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Karlsruhe neutrino mass experiment has been turned on

The Washington Post tells us that the Karlsruhe Tritium Neutrino Experiment (KATRIN) was turned on today.

Karlsruhe looks luxurious. At least the 1715 palace does. Just to be sure, the city is just miles East from the Easternmost point of France.

The experiment costs €60 million i.e. $66 million and has the chance – perhaps the greatest chance among the experiments in the near future – to measure the absolute masses of the neutrinos.

The heart of the experiment is a spectrometer that weighs 200 SI tons. This spectrometer will carefully measure the energy of the electrons that are emitted in a beta decay.

The focus will be on the electrons that have the maximum kinematically allowed energy. In those decays, the smallest amount of energy is left for the neutrinos. By carefully studying the maximum energy of the electrons, within \(1\eV\) or several of them, the experiment will determine the minimum energy of the neutrinos i.e. their rest mass.

Because one wants to have as low-energy electrons as possible, so that the hypothetical neutrino mass is the highest possible fraction of this energy, they picked the tritium (very heavy hydrogen) whose beta-decay leaves a very small amount of energy for the electron and the neutrino. Just to be sure, the total energy of the electron and the antineutrino is \(18.6\keV\) which is very low in comparison with the energies in nuclear physics – we usually talk about several \({\rm MeV}\).

The spectrometer when it was transported 10 years ago.

Note that because the neutrino oscillations have been observed to exist, we know that the neutrinos generally have nonzero masses (which must ultimately come from some non-renormalizable interactions beyond the Standard Model!). However, the oscillations only allow us to determine the differences of the squared mass eigenvalues such as\[

\Delta m_{ij}^2 = m^2_i - m^2_j, \quad i,j=1,2,3

\] and not the absolute shift. The actual lowest eigenvalue of \(m_i^2\) could be much smaller than both differences of the squared masses (and perhaps even zero), or comparable to the lower difference, or comparable to the larger one, or much greater than the differences. We just don't know. If the relevant neutrino mass is significantly larger than \(1\eV\), the experiment should be able to see it. One might guess that this great outcome is unlikely because the square roots of \(\Delta m_{ij}^2\) are closer to \(0.01\eV\) or even \(0.001\eV\). Neutrinos much heavier than \(1\eV\) would mean that all the mass eigenstates are almost degenerate. It's possible but such a degeneracy could be considered unnatural and unlikely.

Even in the optimistic case, there will probably be some uncertainty about the neutrino masses because the experiment will probably not fully settle the question whether the neutrino masses are Dirac, Majorana, or mixed.

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