Neutrino energy

[Moontanman]

How much energy is radiated from the sun in the form of neutrinos? 

 

https://futurism.com/60-years-since-they-were-first-observed-theres-much-left-to-learn-about-neutrinos/

[Vmedvil]

Well, it depends on the amount of Deuterium created each second, before hydrogen fusion to helium two hydrogen atoms fuse to become Deuterium and one proton decays into a neutron this releases a W+ particle which decays into a positron and neutrino, you can calculate the total neutrino release by the rate of transformation of Hydrogen to Deuterium. Its isotope used in hydrogen fusion to helium 4. Secondly, you calculate the neutrino energy-mass which is E= λMC^2, just as any other mass having particle with velocity. These being electron neutrinos released during Hydrogen to Deuterium transformation. 

[swansont]

32 minutes ago, Vmedvil said:

Well, it depends on the amount of Deuterium created each second, before hydrogen fusion to helium two hydrogen atoms fuse to become Deuterium and one proton decays into a neutron this releases a W+ particle which decays into a positron and neutrino, you can calculate the total neutrino release by the rate of transformation of Hydrogen to Deuterium. Its isotope used in hydrogen fusion to helium 4. Secondly, you calculate the neutrino energy-mass which is E= λMC^2, just as any other mass having particle with velocity. These being electron neutrinos released during Hydrogen to Deuterium transformation. 

They will probably have a KE much larger than their mass-energy.

[Vmedvil]

10 minutes ago, swansont said:

They will probably have a KE much larger than their mass-energy.

KE  and Relativistic mass in SR is calculated by γ for high velocity particles, I tend to agree. I used the wrong symbol in the equation, sorry, I mean Gamma not lambda 

 

E= γMC^2 

Edited October 2, 2017 by Vmedvil

[Sensei]

On 28.08.2017 at 9:35 PM, Moontanman said:

How much energy is radiated from the sun in the form of neutrinos?

There is no single answer, different neutrinos, from different reactions which are creating them, have various kinetic energies.

f.e. the most basic fusion reaction is:

[math]p^+ + p^+ \rightarrow D^+ + e^+ + v_e + 0.42 MeV[/math]

This equation can be written in longer form pp->He-2->D

[math]^2_2 He \rightarrow D^+ + e^+ + v_e + 0.647447 MeV[/math]

The more kinetic energy takes positron, the less is left for neutrino, and vice versa.

So average energy emitted to environment around star is 0.21 MeV (or 0.324 MeV) or so, per fusion reaction.

There is emitted 65 billions per cm^2 area at 150 mln km distance from the Sun to the Earth (inverse-square equation, if neutrinos are emitted uniformly in the all directions).

65*10^9 * 324000 eV * 1.602176565*10^-19 J/eV = 0.0033741838 J/cm^2 = 33.74 J/m^2 (divide by time=1s to have power in watts = 33.74 W/m^2.. Regular Sun photons have 1370 W/m^2 for comparison).

Not so much, even if they would be the all captured by the Earth' materials..

 

  

This reaction won't trigger Chlorine-37 based neutrino detector. It requires at least 814 keV.

 

Edited October 2, 2017 by Sensei

[Vmedvil]

1 hour ago, Sensei said:

There is no single answer, different neutrinos, from different reactions which are creating them, have various kinetic energies.

f.e. the most basic fusion reaction is:

p++p+→D++e++ve+0.42MeV

This equation can be written in longer form pp->He-2->D

22He→D++e++ve+0.647447MeV

The more kinetic energy takes positron, the less is left for neutrino, and vice versa.

So average energy emitted to environment around star is 0.21 MeV (or 0.324 MeV) or so, per fusion reaction.

There is emitted 65 billions per cm^2 area at 150 mln km distance from the Sun to the Earth (inverse-square equation, if neutrinos are emitted uniformly in the all directions).

65*10^9 * 324000 eV * 1.602176565*10^-19 J/eV = 0.0033741838 J/cm^2 = 33.74 J/m^2 (divide by time=1s to have power in watts = 33.74 W/m^2.. Regular Sun photons have 1370 W/m^2 for comparison).

Not so much, even if they would be the all captured by the Earth' materials..

 

  

This reaction won't trigger Chlorine-37 based neutrino detector. It requires at least 814 keV.

 

That sounds right, I didn't know the exact numbers.

45 minutes ago, Vmedvil said:

That sounds right, I didn't know the exact numbers.

 

Wait, I just found a problem, that breaks the conservation of momentum, what you just did, you are right the positron would get more of the energy because of its increased mass. They would both get the exact same amount of momentum though.

M₁V₁- M₂V₂ = 0

You misbalanced you equation this time, Sensei. 

Edited October 2, 2017 by Vmedvil

[swansont]

49 minutes ago, Vmedvil said:

 Wait, I just found a problem, that breaks the conservation of momentum, what you just did, you are right the positron would get more of the energy because of its increased mass. They would both get the exact same amount of momentum though. 

Classically KE = p^2/2m, meaning for objects with equal momentum, the less massive one has more KE (more m, less v, and KE depends on v^2). Relativistically the breakdown is more complicated, but the trend does not reverse. So the neutrino should get more energy if the momentum is divided equally.

 

[Sensei]

58 minutes ago, Vmedvil said:

 

Wait, I just found a problem, that breaks the conservation of momentum, what you just did, you are right the positron would get more of the energy because of its increased mass. They would both get the exact same amount of momentum though.

M₁V₁- M₂V₂ = 0

Relativistic momentum is

[math]p=m v \gamma[/math]

And you don't know what is rest-mass of neutrino, in the first place. With constant rest-mass of electron-positron m_e= ~0.511 MeV/c^2

 

Edited October 2, 2017 by Sensei

[Vmedvil]

Just now, swansont said:

Classically KE = p^2/2m, meaning for objects with equal momentum, the less massive one has more KE (more m, less v, and KE depends on v^2). Relativistically the breakdown is more complicated, but the trend does not reverse. So the neutrino should get more energy if the momentum is divided equally.

 

Lol, ya, in any case, he did violate it, So, one of them does get more KE, the neutrino then, man this stuff gets confusing sometimes.

1 minute ago, Sensei said:

Relativistic momentum is

p=mvγ

 

 

well, in that case, M₁V₁γ₁ - M₂V₂γ₂ = 0

[swansont]

1 minute ago, Vmedvil said:

Lol, ya, in any case, he did violate it, So, one of them does get more KE, the neutrino then, man this stuff gets confusing sometimes.

I'm not convinced there is an equal division. I worked on an experiment where we were trying to reconstruct the neutrino momentum spectrum from the momentum of the emitted beta and the recoil daughter (in ³⁸K^m and ³⁷K ) when they emission of the neutrino and beta were in opposite directions. There would be no momentum for the daughter if the momentum were divided equally. 

[Vmedvil]

17 minutes ago, Sensei said:

Relativistic momentum is

p=mvγ

And you don't know what is rest-mass of neutrino, in the first place. With constant rest-mass of electron-positron m_e= ~0.511 MeV/c^2

 

Well, it is around 2.2 ev says the standard model, so lets take them at that.

6 minutes ago, swansont said:

I'm not convinced there is an equal division. I worked on an experiment where we were trying to reconstruct the neutrino momentum spectrum from the momentum of the emitted beta and the recoil daughter (in ³⁸K^m and ³⁷K ) when they emission of the neutrino and beta were in opposite directions. There would be no momentum for the daughter if the momentum were divided equally. 

I dunno, in pair production it looks like this. So, I guess it depends on the direction of the W+?  I mean the W+ is destroyed there is no daughter in this case.

Edited October 2, 2017 by Vmedvil

[Sensei]

11 minutes ago, Vmedvil said:

well, in that case, M₁V₁γ₁ - M₂V₂γ₂ = 0

You don't know what is m₂, what is v₂, and what is γ₂.

You just know m₁=rest-mass of electron/positron, v₁ and γ₁know from observation of longer or shorter traces in f.e. Cloud Chamber in beta decay minus reactions.

So assumption is made that energy is conserved, and thus the rest of energy is taken by neutrino/antineutrino in beta decay plus/minus (and daughter isotope little amount).

 

Example graph that I found on the net:

Bismuth-210 decays via:

Bismuth-210 -> Polonium-210 + e- + Ve + 1.16129 MeV

 

[swansont]

9 minutes ago, Vmedvil said:

 I dunno, in pair production it looks like this. So, I guess it depends on the direction of the W+?  I mean the W+ is destroyed there is no daughter in this case.

But we're talking about decay, not pair production.

1 minute ago, Sensei said:

You don't know what is m₂, what is v₂, and what is γ₂.

You just know m₁=rest-mass of electron/positron, v₁ and γ₁know from observation of longer or shorter traces in f.e. Cloud Chamber in beta decay minus reactions.

So assumption is made that energy is conserved, and thus the rest of energy is taken by neutrino/antineutrino in beta decay plus/minus (and daughter isotope little amount).

 

Example graph that I found on the net:

Bismuth-210 decays via:

Bismuth-210 -> Polonium-210 + e- + Ve + 1.16129 MeV

 

That's another bit. The spectrum is continuous, and the beta can take up almost all of the energy , though it rarely does so. It may be that the most probably energy represents equal momenta, though.

[Vmedvil]

19 minutes ago, swansont said:

But we're talking about decay, not pair production.

Well, but W+ has the same Feynman diagram, so would it have the same path as pair production in its products? They are both bosons.

 

 

From what I understand about Feynman diagrams momentum does have to be conserved in these transactions.

21 minutes ago, Sensei said:

You don't know what is m₂, what is v₂, and what is γ₂.

You just know m₁=rest-mass of electron/positron, v₁ and γ₁know from observation of longer or shorter traces in f.e. Cloud Chamber in beta decay minus reactions.

So assumption is made that energy is conserved, and thus the rest of energy is taken by neutrino/antineutrino in beta decay plus/minus (and daughter isotope little amount).

 

Example graph that I found on the net:

Bismuth-210 decays via:

Bismuth-210 -> Polonium-210 + e- + Ve + 1.16129 MeV

 

Ya, I can only go by what the Standard model says about a electron neutrino rest mass which is around 2.2 eV.

Edited October 2, 2017 by Vmedvil
Typos

[Sensei]

14 minutes ago, Vmedvil said:

Ya, I can only go by what the Standard model says about a electron neutrino mass which is 2.2 eV.

You should read it as "it's not more than 2.2 eV", not "it's exactly 2.2 eV"..

Even on graphics that you sent in 2nd post, there is <2.2 eV (less than sign).

 

[Vmedvil]

2 minutes ago, Sensei said:

You should read it as "it's not more than 2.2 eV", not "it's exactly 2.2 eV"..

Even on graphics that you sent in 2nd post, there is <2.2 eV (less than sign).

 

Ya, that was a typo I fixed it in the edit.

[Sensei]

This could be interesting for you Moontanman

http://www.nature.com/nature/journal/v512/n7515/fig_tab/nature13702_F1.html?foxtrotcallback=true

[Vmedvil]

19 minutes ago, Sensei said:

This could be interesting for you Moontanman

http://www.nature.com/nature/journal/v512/n7515/fig_tab/nature13702_F1.html?foxtrotcallback=true

ya, and the best answer I got for the rest mass of a neutrino was If the neutrino is a Majorana Particle it is between 0.060 eV and 0.161 eV from Fermilab. So, 0.1105 eV  ± 84% which is a huge standard deviation I know.

https://arxiv.org/ftp/arxiv/papers/0903/0903.0899.pdf

Edited October 2, 2017 by Vmedvil

[swansont]

15 hours ago, Vmedvil said:

Well, but W+ has the same Feynman diagram, so would it have the same path as pair production in its products? They are both bosons.

 

There's a recoil involved that isn't part of that end of the diagram. Even for PP, which doesn't occur in free space.

[Vmedvil]

3 minutes ago, swansont said:

There's a recoil involved that isn't part of the diagram. Even for PP, which doesn't occur in free space.

You have a point, ERROR 404.P, Physics not found, Lol

Edited October 3, 2017 by Vmedvil

[Moontanman]

My bad I was trying to compare the energy of neutrinos with the energy of photons being emitted. If you could collect the energy being given off in neutrinos would it compare to what a solar panel collects..    

[Sensei]

43 minutes ago, Moontanman said:

My bad I was trying to compare the energy of neutrinos with the energy of photons being emitted. If you could collect the energy being given off in neutrinos would it compare to what a solar panel collects..    

Like I tried to show in my post above it's between ~22 W/m^2 (~0.21 MeV average per neutrino) to ~34 W/m^2 (~0.324 MeV average per neutrino), while photons give us ~1370 W/m^2, after passing through atmosphere there is left ~1050 W/m^2 (at cloudless noon). Solar panel with efficiency ~16% will be able to give us ~170 W/m^2. 22/170 = ~13% (if the all neutrinos from pp reaction have been captured).

Edited October 4, 2017 by Sensei