Solar fusion, neutrinos and age of solar system

[Sensei]

Hello!

 

Each [latex]m^2[/latex] of surface of Earth is receiving 1367 W (ignoring atmosphere influence).

 

We know inverse-square law:

 

[latex]P = \frac{P_0}{4*\pi*r^2}[/latex]

 

After reversing it, we're receiving:

 

[latex]P_0 = P*4*\pi*r^2[/latex]

 

Distance to the Sun is approximately [latex]r = 150*10^9[/latex] meters.

 

After plugging numbers we get:

 

[latex]P_0 = 1367*4*\pi*(150*10^9)^2=3.8651*10^{26} W[/latex]

 

As you can see this pretty much agree with wikipedia Sun page

http://en.wikipedia.org/wiki/Sun

(Luminosity on the right table)

 

It's energy Sun is emitting every single second.

 

One of possible fusion path is:

 

Fusion of 4 protons to 2 Deuterium:

[latex]p^+ + p^+ \rightarrow D^+ + e^+ + V_e + 0.42 MeV[/latex] (neutrino takes part of energy)

[latex]p^+ + p^+ \rightarrow D^+ + e^+ + V_e + 0.42 MeV[/latex]

[latex]e^+ + e^- \rightarrow \gamma + \gamma + 1.022 MeV[/latex]

[latex]e^+ + e^- \rightarrow \gamma + \gamma + 1.022 MeV[/latex]

 

Fusion of 2 Deuterium to 2 Helium-3:

[latex]p^+ + D^+ \rightarrow _2^3He + \gamma + 5.49 MeV[/latex]

[latex]p^+ + D^+ \rightarrow _2^3He + \gamma + 5.49 MeV[/latex]

 

Fusion of 2 Helium-3 to 1 Helium-4:

[latex]_2^3He + _2^3He \rightarrow _2^4He + p^+ + p^+ + 12.86 MeV[/latex]

 

Sum of energies:

E = 0.42 MeV + 1.022 MeV + 1.022 MeV + 5.49 MeV + 5.49 MeV + 12.86 MeV = 26.304 MeV

 

Conversion to Joules:

[latex]26.304 MeV * 10^6 * 1.602*10^{-19} = 4.2139008*10^{-12} J[/latex]

 

Divide energy Sun has to emit by single fusion cycle to get quantity of reactions per second:

[latex]\frac{3.8651*10^{26} W}{4.2139008*10^{-12} J} = 9.17*10^{37}s^{-1}[/latex]

 

Additionally we can calculate quantity of neutrinos:

[latex]9.17*10^{37}*2 = 1.8344*10^{38}[/latex]

[latex]\frac{1.8344*10^{38}}{4*\pi*(150*10^9)^2}=6.488*10^{14} m^{-2}=64.88 \frac{bln}{cm^2}[/latex]

Which also nicely fits with wiki page neutrino flux 65 bln.

http://en.wikipedia.org/wiki/Neutrino

 

Mass of the Sun is [latex]1.98855*10^{30} kg[/latex]

 

Single atom of Hydrogen has mass = [latex]\frac{\frac{1.007825}{1000}}{6.022141*10^{23}}=1.6735*10^{-27} kg[/latex]

Single atom of Helium has mass = [latex]\frac{\frac{4.0026}{1000}}{6.022141*10^{23}}=6.6465*10^{-27} kg[/latex]

 

If we know mass of object, and what it's made of, we can calculate quantity of atoms.

 

According to this website:

http://en.wikipedia.org/wiki/Solar_core

Sun's core has 34% of mass of the Sun. And it's made of Helium in ~60%, and ~40% Hydrogen.

Phrase "so the innermost portion of the Sun is now roughly 60% helium" can be found on net and Sun wiki page

http://en.wikipedia.org/wiki/Sun

 

[latex]1.98855*10^{30} kg * 0.34 * 0.4 = 2.7*10^{29} kg / mass_H = 1.616*10^{56}[/latex] Hydrogen atoms

[latex]1.98855*10^{30} kg * 0.34 * 0.6 = 4.0566*10^{29} kg / mass_{He} = 6.1*10^{55}[/latex] Helium atoms

 

If core has 34% of mass then in outer regions of the Sun there will be:

[latex]1.98855*10^{30} kg * 0.66 * 0.7346 = 9.6412*10^{29} kg / mass_H = 5.761*10^{56}[/latex] Hydrogen atoms

[latex]1.98855*10^{30} kg * 0.66 * 0.2485 = 3.2614*10^{29} kg / mass_{He} = 4.907*10^{55}[/latex] Helium atoms

 

In total we can find now

[latex]6.1*10^{55}+4.907*10^{55}=1.101*10^{56}[/latex] Helium atoms

 

Initial quantity of Helium atoms according to theories was 27.4%, so:

[latex]1.98855*10^{30} kg * 0.274 = 8.1978*10^{55}[/latex] Helium atoms

 

Time needed to fuse can be calculated by f.e.:

[latex] t = \frac{Q_{current}-Q_{init}}{rate}[/latex]

(Q - quantity)

 

[latex] t = \frac{1.101*10^{56}-8.1978*10^{55}}{9.17*10^{37}}=3.07*10^{17} seconds = 9.7[/latex] bln years

 

Why so large discrepancy from 4.56 bln years?

Where is error?

Abundance of Helium-4 in the core is lower than 60% as all materials are suggesting?

 

And this graph would make Sun even more older than that:

The smaller luminosity (power), the smaller amount of Helium-4 produced per second, and star must be even older..

 

Faster rate of production = more energy emitted to cosmos = vaporization of planets..

 

ps. In older version of calc, I was taking into account also lost of mass due to flares etc. but don't want to be boring you too much..

Edited September 22, 2014 by Sensei

[swansont]

The wikipedia page says that 60% of the innermost region is He, not of the whole sun. I think your outer regions calculation is moot. The core isn't convective, so the He produced there shouldn't get to the outer regions.

 

 

How much of the current mass in the core is He-3? The wikipedia article doesn't say the 60% is all He-4.

[mathematic]

He3 is an intermediate in the H1 to He4 transformation, so the % should be minimal.

[Sensei]

The wikipedia page says that 60% of the innermost region is He, not of the whole sun.

I didn't calculate from whole Sun. But multiplied by 0.34 (34%) first (calculated 60% from just core).

0.34 * 0.6 = 0.204

 

I think your outer regions calculation is moot. The core isn't convective, so the He produced there shouldn't get to the outer regions.

Yes, I am aware of no convection (according to theory/model).

But I don't care about convection in this equation at all..

 

If you start reaction with x particles (no matter if it's star or chemical reaction at home/lab), and then it's growing to x+y after t seconds. x remain constant. y/t is rate of production.

x is initial amount of Helium

x+y is current amount of Helium

 

Meaningless amount of He-4 fused further (according to wiki CNO cycle is 0.8% of energy source at the moment).

 

How much of the current mass in the core is He-3? The wikipedia article doesn't say the 60% is all He-4.

I have no idea.

But 2 He-3 would fuse quickly to He-4 and two protons at the beginning of life of star, so it might be meaningless..

That's why it's so rare isotope (1.34(3)×10^−6)

Edited September 23, 2014 by Sensei

[swansont]

 

I have no idea.

But 2 He-3 would fuse quickly to He-4 and two protons at the beginning of life of star, so it might be meaningless..

That's why it's so rare isotope (1.34(3)×10^−6)

 

Two He-3 in the core are very unlikely to collide, which has to happen for them to fuse. The reaction rate is concentration-dependent. It's probably small enough it can be ignored, but I don't know for sure.

 

The abundance you cite is for earth, where He-4 is (and has been) in constant production via alpha decay. There's no reason to assume that the sun (or any star) has a similar composition.

[Sensei]

Two He-3 in the core are very unlikely to collide, which has to happen for them to fuse. The reaction rate is concentration-dependent. It's probably small enough it can be ignored, but I don't know for sure.

That's true it's unlikely to collide.

 

But as I showed at the beginning of post, it must happen [latex]9.17*10^{37}[/latex] times per second anyway. Otherwise Earth would be able to get 1367 J of energy per meter^2 per second.

 

Helium-4 fusion is even more unlikely to happen, because it requires 3 atoms at the same time (Be-8 has more mass-energy than 2 He-4 alone, and additionally decays to 2 He-4 + 92 keV, very rapidly):

 

[latex]_2^4He + _2^4He + _2^4He \rightarrow _6^{12}C + \gamma + 7.275 MeV[/latex]

 

More likely is fusion between He-3 and He-4 (if your suggestion about larger amount of He-3 in core is true):

 

[latex]_2^3He + _2^4He \rightarrow _4^{7}Be + \gamma + 1.5861 MeV[/latex]

 

Beryllium-7 is unstable and decays to Lithium-7 via electron capture:

 

[latex]_4^7Be + e^- \rightarrow _3^7Li + V_e + 0.861893 MeV[/latex]

 

This is exactly what is triggering Chlorine based neutrino detector (neutrinos from proton-proton have up to 0.42 MeV too low to trigger this kind of detector).

 

The abundance you cite is for earth, where He-4 is (and has been) in constant production via alpha decay.

True.

 

There's no reason to assume that the sun (or any star) has a similar composition.

We're ashes remaining from older generation of stars. He-3 can probably only two (or three) ways to be created: by decay from Tritium, or by fusion between proton and Deuterium.

[swansont]

That's true it's unlikely to collide.

 

But as I showed at the beginning of post, it must happen [latex]9.17*10^{37}[/latex] times per second anyway. Otherwise Earth would be able to get 1367 J of energy per meter^2 per second.

 

And if it has a big cross-section, you don't need much of it. But if it has a small cross-section, you need a lot of it, and then the statement that 60% of the core is Helium doesn't mean it's all He-4, which affects your calculation.

 

I didn't calculate from whole Sun. But multiplied by 0.34 (34%) first (calculated 60% from just core).

0.34 * 0.6 = 0.204

 

You have a section where you calculate the He from the outer part of the sun, and use that number in the total. So yes, you do calculate a number for the whole sun:

 

If core has 34% of mass then in outer regions of the Sun there will be:

[latex]1.98855*10^{30} kg * 0.66 * 0.7346 = 9.6412*10^{29} kg / mass_H = 5.761*10^{56}[/latex] Hydrogen atoms

[latex]1.98855*10^{30} kg * 0.66 * 0.2485 = 3.2614*10^{29} kg / mass_{He} = 4.907*10^{55}[/latex] Helium atoms

 

In total we can find now

[latex]6.1*10^{55}+4.907*10^{55}=1.101*10^{56}[/latex] Helium atoms

[Enthalpy]

I suggest to compare the masses of neutral atoms of hydrogen and helium, without detailing the reactions in between, since this is the net effect of fusion. The only imprecision is between neutral atoms and plasma, approximately 4*13,6eV versus 24,6+54,4eV: only 24,6eV as compared to MeV, and the kinetic energy of the species, 1keV at 10MK.

 

4 neutral hydrogen atoms of 1.00795u become 1 helium of 4.0026u, releasing 0.029u of 1.66e-27kg, that's 4.36e-12J per helium atom. Agreed.

 

3.9e26W need more than (because neutrinos evacuate a part) 1.5e14mol/s helium creation. That's 1.9e28kg/Gyr (billion years).

 

All from Wiki, the core weighing 34% of our Sun's 2.0e30kg being composed of 60% He instead of initial 27.4% has created 2.2e29kg of helium.

 

This would have needed 12Gyr, not 4.6Gyr, agreed as well.

 

----------

 

Explanation attempts:

• Neutrinos evacuate a significant portion of the fusion energy?
• 60% He is a proportion at the center, not a mean value over the core?
• The 60% and 27% at Wiki are wrong or undetailed? Nobody measured them. I suppose their difference results from computations like yours, just more differentiated over the depth.

Edited September 24, 2014 by Enthalpy

[Sensei]

Neutrinos evacuate a significant portion of the fusion energy?

Neutrinos are taken into account. See below "Sum of energies". I am adding just once 0.42 MeV, assuming 50% of energy is taken by neutrino.

 

73.46% Hydrogen and 24.85% Helium from photosphere are pretty accurate measurements (similar content has Jupiter), as they're obtained from spectral lines of light from the Sun.

What is inside is prediction from models, and will remain this way. After all we won't send there any device to check it experimentally.

Edited September 24, 2014 by Sensei

[Sensei]

Using this graph we can calculate energy that Sun emitted in all its predicted by model lifetime.

 

We can draw three figures.

Triangle 1:

[latex]A_1 = \frac{1}{2}*(0.9-0.75)*0.4 = \frac{1}{2}*0.15*0.4 = 0.03[/latex]

Triangle 2:

[latex]A_2 = \frac{1}{2}*(1.0-0.75)*(4.56-0.4) = \frac{1}{2}*0.25*4.16 = 0.52[/latex]

and rectangle:

[latex]A_3 = 0.75*4.56 = 3.42[/latex]

(1.0 represents power of current Sun)

 

Sum of their areas:

[latex]A = A_1+A_2+A_3 = 0.03 + 0.52 + 3.42 = 3.97[/latex]

 

[latex]\frac{3.97}{4.56} = 0.87061 = 87\%[/latex] of idealized energy emission from 1st post.

 

 

[latex]P_{average} = 3.8651*10^{26} W * 87\% = 3.365*10^{26} W[/latex]

Energy emitted for the last 4.56 bln years:

[latex]3.365*10^{26} * 60*60*24*365.25 * 4.56*10^9 = 4.8423243744*10^{43} J[/latex]

So, instead of [latex]9.17*10^{37}[/latex] Helium-4 atoms produced per second there will be average [latex]7.9835^{37}[/latex]

[latex] t = \frac{1.101*10^{56}-8.1978*10^{55}}{7.9835^{37}}=3.52252*10^{17} seconds = 11.13[/latex] bln years

Edited October 3, 2014 by Sensei

[Mordred]

That's the one, I have a luminosity distance to z relation I wanted to verify however may also look at determining age of stars.