# Solar Neutrinos

1. May 27, 2007

### Bret Danfoss

I just read that the mystery of the "missing solar Neutrinos" has been solved.

That is, the "missing" stuff is converted to Muon & Tau Neutrinos (which have been detected at the SNO).

I'm confused .... If Solar Neutrinos are very "non-interactive", then how do enough of them get converted into Muon & Tau Neutrinos?

Could there be other kinds of Neutrinos being detected of similar "mass" that "look-like" Muon & Tau Neutrinos etc?

Thanks.

2. May 28, 2007

### mjsd

via neutrino oscillations

3. May 28, 2007

### Bret Danfoss

Sorry, I don't follow.

Are you saying that they "spontaneously" change species (or something to that effect)?

Thanks.

4. May 28, 2007

### Haelfix

Actually yes, roughly speaking.

Essentially (very loosely speaking) you have a matrix acting on some neutrino state (which is in a mass eigenstate), and this spits out a neutrino in a flavor eigenstate.

Its this mismatch between basis states that is at the heart of neutrino oscillations.

Using this, you can show that say neutrino fields while they propagate in vacuo, have a nonzero chance to basically interfere with themselves and undergo a flavor transition

5. May 28, 2007

### Bret Danfoss

Thanks for that, not that I understand it, but it's a sophisticated response so it must be correct :-)

Let me get this straight: Solar Neutrios come in & about half of them spontaneously "convert" to Muon & Tau Neutrinos?

How is this "spontaneous" conversion factually known to be occuring? .... I realise SNO measures Muon & Tau Neurtinos as well as Electron Neutrinos, but how is it know that the "missing half" spontaneously "converts"? ... That is, not by theoretical assumption, but by physical measurement (i.e. perhaps their produced in the Sun)?

Wouldn't it make more sense to say that the SM is incomplete & perhaps another "kind" of Electron Neutrino "equivalent" is being produced by the Nuclear reactions in the Sun? ..... Alternatively, why can't the "missing Neutrinos" be produced as Muon & Tau Neutrinos to begin with?

Thanks

6. May 28, 2007

### George Jones

Staff Emeritus
It doesn't happen spontaneously, it's an effect that evolves continuously in space and time as a neutrinos travel from the Sun to the Earth.

There are three neutino flavours, electron, muon, tau. There are also three mass states, say m1, m2, and m3, but flavour states and mass states don't correspond.

For example, an electron neutrino is a mixture of all three mass states. The phases of the mass components oscillate in space and time (standard time evolution of quantum theory), and this leads to interference effects. Depending on where it is in space time, what started out as an electron neutrino looks more like a muon neutrino or a tau neutrino. The probabilities for flavour type oscillate back and forth in space and time, so the effects don't happen spontaneously. Neutrino physics predicts the probabilities at the Earth.

In a logic class, knowing that A implies B is true, and that A is true, is enough to say that B is true (modus ponens). An elementary mistake sometimes made in a logic class is to say that becuase B is true, A is true.

However, this is the way science often works. We often have A implies B, where A is a theory and B is prediction. We then try to measure B. If find B, we say that this gives evidence for A. If we find (not B), we say that theory A has been falsified. If we have also A implies C, and we measure both B and C, we gain more conidence in theory A. Etc. (Things are simplified here; things are not this clear-cut.)

In this case, the theory of massive neutrinos and neutrino makes predictions about the numbers of electron, muon, and tau solar neutrino sthat should be found at the Earth. We measure this, and this give evidence for the theory.

Here is http://physicsweb.org/articles/news/8/11/9" for neutrino oscillations. In particular, look at the second-last paragraph.

Last edited by a moderator: Apr 22, 2017
7. May 28, 2007

### malawi_glenn

Here you have a lot of material for neutrinos and specially neutronos in astrophysics:

Last edited by a moderator: May 2, 2017
8. May 29, 2007

### Bret Danfoss

Thanks everyone, much appreciated.

Last edited: May 29, 2007
9. May 29, 2007

### CarlB

Bret, there is another way of explaining neutrino oscillation, one that I think is less confusing.

When we label a state as "electron neutrino" what we mean is that we know that it was associated with an electron in a weak interaction. If you require that a "particle" have a specific and exact mass, like the electron, muon, proton, etc., have, then the electron neutrino is not a particle.

Instead, the three particles with exact masses are $$\nu_1, \nu_2, \nu_3$$. Neutrino oscillation makes a lot of sense if you think of it in terms of these three neutrinos and leave the electron neutrino alone.

According to the laws of quantum mechanics, all the possible ways that an interaction can proceed must be summed over before you take the absolute value and square it to get a probability. For the example of an electron / positron in the sun doing something weak, there are three ways the interaction could proceed, via $$\nu_1, \nu_2, \nu_3$$.

We can't know the actual neutrino species because the mass is so small that it makes no detectable difference in the kinematics. And when we detect the neutrino on this end, it either gets absorbed or flies off to infinity. Either way it goes away and we can't tell what type it was. So we have to treat all three neutrino processes together.

The electron has three different couplings to the three species of neutrinos and it is this set of couplings that are called the "electron neutrino".

Now the three neutrinos have different masses. Differences in mass cause differences in frequency. The mass differences are very small compared to the energy of the reaction so the mass differences make only a little difference in frequency. But over very long distances, the mass difference can make the three neutrino types have enough difference in frequency that it can reach a full pi or 2 pi. It is this phase difference that we interpret as an electron neutrino changing itself into a muon neutrino, etc.

Let me make a simple example with fake numbers but that will give the general idea that is going on here. The first thing to note is that the electron and muon neutrinos are just linear combinations of the "true" neutrinos. Suppose that we have:

$$\nu_e = (\nu_1 + \nu_2)/\sqrt{2}$$
$$\nu_\mu = (\nu_1 - \nu_2)/\sqrt{2}$$

If a neutrino is emitted and we are very close to the electron, it will be received still in the $$\nu_e$$ form. In this form it will couple efficiently to electrons and we will call the reaction an "electron neutrino". But if the neutrino travels far enough, the different frequencies of $$\nu_1$$ and $$\nu_2$$ could add up enough to give a complete sign flip, one relative to the other. Then we would have a combination that interacts efficiently with muons and we would call it a "muon neutrino".

The actual mixing numbers are covered in something called the MNS matrix. There are various places where you learn more about this on the web. There are complications associated with things like differences between how the various neutrinos interact with matter. My favorite write up on it is this one by Alexei Smirnov:
http://www.slac.stanford.edu/econf/C0605151/progfiles/notes/smirnov2.pdf

Carl

10. May 30, 2007

### mjsd

Indeed, at the detector all you really see is either an electron, muon or tauon, and some missing energy (ie. the neutrino) but not directly interacting with the neutrinos themselves. The meaning of an electron/muon/tauon neutrino:
suppose at the source (say the location of hadrons collision) you have an energetic u-quark and d-quark decaying into a W-boson and subsequently the W-boson itself decays into a charged lepton (electron, muon or tauon) plus a neutrino. This neutrino is then allowed to travel only short distance before being "detected" at the detector. At the detector, (basically another interaction involving a W-boson takes place, giving off yet another lepton) one tries to detect the type of the charged lepton involved. If at the detector, we detect a charged lepton that is an electron then we call the neutrino an "electron neutrino". Note that the key here is that it must be a "short journey". If you somehow capture both the source and detector end charged leptons, you will find that they are of the same flavor always. Flavor change in such "short journey" experiments has never been observed.

a good intro to mixing in vacuum and neutrion mass terms, can also try this review written by Boris Kayser
http://arxiv.org/abs/hep-ph/0211134v1

11. May 31, 2007

### Bret Danfoss

Thanks Carl, that's excellent, it helps a great deal.

When you refer to "frequency", I assume you're talking about the wavefunction frequency of the Neutrino?

Cheers

Last edited: May 31, 2007
12. May 31, 2007

### CarlB

Yes.

There's another thing I should mention. The standard model is built on a principle that amounts to assuming that the only terms that can appear in the Lagrangian / Hamiltonian are "renormalizable". The result of this theory was a series of terms that describe the standard model.

Unfortunately, the terms that give the neutrinos mass cannot fit in with this scheme. The world would be kinder and gentler to theoretical physicists if the neutrino were massless. A great book that discusses the problem and various fixes in painstaking detail (see chapter 10) is:

The Standard Model; A Primer
Cliff Burgess & Guy Moore,
Cambridge University Press (2006) \$75
https://www.amazon.com/Standard-Model-Primer-Cliff-Burgess/dp/0521860369

The above, like most physics books, can be read without the need of understanding every calculation.

As discussed in the above book, there is some question of whether or not neutrinos with mass should be included in the "standard model" or not. I think they belong there, and that the renormalization was just a handy way of guessing terms. Experiment has eliminated it, but that doesn't mean that we have to start from scratch. We can keep the Lagrangian with the best experimental evidence so far.

Given that you have three complex degrees of freedom available to define three neutrinos, you can organize those degrees of freedom in various ways. As mentioned above, I think the way that makes sense, from the point of view of the rest of physics, is to diagonalize them according to mass. That is, to treat the physical particles as being the particles with exactly defined masses.

On the other hand, you can also organize those three degrees of freedom according to how they interact weakly with electrons, muons, and taus. Then you get the electron, muon and tau neutrinos. If you take the point of view that neutrino mass does not belong in the standard model, then there is no reason to diagonalize the neutrino degrees of freedom to make mass diagonal. Instead, the natural way of describing the neutrinos is according to how they interact with the electron, muon and tau. And this was how it was always done until it was discovered that neutrinos "oscillate". So there is some historical reason for talking about electron, muon and tau neutrinos.

Last edited by a moderator: Apr 22, 2017
13. May 31, 2007

### Staff: Mentor

Bret, if you're interested, there is a great Nova episode on the discovery of the oscillation and how it explained the missing solar neutrinos. This all happend recently, and was a wonderful example of how scientists had to build better and better and more innovative detectors, and had to be very clever in the hunt for the missing neutrinos. The discrepancy from theory was very disturbing to many scientists (for good reason!), so for the mystery to have been solved and the answer validated through experiment, it was a great relief.

Here is some info about the show:

http://www.pbs.org/wgbh/nova/neutrino/

Looks like it's available on DVD. I think you would enjoy it.

14. May 31, 2007

### Bret Danfoss

Thanks berkeman, I'll look it up.

Carl, would you know what the wavefunction frequencies are for the Neutrinos?

I'd sincerely appreciate hard values if you have them, not just equations.

Last edited: May 31, 2007
15. May 31, 2007

### CarlB

The short answer is that the wave functions have the form $$\exp(i(Et - \vec{p}\cdot \vec{x}))$$, where E is the energy, and Planck's constant and the speed of light have been set to one. To get the frequencies, you have to make a guess at the energy. A typical number might be 1 MeV:
http://www.sns.ias.edu/~jnb/SNviewgraphs/snviewgraphs.html

So the frequency is about $$10^6$$ eV. To convert this into Hz, you divide by Planck's constant $$= h = 4.135 \times 10^{-15}$$ eV-seconds, and you get around $$2.4 \times 10^{20}$$ Hz.

However, what is of more importance for neutrino oscillation is how the frequency changes between the different species. And explaining how this is computed is just a little bit complicated. However, it does illustrate how "virtual particles" work in QM, so I think it's worth going through.

The first thing to note is that the relation between E and p for a given particle is:
$$E^2 = \vec{p}\cdot \vec{p} + m^2 = p^2 + m^2$$
This means that when you switch from one neutrino species to another, you have to change the relationship between the energy E, and momentum p that are carried by the neutrino.

If we were working in pure theory, we could imagine that our experiment was so accurate that we could use conservation of energy and momentum to figure out which neutrino made the trip. (All we would have to do is to compute $$\sqrt{ E^2 - p^2}$$ and see which mass it gave.) But we can't do that. Instead, all three neutrinos contribute to where our dials end up.

A convenient way of detecting a neutrino is to see what it does when it collides with an electron. In doing this, it gives momentum to the electron. This suggests that we make the calculation by assuming that momentum is conserved. To make the calculation, we compute E as a function of p and m:

$$E_m = \sqrt{p^2 + m^2} = p\sqrt{1 + (m/p)^2}$$

Since m is small compared to p, we can approximate this by taking the first term in the expansion of the square root:

$$E_m = p(1 + m^2/(2p^2)) = p + p(m/p)^2 /2$$

This is how the calculation is done in Wikipedia:
http://en.wikipedia.org/wiki/Neutrino_oscillation

Having got $$E_m$$, we can use this to model neutrino oscillation as a beat frequency. But I think this is a little backwards, so let's do it the other way.

Instead of assuming that the measurement conserves momentum, let's assume that our experiment is one that measures energy, and so it conserves energy. Then we want to write $$p_m$$ in terms of E and m:

$$p_m = \sqrt{E^2 - m^2} = E\sqrt{1 - m^2/E^2}$$

Since m/E is very small, we use first order approximations again to get

$$p_m = E(1 - m^2/(2E^2)) = E - E(m/E)^2/2$$

Since to first order, $$E = p$$, so we could write the above in the form $$p_m = E - p(m/p)^2/2$$ which is pretty much the same as the calculation in wikipedia $$E_m = p + p(m/p)^2/2$$, which assumed that momentum was conserved exactly. But the above calculation is in terms of wave vectors (and therefore wave lengths), so I think it is more suited to an explanation for neutrino interference that talks about how they interfere as a function of distance.

The wikipedia calculation is better if you want to think of neutrino "oscillation" instead of neutrino "interference", because it looks at differences in the Et part of the wave function. To convert between oscillation and interference, just use the fact that the neutrinos are traveling very close to the speed of light.

Now in actual fact, the experiment we are running will not be able to measure energy OR momentum so accurately as to ensure that these are conserved. Instead, what we will be doing will be running a whole set of possible experiments with possible energies (or momenta). To compute the overall effect, we would have to assign a probability to each of those possible energies (momenta) and compute how the neutrinos interfere in that particular case.

But you can see from the formulas that the beat frequency (wavelength inteference) does not depend much on small changes in the energy (momentum). Therefore, our analysis that assumed a particular energy (momentum) will give the correct beat frequency (wave length interference) as the actual experiment, which cannot measure things very precisely.

If this doesn't help with the explanation, then at least it may provide a beginning for another person to give a more complete (or more correct) explanation.

Last edited: Jun 1, 2007
16. Jun 4, 2007

### damgo

Hi Carl,

You can certainly have renormalizable neutrino mass terms, for example L_dirac = -m*psi*psibar = -m*(psibar_l*psi_r + psibar_r*psi_l). It's only if neutrinos are Majorana that the simple mass term is nonrenormalizable.

No. Studies at particle accelerators put very tight constraints on the number of ordinary neutrino types -- currently 2.994 +- 0.012. (That is, three. :) ) In particular what we look at are the "invisible" decays of the Z boson. The Z can decay into neutrino-antineutrino pairs (which can't be seen) or other particle-antiparticle pairs (which can be seen). The more types of neutrinos there are, the more often the Z will decay invisibly. So by measuring the ratio of invisible Z decays to visible Z decays, we can spot any extra neutrino flavors.

This bound can be evaded by making the extra neutrino types either extremely heavy or noninteracting with the Z ("sterile".) But these would be very much unlike the ordinary electron, mu, or tau neutrinos.

see eg: http://pdg.lbl.gov/2000/s007.pdf

17. Jun 4, 2007

### CarlB

I think you've got it reversed above, it is the Majorana terms that are renormalizable. If the usual Dirac terms were okay, it wouldn't be an issue. The problem is that all the other massive particles take Dirac masses, rather than Majorana masses.

It's not a subject I have studied carefully because I don't think that renormalizability should be a requirement of a theory. Let me take another look at that chapter.

18. Jun 4, 2007

### damgo

^^^ No, the Dirac term is renormalizable but not the Majorana. Consider: in the case of massive Dirac neutrinos, we have exactly the same situation as with the charged leptons. Since their mass terms are renormalizable, the neutrinos' are as well.

We can see this just by the normal power-counting rules. The Dirac neutrino (and charged lepton) mass term is gotten in GWS electroweak theory via coupling with the Higgs:

L_dirac = f*nu_r_bar*phi*lep_l

where the SU(2) indices on phi (the Higgs field) and lep_l (the left-handed lepton SU(2) doublet) are contracted. The dimensionality of the fields here is 3/2+1+3/2 = 4, so the [Yukawa] coupling f is dimensionless and the term is renormalizable.

For the Majorana term, though, we don't have a separate nu_r field; rather we have to form it by charge-conjugating the left-handed field lep_l, which is an SU(2) doublet. Since we have two SU(2) doublets here, we need at least two copies of the Higgs field to contract them with. (If we contracted them with each other we'd have a non-Higgs mass term of the type forbidden in GSW.) So the simplest mass term is:

L_maj = m_eff * (lep_l^c)bar * lep_l * phi * phi

The fields here have mass dimension 3/2+3/2+1+1 = 5, so the coefficient m_eff has dimension -1. Negative mass dimension tells us that it's non-renormalizable.

I agree that renormalizability isn't an issue here -- in any case, you can get such an effective nonrenormalizable interaction from a renormalizable one via the seesaw mechanism for example. And in fact I don't know anyone who's bothered by it. But people don't like the usual Dirac terms because then you're left with the weird problem of why the neutrino Yukawas are so much tinier than the charged lepton Yukawas, not because it's not renormalizable (it isn't).

19. Jun 4, 2007

### CarlB

Yes, you are right and have written it up nicely. Thanks for the correction. I got confused because I see so many papers talking about "renormalizable Majorana terms". But the context is that this is the man bites dog story.

Carl