In cosmology (Dodelson) , the natural unit is $\hbar=1, c=1, k_B=1 $. There the author writes the critical density of the universe $\rho_{cr}$ as \begin{align*} \rho_{cr}&=1.879 \ h^2\ \times 10^{-29}\ \text{gm/cm$^3$} \\ &= 8.098 \ h^2 \ \times 10^{-11}\ \text{eV$^4$} \end{align*} It is the conversion of units that I am not able to understand.

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    $\begingroup$ The section “Natural units (particle physics and cosmology)” at en.wikipedia.org/wiki/Natural_units shows how mass and length (and therefore mass density) can be measured in eV when $\hbar$ and $c$ are 1. $\endgroup$ – G. Smith Apr 26 '19 at 20:16

You just do the following: $$ \rho_{cr} = 1.879\cdot 10^{-26}\frac{\text{kg}}{\text{m}^{3}} = 1.879\cdot 10^{-26}(\text{kg}\cdot c^{2})\frac{(\hbar c)^{3}}{\text{m}^{3}}\frac{1}{c^{2}(\hbar c)^{3}} $$ Taking into account that $$ \text{kg}\cdot c^{2} = 9\cdot 10^{16}\text{ J} \approx 5.6 \cdot 10^{35}\text{ eV}, $$ $$ \frac{(\hbar c)^{3}}{\text{m}^{3}}\approx 7.7\cdot 10^{-21}\text{ eV}^{3}, $$

you obtain an expression for $\rho_{cr}$ in terms of $\hbar, c$: $$ \rho_{cr} \approx 8.1\cdot 10^{-11}\text{ eV}^{4}\frac{1}{c^{2}(\hbar c)^{3}} $$ In natural units, $c = \hbar = 1$, which gives you the result.

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