happens to all the extra energy in this case just new quarks or will the quarks already there transform into their higher energy state?

When we collide protons together at high energies it is a unnatural thing that causes them to shatter what if you added the energy to it in a way that was more natural.

mfb

No, you cannot just "transform up-quarks to top-quarks" (and similar d->b). This needs the weak interaction, and the processes are quite rare.

This does not make sense. No, I don't get your point. If you mean a heavy baryon, just say heavy baryon. A proton is the only stable baryon as free particle (neglecting a possible proton decay). In addition, neutrons can be stable in nuclei and some other particles might be stable in neutron stars. Everything else decays.

Apart from neutron stars: No. Their mass difference is too big to be compensated by nuclear forces.

TWest

Gotcha, so we would have to create a new force for this to work..... okay you answered my question completely thank you for your time =).


But, lastly actually what determines the decay rate then is it the strength of the field versus energy contained then?

mfb

Quantum field theory.
You can get a good estimate based on the coupling strength (if a decay can occur via the strong interaction, the lifetime is very short), the energy difference between the particle and its decay products (if that is large, lifetime tends to be shorter) and the number of possible decays (more final states => shorter total lifetime). Weak decays are usually slow as they involve a virtual W boson - the top-quark is an exception as it can decay into a real W boson plus a bottom-quark.
Sometimes, parity and spin are important, too.