# Dumbed-down explanation how scientists know the number of atoms in the universe?

It is often quoted that the number of atoms in the universe is 10$^{70}$ or 10$^{80}$.

How do scientists determine this number?

And how accurate is it (how strong is the supporting evidences for it)?

Is it more likely (logically >50% chance) that the numbers are right, or is it more likely that the numbers are wrong?

-
They don't. They estimate. –  OmnipresentAbsence Feb 23 '13 at 22:06
The number of atoms in the universe may be infinite, if the universe is spatially infinite. What can be estimated is the number of atoms in the observable universe. –  Ben Crowell May 28 '13 at 23:17
Stars are fusing four hydrogens into one helium. Older stars fuse up the Periodic Table to iron. A large star will eventually suffer core collapse to a neutron star or a black hole. The number of atoms in the universe has been continuously decreasing since the Big Bang cooled. The "number" of atoms is an estimate assuming mass occurs as atoms, then observed composition abundances. –  Uncle Al Feb 22 '14 at 15:41

The observable universe contains about 100 billion galaxies, each containing on average close to a trillion stars. That is a total of about $10^{23}$ stars. A typical star is like our sun. Sun has a mass of about $2×10^{30}$ kg, which equates to $10^{57}$ atoms of hydrogen per star. A total of $10^{23}$ stars containing $10^{57}$ atoms each gives us a total number of atoms of $10^{80}$.

More detail, including an alternative estimation method based on cosmic microwave background observations, can be found here.

-
But how is it possible that "each containing on average close to a trillion stars" could be verified reasonably enough, for us to say with enough certainty and conviction that there's over 50% chance that it is correct? For all we know, out of this 100 billion galaxy there could be a single galaxy that already has over $10^{23}$ stars... just curious. –  Pacerier Dec 30 '12 at 18:34
@Pacerier - That would be a gross violation of the cosmological principle which says 'Viewed on a sufficiently large scale, the properties of the Universe are the same for all observers.' In other words, our universe is homogeneous. Deep sky galaxy surveys, such as the Sloan Digital Sky Survey, but also cosmic microwave background observations, all confirm the cosmological principle. –  Johannes Dec 30 '12 at 18:39
@Pacerier - what makes you say these numbers are more likely than not correct? It occurs to me that the confidence interval for the numbers of atoms in the universe to be within say $10^{78}$ and $10^{82}$ certainly exceeds 50%. –  Johannes Dec 30 '12 at 18:44
I'd hardly consider "between $10^{70}$ and $10^{80}$" to be a very precise estimate. We could be darn certain that the number of atoms in the universe is between $10^{50}$ (roughly the number of atoms in the earth) and $10^{150}$ (a ridiculously large number). This is a range of $\pm 50$ orders of magnitude, but it is fair to say that we know with virtual certainty that the number of atoms in the universe falls within that range. Somewhere between $10^{70}$ to $10^{80}$ is sufficiently vague as to be accurate, more precise than $10^{50-150}$, and a lot more useful than "unknown". –  KDN Dec 30 '12 at 21:22
@Johannes I think the claim that a galaxy contains on average 1 trillion stars is very bold. Our galaxy is a big one and it contains around 300 billion stars. Otherwise a good answer –  OmnipresentAbsence Feb 23 '13 at 22:06

The cosmological estimation of the number of atoms in the observable universe works as follows: one of the Friedmann equations can be written as $$\dot{a}^2 -\frac{8\pi G}{3}\rho a^2= -kc^2,$$ where the scale factor $a(t)$ describes the expansion of the universe, $\rho$ is the total mass density (radiation, baryonic matter, dark matter, and dark energy) and the integer $k$ is the intrinsic curvature of the universe ($k$ can be 1, 0 or -1). Observations of the Cosmic Microwave Background (CMB) indicate that the spacial curvature $k/a^2$ of the universe is practically zero, so we can set $k=0$. In this case the total density is equal to the so-called critical density $$\rho_\text{c}(t) = \frac{3H^2(t)}{8\pi G},$$ where $$H(t) = \frac{\dot{a}}{a}$$ is the Hubble parameter. The present-day density is then $$\rho_\text{c,0} = \rho_\text{c}(t_0) = \frac{3H_0^2}{8\pi G},$$ with $H_0=H(t_0)$ the Hubble constant. We can write $H_0$ in the following form $$H_0 = 100\,h\;\text{km}\,\text{s}^{-1}\,\text{Mpc}^{-1},$$ with $h$ a dimensionless parameter and $1\;\text{Mpc}=3.0857\times 10^{19}\;\text{km}$ (called a megaparsec). So $$\rho_\text{c,0} = 1.8785\,h^2\times 10^{-26}\;\text{kg}\,\text{m}^{-3}.$$ A detailed analysis of the Cosmic Microwave Background reveals what the density of ordinary matter (baryons) is: according to the latest CMB data, the present-day baryon fraction is $$\Omega_\text{b,0}h^2 = \frac{\rho_\text{b,0}}{\rho_\text{c,0}}h^2 = 0.02205 \pm 0.00028.$$ Notice how accurately this quantity is known. The same data also yield a value of the Hubble constant: $$H_0 = 67.3 \pm 1.2\;\text{km}\,\text{s}^{-1}\,\text{Mpc}^{-1},$$ in other words, $h = 0.673\pm0.012$ so that $$\Omega_\text{b,0} = 0.0487,$$ which means that ordinary matter makes up 4.87% of the content of the universe. We don't actually need the value of $h$ to calculate the baryon density $\rho_\text{b,0}$, because the factor $h^2$ cancels out: we get $$\rho_\text{b,0} = \Omega_\text{b,0}\rho_\text{c,0} = 0.4142\times 10^{-27}\;\text{kg}\,\text{m}^{-3}.$$ About 75% of the baryon density is in the form of hydrogen, and nearly 25% is helium; all other elements make up about 1%, so I'll ignore those. The masses of hydrogen and helium atoms are \begin{align} m_\text{H} &= 1.674\times 10^{-27}\;\text{kg},\\ m_\text{He} &= 6.646\times 10^{-27}\;\text{kg}, \end{align} so the number density of hydrogen and helium atoms is \begin{align} n_\text{H} &= 0.75\rho_\text{b,0}m_\text{H} = 0.1856\;\text{m}^{-3},\\ n_\text{He} &= 0.25\rho_\text{b,0}m_\text{He} = 0.0156\;\text{m}^{-3}, \end{align} and the total number density of atoms is $$n_\text{A} = n_\text{H}+n_\text{He} = 0.02012\;\text{m}^{-3}.$$ Now, the radius of the observable universe is calculated to be $D_\text{ph} = 46.2$ billion lightyears, which is $4.37\times 10^{26}\,\text{m}$ (the subscript 'ph' stands for particle horizon; see this post for a detailed explanation). This is a derived value, which depends on all cosmological parameters; nonetheless, it is accurate to about 1%. The volume of the observable universe is thus $$V = \frac{4\pi}{3}\!D_\text{ph}^3 = 3.50\times 10^{80}\;\text{m}^3.$$ So finally, there are about $$N_\text{A} = n_\text{A}V = 7.1\times 10^{79}$$ atoms in the observable universe.

-
Most interesting indeed - a very thorough and simple explanation. "...detailed analysis of the Cosmic Microwave Background ..." is there a simple summary of how the proportion of baryons is estimated from this? That's quite a colossal paper and a colossal piece of work. Were you involved? –  WetSavannaAnimal aka Rod Vance Nov 4 '14 at 0:03
@WetSavannaAnimalakaRodVance No, I wasn't involved, I'm not even a cosmologist :-) The details are complicated, but it has to do with how baryons influence the propagation of sound waves in the early universe, which eventually are imprinted on the CMB. A great online tutorial is Wayne Hu's site. The key plot can be found here. –  Pulsar Nov 4 '14 at 21:57

## protected by Qmechanic♦Feb 22 '14 at 13:41

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.