The antineutrino in Beta decay

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The discussion centers on the role of the antineutrino in beta decay, specifically how it conserves quantum numbers such as spin and lepton number during the decay process. The W- boson, a virtual particle, decays into an electron and an antineutrino to maintain balance in the decay equation n → p + e^- + \overline{ν}. The necessity of the antineutrino is established through the conservation of spin and lepton number, as the neutrino contributes a lepton number of +1 while the antineutrino contributes -1, ensuring the overall conservation laws are satisfied.

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ajassat
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I understand the physical changes in beta decay that eventually lead to the virtual particle, a W- boson being formed. However I do not understand the physical changes that take place, which eventually lead to an antineutrino being emitted. What changes take place in the W- boson, that make it decay into an antineutrino and electron?

I would also be glad if someone could explain why the physical changes take place?

Thanks in advance
- Adam
 
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ajassat said:
I understand the physical changes in beta decay that eventually lead to the virtual particle, a W- boson being formed.

I'm not sure that I understand what you mean. Also, you do understand that "a virtual particle being formed" is sort of a contradiction. The particle is virtual, not real. It's not formed.
 
Vanadium 50 said:
I'm not sure that I understand what you mean. Also, you do understand that "a virtual particle being formed" is sort of a contradiction. The particle is virtual, not real. It's not formed.

By this do you mean that the antineutrino isn't really there after beta decay?
If the production of an antineutrino is virtual, then would you be able to explain how?

Simply, I would like to understand why we have an antineutrino involved ,regardless of it being real or virtual.

Thanks for attempting the answer the question.
-Adam
 
There is an antineutrino around to conserve spin (and other quantum numbers, like lepton number). If I have the decay n \rightarrow p + e^- + \overline{\nu}, I have a spin-1/2 object on the left hand side and without the neutrino, I have either a spin-0 or spin-1 ensemble on the right. I also have L = 0 on the left and L = 1 on the right. Finally, the energy spectrum of the electron is characteristic of three-body decays, not two body decays. For all of these reasons, one expects an additional particle produced, the antineutrino.

Of course, today those (anti-)neutrinos have been measured.
 
Vanadium 50 said:
There is an antineutrino around to conserve spin (and other quantum numbers, like lepton number). If I have the decay n \rightarrow p + e^- + \overline{\nu}, I have a spin-1/2 object on the left hand side and without the neutrino, I have either a spin-0 or spin-1 ensemble on the right. I also have L = 0 on the left and L = 1 on the right. Finally, the energy spectrum of the electron is characteristic of three-body decays, not two body decays. For all of these reasons, one expects an additional particle produced, the antineutrino.

Of course, today those (anti-)neutrinos have been measured.

This has made me understand why it is necessary to produce an anti-neutrino in beta decay. So what are the spin and lepton values for the neutrino which help to create a balances on both sides of the decay?

Thank you for your time,
-Adam
 
The neutrino is a spin 1/2 particle and a neutrino has lepton number +1 and an antineutrino has -1.
 
malawi_glenn said:
The neutrino is a spin 1/2 particle and a neutrino has lepton number +1 and an antineutrino has -1.

Thanks for this. I now understand how the anti-neutrino helps to balance things out.
 

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