# Why do neutrino oscillations require mass?

1. May 5, 2012

### 96hicksy

I was reading up on chameleon particles today, and this lead me to reading more about neutrino oscillations - I noticed that at one point it said for oscillations to occur, neutrinos must have mass, there was no explanation to this, but it goes against the standard model.
Since I am only young I have not studied things such as neutrino oscillations at a high level, I just research physics in my spare time, but I do want to research it at college and university. So I want to ask you, why do neutrinos require mass for it to happen?
Thanks!

2. May 5, 2012

### Matterwave

Simply put, neutrino oscillations precisely because the flavor eigenstates do not correspond to the mass eigenstates.

With a little more detail, consider the Schroedinger-like equation that Neutrinos obey:
$$H\psi=i\hbar \frac{\partial \psi}{\partial t}$$

H has a term which is proportional to the Neutrino's momentum, and a term which is proportional to the Neutrino's rest mass (if it has any).

If all 3 flavors of Neutrinos had 0 rest mass, then, a particular Neutrino produced with some momentum p will be in an Eigenstate of H (since it has definite momentum p, and definite mass m=0). We know that Eigenstates of H are stationary states and so one would expect the probability for this Neutrino to be a "electron neutrino", for example, to remain constant in time. In order for these probabilities to oscillate, one needs the Neutrino to not be in an Eigenstate of H.

If the 3 different flavor states are not eigenstates of mass, though, then creating an electron neutrino with some momentum p would not be creating a neutrino in an eigenstate of H (because there's a mass term now), and therefore the probabilities can oscillate.

3. May 6, 2012

### Bill_K

Even more simply put, oscillations imply an off-diagonal term in the Hamiltonian, so when the Hamiltonian is diagonalized the eigenvalues will necessarily be unequal, therefore at least one of them is nonzero.

4. May 6, 2012

### Parlyne

It's probably worth pointing out that the necessary condition for oscillations is that the neutrino states have different masses from each other. Oscillations can only occur when different particle states with the same momentum pick up different phases due to free propagation. For a state with momentum $\vec{p}$, the eigenstate of the free Hamiltonian will simply be $\sqrt{p^2+m^2}$, demonstrating that only a difference in mass can cause a difference in $\exp(i\hat{H}t)$.

The oscillation effect, then, will essentially be interference due to these differing phase factors. Of course, to get interference, you need to have a mixed state to begin with, which requires that the neutrino state produced not be an eigenstate of mass, but, rather, a superposition of the mass states.

5. May 7, 2012

### sambristol

96Hicksy

The above posts are correct but not the most straightforward explanation.

Oscillation means something must be changing with time.

Special relativity tells us two things of relevance. Firstly if something travels fast with respect to us we see its time-scale slowed down (its clock seems to run slow), the closer it moves to the speed of light the slower the clock runs and in the limit of a body moving at the speed of light the clock is stopped – nothing can be observed to change. Secondly we know no way of a body with mass to be accelerated to the speed of light, as it would require infinite energy. Special relativity only makes sense if particles with no mass can only travel at the speed of light.

So until a few years ago there was no apparent problem, we didn't know about neutrino oscillation and as far as we knew the neutrino had no mass. We did have one big problem but not from particle physics, it was from astrophysics. The nuclear reactions in the sun produce neutrinos and we could only detect a fraction of the type of neutrinos expected.

When experiments were sufficiently good to show neutrino oscillation it became obvious that the rate of time for a neutrino was not 'frozen' so it could not be travelling at the speed light and therefore it had to have some mass. In addition the 'solar neutrino problem' disappeared.

The 'standard model' does not take a big hit from this. It is, as it says, a model and one that has had many modifications since it was formulated back in the 1960's. The big hit is to special (and by extension) general relativity. It has always been a bit of an embarrassment that relativity singled out just one of the fundamental forces as its key constant is c (the speed of light and hence a product of the electromagnetic force). To those who think about this, it was just about plausible as there were two particles (neutrino and photon) which travelled at the speed of light but with only the photon left now the embarrassment became all too obvious again. But special relativity predicts results which are verified experimentally to quite ridiculous precision so we carry on using it.

Humankind it would seem is happy to use a theory which is useful and gives results verified by experimental evidence even if there are clear logical problems with the theory. Some examples are Calculus – Newtons formulation was logically flawed (it involved dividing 0 by 0) and the whole mess took almost 200 years to sort out (the theory of limits) or Fourier analysis – just plain wrong in its original form only corrected over 100 years later but in both cases everyone used used them because they got results that agreed with experiment.

There are many other examples , so you see physics is 'a work in progress'. It is to future generations of physicists, such as you, we leave these issues, good luck – we were not clever enough to solve them.

Regards

Sam

6. May 7, 2012

### Parlyne

I'm afraid that this explanation doesn't really capture the issue. If I followed your reasoning, I would conclude that massless particles can't be involved in oscillations at all. But, this is simply not correct. If, for instance, we had two neutrino species - one with 0 mass and one with non-zero mass - there could still be oscillations if the neutrinos were physically produced in mixed states. In point of fact, present measurements only guarantee that two of the three neutrino mass states have non-zero mass. What we know for certain, however, is that the masses of all three states differ from each other.

You seem to be under the impression that the speed of light somehow comes into relativity because it is a property of electromagnetism; when, in fact, it's the other way around. c shows up in EM specifically because EM is a relativistic theory. If we were to formulate the other forces in everyday units, there would be factors of c in those equations, as well. In fact, you would find that, were it possible to have a free gluon, gluons would also travel at c.

The fact the photon is the only freely propagating particle which travels at c is in no way an embarassment. It is a fully understood property of the theory. Nothing more, nothing less.