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 \nu_{1} and \nu_{2} is denoted \Delta m^2_{12}.

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 \Delta m^2_{sol}.

Am I right in thinking that whenever I see \Delta m^2_{sol} it will always mean \Delta m^2_{12} 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 \Delta m_{sol}^{2} refers to \Delta m_{12}^{2} and not some other mass^2 difference ?

What I mean is, the sun's reactions produce \nu_{e} 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 \nu_{\mu} or \nu_{\tau} since they lack the energy required to produce the relevant charged lepton. That means you don't know whether they are turning mostly to \nu_{\mu} or \nu_{\tau}.

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

\rvert \nu_{e} \rangle = cos\theta_{12}cos\theta_{13} \rvert \nu_{1} \rangle + <br /> sin\theta_{12}cos\theta_{13} \rvert \nu_{2} \rangle + <br /> sin\theta_{13}e^{-i\delta} \rvert \nu_{3} \rangle

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

\rvert \nu_{e} \rangle = cos\theta_{12} \rvert \nu_{1} \rangle + <br /> sin\theta_{12} \rvert \nu_{2} \rangle

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

 

[AdrianTheRock]

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

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

 

[Doofy]

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

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

is it still a valid analysis given that we now know that theta_{13} 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 \theta_{13}, not \theta_{12}.