Which force mediates a reaction in which a pion and a deuteron form two protons?

Which fundamental interaction is responsible?

  • Strong

    Votes: 2 66.7%
  • Weak

    Votes: 1 33.3%
  • Electromagnetic

    Votes: 0 0.0%

  • Total voters
    3
  • Poll closed .
alby
Messages
4
Reaction score
0
Which fundamental force mediates the reaction \pi^+ + D \rightarrow p + p?

My initial assumption was the strong force because states only feature quarks (i.e. there are no leptons) but looking at it again I'm not 100% sure.

Do I need to include colour factors? Could it be a neutral current Z without lepton pair production? Or electromagnetic, photon absorbed by proton?
 
Physics news on Phys.org
am sure that the strong interaction mediates the reaction in which a pion and a deuteron form two protons.and yes u should include a color factor.the deuteron is made of a proton and neutron.the positive pion interacts with the neutron and leads to the formation of a proton .is a simple way a positve pion is made of up and anti down,and the neutron is made of ddu.so the down and anti down annihilate which leaves uud which is a proton.
u can also read this paper made specifically for pre-university level to understand the pion exchange mechanism which include color charge and gluones.http://teachers.web.cern.ch/teachers/archiv/HST2002/feynman/Pion%20exchange.pdf"
 
Last edited by a moderator:
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...

Similar threads

Back
Top