I Neutrino flavour eigenstates and expansion of the universe

[Carlos L. Janer]

Neutrinos were flavor eigenstates at the time of their decoupling from baryonic matter. Since they were not pure mass eigenstates, how do you take this fact into account if you try to study how they evolved as the universe expanded?

Could we determine if the heaviest neutrino could be non relativistic at present times?

Can we even try?

 

[mfb]

The flavor eigenstates are superpositions of the mass eigenstates, and you can simply follow the mass eigenstates over time.
The heaviest neutrino has a mass of at least 50 meV (mixing) and of at most ~70 meV (cosmological constraints), at its present temperature of 2 K = 0.17 meV these neutrinos are moving at a few percent the speed of light.

 

[Carlos L. Janer]

he flavor eigenstates are superpositions of the mass eigenstates, and you can simply follow the mass eigenstates over time.
The heaviest neutrino has a mass of at least 50 meV (mixing) and of at most ~70 meV (cosmological constraints), at its present temperature of 2 K = 0.17 meV these neutrinos are moving at a few percent the speed of light.

Thanks for the info.

There's something I don't get, though. Should the heavier neutrinos not be unstable and decay into the lightest one? If the're not flavour eigenstates I do not see what could prevent that from happening.

 

[mfb]

What would be the corresponding decay process?

 

[Carlos L. Janer]

EM? Two low energy photons? Not saying I'm right, just asking why I'm wrong.

 

[Vanadium 50]

Should the heavier neutrinos not be unstable and decay into the lightest one?

The lifetime is so long that there may not have been enough time for a single neutrino in the universe to decay.

 

[Vanadium 50]

Two low energy photons?

Violates conservation of angular momentum.

 

[Carlos L. Janer]

The lifetime is so long that there may not have been enough time for a single neutrino in the universe to decay.

Does your assertion indicate that you have a decay process in mind? I'm asking, not pretending to know. What would the point in that be?

 

[Carlos L. Janer]

Violates conservation of angular momentum.

Why do the two photons have to be exactly equal and propagate in opposite directions?

I keep telling you that if I'm asking is because I don't know.

 

[Vanadium 50]

Why do the two photons have to be exactly equal and propagate in opposite directions?

At what point did I say that? It's bad enough that you come here pushing fringe theories, but don't put words in my mouth.

 

[Carlos L. Janer]

At what point did I say that? It's bad enough that you come here pushing fringe theories, but don't put words in my mouth.

I honestly don't have any idea what you are talking about.

Whenever I have something to say I just say it straight away. Life is already complicated enough to try to double guess what other people may have in their minds and are, allegedly, not willing to tell.

I think straight, I walk straight and I talk straight. Do not try to find any hidden message because there is none.

 

[PeterDonis]

I honestly don't have any idea what you are talking about.

Here's what you said:

Why do the two photons have to be exactly equal and propagate in opposite directions?

@Vanadium 50 did not say the two photons had to be exactly equal and propagate in opposite directions. Your question, just quoted, implies that he did say that. That's why he objected to it.

 

[kimbyd]

EM? Two low energy photons? Not saying I'm right, just asking why I'm wrong.

Photons only interact with charged particles. Neutrinos aren't charged.

Such a decay process, if it were to exist, would require either W or Z boson intermediaries, and would thus be suppressed dramatically by the very small mass differences between neutrino flavors.

Btw, I found this paper on neutrino decay:
https://arxiv.org/abs/1208.4600

 

[Carlos L. Janer]

Photons only interact with charged particles. Neutrinos aren't charged.

Well, this reference considers posible neutrino electromagnetic interactions. I's beyond me, at the moment, but I will post the reference in case you can tackle it (of course, provided that you're interested in the subject):
https://arxiv.org/abs/1403.6344

 

[PeterDonis]

this reference considers posible neutrino electromagnetic interactions

None of which have been observed; as the introduction to the paper says, we have no evidence for the existence of neutrino electromagnetic interactions. So we also have no evidence that heavier neutrinos could decay into lighter ones by such a process.

 

[Carlos L. Janer]

None of which have been observed; as the introduction to the paper says, we have no evidence for the existence of neutrino electromagnetic interactions. So we also have no evidence that heavier neutrinos could decay into lighter ones by such a process.

I am not even sure that I should be answering your remarks.

If there are neutrino electromagnetic decays, the outgoing photons would be of very low energy and, therefore, extremely dificult to detect.

So, what are your implying? Do you think that both authors, Carlos Giunti and Alexander. I Studenikin are a couple of incompetent researchers? Is it an heresy to question the validity of the Standard Model of Particle Physics and look for Physics beyond it? Should their paper citations be banned from this forum? Have you asked your colleague Orodruin, who works in neutrino physics, if all his papers conform to the scientific consensus on Fundamental Physics?

Why are you such an adamant defender of the SM of Particle Physics and de LambdaCDM-model, what do you like about fine tuning so much? They are, undoubtedly, the best theories we have. But they're not good enough.

 

[mfb]

Btw, I found this paper on neutrino decay:
https://arxiv.org/abs/1208.4600

They only discuss experimental lower bounds on the lifetime, but don't discuss new physics models that could lead to these decays.

If there are neutrino electromagnetic decays, the outgoing photons would be of very low energy and, therefore, extremely dificult to detect.

meV, above to the cosmic microwave background. A large source of meV photons would be notable.

We cannot rule out neutrino/photon interactions, and at loop-level we have them even in the SM, but they have to be extremely weak. And that is elastic scattering - I still don't see how you would get a decay.

 

[PeterDonis]

If there are neutrino electromagnetic decays, the outgoing photons would be of very low energy and, therefore, extremely dificult to detect.

If such a decay took place now, yes, this is correct. But if, as you hypothesize, a heavier neutrino decayed into a lighter neutrino by such a mechanism, we wouldn't have to detect the photons to know it took place; we could just detect the change in the neutrino itself.

Also, as @mfb points out, if such decays had taken place in the early universe we would see the EM radiation from them, because it would have a significantly different spectrum from the CMBR and would be more intense. So if we can detect the CMBR, we would be able to detect the radiation from a significant number of such decays.

what are your implying?

That, as the paper you linked to explicitly says, we have no evidence of neutrino electromagnetic interactions. And that, as noted above, if such interactions had taken place to a significant extent in the early universe, we would expect to have evidence of them.

As for the rest of your post, I have not made any of the claims you attribute to me.

 

[Carlos L. Janer]

If such a decay took place now, yes, this is correct. But if, as you hypothesize, a heavier neutrino decayed into a lighter neutrino by such a mechanism, we wouldn't have to detect the photons to know it took place; we could just detect the change in the neutrino itself.

I am not aware of any calculation of heavy neutrinos lifetime in that paper. But anyhow I am pretty sure that the relic neutrino bakgroud emission will never be detected.

Also, as @mfb points out, if such decays had taken place in the early universe we would see the EM radiation from them, because it would have a significantly different spectrum from the CMBR and would be more intense. So if we can detect the CMBR, we would be able to detect the radiation from a significant number of such decays.

Wrong! The EM radiation temperature would have been tiny because it depends on the neutrinos mass differenes and I hope that I do not have to remind you that the photon-matter decoupling happened at z=1100 (T=3000K). That radiation wavelength at present times (caused by the hypothetical heavy neutrinos decay) might even be larger that the size of our observable universe.

I have not made any of the claims you rather hysterically attribute to me

What on Earth do you think it gives you the right to insult me!

 

[mfb]

But anyhow I am pretty sure that the relic neutrino bakgroud emission will never be detected.

PTOLEMY tries to see the cosmic neutrino background.

Wrong! The EM radiation temperature would have been tiny because it depends on the neutrinos mass differenes and I hope that I do not have to remind you that the photon-matter decoupling happened at z=1100 (T=3000K). That radiation wavelength at present times (caused by the hypothetical heavy neutrinos decay) might even be larger that the size of our observable universe.

Be careful with statements like that. You are wrong here.

The absolute mass difference might be small, but relative to the neutrino masses it is large (O(1)). No matter when the process happens, the photons would carry a significant fraction of the neutrino energy. Their energy goes down at the same rate as the neutrino energy (as long as they are relativistic). The photons today would have meV energies (~0.1 meV to few meV depending on the timescale), similar to the neutrinos, at roughly the same temperature as the CMB or hotter.
If the process happens faster than recombination, the photons might get buried in the CMB, but that case would need a more careful analysis.

 

[Carlos L. Janer]

The absolute mass difference might be small, but relative to the neutrino masses it is large (O(1)). No matter when the process happens, the photons would carry a significant fraction of the neutrino energy. Their energy goes down at the same rate as the neutrino energy (as long as they are relativistic). The photons today would have meV energies.
If the process happens faster than recombination, the photons might get buried in the CMB, but that case would need a more careful analysis.

OK, I get it. You're right.

 

[Carlos L. Janer]

PTOLEMY tries to see the cosmic neutrino background.
Be careful with statements like that. You are wrong here.

The absolute mass difference might be small, but relative to the neutrino masses it is large (O(1)). No matter when the process happens, the photons would carry a significant fraction of the neutrino energy. Their energy goes down at the same rate as the neutrino energy (as long as they are relativistic). The photons today would have meV energies (~0.1 meV to few meV depending on the timescale), similar to the neutrinos, at roughly the same temperature as the CMB or hotter.
If the process happens faster than recombination, the photons might get buried in the CMB, but that case would need a more careful analysis.

It's getting late and I'm tired. I'm going to bed. It was nice talking to you.

 

[kimbyd]

We cannot rule out neutrino/photon interactions, and at loop-level we have them even in the SM, but they have to be extremely weak. And that is elastic scattering - I still don't see how you would get a decay.

To examine this a little bit, you can contrast it against the decay of the muon, which typically looks like:

\mu \rightarrow \nu_\mu + \nu_\bar{e} + e

This is because lepton flavor is conserved in the standard model. In order to get a neutrino to decay into a lighter neutrino, you'd need, at the very least, for this decay to be possible:

\mu \rightarrow e + \gamma

It might be heavily suppressed, but it should at least happen with some frequency. I don't think any such decay has ever been observed.

 

[PeterDonis]

What on Earth do you think it gives you the right to insult me!

On consideration, you're right, some of what I said that you quoted was uncalled for. I have edited the post of mine from which you quoted to remove that part. But I have let stand the plain statement that I did not make the claims you attributed to me, since it is true.

 

[mfb]

It might be heavily suppressed, but it should at least happen with some frequency. I don't think any such decay has ever been observed.

MEG searched for it, and set an upper limit of 4*10^-13. One of the best branching fraction limits ever set for any particle, possibly the best after an obscure neutron decay*.* 10^-26 for a charge changing decay mode. The trick here is the huge number of neutrons in nuclei in the detector, where such an obscure decay would be possible, but the dominant decay of free neutrons is not available.

 

[snorkack]

Would it be legal for a neutrino to decay into three lighter neutrinos? Spin could be conserved, and so could flavour numbers...

 

[kimbyd]

Would it be legal for a neutrino to decay into three lighter neutrinos? Spin could be conserved, and so could flavour numbers...

No. What you're suggesting would be:

\nu_\mu \rightarrow \nu_\mu + \nu_e + \nu_\bar{e}

You have to have a muon neutrino on both the left and right hand side of the equation. But obviously that can't be because some energy needs to be carried away by the electron neutrino/anti-neutrino pair.

 

[snorkack]

No. What you're suggesting would be:

\nu_\mu \rightarrow \nu_\mu + \nu_e + \nu_\bar{e}

You have to have a muon neutrino on both the left and right hand side of the equation. But obviously that can't be because some energy needs to be carried away by the electron neutrino/anti-neutrino pair.

But what is the mass eigenstate of the muon neutrino on left and right side of the equation?
If oscillation of a propagating neutrino allows a neutrino to change flavour while keeping its rest mass unchanged, can there be interactions in which a neutrino leaves its flavour unchanged but changes its rest mass, transferring energy and momentum to other participants of interaction?

 

[kimbyd]

An interaction like this can only involve flavor eigenstates. Thus it's impossible.

 

[Vanadium 50]

Against my better judgment, I am re-entering this. I hope very much that I will not be misrepresented by other participants.

The particles nu_e, nu_mu and nu_tau do not exist as physical particles. Talking about them as if they are can only lead to confusion. The physical particles are the mass eigenstates, nu_1, nu_2 and nu_3. Decays like $\nu_3 \rightarrow \nu_1 + \overline{\nu_2} + \nu_2$ can happen if they are kinematically allowed, i.e. in this case $m(\nu_3) > 2m(\nu_2) + m(\nu_1)$.

The decay rate will go roughly as $m_\nu^5/M_W^4$. This is very, very small. For a mass of 50 meV, it's of order 10^33 or 10^34 years.

If these decays are kinematically blocked, you can still have penguin decays like $\nu_3 \rightarrow \nu_2 + \gamma$ but these will be even more suppressed.

 

[kimbyd]

The particles nu_e, nu_mu and nu_tau do not exist as physical particles. Talking about them as if they are can only lead to confusion. The physical particles are the mass eigenstates, nu_1, nu_2 and nu_3. Decays like $\nu_3 \rightarrow \nu_1 + \overline{\nu_2} + \nu_2$ can happen if they are kinematically allowed, i.e. in this case $m(\nu_3) > 2m(\nu_2) + m(\nu_1)$.

I think you'll find that the flavor numbers don't match on the left and right sides of that equation, making it questionable at best. This is why such interactions are always written in terms of the flavor eigenstates.

 

[Vanadium 50]

I think you'll find that the flavor numbers don't match on the left and right sides of that equation

Neither the left side nor the right side is in a flavor eigenstate. So how can that possibly be true?

Furthermore, the weak interaction does not conserve flavor.

 

[snorkack]

Furthermore, the weak interaction does not conserve flavor.

What precisely requires a muon to emit two neutrinos to get rid of its muonic flavour? Like an alternative
μ^-→e^-+e^-+e⁺
Spin is conserved, lepton number is conserved... what´s not conserved is lepton flavour. What´s wrong with neutrinoless muon decay?

 

[mfb]

It can happen. It is just extremely unlikely due to the tiny neutrino masses. See post 30. Something like $\displaystyle \frac{m_\nu^4}{m_W^4} = 10^{-50}$ branching fraction, give or take a few orders of magnitude.

Mu3e wants to set 10^-16 as upper limit in the next years. Well, they would prefer finding the decay, but then we need new physics.

 

[kimbyd]

It can happen. It is just extremely unlikely due to the tiny neutrino masses. See post 30. Something like $\displaystyle \frac{m_\nu^4}{m_W^4} = 10^{-50}$ branching fraction, give or take a few orders of magnitude.

I don't think that makes sense as an explanation, because in that instance this decay is also possible:
\mu \rightarrow e + \gamma

Since the photon has zero mass, it is kinematically favored over the neutrino decays.

The Wikipedia article points out that this decay is possible through neutrino oscillation of a virtual neutrino, but this is highly unlikely, probably due to the short amount of time involved for the decay interaction.

 

[Carlos L. Janer]

I don't think that makes sense as an explanation, because in that instance this decay is also possible:
\mu \rightarrow e + \gamma

Since the photon has zero mass, it is kinematically favored over the neutrino decays.

The Wikipedia article points out that this decay is possible through neutrino oscillation of a virtual neutrino, but this is highly unlikely, probably due to the short amount of time involved for the decay interaction.

What makes no sense to me is to consider the propagation of the quantum superposition of three particles in a classical curved space-time. Unfortunately, I am not aware that we have such a theory. Using QFT formalism in a curved space-time is contradictory (there's no Poincaré invariance). I don't know how it could possibly work.

 

[kimbyd]

What makes no sense to me is to consider the propagation of the quantum superposition of three particles in a classical curved space-time. Unfortunately, I am not aware that we have such a theory. Using QFT formalism in a curved space-time is contradictory (there's no Poincaré invariance). I don't know how it possibly work.

For the most part, the curvature of space-time is low enough that the fact that it isn't flat space-time is irrelevant. Just do the quantum examination for the case where space-time is flat and you'll get close to correct.

This breaks down if you're close to a black hole, but works fine in most other cases.

 

[mfb]

The Wikipedia article points out that this decay is possible through neutrino oscillation of a virtual neutrino, but this is highly unlikely, probably due to the short amount of time involved for the decay interaction.

It is not really an amount of time involved in internal lines in a Feynman diagram.

You say the same thing in different words. All the flavor violating decays are extremely unlikely because the neutrinos are so light.

 

[Carlos L. Janer]

For the most part, the curvature of space-time is low enough that the fact that it isn't flat space-time is irrelevant. Just do the quantum examination for the case where space-time is flat and you'll get close to correct.

This breaks down if you're close to a black hole, but works fine in most other cases.

Are you sure about that? I's a quotation I've been given several times. However, your tracing the trajectory and behaviour of relic neutrinos, that decoupled a long time ago from barionic mass and have been contribuiting to the universe expansion ever since. That does not sound right to me.

 

[Carlos L. Janer]

For the most part, the curvature of space-time is low enough that the fact that it isn't flat space-time is irrelevant. Just do the quantum examination for the case where space-time is flat and you'll get close to correct.

Could you give me some references where I could check under what assumptions that aproximation is valid?

 

[kimbyd]

It is not really an amount of time involved in internal lines in a Feynman diagram.

You say the same thing in different words. All the flavor violating decays are extremely unlikely because the neutrinos are so light.

That doesn't make sense to me. It could potentially explain why the $e^- + e^+ + e^-$ decay is disfavored, because the neutrinos are so much lighter than the electrons the decay to neutrinos will be far more common. That doesn't explain why the $e^- + \gamma$ decay is disfavored, as the photon is even less massive.

Where do you get the statement that it's the small neutrino mass that is important in these decay rates?

 

[kimbyd]

Could you give me some references where I could check under what assumptions that aproximation is valid?

Unfortunately I'm not aware of any references offhand. You'd likely be able to search for them about as well as I can.

But the picture is pretty simple: quantum superpositions only really matter in two general cases:
1. There's some kind of discontinuity, such as an event horizon. What happens if one part of the superposition crosses, but the other does not? Thus there are likely issues with the flat-space approximation near black holes.
2. The quantum system produces significant space-time curvature. In this case, because we can't properly describe how a superposition of states impacts space-time curvature, we can't say how such a system gravitates. This isn't really relevant for neutrinos because their density isn't high enough for any significant gravitational effect. It would have been high enough in the early universe, but back then the neutrinos were thermalized enough that they could be treated classically.

 

[Carlos L. Janer]

Unfortunately I'm not aware of any references offhand. You'd likely be able to search for them about as well as I can.

But the picture is pretty simple: quantum superpositions only really matter in two general cases:
1. There's some kind of discontinuity, such as an event horizon. What happens if one part of the superposition crosses, but the other does not? Thus there are likely issues near black holes with the flat-space approximation.
2. The quantum system produces significant space-time curvature. In this case, because we can't properly describe how a superposition of states impacts space-time curvature, we can't say how such a system gravitates. This isn't really relevant for neutrinos because their density isn't high enough for any significant gravitational effect. It would have been high enough in the early universe, but back then the neutrinos were thermalized enough that they could be treated classically.

Ok, I get what you say and I buy it.

However, you're ruling out yourself the possibility for the more massive neutrinos to quickly decay (by an interaction that is beyond the Standard Model) to the lightest one and I thought it was the idea that you were trying to explore with mfb and the reason you were having this conversation with Vanadium50 and mfb.

 

[mfb]

That doesn't make sense to me. It could potentially explain why the $e^- + e^+ + e^-$ decay is disfavored, because the neutrinos are so much lighter than the electrons the decay to neutrinos will be far more common. That doesn't explain why the $e^- + \gamma$ decay is disfavored, as the photon is even less massive.

Where do you get the statement that it's the small neutrino mass that is important in these decay rates?

Draw the Feynman diagram for $\mu \to e \gamma$. It has a neutrino oscillating to a different neutrino type.
The photon mass doesn't matter ($\mu \to Z e$ would be rare as well if it would be possible kinematically). The process is rare due to the flavor violation.

 

[kimbyd]

However, you're ruling out yourself the possibility for the more massive neutrinos to quickly decay (by an interaction that is beyond the Standard Model) to the lightest one and I thought it was the idea that you were trying to explore with mfb and the reason you were having this conversation with Vanadium50 and mfb.

No, I don't think we can rule out a BSM decay. I thought I had been explicit that such a decay, if it exists, would be beyond the standard model, because it would require a lack of conservation of flavor. My understanding is that the ways in which flavor is not conserved in the standard model just don't apply to these sorts of interactions.

Your statement that this could be recovered by looking at the reaction in terms of mass eigenstates rather than flavor eigenstates does nothing but obfuscate the difficulty. Mass isn't a conserved quantity, so it doesn't even make sense to write down an interaction equation in terms of mass eigenstates. We write such interactions in terms of flavor eigenstates because flavor is mostly conserved.

 

[kimbyd]

Draw the Feynman diagram for $\mu \to e \gamma$. It has a neutrino oscillating to a different neutrino type. The photon mass doesn't matter. The process is rare due to the flavor violation.

That's what I'm trying to say: the low neutrino mass isn't the dominant factor in suppressing such interactions.

 

[mfb]

That's what I'm trying to say: the low neutrino mass isn't the dominant factor in suppressing such interactions.

But... it is!

Flavor violation is rare because the neutrinos are light.

 

[PeterDonis]

There's some kind of discontinuity, such as an event horizon. What happens if one part of the superposition crosses, but the other does not?

An event horizon is not a discontinuity locally; it's just a lightlike surface, which locally looks like any other lightlike surface. Quantum field theory works fine locally across lightlike surfaces.

 

[kimbyd]

But... it is!

Flavor violation is rare because the neutrinos are light.

Why is that?

You could argue that the flavor violation is rare because neutrino oscillation happens slowly due to the small differences in masses between neutrinos, but that only explains part of it. The more important question, to me, is why the neutrino oscillation is necessary at all to produce the flavor violation.

 

[Vanadium 50]

This is a B level thread. I don't think any question whose answer starts "write down the Feynman diagram" can be answered at the B level.