2
$\begingroup$

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 nuclesynthesis?

$\endgroup$
  • $\begingroup$ Related: physics.stackexchange.com/q/47941/2451 , physics.stackexchange.com/q/174820/2451 and links therein. $\endgroup$ – Qmechanic Nov 12 '17 at 18:13
  • $\begingroup$ There is a simple way. (A coincidence? I think not.) Assume the radius of the universe relates to its mass as in the Schwarzschild solution and also that the radius in light years equals the Hubble time. Then you get $M=\dfrac{c^3}{2HG}$. With the current value of H=73.8 you get $\text{N}=5\cdot 10^{79}$ nucleons. $\endgroup$ – safesphere Nov 21 '17 at 20:12
  • $\begingroup$ Most of the matter is not in stars but in intergalactic and interstellar mediums. So using stars to count nucleons should giv a significantly lower value. $\endgroup$ – M111 Nov 25 '17 at 21:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.