Understanding Quark Interactions in Electron-Proton Fusion

[taylordnz]

when electrons and protons fuse to canel out each others charge it seems my equations dotn add up, i need to know what quarks are in electrons.

 

[chroot]

Originally posted by taylordnz
when electrons and protons fuse to canel out each others charge it seems my equations dotn add up, i need to know what quarks are in electrons.

The inverse beta decay is

p + e^- \rightarrow n + \nu_e

- Warren

 

[taylordnz]

but what i mean if it dosent have any quarks is what particles does an electron contain. (great equation but i need it also ned diffrent particles in the electron)

 

[meteor]

There's a hypothetical particle called leptoquark that should be an hybrid between a lepton (the electron is a lepton), and a quark, but it's just that, hypothetical

 

[mormonator_rm]

An electron and a proton couple via a massive vector boson, resulting in the neutron and a neutrino, just as chroot showed. The underlying interaction with the boson and the proton causes an up quark to be converted into a down quark, reversing the isospin magnitude of the affected nucleon. A neutrino remains from the electron interacting via W-. So the basic flavor change gives us a sort of equivalence statement;

e- --> W- + v~e

and

W- --> d + -u

where e- is the electron, W- the weak boson involved, v~e the electron-neutrino, d the "down" quark, and -u is the antiquark to the "up" quark. So;

d + -u <--> W- <--> e- + v~e

is sort of an equivalence statement, but it is probably best to deal with this in terms of quantum numbers, rather than particle addition. You can set up the problem so that Q, I~3, Y, and L are conserved. Q is the electric charge, I~3 is the eigenvalue of isospin, Y is the hypercharge, and L is the lepton number. Y is a composite that includes only b, the baryon number, in this case. These quantities are related by the equations;

Q = I~3 + Y,
Y = b/2

and

Q = I~3 + L/2

For the proton, Q = 1, I~3 = 1/2, b = 1. For the neutron, Q = 0, I~3 = -1/2, and b = 1. For the W- boson, Q = -1, I~3 = -1, and b = 0. For the e-, Q = -1, I~3 = -1/2, and L = -1. For the v~e, Q = 0, I~3 = 1/2, and L = -1. Try it out, and you'll find that it should conserve all of these quantities; if it doesn't, then I made a mistake somewhere that I haven't seen as yet, so let me know if it breaks down somewhere.