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MSW effect for Solar Neutrinos

  1. Jul 14, 2015 #1

    ChrisVer

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    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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  3. Jul 14, 2015 #2

    Orodruin

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    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.
     
  4. Jul 14, 2015 #3

    ChrisVer

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    Sorry "##\nu_{1m}##" was a typo of mind (it doesn't correspond to the figure) and I meant ##\nu_{2m}##.
     
  5. Jul 14, 2015 #4

    Orodruin

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    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##.
     
  6. Jul 14, 2015 #5

    ChrisVer

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

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

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

    By continuity then, the inbetween values will have a dominant [itex]\nu_{2m}[/itex]...the opposite holds for [itex]n_e <n_R[/itex] until it reaches the [itex]\theta_m= \theta \approx 33^o[/itex].
     
    Last edited: Jul 14, 2015
  7. Jul 14, 2015 #6

    Orodruin

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    You can remove the minus sign from the ##\theta_m = \pm \pi/4##, otherwise yes.
     
  8. Jul 14, 2015 #7

    ChrisVer

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    OK I think I got it, thanks :biggrin:
     
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