# Neutrino confusion

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1. Apr 19, 2015

### ryanuser

Hi
Why are there three types of neutrinos? Why isn't there just one neutrino just like there is only one electron? (I mean electron doesnt have different flavours to it, same with tau and muon.)
My text book touches on this very breifly and confuses me with "if there were only one type of neutrino and antineutrino, equal numbers of electrons and muons would be produced."
????

2. Apr 19, 2015

### Orodruin

Staff Emeritus
The neutrinos are partnered with the charged leptons. So the correct question would be "why are there not three neutrinos just as there are three charged leptons?" Except that there is three charged leptons so there really is no problem.

3. Apr 19, 2015

### ryanuser

You have not understood my question. Neither your 'correct question' is what I intended to ask.
Why are neutrinos partnered with leptons? Why we are not having a single particle called neutrino?

4. Apr 19, 2015

### Orodruin

Staff Emeritus
Because this is not what we observe in nature.

But my point is that your question in itself is a bit confusing. You seem to be fine with charged leptons coming in three different flavours, why would neutrinos be different?

If you want to be more theoretical, I could tell you that it is required for the Standard Model to be gauge invariant, but I doubt this is enlightening you much. In fact, the neutrino and the left-handed electron to some extent form an entity which is far more fundamental then what we typically call "an electron", namely an SU(2) doublet under the SM gauge group. Without the neutrino there for each family of charged leptons, the theory would not be consistent.

5. Apr 19, 2015

### ryanuser

6. Apr 19, 2015

### ChrisVer

To go further and talk about experiments also. The light neutrinos types has been measured (most recently by LEP) to be $N_\nu = 2.984 \pm 0.008$ (so roughly around 3... When combined with previous lower-energy data this result goes to $N_\nu = 2.92 \pm 0.05$.

In LEP you collided electrons with positrons and produced Z-bosons. The Z-bosons can decay into $Z \rightarrow q \bar{q}$ (hadrons), $Z \rightarrow l \bar{l}$ (lepton-antilepton) and $Z \rightarrow \nu \bar{\nu}$ neutrinos-antineutrinos.
So you are going to have a "visible" width as a result of the hadrons and leptons, and an "invisible" width due to the light neutrinos each contributing a Standard Model (SM) $\Gamma_\nu$ width so $\Gamma_{inv}= N_\nu \Gamma_\nu$. You can find the invisible width by subtracting the visible Z-width from the total Z-width. The result for the number of light neutrinos in the SM is then given by:
$N_\nu = \frac{\Gamma_{inv}}{\Gamma_\nu} = \frac{\Gamma_{inv}}{\Gamma_{l}} \frac{\Gamma_l}{\Gamma_\nu}$
To make it more model-independent one you use the SM ratio of neutrinos' to the leptons' widths: $\frac{\Gamma_\nu}{\Gamma_l} = 1.991 \pm 0.001$ instead.

The fittings of this type: http://www.fzu.cz/sites/default/files/images/gallery/zlineshape.gif
are the ones that can also help you see that N=3 is closest to the measured data (this plot is from DELPHI experiment of LEP)

This is true for light neutrinos... if there are other neutrinos out there, they have to be more massive.

Last edited: Apr 19, 2015
7. Apr 19, 2015

### ChrisVer

Also there is one electron-neutrino, one muon-neutrino and one tau-neutrino, as there is one electron, one muon and one tau. The difference between the neutrinos and the charged leptons is that neutrinos can oscillate between different flavors.

I don't understand what your book says, which book are you reading? in what context does it say that you have equal numbers of muons/electrons produced?

8. Apr 20, 2015

### ChrisVer

A further notice one can make on this, again indirectly striking the idea of the existence of only one neutrino, would be that if there was one neutrino then the electron and muon neutrinos would have to be the same.

The "muon neutrino" initially called "neutretto" was found (in 1948) when the spectrum of the electron product of muon decay was found to be continuous (so there should be two extra neutrinos/or invisible particles in the products):
http://arxiv.org/ftp/physics/papers/0503/0503172.pdf
$\mu \rightarrow e^- + \text{2~invisibles}$
I think this 3 products for continuous spectrum instead of 2 should be clear to you- it's the same reasoning of energy-momentum conservation in two bodies vs three bodies which led Pauli in introducing something that cannot be detected.

However, in 1959 Pontecorvo investigated whether the neutrinos emitted together with electrons (in beta-decays) were the same as the neutrinos emitted in pion decays.
If electron neutrino and muon neutrino were the same particle, then the reactions:
$\nu_\mu + n \rightarrow \mu^- + p~~,~~ \bar{\nu}_\mu + p \rightarrow \mu^+ + n$
and
$\nu_\mu + n \rightarrow e^- + p~~,~~ \bar{\nu}_\mu + p \rightarrow e^+ + n$

would have the same rate, because the later can also be done by electron (anti)neutrinos, otherwise the last two should not be observed at all.
At Brookhaven AGS using 15GeV protons hitting a Beryllium target we created secondary pions/kaons beams which produced an almost pure $\nu_\mu$ beam (charged pions were given enough time to decay dominantly into muons and neutrinos - electrons decay mode is helicity suppressed). Then an iron wall (13.5m thickness= enough to absorb up to 17GeV muons) was used as a shield for hadrons and muons, and then 10 modules of spark chambers were installed weighting 1ton each. Muons and electrons were discriminated by their tracks: the muons left straight lines whereas the electrons caused electromagnetic showers.
In total the result was that we had 29 muon-like and 6 electron-like events observed (so not the same rate for those interactions), showing that $\nu_\mu \ne \nu_e$. The same thing was shortly afterwards tested by CERN, reproducing/confirming the same results with better statistics.
The extra electron neutrinos were expected to come from the Kaon decays: $K^+ \rightarrow e^+ \nu_e \pi^0$.
The distinction was made clear by 1962.

From my point of view, oscillations for neutrinos won't play a role in this detection distances so they can't explain the difference.

(I also got helped in writing this information)

Last edited: Apr 20, 2015
9. Apr 20, 2015

### Staff: Mentor

As the point was showing that you have more than one neutrino type, you don't have to worry about oscillations. A single neutrino type could not oscillate to other types.

The distance is so short that neutrino oscillations are negligible (but back then no one knew about them).
Kaon decays look plausible. I don't know what the background from atmospheric neutrinos or cosmic rays would be.

10. Apr 20, 2015

### ChrisVer

Yes this is a further comment made by me, since we now know for neutrino oscillations. The thing is that in the experiment's idea you take the pions products that give:
$\pi^+ \rightarrow \mu^+ + \nu_\mu + \pi^0$
...you let the $\nu_\mu$ beam travel the distance of 13.5m in the wall (+ some extra distance between 0 and 21 meters of air between the target and the wall, depending on where you the pions decay) , and you consider that it won't change in this interval (because they didn't know of oscillations back then).
If it changed into electron neutrino, then you would indeed measure an electron-like signal.
However, as you and I mentioned, the distances of 13.5- 35m is not enough for such a neutrino flavor change (so the experiment's consideration is safe from neutrino oscillations). Also this scenario gets further suppressed if you add to the probability of the flavor transition, the probability for the resulted electron neutrino to interact.

As for the Kaon, it's what I found in my source... I am not sure whether the background neutrinos would affect the experiment in any way or how...

Last edited: Apr 20, 2015
11. Apr 20, 2015

### Staff: Mentor

That should be at least the same as the probability for muon neutrinos.

You certainly get muons from the atmosphere, but those all move downwards under various angles. I guess the atmospheric neutrino background is negligible (those events go in all directions uniformly). A good angular resolution helps to suppress those backgrounds.

12. Apr 26, 2015

### ChrisVer

True, but the overall probability is "doubled" (not as a quantity but as an idea) suppressed by the probability of the muon-electron transition + the interaction.
The differential event rate at a detector for the neutrino oscillations should be something like:

$\frac{dN^\alpha}{dE dx} = K_y \sum_\beta \Phi_\beta (E,x) ~P_{\beta \alpha} (E,x) ~\sigma_\alpha (E)$
where $x=L/E$, $\sigma_\alpha$ the total interaction cross section for a neutrino of flavor $\alpha$, $P_{\beta \alpha}$ the probability for $\nu_\beta \rightarrow \nu_\alpha$ and $\Phi_\beta$ a flux-dependent term of the $\beta$ neutrinos (depends on $\frac{d^2 \phi_\beta}{d \ln E d \cos \theta}$). $K_y$ is some detector dependent factor.

http://arxiv.org/pdf/hep-ph/0603264.pdf (eq.12)

I think for no oscillations, the formula would be somehow the same (maybe with some value differences for the flux) but with missing $P_{\beta \alpha}$ or better said with $P_{\beta \alpha} = \delta_{\beta \alpha}$.