Undergrad How to derive number of nucleons in Universe?

[jcap]

I understand that there are roughly $N=10^{79}$ nucleons in the visible Universe. This number comes from adding up the nucleons of $100$ billion stars in $100$ billion galaxies in the visible Universe i.e.
$$N=\frac{10^{30}}{10^{-27}}. 10^{11}.10^{11}=10^{79}$$
where mass of sun is $10^{30}$ kg and mass of proton/neutron is $10^{-27}$ kg.

Is there a simple way of deriving $N$ using the fundamentals of Big Bang Nucleosynthesis?

 

[phyzguy]

No, I don't think so. Big Bang Nucleosynthesis only re-arranged nucleons, it didn't change the total number of nucleons. It is possible to estimate this number from the Λ-CDM standard model of cosmology. In this model, the current density of the universe is closely equal to the critical density, which is about 8.6E-27 kg/m^3. About 5% of this is baryons, and the comoving radius of the observable universe is about 46 billion light-years or about 4E26 m. Multiplying this out gives:

N = 4 * π * (4E26 m)^3 * 8.6E-27 kg/m^3 * .05 / 1.6E-27 kg/nucleon = 2E80

This is close to what you said. These are order of magnitude estimates anyway. One difference is that most of the baryons are not in stars, most are in interstellar gas.

 

[kimbyd]

I understand that there are roughly $N=10^{79}$ nucleons in the visible Universe. This number comes from adding up the nucleons of $100$ billion stars in $100$ billion galaxies in the visible Universe i.e.
$$N=\frac{10^{30}}{10^{-27}}. 10^{11}.10^{11}=10^{79}$$
where mass of sun is $10^{30}$ kg and mass of proton/neutron is $10^{-27}$ kg.

Is there a simple way of deriving $N$ using the fundamentals of Big Bang Nucleosynthesis?

First, it is impossible to derive $N$ using fundamental arguments at the moment. As phyzguy mentioned, $N$ only shuffles them around a bit. The number is determined instead by the asymmetry between matter and anti-matter. Since we don't yet have a good theoretical understanding of the causes of this symmetry, it's impossible to derive from first principles.

That said, the calculation you laid out doesn't give us a very accurate picture of how many nucleons there are in the observable universe.

Our most precise measurements of the number of nucleons in the universe stems not from observations of galaxies, but from the cosmic microwave background.

The problem with measuring nucleons from galaxies is that only a fraction of the nucleons are visible. Most nucleons exist in diffuse interstellar and intergalactic gases that are very difficult to detect.

But before the CMB was emitted, the entire universe was a plasma. Within the plasma, the photons and electrons interacted very strongly, and when that plasma cooled to the point that it turned into a gas, those photons streamed freely through the universe with very little interference. Thus measuring the properties of the CMB gives us an extremely accurate measurement of the number of electrons in the universe. Then, from Big Bang Nucleosynthesis, we know the ratio of protons to neutrons quite precisely (both from measuring the primordial abundances of light elements, and from theory).

From the above, we know that the Baryon density of the universe is (assuming I didn't make a mistake):
$$8.50 \times 10^{-29} kg/m^3$$

The above number is accurate to about 1%. You can multiple the above number by the volume of the observable universe to get an estimate of the total mass, and then divide that by the mass of a nucleon to get the total number. But the density is usually quoted because it's the more physically-relevant number (any process of generating the matter/anti-matter symmetry would predict a density, not a total number, as the volume depends upon many other factors related to the rate of expansion and the other contents of the universe).