Wednesday, May 28, 2014

temperature in a FLRW-universe

what temperature would you associate with gravity in a FLRW-universe and how would that temperature compare to the actual temperature or the Planck-temperature?

3 comments:

  1. i never felt as uncertain as this when answering: one can write down the first law of thermodynamics as $\theta\mathrm{d}A=c^2\mathrm{d}M=T\mathrm{d}S$ with area $A$, mass $M$ and entropy $S$. $\mathrm{d}A/\mathrm{d}S=4l_p^2$ is given by the BH-entropy with the Planck-length $l_p=\sqrt{\hbar G/c^3}$, and $\theta$ can be determined using $\theta=c^2\mathrm{d}M/\mathrm{d}A$, which gives $\theta=c^2\rho_\mathrm{crit}\chi_H/2$ with the length scale $\chi_H=c/H_0$ and the critical density $\rho_\mathrm{crit}=3H_0^2/(8\pi G)$. then, the temperature follows as $T = 3/(4\pi)c\hbar/\chi_H$.

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  2. it's possible to bring this result into a more "accessible" form by expressing the Hubble-radius $\chi_H$ in terms of the Planck length $l_p$. then, the temperature $T$ in units of the Planck-temperature $T_p = \sqrt{c^5\hbar/G}$ is just proportional to $l_p/\chi_H$, which has a numerical value of $\sim 10^{-60}$!

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  3. one more thing that blows my mind: with a numerical value of the Planck temperature of $10^{30}$ Kelvin one would obtain a value of $10^{-30}$ Kelvin for the FLRW-temperature... and the actual CMB-temperature would be right in the middle (on a logarithmic scale)!

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