# Energy/Momentum transfer from neutrinos to black holes

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1. Jan 16, 2015

### Larry Pendarvis

Suppose you have a source of electron antineutrinos, and you arrange your apparatus so that a billion billion billion of them collide directly with a black hole. In principle, you could measure the change in momentum and energy from that occurrence.
Suppose you did that the next day. According to current theory, would you always get the same result?
Suppose then you back off a few miles and do the experiment again, with the same number of neutrinos. Would your measurements give the same energy and momentum change as the first time?

2. Jan 16, 2015

### Staff: Mentor

There is no evidence that the laws of physics or fundamental constants would change over time. Why do you expect a difference? The black hole might be a tiny bit larger or smaller, or move in a different way, depending on what happened in the meantime.
If all of your neutrinos hit, why do you expect a difference? This is just conservation of energy and momentum.

3. Jan 16, 2015

### Larry Pendarvis

That is exactly what I would expect.
In that case, would you say that we have measured the energy and the momentum of those neutrinos, and therefore the rest mass; or only some average that is indicative of a definite rest mass; or only something useless because there IS no definite rest mass?

4. Jan 17, 2015

### Orodruin

Staff Emeritus
As you have been told repeatedly, electron neutrinos do not have a definite rest mass. What is an electron neutrino is a particular superposition of the neutrinos that do have definite masses, the neutrino mass eigenstates. Any experiment you do by creating "electron neutrinos" will measure some effective average of the neutrino mass eigenstate masses. This does not mean that the electron neutrino has a definite mass.

If you do an experiment that is accurate enough to separate the neutrino mass eigenstates so that they no longer are in a coherent superposition, then you would consider the mass eigenstates separately, much in the same way as different quarks are considered separately while having some off-diagonal flavour interactions due to quark mixing. There really is not anything else to this apart from the neutrino masses being so close together that this leads to coherent interference with the mass eigenstates acquiring different phases during propagation.

In order to start understanding at a more basic level, I suggest studying quantum field theory and reading up on the theory behind neutrino masses and oscillations. Some relevant references are:
Giunti, Kim, Lee, Phys.Lett. B274 (1992) 87-94
Akhmedov, Kopp, arXiv:1001.4815

In the end, you can only get so far by popularised versions of the actual physics and it is often dangerous to take these explanations beyond their context.

5. Jan 17, 2015

### Larry Pendarvis

What kind of measurement are you imagining here? If you measure the spin, you will get a definite spin. If you try to measure the rest mass, I should think that each measurement will give one definite result, but if you do many such measurements you would get various results because each measurement is of a superposition. Thus your measurement would correspond to one mass eigenstate one time, and another some other time. Is that what you are saying? I can't think you are saying that any measurement of one neutrino would give more than one eigenstate by separating them.

6. Jan 17, 2015

### Orodruin

Staff Emeritus
The electron neutrino does not have a definite mass as it is a superposition of the three neutrinos that do. The flavour states, and thus neutrino oscillations, arise due to the fact that these three neutrinos have so similar masses that they tend to keep coherence for a very long time and/or distance. You could decide to talk about neutrinos as the states of definite masses just as you do in the quark sector but it is not convenient to do so just because of the fact that the massive neutrino states keep coherence for a long time. This is the only reason we tend to talk about flavour states, nothing else.

If you did create electron neutrinos (which is by definition what you do in a weak charged current interaction involving an electron or positron) what is created is a linear combination of the different mass eigenstates. This linear combination is generally coherent, i.e., the masses are so close that the energy uncertainties in the other particles involved are not small enough to resolve it. If you could create a device that measured neutrino masses with such a precision that you could separate the mass eigenstates, you would sometimes produce a $\nu_1$ and sometimes a $\nu_2$ (and, very rarely, a $\nu_3$) in conjunction with the electron/positron weak CC interaction. If you have that kind of precision, you would not be talking about an electron neutrino, you would be talking about the branching ratios to the mass eigenstates, just as you do in the quark sector with the charm quark having some probability of weak decay into a strange quark and some smaller probability of weak decay into a down quark. In the quark case, the branching ratios are given by the Cabibbo angle, in the neutrino case, it is (mainly) the lepton mixing angle $\theta_{12}$.

Edit: LaTeX typo.

Last edited: Jan 17, 2015