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Take an squared area with (10²)² m² in front of the sun. What is the bigger number of particles crossing an area: the number of photons or the number of neutrinos?

Just for clarification: you can calculate the values on earth, if you want (you'd need a distance to be able to calculate). You can use simplifications if they are needed and justified. If my area is limiting the wavelength you can as well make the math with an arbitrary (but a completely detailed and specified area in that case).

And make sure to put some sources of the data you've taken.

I'm asking this question after coming across the upper bound of the photon mass.

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  • $\begingroup$ There are far more photons than neutrinos. That should be obvious from the peaks of their spectra. $\endgroup$
    – CuriousOne
    Commented Oct 29, 2015 at 20:29
  • $\begingroup$ Estimates for both quantities can be found in Mukhanov's book on cosmology. $\endgroup$
    – Danu
    Commented Oct 29, 2015 at 20:29
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    $\begingroup$ @Vendetta That is indeed Mukhanov's (only) book on cosmology. $\endgroup$
    – Danu
    Commented Oct 29, 2015 at 20:36
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    $\begingroup$ Wait! I know this one, is it $7 \times 10^{10} neutrinos\:\mathrm{ cm^{-2} s^{-1}}$? en.wikipedia.org/wiki/Solar_neutrino#Observed_data $\endgroup$
    – Gert
    Commented Oct 29, 2015 at 21:27
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    $\begingroup$ Are you looking for just solar photons and solar neutrinos, or all photons and all neutrinos, or something else? $\endgroup$
    – user10851
    Commented Oct 29, 2015 at 22:52

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That the numbers of photons far, far outweighs the number of neutrinos can also be determined as follows.

One of the main contributors of energy production in the Sun's core is the proton-proton chain reaction, which produces one neutrino and about $26.7\:\mathrm{MeV}$ of energy per helium atom produced. This colossal amount of energy is gradually partitioned into smaller and smaller packets as it diffuses from the core to the surface of the Sun, which it leaves mainly as visible light photons.

The energy of an average VIS light photon however is only about $1.5\:\mathrm{eV}$, so millions of times smaller than $26.7\:\mathrm{MeV}$, ergo the number of photons leaving the Sun must be many times that of neutrinos leaving the Sun. And assuming isotropic (uniform) distribution of solar radiation, the number of solar photons reaching Earth must far outnumber the number of solar neutrinos reaching Earth.

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