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 $\theta_{12}$.