I was reading various reviews on neutrino physics but I couldn't understand the following to complete satisfaction.

How is $\theta_{12}$ identified with the Solar mixing angle and $\Delta m^2_{21}$ the Solar mass-squared difference?

The formulae for three-flavor oscillation probabilities, in general, depends on all three mixing angles $\theta_{12},\theta_{23}$ and $\theta_{13}$:

$$P_{\alpha\rightarrow\beta}=\delta_{\alpha\beta} - 4 \sum_{i>j} {\rm Re}(U_{\alpha i}^{*} U_{\beta i} U_{\alpha j} U_{\beta j}^{*}) \sin^2 \left(\frac{\Delta m_{ij}^2 L}{4E}\right) \\ + 2\sum_{i>j}{\rm Im}(U_{\alpha i}^{*}U_{\beta i}U_{\alpha j} U_{\beta j}^{*}) \sin\left(\frac{\Delta m_{ij}^2 L}{2E}\right).$$

It's not clear to me why only one mixing angle be associated with solar neutrino oscillation. An answer starting from the three-flavour oscillation formulae will be helpful for my understanding.

  • $\begingroup$ Related: physics.stackexchange.com/questions/270727/… $\endgroup$ – dmckee --- ex-moderator kitten Jun 28 '17 at 16:54
  • $\begingroup$ Too bad there is not official mechanism to combine good answers into a comprehensive one because with all the answers to that very same question, each stressing particularly well a specific aspect of it, we would have the mother of all answers on the subject!!! $\endgroup$ – user154997 Jul 1 '17 at 19:17

The most important effect in estimating the anticipated production of neutrino types out of the sun is the MSW effect, resonant conversion in matter, and the estimate is a complicated business, but apparently you are not interested in that, in this question.

The point, well made before in @PascExchange 's answer, is sheer numerical accident. Since L, the distance from the sun, is so enormous, the oscillatory factors (see below) in the transition formula in vacuum are dominated by the smallest $\Delta m^2$s: in this case, the 1-2 mass-squared-difference is two orders of magnitude smaller than the other two.

This fortunate accident makes 1-3 and 2-3 oscillation frequencies a hundred times higher, so the corresponding trig functions average out to zero. An intermediate expression reflecting this fact is oscillation equations

Proceed to neglect the very small $\theta_{13}$, and you are there, as only $S_{12}$ survives.

  • $\begingroup$ "Proceed to neglect the very small $\theta_{13}$, and you are there, as only $S_{12}$ survives." Why do you say that? Even when $\theta_{13}$ is neglected, your equations (30)-(40) still contain $\theta_{23}$. $\endgroup$ – SRS Jul 14 '18 at 13:00
  • $\begingroup$ Adding (39) to (40) in that limit to subtract from 1, yields the persistence probability characteristically depending on $\theta_{12}$, and not $\theta_{23}$, as implied. $\endgroup$ – Cosmas Zachos Jul 14 '18 at 13:51

In order to answer this question let me go back a bit in the discussion about solar neutrinos. The process, which keeps our sun alive can basically be described as $$p+p\to\ ^2_1\text{H}+e^++\nu_e$$ $$D+p\to\ ^3\text{He}+\gamma$$ The important fact for our current discussion nonetheless is, that during this process we produce electron neutrinos $\nu_e$.

Using the Luminosity, the distance between the sun and the earth and as the energy being released by one single such fusion process we can calculate the expected neutrino flux on earth. Various experiments, covering different Neutrino energies and allowing to detect $\nu_\mu$ and $\nu_e$ now showed, that the there is only half of the expected amount measured. The Sudbury Neutrino Observatory (SNO) now included the capability to measure $\nu_\tau$ by using heavy water instead of normal water, where-after the expected flux was measured. Thus we can conclude that the lack is actually an observation of $\nu_e \to \nu_\mu\to \nu_\tau$ aswell as $\nu_e\to \nu_\tau$ oscillation. To understand why we still claim that we get one squared mass difference out of that can be understood when consulting the particle physics booklet. Here we find the following values for the squared mass differences between the two generations $\Delta m_{12}^2=(7.53\pm 0.18)\cdot 10^{-5}$ and $\Delta m_{23}^2=(2.52\pm 0.05)\cdot 10^{-3}$. We realize that the difference $\Delta m_{12}$ is quite small compared to $\Delta m_{23}$ and therefore we can only distinguish between the first two, respectively the second and third generation and therefore we can only regard the $\mu_e\to \mu_\tau$ part.


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.