Neutrino Oscillation: Mass Differences

[Doofy]

In neutrino oscillation the probability a neutrino changing its flavour depends on the difference between the squares of the masses of the neutrino mass eigenstates. For example, the squared-mass difference between the mass states [itex]\nu_{1}[/itex] and [itex]\nu_{2}[/itex] is denoted [itex] \Delta m^2_{12} [/itex].

However, I keep reading stuff that refers to the neutrino source used in the experiment when it talks about the mass difference, for example, in solar neutrinos it is [itex] \Delta m^2_{sol} [/itex].

Am I right in thinking that whenever I see [itex] \Delta m^2_{sol} [/itex] it will always mean [itex] \Delta m^2_{12} [/itex] etc.?

 

[AdrianTheRock]

I can't immediately locate a definitive answer, but I think you are right for Δm²_sol. But one textbook I have uses the definition

Δm²_atm = m₃² - 1/2 (m₁² + m₂²)​

 

[Doofy]

I can't immediately locate a definitive answer, but I think you are right for Δm²_sol. But one textbook I have uses the definition

Δm²_atm = m₃² - 1/2 (m₁² + m₂²)​

I don't suppose you know why it is that [itex] \Delta m_{sol}^{2} [/itex] refers to [itex] \Delta m_{12}^{2} [/itex] and not some other mass^2 difference ?

What I mean is, the sun's reactions produce [itex] \nu_{e} [/itex] and fewer of them arrive at Earth than expected, implying oscillation is happening. However, they only have a few MeV of energy, so when these solar neutrinos reach a detector, they cannot undergo CC interactions as [itex]\nu_{\mu} [/itex] or [itex]\nu_{\tau} [/itex] since they lack the energy required to produce the relevant charged lepton. That means you don't know whether they are turning mostly to [itex]\nu_{\mu} [/itex] or [itex]\nu_{\tau} [/itex].

Am I right in thinking that, since you can express [itex] \nu_{e} [/itex] as

[itex] \rvert \nu_{e} \rangle = cos\theta_{12}cos\theta_{13} \rvert \nu_{1} \rangle +
sin\theta_{12}cos\theta_{13} \rvert \nu_{2} \rangle +
sin\theta_{13}e^{-i\delta} \rvert \nu_{3} \rangle [/itex]

you can approximate [itex] sin\theta_{13} = 0 [/itex] and [itex] cos\theta_{13} = 1 [/itex] so that you just deal with

[itex] \rvert \nu_{e} \rangle = cos\theta_{12} \rvert \nu_{1} \rangle +
sin\theta_{12} \rvert \nu_{2} \rangle [/itex]

and just neglect any oscillation to [itex]\nu_{\tau} [/itex], ending up with a two-neutrino treatment where the only parameters you have are [itex] \Delta m_{12}^{2}, \theta_{12} [/itex]?

 

[AdrianTheRock]

Yes, that's exactly why [itex]\Delta m^2_{sol}[/itex] means [itex]\Delta m^2_{12}[/itex].

With atmospheric neutrinos you are starting with [itex]\nu_\mu[/itex], so even with the approximation [itex]\theta_{12} = 0[/itex] you still have to take account of the [itex]\nu_3[/itex] state.

 

[Doofy]

Yes, that's exactly why [itex]\Delta m^2_{sol}[/itex] means [itex]\Delta m^2_{12}[/itex].

With atmospheric neutrinos you are starting with [itex]\nu_\mu[/itex], so even with the approximation [itex]\theta_{12} = 0[/itex] you still have to take account of the [itex]\nu_3[/itex] state.

is it still a valid analysis given that we now know that [itex]theta_{13}[/itex] is non-zero though?

 

[AdrianTheRock]

Given the relatively low levels of precision currently available in experimental measurements, I imagine it's still a reasonable approximation.

BTW apologies for the typo in my previous post, I did of course mean [itex]\theta_{13}[/itex], not [itex]\theta_{12}[/itex].