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"Three Neutrino Families"

7 Comments -

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Blogger stevelinton said...

Could there be fourth neutrino type that was at least as massive as a Z0? Presumably then the Z0 would be unable to decay into it and so it would not show up on this plot?

10:19 AM, December 17, 2007

Blogger Bee said...

Hi Steve:

In principle, yes. As we've written

its mass should be below half the mass of the Z (the Z should be able to decay into the particle-antiparticle pair) for the method to work

Such a very heavy neutrino might mess up some other data though, astrophysics/cosmology wise. I am not exactly sure about the constraints from this.

Best,

B.

10:34 AM, December 17, 2007

Anonymous Coin said...

But wouldn't a fourth neutrino type imply an entire fourth generation of quarks/leptons/etc, which one assumes would create all kinds of difficult consequences?

Or is there some way we could possibly posit a fourth neutrino family, without having to add an entire new row onto that lepton periodic table?

5:18 PM, December 17, 2007

Blogger Bee said...

Hi Coin:

Well, if they are sufficiently heavy, I'd think it was possible. Besides this people like to play around with the mentioned 'sterile' neutrinos that don't carry flavor and thus wouldn't be bound to charged leptons. These could even be lighter than 45 GeV since the Z wouldn't decay into them. E.g. some people have played around with 'hot' dark matter made of sterile neutrinos I think somewhere in the MeV range. I can't say though I find these speculations neither appealing nor plausible. Best,


B.

5:25 PM, December 17, 2007

Blogger Doug said...

Hi Stefan and Bee,

Your 'resonance' wiki reference with the helical bouncing spring reminds me of the likely helical trajectories for QM particles such as neutrinos and GR celestial bodies.

See this example from Cleon Teunissen, Nederlands
'Nonsymmetric velocity time dilation' [bottom of web page] Picture 7. Animation
A straight worldline and a helical worldline.

6:30 PM, December 17, 2007

Anonymous mono said...

Hi Bee and Stefan, thanks for this interesting post. A question: you say

"But this means, in the language of resonances, that the Z is quite strongly damped..."

I don't understand what the word "this" refers to, i.e., what aspect of this decay process can be thought of as corresponding to a strong damping, in the language of resonances. Could you elaborate?

12:02 AM, December 19, 2007

Blogger stefan said...

Hi mono,


the Z has many different "decay channels", i.e. options to decay in various other particles. As a result, the Z decays rapidly, and its energy is "dissipated" - it will distributed over the decay products.

In a vibrating system, damping means dissipation of the energy which has been applied to the system to set it into vibration. The stronger the damping, the faster the dissipation of the energy, and the sooner the vibrations comes to a stop.

So, the Z with its many efficient decay channels and fast decay is analogous to a strongly damped resonating system - including the characteristic feature of a broadened peak.

Maybe it helps to compare the Z peak to the J/Psi or Upsilon states - these states have a longer lifetime, and as a consequence, they correspond to much sharper lines - as for a resonance with small damping.

Hope that helps, Stefan

6:28 PM, December 19, 2007

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