Neutrinos produced in the interaction of very high energy cosmic rays with the CMB are produced in the proportion $\nu_\mu:\nu_e:\nu_\tau=2:1:0$

because of the decays produced after the cosmic ray and the photon. If they are produced far enough from the Earth, the proportion when they get here at the Earth is 1:1:1. I wanted to know why does the neutrino mixing produced a 1:1:1 proportion after a long-enough time.


The measured flavor content of the beam as a function of distance from the source depends on (a) the initial composition of the beam and (b) some set of transition probabilities $P_{\alpha,\beta}$ from flavor $\alpha$ to flavor $\beta$.

Let's consider some properties of those probabilities:

  • They are time reversal invariant. That is for a given set of physical conditions $P_{\alpha,\beta} = P_{\beta,\alpha}$, so there can exist an equilibrium in which the three flavors remain in the same proportion as time passes.

  • Second the individual probabilities mentioned above are constructed as squares of sums of terms proportional to $$P_{\alpha,\beta} \propto \left| \sum_{i,j} \sin \left(\frac{\Delta m_{ij}^2 \,L}{4E}\right) \right|^2\;,$$ where $\Delta m_{ij}^2$ is the difference of squared masses between states $i$ and $j$, $L$ is the propagation distance since the source and $E$ is the neutrino energy. For a given range and energy this means you expect that transition probabilities would rise and fall along the track, but ...

  • The neutrinos necessarily have a range of energies, so the peaks and valleys in the probability distributions occur at different ranges for different energies. For visualization purposes notice that $\lambda(E) = 8\pi E/\Delta m^2$ is a smooth function of energy over some range. The result is that for the whole beam incident on a energy-sensitive detector $$ P_{\alpha,\beta}^\text{beam} = \int \mathrm{d}E\, P_{\alpha,\beta} \;, $$ is a smeared out by the sum over many wavelengths.

  • Finally, neutrinos are massive particles, so there should be some dispersion, thought that is a complicated issue in and of itself.

The overall results is that even for precisely defined energy bands ranges $L$ at which one transition dominates are rare, and when you consider a population with a wide range of energies you have no expectation of finding a particular flavor dominate.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.