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Solar Neutrinos - Calculation of the solar neutrino flux on Earth

Photon flux calculation below

Proton-proton fusion chain process:
Two protons fuse to a deuterium nucleus and release a positron and a neutrino. The positron quickly annihilates with an electron and produces a high energy photon (gamma ray). This photon is bouncing between the dense matter in the solar interior, can also be absorbed and be re-emitted in random directions as one or more photons (radiative transport).

The energy transported in this process needs a few millions(!) of years to reach the outer parts of Sun (due the dense interior of Sun) in form of many low energy photons (visible light). In the outer part of the Sun the photons are more efficiently absorbed, therefore radiative transport is replaced by convection to transport the energy to the solar surface. 

p + p -> 2H + e+ + νe
p + 2H -> 3He + gamma
3He + 3He -> 4He + p + p

Calculation of the solar neutrino flux:

Units and conversions:
1 eV = 1.60217733*10-19J
1J = 1Nm
1N = 1kgm/s2
1u = 1.6605402*10-27kg

1 eV = 1.60217733*10-19 J = 1.60217733*10-19 kgm2/s2
1m2/s2 = 6.241506*1018 eV/kg

c2 = (3*108m/s)2 = 9*1016m2/s2 = 9*1016 * 6.241506*1018 eV/kg => 
5.609586*1035 eV/kg = 931.494334 MeV/u


In the proton-proton fusion chain process, 4 protons fuse into 1 Helium atom. The mass of 4 protons is not equal to the mass of one Helium atom - the difference is converted into neutrinos: 
dm = (4*mp - mHe) = (4*1.00794 - 4.002602) = 0.029158 u
Now we can calculate the energy of one neutrino: 
Eν = dm*c2 = 0.029158u * 931.494334 MeV/u = 27.160511 MeV
We need to convert the solar-constant for our final formula: 
Solar-constant: 1370W/m2
1W = 0.238846 cal/s ; 1cal = 2.613194*1019 eV
1W = 6.241506*1018 eV/s
1370W/m2 = 8.550863*1021 eV/(m2s) = 8.550863*1011MeV/(m2s)
With this final formula we get the solar neutrino flux at Earth: 
n = S * 2 / Eν = 8.550863*1011MeV/(m2s) * 2 / 27.160511MeV = 6.296541*1014 /(m2s)


It follows from the above:
~6·1014 solar neutrinos per square meter per second
600000 billions neutrinos per square meter per second
6000 billions neutrinos per square decimeter per second
60 billion neutrinos per square centimeter per second
or if we use our hand for comparison: 6000 billions neutrinos pass thru your open hand per second
or if we use one thumbnail for comparison: 60 billion neutrinos pass thru your thumbnail per second

Photons - Calculation of the solar photon flux on Earth

Some definitions: 

1 W	= 1 J/s
1 eV	= 1.602*10-19 J
1J	= 1/(1.602*10-19)eV = 6.242*1018eV
=> 1 W	= 6.242*1018eV/s
=> 1 W/m2 = 6.242*1018 eV/(m2*s)

h = 6.626068*10-34 kg*m2/s (Planck's constant)
with [erg] = kg*m2/s2
and 1 erg = 10-7 J
h = 6.6*10-27 erg*s
The sun emmits a broad range of wavelengths, but 556nm is considered as the central wavelength of suns light. 
f	= c / 556nm = 5.392*1014 Hz
1 erg	= 6.242*1011eV

Esingle-photon = f * h = 3.573*10-12 erg = 3.573*10-12 erg * 6.242*1011eV/erg = 2.230 eV/photon
So we calculate the number of photons @ 556nm for the power of 1W/m2: 
P1W/m2@556nm = 6.242*1018 eV/(m2*s) / 2.230 eV/photon = 2.799*1018 photons/(m2*s)
Multiplying P1W/m2@556nm with the power of sunlight at Earth distance (1370W/m2) 
P1370W/m2@556nm = 2.799*1018 photons/(m2*s) * 1370 = 3.835*1021 photons/(m2*s)
we get the photon flux at Earth with roughly 3.8*1021 photons / (m2*s)

References:
R. Frey - Lecture Notes for Astronomy 321, W 2004
J. R. Reid - Space Weather

Last-Modified: Thu, 23 Feb 2006 19:49:30 GMT

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