Wednesday, January 22, 2014

relativistic natural system of units

can you define fundamental scales for length, time and mass from the speed of light $c$, the gravitational coupling constant $G$ and the cosmological constant $\Lambda$? please compare size, age and density of the Universe to these fundamental scales.

2 comments:

  1. the density parameter $\Omega_\Lambda$ of the cosmological constant $\Lambda$ is $\Lambda/3/H_0^2$ because $\Lambda$ has units of squared inverse time, and the proximity of $\Omega_\Lambda$ to one means that $\Lambda$ is about an inverse squared Hubble time. $c$ has units of length divided by time and $G/c^2$ has units of length over mass, because as the gravitational coupling constant it assigns a length scale to the gravitational field of a mass. setting up an ansatz $\Lambda^\alpha\times c^\beta\times (G/c^2)^\gamma = l^{\beta+\gamma}\times t^{-2\alpha-\beta}\times m^{-\gamma}$ with units of length, time and mass to be determined: for the length scale $l$ one obtains $\alpha = -1/2$, $\beta=1$ and $\gamma=0$, yielding $l = c/\sqrt{\Lambda}=c/H_0/\sqrt{3\Omega_\Lambda}$, i.e. the Hubble distance with a small correctional factor. similarly, $t = 1/H_0/\sqrt{3\Omega_\Lambda}$ and $m=c/H_0\:c^2/G\:/\sqrt{3\Omega_\Lambda}$, implying that the mass scale $m$ corresponds to a mass whose length scale is the Hubble length. in contrast to Planck-units, the Universe looks very natural in this system of units.

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  2. from these scales one can derive a density scale $\rho=m/l^3=8\pi\:\Omega_\Lambda\:\rho_\mathrm{crit}$ with $\rho_\mathrm{crit}=3H_0^2/(8\pi G)$, again with a factor of "order unity".

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