Thursday, September 19, 2019

cosmology - How do Hubble Units work?


I'm reading through a document - Briel et al (1992) (pdf) - and I've come across Hubble units with which I'm not familiar. How do you interpret something like $$h^{1/2}_{50} \:\mathrm{cm}^{-3}$$ or similar constructs? This one appears to be the square root of some unit-less value per cubic centimeter.



Answer



It appears that the results of this paper are dependent on the precise value of the Hubble constant, but the only effect of this is to scale the different measured quantities by different amounts. The initial approach is to simply assume a reasonable value, which they take as $$H_0=50\:\mathrm{km\:s^{-1}Mpc^{-1}},$$ but a slightly more sophisticated approach is to incorporate the actual value of $H_0$ into a dimensionless constant which scales everything else. This is exactly what $\newcommand{\hf}{h_{50}}\hf$ is: $$ \hf:=\frac{H_0}{50\:\mathrm{km\:s^{-1}Mpc^{-1}}}, $$ and it is dimensionless.


Thus, $\hf^{1/2}\:\mathrm{cm}^{-3}$ is a number density (as stated in p. L33, final line of first paragraph, 'core electron density'), $\hf^{-1}\:\mathrm{Mpc}$ is a length ('core radius'), $\hf^{-2}\:\mathrm{erg\:sec^{-1}}$ is an energy flow ('luminosity of the source') and so on.



This usage is explicitly confirmed by e.g. arXiv:astro-ph/9504015.


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