# How does the Planck satellite's detailed map of the CMB lead to a value of the Hubble constant, $H_0$?

Cosmologists have put great faith, it seems, on the cosmological model that led to a value (of 67) of the Hubble 'constant', after carefully peering at Planck's map of the cosmic microwave background....

But, I have nowhere read an article or (journal) paper, detailed or not, about how a close observation and analysis of the CMB leads to a value for Hubble's constant....

Does anybody know how Planck's CMB data/map led to a calculation of Hubble's constant?

Does anyone know the logic behind this?

Or a link? (Even if behind a paywall, or something....)

• Have you read the papers by the Planck Collaboration? They're very detailed Commented Mar 3, 2021 at 11:38

In cosmology, there are three "things" in the universe: radiation, matter, and dark energy. At various points in the universe's history, each of these things have dominated the universe's evolution. We call these periods "radiation-dominated", "matter-dominated", and "dark energy dominated" (this last period also goes by other names since dark energy isn't well understood). The three are different because they scale differently with volume. In particular, if you have a particle in the box and double the size of the box, the volume goes up by a factor of 8, and the density goes down by the same factor. Mathematically we say $$\rho(m) \propto a^{-3}$$, where $$\rho$$ is the density, and $$a$$ is the so-called scale factor. With radiation, not only does the density go down, there is also a redshift. We write $$\rho(r) \propto a^{-4}$$. Finally with dark energy, if it's described by a cosmological constant, then it is a constant - its density does not change. $$\rho(\Lambda) \propto a^{0}$$.
$$\frac{H^2}{H_0^2} = \Omega_{0,R} a^{-4} + \Omega_{0,M} a^{-3} + \Omega_{0,k} a^{-2} + \Omega_{0,\Lambda}$$
Here the $$\Omega$$'s are the present-day density of radiation, matter and dark energy, and $$\Omega_{0,k}$$ is the contribution due to the curvature of the universe.