What is the correct solar neutrino reaction equation and how was it discovered?

[Buzz Bloom]

Hi mfb:

The following messages were in the now closed thread

"Neutrinos-Antineutrinos in the universe".​

Bill_K said (msg #2):

Most of the neutrinos we detect are solar neutrinos coming from

p + P → d + e⁺ + ν.​

I said (msg #5):

Is the "P" here a typo where "p" was intended? If not, what does "P" mean?​

You said (msg #7):

It was a two year old typo (and as the thread is very old, I'll close it, feel free to discuss in other threads or open a new one).​

Actually, it was not correct either because neutrinos from that particular reaction are very low-energetic and hard to detect, that was achieved for the first time in 2014.​

Would you please post the new correct solar reaction equation, and if you know one, give a reference to an explanation about how the new reaction formula was discovered?

Thanks for your help,
Buzz

 

[ChrisVer]

If I recall well, the neutrinos that were detected were from Be8...
http://t2k-experiment.org/wp-content/uploads/colorbahcallserenellibs05OP.gif
Of course the pp reaction gives the most of the flux, but those neutrinos have small energies and that's why they were difficult to detect...

 

[Buzz Bloom]

Hi ChrisVer:

The solar neutrino energy spectrum diagram you cited is quite interesting. Can you help me understand how to interpret the three isolated vertical lines as spectrums? They are labeled:

⁷Be at ~0.39 MeV
⁷Be at ~0.87 MeV
pep at ~1.4 MeV.​

I was also able to find at https://en.wikipedia.org/wiki/Solar_neutrino the following diagram showing the chains of Solar reactions.

Can you tell me what the asterisk "*" means in the two boxes at the lower right attached to ⁸Be? My guess is that it means ⁸Be is radioactive.
Also, why in the lowest box is there no γ released in what seems to be a fission reaction?

Thank you for you post,
Buzz

 

[e.bar.goum]

Hi ChrisVer:

The solar neutrino energy spectrum diagram you cited is quite interesting. Can you help me understand how to interpret the three isolated vertical lines as spectrums? They are labeled:

⁷Be at ~0.39 MeV
⁷Be at ~0.87 MeV
pep at ~1.4 MeV.​

I was also able to find at https://en.wikipedia.org/wiki/Solar_neutrino the following diagram showing the chains of Solar reactions.
View attachment 85451
Can you tellme what the asterisk "*" means in the two boxes at the lower right attached to ⁸Be?

Thank you for you post,
Buzz

The ⁷Be neutrinos come from ⁷Be decay via electron capture - $e + ^7Be \rightarrow \nu_e + ^7Li$, which has a 90% branching ratio to 0.87 MeV, and a 10% branch to 0.39 MeV. The pep reaction comes from $p+e+p \rightarrow d+ \nu_e$. They're isolated vertical lines as the reactions are two body processes, rather than the rest, which are three body - e.g. $^8B \rightarrow ^8Be + e^+ + \nu_e$ - this is just energy/momentum conservation.

the "*" just indicates that $^8Be$ is populated in an excited state (technically you'd say it was in a resonant state, but that's beside the point)

This might be helpful: http://www.slac.stanford.edu/econf/C990809/docs/suzuki.pdf

 

[Buzz Bloom]

Hi e.bar.goum:

I much appreciate your answering my questions.

Just to make sure I understand the vocabulary:
1) A two body process is a reaction the produces only two output particles. This kind of process always produces a discrete rather than a continuous neutrino energy spectrum.
2) The excited ⁸Be* nucleus has either (a) exactly twice the energy of the ⁴He nucleus, or (b) a bit more than that. If (a), then the two ⁴He nuclei will move apart due to the repulsive force from their positive charges. If (b) then they will move apart somewhat faster to convert the extra energy available into kinetic energy. In either case, there is no residual energy to be carried away by a photon.

If I have misinterpreted, I would much appeciated being corrected.

Thanks for your post and for the citation of the Suzuki et. al. paper.
Buzz

 

[e.bar.goum]

Hi e.bar.goum:

I much appreciate your answering my questions.

Just to make sure I understand the vocabulary:
1) A two body process is a reaction the produces only two output particles. This kind of process always produces a discrete rather than a continuous neutrino energy spectrum.
2) The excited ⁸Be* nucleus has either (a) exactly twice the energy of the ⁴He nucleus, or (b) a bit more than that. If (a), then the two ⁴He nuclei will move apart due to the repulsive force from their positive charges. If (b) then they will move apart somewhat faster to convert the extra energy available into kinetic energy. In either case, there is no residual energy to be carried away by a photon.

If I have misinterpreted, I would much appeciated being corrected.

Thanks for your post and for the citation of the Suzuki et. al. paper.
Buzz

1. Yes. ETA: I mean "two body decay" here, rather than "two body reaction".
2. In essence, yes. For there to be a photon, you would either have to have 8Be de-excite via photon emission before breaking up (which won't happen fast enough, considering the lifetime of 8Be resonant states), or 4He be populated into an excited state which then decays via photon emission. There won't be the energy for this, and even then, excited states of 4He are above the proton emission threshold. The mass difference between 8Be and 2*4He is such that at minimum, the 4He particles will have 92 keV of relative energy (and some total kinetic energy depending on the kinetic energy of the 8B. If you put this all into the centrer-of-mass frame, the 8B is at rest). If you populate 8Be into the first resonant state (at 3.03 MeV) the alphas will have 3.03+0.092 MeV of relative energy. In the center-of-mass frame, the alphas will each have half the total energy.

 

[Buzz Bloom]

Hi: e.bar.goum:

I much appreciate your elboration and clarification.

There is a typo you may want to correct in

excited states of 4He are above the proton emission threshold.

"proton" -> photon".

Thanks for your post,
Buzz

 

[e.bar.goum]

Hi: e.bar.goum:

I much appreciate your elboration and clarification.

There is a typo you may want to correct in

"proton" -> photon".

Thanks for your post,
Buzz

No, I definitely meant proton. http://www.tunl.duke.edu/nucldata/figures/04figs/04_02_1992.gif

 

[Buzz Bloom]

Hi e.bar.goum:

I don't get the meaning of

excited states of 4He are above the proton emission threshold

I looked at the GIF diagram, and I think I undesrtand that the large display in the middle is about about the excited energy states of ⁴He.
I interpret the top and bottom entries in this display as meaning that the max excited energy state is 29.89 Gev, and the min is 20.21Gev. Even if this is all correct, I still do understand how it relates to: "are above the proton emission threshold", and how that is related to

4He be populated into an excited state which then decays via photon emission. There won't be the energy for this

(The underline is mine.)​

I hope you will be able to help me through my confusion.

Thank you for the discussion,
Buzz

 

[e.bar.goum]

Hi e.bar.goum:

I don't get the meaning of

​

I looked at the GIF diagram, and I think I undesrtand that the large display in the middle is about about the excited energy states of ⁴He.
I interpret the top and bottom entries in this display as meaning that the max excited energy state is 29.89 Gev, and the min is 20.21Gev. Even if this is all correct, I still do understand how it relates to: "are above the proton emission threshold", and how that is related to

(The underline is mine.)​

I hope you will be able to help me through my confusion.

Thank you for the discussion,
Buzz

That is an energy level diagram, yes, but the units are in MeV, and the max state is not neccesarily the max state in principle, just the highest measured state.

The proton emission threshold is shown at 19.815 MeV, just to the left of the level diagram. This is the energy level at which 4He has enough energy to decay to 3H + p, rather than via gamma emission to the ground state.

This is relevant because there are two ways to get photon in this reaction - from 8Be or 4He. 8Be just doesn't last long enough for an EM process to occur (the lifetime of the 2+ and 4+ state are a fraction of a zeptosecond), which leaves us with 4He, now, even if you had the energy for it, 4He will decay via proton emission instead (or n or D for higher states).

Does this clear things up?

 

[Buzz Bloom]

Hi e.bar.goum:

even if you had the energy for it, 4He will decay via proton emission instead

Does this clear things up?

I think I almost get it now. If the 8Be -> 4He + 4He reaction produced an excied 4He nucleon with sufficient energy to emit a photon (20.21 MeV), it would instead emit a proton (or if sufficiently greater, something more massive).

Does it also mean that it is known experimentally that the above (8Be) reaction never does excites either of the two produced 4He nuclei to as much as 19.815 MeV since such a photon is never observed?

Thank you for patience in answering my questions,
Buzz

 

[e.bar.goum]

Hi e.bar.goum:

I think I almost get it now. If the 8Be -> 4He + 4He reaction produced an excied 4He nucleon with sufficient energy to emit a photon (20.21 MeV), it would instead emit a proton (or if sufficiently greater, something more massive).

Yes.

Does it also mean that it is known experimentally that the above (8Be) reaction never does excites either of the two produced 4He nuclei to as much as 19.815 MeV since such a photon is never observed?

Thank you for patience in answering my questions,
Buzz

I don't know about experimentally (it's a well studied reaction, but I don't know off hand), but we can work this out pretty easily, there's no need for experiment:

Let's consider 8B decaying in it's rest frame.

The mass difference between 8B and 8Be is 17.98 MeV/c^2. So, at most you've got 17.98 MeV - E_electron - E_neutrino of available excitation in 8Be. The alphas then have at most half of that + 0.092 MeV available for excitation (in their center of mass frame). Not enough to excite 4He. Hence my first comment about not having enough energy.

 

[Buzz Bloom]

Hi e.bar.goum::

The mass difference between 8B and 8Be is 17.98 MeV/c^2. So, at most you've got 17.98 MeV - Eelectron - Eneutrino of available excitation in 8Be. The alphas then have at most half of that + 0.092 MeV available for excitation (in their center of mass frame). Not enough to excite 4He. Hence my first comment about not having enough energy.

Sometimes the experience of a senior moment makes me feel stupid. I should have been able to work all that out from the information you gave me.

Thanks very much for your patience,
Buzz

 

[my2cts]

More degrees of freedom than restrictions from energy and momentum conservation.

 

[Buzz Bloom]

Hi my2acts:

More degrees of freedom than restrictions from energy and momentum conservation.

I am not sure I fully understand this.

Do you mean that the degrees of freedom which are involved in restricting the emission of a photon in the sequence of the following two reactions are more than just conservation restrictions?

⁸B → ⁸Be* + e⁺ + ν_e
⁸B_e* → ⁴He + ⁴He​

If that is correct, would you please post something specific about these other restrictions?

Thanks for your post,
Buzz

 

[my2cts]

Hi my2acts:
I am not sure I fully understand this.

Do you mean that the degrees of freedom which are involved in restricting the emission of a photon in the sequence of the following two reactions are more than just conservation restrictions?

⁸B → ⁸Be* + e⁺ + ν_e
⁸B_e* → ⁴He + ⁴He​

If that is correct, would you please post something specific about these other restrictions?

Thanks for your post,
Buzz

This was an answer to the question as to why a three particle emission resulted in a continuum spectrum. The post seems to have been deleted.

 

[jtbell]

This was an answer to the question as to why a three particle emission resulted in a continuum spectrum. The post seems to have been deleted.

The poster re-posted that question in a separate thread here:

https://www.physicsforums.com/threads/how-two-body-decay-energy-spectrum-is-discrete.821999/

 

[ChrisVer]

This was an answer to the question as to why a three particle emission resulted in a continuum spectrum. The post seems to have been deleted.

You can also find the answer in:
https://www.physicsforums.com/threads/experiments-to-distinguish-3-body-and-2-body-decays.819692/
posts #6,7,8

restricting the emission of a photon

Let me understand better what you are asking, by answering this question of mine:
Why do you want a photon to be emitted?

 

[Buzz Bloom]

Hi ChrisVer;

Why do you want a photon to be emitted?

I have no desires about photon emission one way or or another. I asked the question out of curiosity, seeking to improve my very limited understanding of QM phenomena. I was aware that in many reactions photons are emitted and I was currious about why they were not in the reaction I asked about,

You can also find the answer in:
https://www.physicsforums.com/threads/experiments-to-distinguish-3-body-and-2-body-decays.819692/
posts #6,7,8

The posts in this new thread are very instructive. Thanks for the link.

Regards,
Buzz

 

[ChrisVer]

I was aware that in many reactions photons are emitted and I was currious about why they were not in the reaction I asked about,

Photons are emitted after electromagnetic interactions...or in some cases when excited nuclei are formed and transit to the ground state (again by electromagnetic interactions)...
In beta decays (or weak interactions) and strong interactions you don't have a photon emission.

 

[Buzz Bloom]

Hi ChrisVer:

All of this discussion since post #15 has been very enlightening.

or in some cases when excited nuclei are formed and transit to the ground state (again by electromagnetic interactions)...

The scenario in the quote was the type of the reaction ⁸Be* → ⁴He + ⁴He that I thought might produce a photon. And the following explantion seemed to me to be adequate regarding photon emission: (1) insufficient energy for a photon and (2) even if there were sufficient energy, a proton would be emitted instead. The discusions of the two vs three particle decays and the discrete vs continuous spectrums explained further, regarding a hypothethetic similar reaction, that even if there were (1) sufficient energy to emit a photon, and (2) not enough energy to emit a proton, there still might not be a photon emitted.

Thanks for your discussion,
Buzz

 

[ChrisVer]

Finally you asked about the pp reaction experiment?

First of all, the initial Solar Neutrinos were measured with the help of $_{17}^{37}Cl$... In particular you had $Cl$ in large quantities within a liquid, and you were looking at events:
$\nu + _{17}^{37}Cl \rightarrow e^- + _{18}^{37}Ar$.
This reaction needs at least $0.81~MeV$ free energy to happen and that's why pp neutrinos were not detectable. As a final result you were looking for the characteristic beta-decay of $Ar$ which you identified as a neutrino interaction...

Another way to build your detector is instead of using some tons of Cl in your experiment, to use some tons of $_{31}^{71}Ga$. The interaction is:
$\nu + _{31}^{71}Ga + 0.23~MeV \rightarrow _{32}^{71}Ge + e^-$
So with Gallium you only need 0.23MeV neutrinos. The Gallium experiment can be sensitive to neutrinos coming from the pp interaction.