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Neutrinos change flavor, energy transfer

by rrogers
Tags: energy, flavor, neutrinos, transfer
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rrogers
#1
Jan18-14, 09:31 AM
P: 80
I have a similar question. Since the neutrino's has a rest mass I can observe it in a SR frame where it is at rest. Now if I measure the mass (say by energy absorption inside a light trap) at one time. To the degree necessary to distinguish it's state from the other neutrino states I get a number and corresponding energy. Now if I measure it later when it has morphed into another flavor; where is the extra/deficit energy? Or is it, that once I determine the mass the flavor is fixed forever?
My knowledge of QM is such that the link "this thread" is difficult. Is there a possibility of a simple answer to my experiment? I do realize this sort of thing apparently happens elsewhere in QM but I don't understand that either and there seem to counterbalancing energy carriers in the cases I know about.
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mfb
#2
Jan18-14, 09:55 AM
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Please start a new thread if the old one is so old.

Usually neutrinos are created as flavor eigenstates (example: "electron neutrino"). This is a mixture of the three mass eigenstates - there is no reference frame where all three are at rest. You can pick a frame where one mass eigenstate is at rest. The individual mass eigenstates have no interesting time-evolution, the mixture of all 3 generates the interesting effects (=distance-dependent mixing).

Now if I measure it later when it has morphed into another flavor; where is the extra/deficit energy?
There is no extra or deficit energy. You always get back the energy you used to produce the particle. Note that "the mass of an electron neutrino" is a very problematic concept.
ChrisVer
#3
Jan18-14, 10:44 AM
P: 751
measuring neutrinos is kind of difficult, so is their masses (that's why we are still on boundaries and not having exact masses for them). What we have measured is the differences of their squared masses, so we are talking for them being massive.
Now as already said, the flavor eigenstates are not the same as the mass eigenstates. That also means that you cannot diagonalize simultaneously the mass and the flavor (their interactions) matrix terms in the Lagrangian. So you have to choose a basis (mass or flavor). Once you have the one, the other can be written as a linear combination of the first (in the same way you work with QM once you take a basis eg position and write it as a linear combination of things that it doesn't commute with eg momentum via fourier transform).

So you measure the mass? you can't distinguish what flavor your neutrino is- it's a linear combination. Same for flavors. So you will always write:
[itex]v_{a}= Σ_{i} U_{ai} u_{i}[/itex]
where [itex]v_{a}[/itex] are the flavor e.states, and [itex]u_{i}[/itex] are the mass e.states, and U being the mixture matrix. So what you measure is the electron neutrino, this will be:
[itex]v_{e}= U_{e1} u_{1}+U_{e2} u_{2}+U_{e3} u_{3}[/itex]

rrogers
#4
Jan18-14, 03:17 PM
P: 80
Neutrinos change flavor, energy transfer

Okay, this fits in with what little I know. I guess E=mc^2 isn't quite true in any case since a photon has energy but no mass. It fits into a different place in the stress-energy tensor. So if I place a neutrino in a trap, say between two masses, and then weighed the total the weighing would vary in time. Somewhat like an interference pattern in time.
Apologies for such an absurd measurement :)
Thanks again
ChrisVer
#5
Jan18-14, 03:43 PM
P: 751
I am sorry, I am not sure I can understand what you have in your mind by trap. I don't know if someone more specialized in neutrino measurements can help more, but according to my understanding this is impossible.
You can't trap neutrinos, because they can interact only via weak interactions, and that is uncommon to happen. But suppose you can (let us say) capture a neutrino, and keep it localized in a vollume/box of V. And you tried to weight it. Why do you say its weight would vary?
The total energy doesn't change.
rrogers
#6
Jan18-14, 05:42 PM
P: 80
Well I think it might be conceivable in something like a Lagrangian point; gravity is hard to ignore.
"Why do you say its weight would vary?"
Well what is meant by a neutrino's "mass"? I guess this is the heart of my problem.
Surely we don't have a case where inertial mass is different than gravitational mass?
ChrisVer
#7
Jan18-14, 07:14 PM
P: 751
in general, the Standard Model is not built on consideration of gravity.
So in general, we neglect gravity in physics under GUT energy scale, because it's very weak.
Neutrinos having mass, in general, means that neutrinos' left and right components (in the Weyl representation of Dirac spinors) interact with the Higg's vev and gain mass, through an interactive term in the Lagrangian called "Yukawa term". This thing leads in some extended theories of standard model (for example theories of very heavy right handed neutrinos, with mass equal [itex]M_{GUT}\approx 10^{16}GeV\approx 10^{-11}kg [/itex] or [itex]10 [/itex] nanogram-something enormous for pointlike particle). Why is their mass important? For some reasons, probably someone could write a list. Rest mass is important because it helps you in dealing kinematically with interactions... another reason could be that it help you add it up to the total mass of the universe... maybe even more.

But if you expect for a neutrino to interact gravitationally, I would suggest you'd have to wait a lot-leaving aside the fact that we don't know "how"-. Neutrinos for bad or good luck are hardly ever interacting weakly, and that's why they are so difficult to deal with. Now expecting to detect a force that would be some orders of magnitude less "strong" than the weak interaction - you'd need enormous amount of energies to look at it.

I guess the biggest natural mechanism of neutrino interactions in our universe, is the mechanism which initiates the Supernovae (neutrinos create a shockwave strong enough to expand the layers of the star). And that is not done gravitationally but via weak interactions.
mfb
#8
Jan19-14, 12:58 PM
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A captured neutrino would probably be in a mass eigenstate - otherwise I don't see how you want to capture it (even in theory). Then you can measure its mass as often as you like (again, just in theory...), it will keep the same mass.
rrogers
#9
Jan19-14, 01:44 PM
P: 80
Okay I give. I will do some personal research :) I think the above, and other posts, provide enough guidance to guide my ADD mind in some consistent direction. The meaning of energy conservation, particle eigenstates and evolution, and the relevant Lagrangian. I am really very weak in QM; a couple of classes and about 10 books I have read the first couple of chapters in.
snorkack
#10
Jan20-14, 03:58 AM
P: 381
Quote Quote by mfb View Post
Please start a new thread if the old one is so old.

Usually neutrinos are created as flavor eigenstates (example: "electron neutrino"). This is a mixture of the three mass eigenstates - there is no reference frame where all three are at rest. You can pick a frame where one mass eigenstate is at rest.
How can you create a neutrino that does not also have a definite rest mass?
jtbell
#11
Jan20-14, 06:49 AM
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The other products of the interaction that produces the neutrino have correspondingly indefinite momentum and energy.
mfb
#12
Jan20-14, 12:39 PM
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Quote Quote by snorkack View Post
How can you create a neutrino that does not also have a definite rest mass?
Radioactive decays do this all the time as they produce flavor eigenstates.
snorkack
#13
Jan20-14, 03:58 PM
P: 381
Quote Quote by mfb View Post
Radioactive decays do this all the time as they produce flavor eigenstates.
But how do the products of the decay acquire uncertainty of their own mass?
mfb
#14
Jan20-14, 04:05 PM
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They are produced in a superposition of the mass eigenstates.
This is just quantum mechanics. Superpositions are nothing special.
snorkack
#15
Jan20-14, 04:28 PM
P: 381
Quote Quote by mfb View Post
They are produced in a superposition of the mass eigenstates.
This is just quantum mechanics. Superpositions are nothing special.
Do (stable nuclei, ground state) products of beta decay possess multiple mass eigenstates to match the multiple mass eigenstates on neutrino?
mfb
#16
Jan20-14, 04:42 PM
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No, there is no need to have "matching masses" in any way (the masses don't match anyway, there is energy released in the process). Different kinetic energies, corresponding to those different mass eigenstates, are fine, as just energy has to be conserved.
jtbell
#17
Jan20-14, 06:06 PM
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Quote Quote by mfb View Post
as just energy has to be conserved.
...and momentum, of course.

The kinetic energies and momenta of the beta-decay electron (or positron) and remaining (recoiling) nucleus are slightly different for different neutrino masses. For that matter, so is the kinetic energy and momentum of the outgoing neutrino.

Therefore, for three different neutrino masses, the entire final state of the decay (neutrino, electron and nucleus) is a superposition of three states with (very) slightly differing neutrino mass, and with (very) slightly differing kinetic energies and momenta for the electron and nucleus.
ChrisVer
#18
Jan20-14, 07:52 PM
P: 751
The problem with the mass and flavor eigenstates can easily be seen after you look at the corresponding Lagrangian containing the interaction and the mass terms- all these written in matrix forms.
There you have matrices multiplying your states. You can do a unitary transformation on your states. That unitary transformation can diagonalize either the interaction matrix (so the states are eigenstates of flavor) or the mass matrix (so the states are eigenstates of the mass). The problem is that you cannot diagonalize those two simultaneously for the neutrinos, so the states that you can acquire cannot be simultaneously eigenstates of both.

This is not something new for someone who has studied classical quantum mechanics either. When you have two operators A,B they can have the same eigenstates if they commute. The commutation means that you can choose a basis such that (in that basis) the matrix representations of A and B will be diagonal. When you can't do that, and [A,B] is not trivial, you can only write one's eigenstates as a superposition of the other's.


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