Neutrinos and neutrino interactions

1. Oct 13, 2004

Andrew Mason

I have two questions about neutrinos that I haven't been able to figure out:

1. The evidence seems to indicate that neutrinos have rest mass. Since it is invisible to charge and electromagnetic fields and, apparently, gluon fields, how does it interact with matter at all? I think it may have something to do with it having a magnetic moment, but I also find the existence of a magnetic moment puzzling if it cannot interact with charges.

2. If the neutino lacked rest mass it would have to travel at the speed of light. Since it could not be an electromagnetic wave/particle (eg. photon) what kind of 'wave' could it be? (the lack of an answer may imply that it has rest mass).

AM

2. Oct 13, 2004

vanesch

Staff Emeritus
It can (and does) interact through the weak interaction. That's the only way to see it, and because the weak interaction is ... weak, it is difficult to see them.

cheers,
Patrick.

3. Oct 14, 2004

zefram_c

1. In the Standard Model, neutrinos interact with normal matter through the weak interaction, mediated by Z and W. Theoretically, since neutrinos do not interact with the photon, they have no magnetic moment. However, there have been recent claims that neutrinos might have a magnetic moment: http://hbar.stanford.edu/sturrock/peter/PASpreprints.html

2. There is no requirement for something to be a wave to be able to travel at the speed of light. For example, gluons are also massless, but there is no analogue to Maxwell's equations for them that I am aware of, and hence no wave theory of gluons. Besides, according to QM, there are no such things as waves or particles as far as fundamental entities are concerned.

4. Oct 14, 2004

vanesch

Staff Emeritus
Well, there is of course the classical SU(3) gauge theory, no ? That's a classical field theory, which, after quantization, becomes QCD.

cheers,
Patrick.

5. Oct 14, 2004

humanino

I agree with Patrick, and I want to add that in QCD, there is also a separation of the color field into "electrical" and "magnetical" parts, called chromoelectric and chromomagnetic fields. This is very convenient as a separation when one uses hamiltonian formailsm. They satisfy equations similar to Maxwell equations ! See for instance QCD by A.M.Smilga. Also, when one deals with the quark-gluon plasma, this is useful too : Topics in the Transport Theory of Quark-Gluon Plasma

Last edited: Oct 14, 2004
6. Oct 14, 2004

Andrew Mason

I can begin to see how the weak interaction explains a neutrino - neutron interaction as in $\nu_e + d \rightarrow p + p + e^-$. That just seems to be the reverse of beta decay. But does the weak interaction explain neutrino-electron scattering in which there does not appear to be any weak interaction? - just a transfer of momentum resulting in the same particles: $\nu_e + e^- \rightarrow \nu_e + e^-$

AM

7. Oct 14, 2004

vanesch

Staff Emeritus
Of course it does. By exchange of a Z0 or by exchange of a W+/-.
For instance you can have \nu_e + e^- -> W^- -> \nu_e + e^- (s-channel) and also a Z0 in the t channel.

cheers,
Patrick.

8. Oct 14, 2004

zefram_c

Thanks, this is new to me. I was half expecting that something like this exist, hence my phrasing "no theory... that I am aware of"

Going back to neutrinos, their interaction with particles of regular matter can also be seen from the difference in oscillations of solar neutrinos that reach the detector from directly overhead versus those that pass through the Earth to reach it from below.

9. Oct 14, 2004

Andrew Mason

There is a good paper on this at: http://xxx.lanl.gov/PS_cache/nucl-ex/pdf/0204/0204009.pdf

I don't pretend to grasp all of the detail here but the conclusion is reasonably clear [bracketed terms are added by me]:
"In summary, SNO has measured the day-night asymmetries of the CC [charged current], NC [neutral current], and ES [electron scattering] reaction rates. From these results the first direct measurements of the day-night asymmetries in the $\nu_e$ flux and the total $\nu$ flux from the Sun have been deduced. A global fit to SNO’s day and night energy spectra and data from other solar neutrino experiments strongly favors the LMA [Large Mixing Angle] solution in a 2-flavor MSW [Mikheyev Smirnov Wolfenstein - matter enhanced oscillation] neutrino oscillation analysis."​

10. Oct 15, 2004

Andrew Mason

Thanks for this explanation.

Question:
Since the forces and interactions between fundamental particles are necessarily mediated by particle exchanges (bosons or gluons), am I correct in thinking that there cannot be such a thing then as a contact collision between fundamental particles? (ie the exchange of momenta between two moving fundamental particles is always achieved by the exchange of virtual particles not by some concept of 'direct contact')?

AM

11. Oct 16, 2004

vanesch

Staff Emeritus

(btw: gluons are bosons)

The "old" theory of weak interactions by Fermi _was_ a "contact interaction" and it works quite well on tree level ; only it is not renormalizable. Currently, the view is indeed that you always need a boson exchange.
The reason is that to have a fermionic "contact interaction", you'd need a vertex with 4 fermions, which leads automatically (as in Fermi's theory) to a non-renormalizable term when working in 4-dimensional space-time. Not that that is so terribly bad as it used to be. But currently it is not considered (in the standard model and most of its extensions I'm aware of - which is certainly not exhaustive).

However, spin-1 bosons CAN have a contact interaction (a 4-vertex). In QCD for instance, there is a 4-vertex with 4 gluons.

cheers,
Patrick.

12. Oct 16, 2004

Andrew Mason

While Maxwell's equations show that the speed of light is determined by electromagnetic properties of free space, relativity makes the speed of light something more than just a property of electromagnetism. The principle of relativity implies that massless particles must travel at the speed of light.

$E = mc^2$ with the addition of kinetic energy gives: $E^2 = (m_0c^2)^2 + (pc)^2$

So if $m_0 = 0$ then E = pc

Since $p = m_0v\gamma$ where $\gamma = \frac {1}{\sqrt{1-v^2/c^2}}$
if v is less than c, and $m_0 = 0$, p is necessarily 0.

From the above, we see that a massless particle traveling at a speed less than c could not have any momentum or energy in any inertial frame. If v=c the denominator is 0 so the momentum, p, cannot determined by the Lorentz transformation. This is the only situation in which p can be nonzero. Since massless particles do have nonzero momentum and energy, they must travel at the speed of light with respect to all intertial frames of reference.

AM

13. Oct 16, 2004

Tom Mattson

Staff Emeritus
I suppose that's one way to interpret it. Another way would be to recognize that p=&gamma;mv simply does not apply to massless particles. QM tells us exactly what the photon's momentum is: p=hc/&lambda;.

14. Oct 16, 2004

Andrew Mason

(I think you meant p=h/λ). The momentum of a photon is a consequence of relativity. The conservation of momentum requires the photon to have momentum: since the emission of energy from a mass, m1, decreases the m1 by $E/c^2$, and when absorbed by mass m2, increases the mass of m2 by $E/c^2$ the photon must carry momentum p = E/c. Otherwise the center of mass of the inertial system of m1 and m2 moves which would violate conservation of momentum.

Quantum Mechanics simply expresses that momentum in terms of wavelength or frequency. (ie. the momentum $p = E/c = \frac{h\nu}{c} = \frac{h}{\lambda}$). Relativity explains it.

I think it is pretty clear that a massless particle cannot move at a speed less than c. Put another way, a massless particle having a speed of less than c in free space would have no momentum or energy. Your example of the photon, of course, is of a massless particle travelling at c, not less than c.

AM

15. Oct 17, 2004

Tom Mattson

Staff Emeritus
Of course, you're right.

I agree. When I first responded, I wasn't sure of the validity of using the relation p=&gamma;mv to make inferences on what massless particles can and cannot do, but I've since worked it out for myself.