why in the case of MINOS experiment (for neutrino mass detection) the squared mass difference of neutrino 3 and 2 is a absolute value(mod) compared with KamLand experiment?


The experiments we've done so far are sensitive to the difference of the squares of the masses $\Delta m_{ij}^2 = m_i^2 - m_j^2$, not the square of the differences $\left( m_i - m_j \right)^2$. As such they can be negative. At first glance it appear that we should be sensitive to the sign of these values, but they come into the experimental observables squared. That is, we're actually sensitive to the square of the difference of the squares of the masses $\left(\Delta m_{ij}^2\right)^2$.

Our measurements so far give us two values, a small one (first found in solar neutrino studies) and a large one (first identified in atmospheric studies). \begin{align*} \Delta m_\text{small}^2 = \Delta m_\text{sol}^2 &\approx 7.6 \times 10^{-5}\,\mathrm{eV}^2 \\ \Delta m_\text{large}^2 = \Delta m_\text{atm}^2 &\approx 2.4 \times 10^{-3}\,\mathrm{eV}^2 \\ \end{align*} The third mass difference must be very similar to the one listed here as "large".

The numbering of the mass states is somewhat arbitrary, because we never observe them in an experiment (we always observe weak interaction, which select for the flavor states), so by convention we assign the label 1 to the lighter state that participates in the small mass difference, and the label 2 to the other state participating in the small mass difference. This sets the sign of $\Delta m_{12}^2$ to be positive: $m_1 < m_2$.

Finally, there are two ways the actual masses could be ordered:

  • $m_1 < m_2 < m_3$ which is called the "normal hierarchy"
  • $m_3 < m_1 < m_2$ which is called the "inverted hierarchy"

Because we don't know which actually obtains, the sign of the large mass difference is not actually known.

Thus we write $$ \Delta m_{21}^2 = \Delta m_\text{sol}^2 \approx 2.4 \times 10^{-3}\,\mathrm{eV}^2\,, $$ but one of \begin{align*} \left| \Delta m_{32}^2 \right| &= \Delta m_\text{atm}^2 \approx 7.6 \times 10^{-5} \,\mathrm{eV}^2 \quad \text{or}\\ \left| \Delta m_{31}^2 \right| &= \Delta m_\text{atm}^2 \approx 7.6 \times 10^{-5} \,\mathrm{eV}^2\,. \end{align*}

There is a pretty good chance that $\text{NO}\nu \text{A}$ (currently running from Fermilab to Soudan mine in Minnesota) will return a usable result on which hierarchy obtains, and these absolute error bars will be removed from the next generation of posters.

  • $\begingroup$ It is probably obvious, but you confused the $\Delta m_{21}^2$ value with the $\Delta m_{32}^2$/ $\Delta m_{31}^2$ value at the close of your answer (solar is $\sim 7.5 \, \times 10^{-5} \,\mathrm{eV}^2$. $\endgroup$ – Dalton Bentley Jun 7 '20 at 18:25

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