# Solar neutrinos problem

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The solar neutrinos problem has represented one of the most important issues in the history of physics. It concearned a discrapancy between the number of neutrinos which reach the Earth from the Sun predicted by the physics models and the effective number of neutrinos observed in experiments.

-Flow expected

Solar neutrinos with photons are created in Sun's core thank to the nuclear reaction between Hydrogen cores, which form Helium; In this type of reaction is created an amount of neutrinos. The flow of neutrinos could be calculated by the brightness of the Sun, which is propotional to the amount of energy released from it.

n ν = 2 L Q − 〈 q ν 〉 {\displaystyle n_{\nu }=2{\frac {L}{Q-\langle q_{\nu }\rangle }}}

where L is the brigthness, Q is the energy created with the reactions and qv is the medium energy of a neutrino.〈

L = 3 , 864 ⋅ 10 33   e r g s {\displaystyle L=3,864\cdot 10^{33}\ {\frac {erg}{s}}} , Q = 26 , 7   M e V {\displaystyle Q=26,7\ MeV} e 〈 q ν 〉 = 0 , 6   M e V {\displaystyle \langle q_{\nu }\rangle =0,6\ MeV} , we found that:

n ν = 1 , 851 ⋅ 10 38   n e u t r i n i s {\displaystyle n_{\nu }=1,851\cdot 10^{38}\ {\frac {neutrini}{s}}}

The number of neutrinos which reach the Earth, considering an uniform emission in all the directions is:

Φ ν = n ν 4 π R 2 {\displaystyle \Phi _{\nu }={\frac {n_{\nu }}{4\pi R^{2}}}}
where R is the distance Sun-Earth
Φ ν = 6 , 588 ⋅ 10 10   n e u t r i n i s × c m 2 {\displaystyle \Phi _{\nu }=6,588\cdot 10^{10}\ {\frac {neutrini}{s\;\times \;cm^{2}}}}
In the experiments the amount of neutrinos revealed is only 1/3 of the all we have expected. Approximately the 2/3 of neutrinos miss..

-The solution
The solution of the problem came from a modification of the Standard model of particles: Neutrinos could change their "flavor" in a process called neutrinos oscillation.
The flavor charge is a set of quantum number which characterize quarks and leptons; for which, for example a muon neutrino is different from an electron neutrino...
So, on the road from the Sun to the Earth, neutrinos change their flavor, from electronic they could become tauonic or muonic ones.
In Experiments only the electron ones can be detected, it explains the discrepancy between theory and observation. The oscillation happens only for a brief period of time.
The leptons (electrons, tauons, muons and respective neutrinos) have a Lepton quantum number L (L=1) of flavor. They also have a weak isospin which is Tz=-1/2 for (e, t and m) and Tz=1/2 for neutrinos and a weak hypercharge, Yw=-1 for charged leptons and Yw=1 for neutrinos.

## Esperimenti

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!

Moderator Note

Is there a question here, or a point of discussion?

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@Heis3nberg It's not the simplest subject. I do not see any related question apart from the historical facts.

To verify this hypothesis one avenue envisaged is to compare the flux of neutrinos detected between day and night. Indeed, at night, solar neutrinos, to be detected, must cross the whole of the Earth, and should therefore be able to oscillate by interaction with matter, therefore at a rate different from the oscillation in vacuum, which should cause a flux detected different from that detected during the day. In addition, such an experiment would allow us to better understand the periods of these oscillations, which would improve our understanding of the phenomenon. But currently, the detectors are not sensitive enough to allow this measurement. I do not understand everything to get this equation, but I found this form of expression where the probability equation becomes:

P=|νe|ν(t)|2|νe|ν2m(t)|2=|νe|ν2|2sin2θ13

A deviation from the value of the sin²θ probability is an indication of the presence of oscillations.

Quote

Many experiments have searched for oscillation of electron anti-neutrinos produced at nuclear reactors. No oscillations were found until the detector was installed at a distance 1–2 km. Such oscillations give the value of the parameter θ13. Neutrinos produced in nuclear reactors have energies similar to solar neutrinos, of around a few MeV. The baselines of these experiments have ranged from tens of meters to over 100 km (parameter θ12). Mikaelyan and Sinev proposed to use two identical detectors to cancel systematic uncertainties in reactor experiment to measure the parameter θ13.

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1 hour ago, swansont said:
!

Moderator Note

Is there a question here, or a point of discussion?

a point of discussion, there are a lot of things about the neutrinos physics on which we can discuss

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2 minutes ago, Heis3nberg said:

there are a lot of things about the neutrinos physics on which we can discuss

Given that you’re the OP, how about picking one

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5 hours ago, Heis3nberg said:

The solar neutrinos problem has represented one of the most important issues in the history of physics. It concearned a discrapancy between the number of neutrinos which reach the Earth from the Sun predicted by the physics models and the effective number of neutrinos observed in experiments.

-Flow expected

Solar neutrinos with photons are created in Sun's core thank to the nuclear reaction between Hydrogen cores, which form Helium; In this type of reaction is created an amount of neutrinos. The flow of neutrinos could be calculated by the brightness of the Sun, which is propotional to the amount of energy released from it.

n ν = 2 L Q − 〈 q ν 〉 {\displaystyle n_{\nu }=2{\frac {L}{Q-\langle q_{\nu }\rangle }}}

where L is the brigthness, Q is the energy created with the reactions and qv is the medium energy of a neutrino.〈

L = 3 , 864 ⋅ 10 33   e r g s {\displaystyle L=3,864\cdot 10^{33}\ {\frac {erg}{s}}} , Q = 26 , 7   M e V {\displaystyle Q=26,7\ MeV} e 〈 q ν 〉 = 0 , 6   M e V {\displaystyle \langle q_{\nu }\rangle =0,6\ MeV} , we found that:

n ν = 1 , 851 ⋅ 10 38   n e u t r i n i s {\displaystyle n_{\nu }=1,851\cdot 10^{38}\ {\frac {neutrini}{s}}}

The number of neutrinos which reach the Earth, considering an uniform emission in all the directions is:

Φ ν = n ν 4 π R 2 {\displaystyle \Phi _{\nu }={\frac {n_{\nu }}{4\pi R^{2}}}}
where R is the distance Sun-Earth
Φ ν = 6 , 588 ⋅ 10 10   n e u t r i n i s × c m 2 {\displaystyle \Phi _{\nu }=6,588\cdot 10^{10}\ {\frac {neutrini}{s\;\times \;cm^{2}}}}
In the experiments the amount of neutrinos revealed is only 1/3 of the all we have expected. Approximately the 2/3 of neutrinos miss..

-The solution
The solution of the problem came from a modification of the Standard model of particles: Neutrinos could change their "flavor" in a process called neutrinos oscillation.
The flavor charge is a set of quantum number which characterize quarks and leptons; for which, for example a muon neutrino is different from an electron neutrino...
So, on the road from the Sun to the Earth, neutrinos change their flavor, from electronic they could become tauonic or muonic ones.
In Experiments only the electron ones can be detected, it explains the discrepancy between theory and observation. The oscillation happens only for a brief period of time.
The leptons (electrons, tauons, muons and respective neutrinos) have a Lepton quantum number L (L=1) of flavor. They also have a weak isospin which is Tz=-1/2 for (e, t and m) and Tz=1/2 for neutrinos and a weak hypercharge, Yw=-1 for charged leptons and Yw=1 for neutrinos.

## Esperimenti

Why is it that we can only detect electron neutrinos?

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One factor is that the muon and tau particles that would be produced by the muon and tau neutrinos are more massive than the electron, so more energy is required in the interactions.

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2 hours ago, exchemist said:

Why is it that we can only detect electron neutrinos?

Neutrinos are undetectable, they interact only with weak force, scientists exploit the Beta decay reactions to detect them: I'm going to post something about the way to reveal neutrinos, but you need to know that in Beta decay reaction only electron neutrinos participate, not the muon and tau ones.

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18 hours ago, Heis3nberg said:

Neutrinos are undetectable, they interact only with weak force, scientists exploit the Beta decay reactions to detect them

So they are, in fact, detectable.

Quote

but you need to know that in Beta decay reaction only electron neutrinos participate, not the muon and tau ones.

The ones we use, because of the neutrinos we detect. Muon and tau neutrinos do similar reactions.

The detection of the muon neutrino got the Nobel prize in 1988

(edit: it's trivially true that they don't participate in a reaction where an electron is emitted, because that violates lepton number conservation)

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1 hour ago, swansont said:

So they are, in fact, detectable.

The ones we use, because of the neutrinos we detect. Muon and tau neutrinos do similar reactions.

The detection of the muon neutrino got the Nobel prize in 1988

Do muon and tau neutrinos also participate in beta decay reactions, then? Or, when you say "similar", do you mean different from beta decay but in some way analogous?

Or is it that they do, but to a much lesser degree, due to their greater mass (which you referred to earlier)?  Is this a cross-section effect?

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16 hours ago, exchemist said:

Do muon and tau neutrinos also participate in beta decay reactions, then? Or, when you say "similar", do you mean different from beta decay but in some way analogous?

Or is it that they do, but to a much lesser degree, due to their greater mass (which you referred to earlier)?  Is this a cross-section effect?

“beta decay reactions” are ones that swap a proton and neutron. In a decay, both the particle and its associated neutrino (one of them is an antiparticle) is released. In the detection reaction, the particle is absorbed and the neutrino emitted

If a reaction needs e.g. 10 MeV to occur and your source is giving you 5 MeV particles, you won’t cause the reaction. One would need to check the specifics of the solar neutrino spectrum and the specific reaction to see how much of an issue this is. The muon mass is > 100 MeV  and tau mass is ~1776 MeV, so they definitely require added energy. A bare neutron is only a few MeV more massive than a proton, and the energy difference in a nuclear shell is of similar magnitude.

edit: from this document from SLAC it looks like the solar neutrinos are all under 20 MeV (p.208, graph on 209)

So if they changed to muon neutrinos along the way, they don't have enough energy to create muons in a n—>p reaction

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7 hours ago, swansont said:

“beta decay reactions” are ones that swap a proton and neutron. In a decay, both the particle and its associated neutrino (one of them is an antiparticle) is released. In the detection reaction, the particle is absorbed and the neutrino emitted

If a reaction needs e.g. 10 MeV to occur and your source is giving you 5 MeV particles, you won’t cause the reaction. One would need to check the specifics of the solar neutrino spectrum and the specific reaction to see how much of an issue this is. The muon mass is > 100 MeV  and tau mass is ~1776 MeV, so they definitely require added energy. A bare neutron is only a few MeV more massive than a proton, and the energy difference in a nuclear shell is of similar magnitude.

edit: from this document from SLAC it looks like the solar neutrinos are all under 20 MeV (p.208, graph on 209)

So if they changed to muon neutrinos along the way, they don't have enough energy to create muons in a n—>p reaction

Thanks, that's very informative.

But according to the link, neutrinos are detected, not by beta decay, as @Heis3nbergsaid,  but by emission of Cerenkov radiation, in which, apparently, both electron and muon neutrinos can show up but not, for some reason tau neutrinos. They don't say why, but I can imagine that it would be less likely a particle with such a large mass will be accelerated to speeds greater than the local phase velocity of light in the detection medium.

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38 minutes ago, exchemist said:

Thanks, that's very informative.

But according to the link, neutrinos are detected, not by beta decay, as @Heis3nbergsaid,  but by emission of Cerenkov radiation, in which, apparently, both electron and muon neutrinos can show up but not, for some reason tau neutrinos. They don't say why, but I can imagine that it would be less likely a particle with such a large mass will be accelerated to speeds greater than the local phase velocity of light in the detection medium.

In the document I linked to they discuss this, on p 210. They use multiple methods, depending on which experiment you're looking at

"These experiments have used four different techniques (target materials) and therefore have different energy thresholds—233 keV for the Gallium experiments (SAGE and GALLEX), 814 keV for the Chlorine experiment (Homestake) and a few MeV for the water Cherenkov experiments (Kamiokande and Super-Kamiokande; 5.5 MeV for the latest analy-sis threshold of Super-Kamiokande),and,we expect,5MeV for the heavy water Cherenkov experiment (SNO). It is an advantage that these experiments have dif-ferent energy thresholds and therefore have sensitivity to different regions of the solar neutrino spectrum."

The chlorine and gallium experiments are the beta-decay technique

"the neutrino enters and then leaves the detector after having transferred some of its energy and momentum to a target particle. If the target particle is charged and sufficiently lightweight (e.g. an electron), it may be accelerated to a relativistic speed and consequently emit Cherenkov radiation"

I don't know if a muon or tau neutrino would undergo this interaction with an electron. The entry suggest that they can, but you can't tell what kind of neutrino was involved. (this is beyond my schooling)

The other interaction says "a high-energy neutrino transforms into its partner lepton (electron, muon, or tau).[7] However, if the neutrino does not have sufficient energy to create its heavier partner's mass, the charged current interaction is unavailable to it."

So here we know there is not enough energy to do this, since the requirement is bigger than 100 MeV

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