# Neutrino detectors

1. Apr 10, 2015

### ChrisVer

What would be the signature in a neutrino detector of these interactions?
$\nu_l q \rightarrow q^\prime X$
and
$\bar{\nu}_l e \rightarrow \bar{\nu}_l l^-$
with $q$ a nucleon's quark and $l^-$ a lepton?

In general I would say that the signal would be that of the resulting lepton and maybe hadron jet, but I am not sure if the neutrino detectors work in the same way...

For example, in an atomic environment, the special example of interaction $] \bar{\nu}_e e \rightarrow \bar{\nu}_e e$ can result in ionizations...

2. Apr 10, 2015

### Orodruin

Staff Emeritus
This depends a lot on the neutrino energy. Remember that neutrinos (in beam experiments) are not as energetic as the LHC. You are getting neutrinos in the GeV range at most. Very often, the interactions are elastic or quasi elastic, again depending on energy. At most you get a small hadron shower, but I would not call it a jet in the LHC sense. I would say the most successful detector technology so far in neutrino physics has been the water/ice Cherenkov (ie SuperK, SNO, IceCube), which is based on detecting the Cherenkov light of the reaction products.

3. Apr 11, 2015

### Staff: Mentor

Stellar neutrinos don't have enough energy to produce new particles (apart from electrons).
Atmospheric and astrophysical neutrinos can have enough energy, but they are very rare - they can produce showers, and IceCube sees many of those via the Cherenkov light of the shower particles.

4. Apr 11, 2015

### Orodruin

Staff Emeritus
I still think the coolest thing is how IceCube participates in the SNEWS project. It cannot detect individual supernova neutrino events due to the low energy, but would detect a supernova by an increase in the background for all PMs as if the ice was glowing.

5. Apr 11, 2015

### ChrisVer

So let me take each one of the two interactions seperately...

The "inverse beta decay" which would be the $\bar{\nu}_e u \rightarrow e^+ d$ or $\bar{\nu}_e p \rightarrow e^+ n$
is going to give as a measurable/detectable result a positron and a neutron. The positron will annihilate with an electron and give the known two gammas with a peak around $0.5MeV$. The neutron on the other hand (from what I saw) is going to be absorbed by a nucleus and undergo another gamma (nuclear) transition, which will itself give a peak at the corresponding energy (I guess this depends on the absorbing nucleus type). So the signature will be these 3 gamma photons, right?
Also I wonder whether the neutrino should be an antineutrino in the left-side of the interaction or I can turn it into neutrino (but then I think it can't interact with the quarks? or something is wrong with my feynman diagram fermionic arrows)
In case of an elastic scattering (let's say with a Z boson interaction), the result won't be measurable/detectable... It would be like $\bar{\nu}_e p \rightarrow \bar{\nu}_e p$ and only the proton's momentum might change... but I don't think we can detect it in any way, can we?

Then the other interaction with leptons... obviously there are two types again, the elastic scattering (which can lead to ionization depending on the incoming neutrino energy) and this can give a signal , as well as a change of the neutrino and lepton flavor $\bar{\nu}_e e \rightarrow \bar{\nu}_\mu \mu$, in that case the only possible detection would be by looking for muons (or taus) at the end -like muon "orbiting" the nucleus-? still I think neutrino detectors won't be able to get this signal.

6. Apr 11, 2015

### Orodruin

Staff Emeritus
There are several different things you can look for in the inverse beta decay reaction. Again, many neutrino detectors would look for the Cherenkov light from the positron rather than the annihilation. When it comes to muons, they make very nice Cherenkov rings in your detector.

Also, the detection method may depend on the target in the detector. You should be able to find some techniques for solar (very low energy) neutrinos by checking out the methods used by the SNO experiment.

7. Apr 11, 2015

### Staff: Mentor

The proton recoil (or, more likely, electron or nucleus recoil) could still lead to scintillation. Dark matter detectors use similar detection methods, and the neutrino background will become important in a few years.

8. Apr 12, 2015

### Orodruin

Staff Emeritus
Just adding for completeness since it has not been mentioned yet. One of the more popular technologies among experimentalists today is that of a liquid argon time projection chamber. You can read a bit about it here: http://www-lartpc.fnal.gov/summary.htm [Broken] - although, naturally, they might be a bit biased in saying that it is the future technology.

Last edited by a moderator: May 7, 2017
9. Apr 12, 2015

### ChrisVer

10. Apr 12, 2015

### Staff: Mentor

11. Apr 12, 2015

### Orodruin

Staff Emeritus
12. Apr 12, 2015

### Staff: Mentor

The exclusion limits got better since 2012. For the low-mass region, see this plot taken from the CRESST paper discussed there. Just an order of magnitude to go in the 7-10 GeV mass range. Include experiments currently running or under construction and some neutrino events should appear. There was a talk at the Moriond conference how to exceed searches below that limit by doing more careful analyses in terms of energy spectrum or directional detectors. Oh wait, the topic of this thread were neutrinos ;), they will be a relevant background soon.

13. Apr 12, 2015

### Orodruin

Staff Emeritus
Today's signal, tomorrow's ... Well, you know.

I almost went to Moriond this year, but it conflicted with my teaching schedule. :(
The internet connection there is crappy anyway

14. Apr 13, 2015

### ChrisVer

In a case where one would think about the neutrino-nucleon interaction:
The cross section is given by $\sigma = 10^{-38}~cm^2 \Big(\frac{E_\nu m_N}{GeV^2} \Big)$

In that case we can find for earth (assuming it as a homogeneous sphere $\rho =5.5 ~gm/cm^3$ and $R=6.4~km$) that the mean free path for the neutrinos is:

$2R=^{\!}\lambda = \frac{1}{\sigma \rho} \frac{1g}{N_A} \Rightarrow E \sim 30~TeV$

This means that the neutrinos of energies above this energy threshold will undergo a neutrino-nucleon interaction while crossing through the Earth. Right?
Experimentally, would that cause a variation of the neutrino signals throughout the day? I've seen this effect of earth shielding can help to extrapolate the cross section of UHE neutrino-nucleon interaction, but I don't seem to understand how...

Physically what will happen depends on the neutrino type...
for electron-neutrino we will have: $\nu_e p \rightarrow e^- n$ and the electron will surely be trapped within the earth (so we won't find it?).
For muon-neutrino we will have $\nu_\mu q \rightarrow \mu^- q'$ and the muon will decay $\mu \rightarrow e \nu_\mu \bar{\nu}_e$. I think the muon can penetrate the rest of the earth and so we can observe it?
For tau-neutrino we will have the creation of a tau fermion, which decays dominantly into pions (charged and neutral). The charged pions can also end up in muons ,which the neutral will give photons. I think the detection can be made in the same way as the muon neutrinos (the photons won't be measured).

Last edited: Apr 13, 2015
15. Apr 13, 2015

### Staff: Mentor

I know, there are studies that have Higgs as background contribution.

@ChrisVer: 30 TeV is lower than I remembered, the value should be somewhere at ~1000 TeV (which means the most high-energetic events of IceCube have significant shielding on one side).
Well, it is a matter of probability.
IceCube is the only current detector large enough to find those events with reasonable frequency, and it is directly at the south pole - no variations.

An electron gives an electromagnetic shower close to the interaction point of the neutrino.
Muons as minimally ionizing particles give long tracks (roughly 1km/TeV at typical energies), at very high energy (many TeV) they start to emit some bremsstrahlung so they won't penetrate too much rock. Their decay is negligible - they don't travel long enough to make that relevant.
Charged pions produce hadronic showers long before they would get a chance to decay. Photons become part of the hadronic shower then.

16. Apr 13, 2015

### ChrisVer

*I made mistake for the radius of the earth , I wrote 6.4km, obviously I mean 6400km -it's bigger than a city*

Well one has to solve the equation I wrote for $E_\nu$.

$E_\nu (GeV)= \frac{10^{38}}{5.5 \times 12800 \times 6.022 \times 10^{23}} \frac{cm}{km} \frac{GeV}{m_N}$

where $N_A = 6.022 \times 10^{23}$ the avogadro number.

For $m_N \sim 1GeV$, $\frac{cm}{km} = 10^{-5}$ and doing the calculations you reach:

for your convenience the non-latex calc: (10^33)/(5.5*12800*6.022 *10^(23))

$E_\nu = 2.35 \times 10^{4} GeV = 23.5~TeV$

I don't know if you find something wrong in the above...(?)

As for the muon I think you are right...
Hadronic showers within the earth won't give anything detectable either...

17. Apr 13, 2015

### Staff: Mentor

Hmm, I can confirm your calculation for the given cross-section, although that seems to be a bit on the high side that does not explain a factor of 30. The 1000 TeV came from this amusing article

This presentation has interesting slides, like 24 and 31.