Beta decay Charge conservation

jc09
Messages
42
Reaction score
0
In beta decat the u quark fromthe proton and the d quark from the neutron interact to produce a electron and it's neutrino. I'm wondering about the charge conservation of this process because a neutron has zero charge and a proton has charge of 1. How do we end up with a electron charge of -1 at the end.
 
Physics news on Phys.org
A neutron decays into a proton, an electron, and a neutrino.
 
jc09 said:
In beta decat the u quark fromthe proton and the d quark from the neutron interact to produce a electron and it's neutrino. I'm wondering about the charge conservation of this process because a neutron has zero charge and a proton has charge of 1. How do we end up with a electron charge of -1 at the end.

You should look for a Feynman diagram of this process showing the quark lines
 
If a neutron has zero charge and produces a proton charge one and a electron charge -1 and a neutrino charge 0 why does it need a W- to mediate this
 
In \beta^- decay, a down quark (charge -1/3) converts to an up quark (charge +2/3), an electron (charge -1) and an antineutrino (charge 0). The total charge is -1/3 before and after. Introducing a virtual W boson, we have a two-step process:

d \rightarrow u + W^-

W^- \rightarrow e^- + \bar \nu_e
 
Last edited:
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}...
Back
Top