# Interactions between quarks and neutrinos

1. Jul 17, 2006

### Signifier

I am wondering how to calculate how far a neutrino would have to pass through a substance for it to have a probability P of interacting at least once.

Water, for instance, has a density of 1 g / cm^3; using Avogadro's number I think this means that there is about 6.02 x 10^29 protons and/or neutrons per 1 m^3; as a proton or a neutron is made up of three quarks, this means that for water there is about 1.806 x 10^30 quarks per m^3. I know that a neutrino has to get very close (~10^-18 m) to a quark to interact with it, and that very few of the neutrinos that make it into the radius of some arbitrary quark actually do interact with it (~1 in 10^12).

How could I use this information to (VERY) roughly estimate how far some random neutrino would have to pass through water before it would interact with a quark? Or, say, how long it would have to pass before it had a 60% chance of interacting?

Any help, hints or guidance to solving this problem would be appreciated. I have read (here, for example: http://www.physics.usyd.edu.au/hienergy/forces_and_neutrinos.html [Broken]) that a neutrino would have to travel ~9.5 x 10^17 m through pure water before interacting, but I haven't been able to find any explanation of how this may be reasoned out.

Last edited by a moderator: May 2, 2017
2. Jul 17, 2006

### CarlB

It very much depends on the energy and type of the neutrino. Hopefully someone will provide a formula.

Carl

3. Jul 17, 2006

### Staff: Mentor

You need to find out the total interaction cross section with nucleons, for the particular type of neutrino you're dealing with. I don't have it at my fingertips, but that's the term to search for. At high energies, it increases linearly with energy. Try rummaging through the Particle Data Group's web site.

This will be a number, $\sigma$, with units of area, probably cm^2. To calculate the number N of neutrinos that interact, out of a beam that initially contains $N_0$ neutrinos:

$$N = N_0 n \sigma L$$

where n is the number of nucleons per unit volume (use cm^3 if the cross section is in cm^2) and L is the distance (cm) the neutrinos travel through the target.

This formula is an approximation which is valid for very weak interactions, which is certainly the case for neutrinos.

I think you can safely neglect interactions with electrons, but you might want to look up the total interaction cross-section for neutrinos with electrons, while you're at it.

4. Jul 22, 2006

### Signifier

May I ask how you derived that formula? (I am not very good at statistics). Or is it an empirical formula? It seems to produce fairly good results...

Thank you!

5. Jul 23, 2006

### Staff: Mentor

That formula defines the cross section $\sigma$. If the interaction is very weak, one would expect the number of neutrinos absorbed to be proportional to the thickness of target material (L), the number density of target particles (n), and the number of incoming neutrinos. The proportionality constant is $\sigma$.

6. Jul 23, 2006

### Astronuc

Staff Emeritus
From NEUTRINO DETECTION EXPERIMENTS - http://wwwlapp.in2p3.fr/neutrinos/anexp.html

37Cl -> 37Ar Homestake Mine, South Dakota
98Mo -> 98Tc Henderson Mine, Colorado
71Ga -> 71Ge SAGE, URSS (Russia?) and Gallex, Italy.

$\nu_e\,+\,d\,\rightarrow\,p\,+\,p\,+\,e^-$ Sudbury, Ontaria (according to hyperphysics, but the above site (in2p3) has "p + n" as the product of neutrino interaction with a deuteron. ( http://wwwlapp.in2p3.fr/neutrinos/neutimg/nexp/solar_chimie.gif )

Other reactions are possible, but jtell provided the answer. The microscopic cross-section is derived from experiments, and is dependent upon the target. Clearly, some nuclei allow for interaction of neutrinos and the quarks in their nucleons, or actually neutrons.