# MSW effect for Solar Neutrinos

1. Jul 14, 2015

### ChrisVer

Can someone help me understand the following figure?

It shows how the conversion of neutrons (electron neutrinos=red, muon neutrinos=blue) is happening through the Sun for different cases of the core's density (relative to the resonance density).

I only understood the top figure, which tells me that I get an electron neutrino produced that is mainly in the $\nu_{1m}$ state, and because of matter effects, when it reaches the resonance ($\Delta m_m =min$) we get each $\nu_{im}$ being 50-50 of neutrinos. Then the $n$ keeps dropping and so the muonic component of $\nu_{1m}$ gets larger until $\nu_{1m} \rightarrow \nu_1$ (vacuum) with the small mixing angle (here zero/no oscillations only matter effects).

However in the rest diagrams, I don't understand why the initial neutrino composition is such.

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2. Jul 14, 2015

### Orodruin

Staff Emeritus
This is not what the top figure shows, it show the electron neutrino produced mainly in the $\nu_{2m}$ state. As the neutrino evolves adiabatically, when the state exits the Sun, it is essentially in the $\nu_2$ state, which only has a subdominant component of electron neutrino ($\theta_{12} \simeq 33^\circ$).

The initial matter angle depends on the relation between the neutrino energy and the matter (electron) density. If the energy is on-resonance, the electron neutrino would be 50-50 $\nu_{1m}$ and $\nu_{2m}$. Anywhere above resonance, it is mainly $\nu_{2m}$ and below resonance it goes towards the vacuum composition.

3. Jul 14, 2015

### ChrisVer

Sorry "$\nu_{1m}$" was a typo of mind (it doesn't correspond to the figure) and I meant $\nu_{2m}$.

4. Jul 14, 2015

### Orodruin

Staff Emeritus
Also, I generally prefer figure 2 of this paper for understanding the flavour composition dependence on the matter potential, but admittedly I am biased for obvious reasons. The solar resonance is that which appears around $VE \simeq 10^{-5}\ \rm{eV}^2$.

5. Jul 14, 2015

### ChrisVer

I'm trying to see those transitions from formulae...
You get the initial electron-neutrino composition from:
$|\nu_e> = \cos (\theta_m) | \nu_{1m}> + \sin (\theta_m) |\nu_{2m}>$
With
$\sin (2\theta_m) = \frac{\sin (2\theta)}{C}$
and
$C= \sqrt{[A_{CC}-\cos(2\theta)]^2+ \sin^2(2\theta)}$
$A= \frac{2 \sqrt{2}G_F n_e E}{\Delta m^2}$?

So for $n_e =n_R$ you get $\sin^2 2 \theta_m =1 \Rightarrow \theta_m = \pm \frac{\pi}{4}$ and so the $\nu_e \approx \frac{1}{\sqrt{2}}[|\nu_{1m}>+ | \nu_{2m}>]$?

Then for $n_e \gg n_R$ the $C\approx A_{CC} \gg 1$ and so $\sin 2 \theta_m \approx 0 \Rightarrow \theta_m \approx 0,\frac{\pi}{2}$ and for 0: $|\nu_e> \approx |\nu_{1m}>$ while for $\pi/2$ $|\nu_e> \approx |\nu_{2m}>$? *that's reversed*

By continuity then, the inbetween values will have a dominant $\nu_{2m}$...the opposite holds for $n_e <n_R$ until it reaches the $\theta_m= \theta \approx 33^o$.

Last edited: Jul 14, 2015
6. Jul 14, 2015

### Orodruin

Staff Emeritus
You can remove the minus sign from the $\theta_m = \pm \pi/4$, otherwise yes.

7. Jul 14, 2015

### ChrisVer

OK I think I got it, thanks