# What is the de Sitter temperature as of now?

It is claimed that de Sitter temperature is $$T=\frac{1}{2\pi}H,$$

where $H$ is the Hubble constant. I presume it is expressed in natural units with which I am not familiar. So what it will be in Kelvins? Is it higher than the temperature of CMB? If so, is there any heat exchange between the both?

The unit of $H$ is the inverse of time denoted by $[\mathrm T^{-1}]$. The dimension of a temperature is denoted by $[\Theta]$. To find the numerical value of $T$ in kelvin, one should find a combination of $c:[\mathrm{LT^{-1}}]$, $\hbar=[\mathrm{ML^2T^{-1}}]$, $\mathcal G:[\mathrm{M^{-1}L^3T^{-2}}]$ and $k_{\mathrm B}:[\mathrm{ML^2T^{-2}\Theta^{-1}}]$ ($[\mathrm L]$ and $[\mathrm M]$ represent length and mass respectively) such that $$c^x \;\hbar^y \;\mathcal G ^z\; k_{\mathrm B}^u\times H$$ has the dimension of a temperature ($x$, $y$, $z$ and $u$ are unknown). This gives the system $$\left\{\begin{array}{rcc} y-z+u&=0&\quad[\mathrm M]\\ x+2y+3z+2u&=0&\quad[\mathrm L]\\ -x-y-2z-2u&=1&\quad[\mathrm T]\\ -u&=1&\quad[\Theta] \end{array}\right.$$ The solution is $$\left\{\begin{array}{cl} x&=0\\ y&=1\\ z&=0\\ u&=-1 \end{array}\right.$$ We obtain thus $$T=\frac{\hbar}{2\pi k_{\mathrm B}}H.$$ The value of $H$ is $H=67.8\, \mathrm{km.s^{-1}.Mpc^{-1}}=2.194\times 10^{-18}\,\mathrm{m.s^{-1}}$. We find a temperature of $T=2.67\times10^{-30}\,\mathrm K$.