The neutrino spin

I haven't seen that, but I am not familiar with the physics behind it.But that's how it's defined. You can talk about angular momentum that is not intrinsic, but that is by definition, not spin.Yes, that is how it is defined, but it is still spin.
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Is there any experimental evidence that neutrino spin is equal to the electron spin ( I think h/4pi, right?)

If it is just an assumed intrinsic property,can you explain on what grounds it is assumed as such, is it juas a matter of parity?
 
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  • #3
davenn said:
I read those (an many more) articles, do they contain experimental data? I must have missed them.I am asking if there is any concrete experimental data that confirms that it has indeed some spin and that its magnitude is h/4pi
 
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  • #4
there probably is, but I haven't found any info in my looking

that link I gave ( and many others I read) identifies the type of particle the neutrino is ... that info was gained by experimentation
it is identified in being part of the same group as the electron ie the lepton, ergo it has the same spin as the electron ...
The neutrino has half-integer spin (ħ⁄2)


not sure what else you want ?


maybe some of our particle physicists can take it further ?
@ZapperZ


Dave
 
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  • #5
davenn said:
there probably is, but I haven't found any info in my looking
not sure what else you want ?Dave
Thanks , Dave, I was just looking for some experimental evidence, or at least some indirect confirmation. I know the Standard Model assumes all fermions, leptons have spin 1/2. But suppose (ad absurdum) neutrino has no spin, what happens, what is the problem?
 
  • #6
alba said:
.I am asking if there is any concrete experimental data that confirms that it has indeed some spin and that its magnitude is h/4pi

It is by conservation of angular momentum.
Also the helicity of the neutrinos has been measured (eg the Goldhaber et al experiment)
 
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  • #7
Accurate measurements are available of neutron decay.
My guess, as I have not analysed the paper, is that these would have shown up - Nobel prize winning - deviations from the present electroweak theory if the neutrino would not be (nearly) massless with spin 1/2.
http://neutron.physics.ncsu.edu/UCNA/protected/55_2_0119.pdf
 
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  • #8
A different spin would lead to different angular distributions of the decaying particles, something measured in the experiments of Wu and Goldhaber, and later much more precisely with particle accelerators.
 
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  • #9
If the neutrino were spin-0, it would mean angular momentum is not conserved.

If the neutrino were spin-3/2, [itex]\frac{B(\pi \rightarrow \mu \nu \overline{\nu})}{B(\pi \rightarrow e \nu \overline{\nu})}[/itex] would be some number other than the measured ~12,000.
 
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  • #10
mfb said:
A different spin would lead to different angular distributions of the decaying particles, something measured in the experiments of Wu and Goldhaber, and later much more precisely with particle accelerators.
Why can't spin be added during the process?
 
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Vanadium 50 said:
If the neutrino were spin-0, it would mean angular momentum is not conserved.

If the neutrino were spin-3/2, [itex]\frac{B(\pi \rightarrow \mu \nu \overline{\nu})}{B(\pi \rightarrow e \nu \overline{\nu})}[/itex] would be some number other than the measured ~12,000.
What I meant is that the spin musn't necesseraly be an intrinsic property. Like in a billiard ball any spin can be generated by the cue or by a collision, the value of the spin +1/2 or -1/2 (or any other value) could be determined only by the process. This would also eliminate the akward assumption of an antineutrino antiparticle of itself. Does the linked experiment rule out this possibility? Ihave not the expertise to reach any conclusion.

Thanks for your help. anyway
 
  • #12
alba said:
What I meant is that the spin musn't necesseraly be an intrinsic property.

ahhh but it is
 
  • #13
@Vanadium 50: I think you just want one neutrino in the decay.
alba said:
Why can't spin be added during the process?
What do you mean by "added"?
Angular momentum is conserved.

alba said:
This would also eliminate the akward assumption of an antineutrino antiparticle of itself.
Assumption of what? It is possible that neutrinos are their own antiparticles, but they don't have to be - measurements are not precise enough yet to distinguish between those possibilities.
 
  • #14
mfb said:
I think you just want one neutrino in the decay.

You're right. The one I want is [itex]\frac{B(\pi \rightarrow \mu \nu)}{B(\pi \rightarrow e \nu)}[/itex]

(Bonus points for guessing which reaction I was starting to write down when I thought better of it - and then blew it)
 
  • #15
alba said:
What I meant is that the spin musn't necesseraly be an intrinsic property.

But that's how it's defined. You can talk about angular momentum that is not intrinsic, but that is by definition, not spin.
 
  • #16
alba said:
What I meant is that the spin musn't necesseraly be an intrinsic property. Like in a billiard ball any spin can be generated by the cue or by a collision, the value of the spin +1/2 or -1/2 (or any other value) could be determined only by the process. This would also eliminate the akward assumption of an antineutrino antiparticle of itself. Does the linked experiment rule out this possibility? Ihave
By the way, have you ever seen a billiard ball "spinning" at half-integer angular momentum?
 

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