# How quickly do neutrinos change flavor?

DOE’s Fermilab has switched on its newly upgraded neutrino beam. This is in preparation for the NOvA experiment, which will study neutrinos using a 200-ton particle detector at Fermilab and a 14,000-ton detector in northern Minnesota. Story here.

This experiment is to test "flavor changes" in neutrinos. What does current theory predict in terms of flavor changes? Ie. does a electron neutrino change to a muon neutrino every so many milliseconds?

• Did you read the tag-wiki for neutrinos or the wikipedia page? We've only been unambiguously observing neutrino oscillations since circa 1990. It's not like the fact that this happens is news or anything. There are already 18 questions concerning neutrino oscillations on the site. – dmckee Sep 20 '13 at 16:50
• From the tag: "Current efforts are focused on determining the parameters of the mixing matrix (two mass differences and two mixing angles are known), searching for evidence of CP violation in the neutrino sector, and determining if the neutrinos are Dirac or Majorana particles." My question relates to if there are any actual specific predictions regarding the number of changes they will see at the two detectors or if this is more of a blind experiment to build a theory around? – AnimatedPhysics Sep 20 '13 at 17:13

## 1 Answer

You may check the Wikipedia page for neutrino oscillations

https://en.wikipedia.org/wiki/Neutrino_oscillations

where you may also find the derivation of the formula for the neutrino oscillations in the case of two flavors (in which case the oscillations are harmonic and simple)

$$P_{\alpha\rightarrow\beta, \alpha\neq\beta} = \sin^{2}(2\theta)\, \sin^{2} \left(\frac{\Delta m^2 L}{4E}\right)\quad \mathrm{(natural\,units)}$$

Numerically, in SI units, that is

$$P_{\alpha\rightarrow\beta, \alpha\neq\beta} = \sin^{2}(2\theta) \, \sin^{2}\left( 1.267 \frac{\Delta m^2 L}{E} \frac{\rm GeV}{\rm eV^{2}\,\rm km}\right).$$

You may see that the probability depends on the length $L$ in kilometers – you may write $L=ct$ using the time $t$ as well, if you wish, $c$ is the speed of light (their speed is undetectably different from $c$). However, the frequency also depends on the neutrino energy $E$ expressed in ${\rm GeV}$ above.

The factor $\Delta m^2$ is the difference between the eigenvalues of $m^2$. For the three species, these are $$\Delta m_{12}^2 = 7.59\times 10^{-5} {\rm eV}^2$$ $$\Delta m_{32}^2 \approx \Delta m_{13}^2 = 2.32 \times 10^{-3}{\rm eV}^2$$ Two of the neutrino mass eigenvalues are very close to each other. This results in very slow oscillations (solar oscillations, $12$). The remaining third eigenvalue is further from them and it results in faster oscillations (seen in atmospheric oscillations, $23$).

At any rate, the units in the formula above are designed so that $E$ is of order one (i.e. GeV). Because $\Delta m^2$ is 3-5 orders of magnitudes smaller than 1, $L$ of the wavelength is of order between thousands and millions of kilometers which translates roughly to something between milliseconds and seconds. You may calculate the precise frequency for yourself – the periodicity is proportional to the neutrino energy.

Most of these things have been known for a few decades. What was observed in the last year was a matrix element that allows the $13$ transmutation "directly". The previous solar and atmospheric oscillations de facto measured the angles and mass differences $12$ and $23$ only. The accuracy of all the parameters is rather close to one percent these days.

• What would be the frequency in the rest frame of the neutrino? Does that question even make sense? – user56903 Jul 13 '16 at 9:32
• Dear @Dirk, if you knew you have an oscillating neutrino at rest, the frequency would be $hf = (m_j-m_k)c^2$ - like for any 2-level system (where the momentum is set to zero and disappears). But that would be a state extremely hard to produce, and it would be different from the very fast neutrinos discussed above which are near $c$. You know, the funny thing about the superpositions discussed in my answer are very far from eigenstates of $v-c$ or Lorentz $\gamma$: the deviation of $v$ from $c$ is highly undefined in the state. – Luboš Motl Jul 13 '16 at 12:55
• For example, the Lorentz gamma factor is $E / m_0$. Here, the $E$ is roughly well-defined even when the neutrino isn't a mass eigenstate. But because it's not a mass eigenstate, $m_0$ is in a superposition of two very different eigenvalues. This also meansa superposition of very different amounts of boosts, and a superposition of very different ideas what is the "rest frame of the object". – Luboš Motl Jul 13 '16 at 12:57