# What is the age of a universe with positive, negative, and zero curvature?

I am trying to calculate the age of universes with different curvatures using the Hubble constant and Friedmann equation.

What does it mean when we say that the universe started out at equipartition at the Planck time? I was told the statement implied that the energy density in matter and radiation are of order the Planck density, and the temporal and spatial curvature radii are of order the Planck length. But what can we get about the Hubble parameter $$H(z)$$, where $$z$$ is the redshift, from this statement?

• If you're just asking about calculating the age of the universe that is done in How does the Hubble parameter change with the age of the universe?. Commented Dec 17, 2018 at 5:49
• There is some cosmological calculators for that. You can find them by reasearch. Commented Dec 17, 2018 at 6:10

The spatial curvature $$K$$ of a universe depends on its energy densities according to the first Friedmann Equation. Given the energy densities the age of the universe is model dependent, see 4.6 Simple Universe Models.
The Friedmann Equations start out at $$t = 0$$. The Big Bang and things like the Equipartition Theorem are not related to them nor is the Hubble Parameter $$H$$. I'm not sure why you want to combine $$H=\dot{a}/a%$$ with the cosmological redshift $$1+z=a_{now}/a_{then}$$. Both, $$H$$ and $$z$$ are not related somehow to the Planck epoch which is subject of a future theory of Quantum Gravity.
• I see. Thank you for your reply！I am trying to follow the lookback time derivation in Carroll (ned.ipac.caltech.edu/level5/Hogg/Hogg10.html), and calculate the time taken for a positively-curved universe to collapse, and the time to reach 3 K for a flat/negatively-curved universe. In order to use this equation, I wan tto choose a form for E(z) = $\sum_{i(c)}^{} \Omega_{i0}(1+z)^{n_{i}}$, which depends on whether the universe is matter/radiation-dominated. In this case, I am not sure what to use for E(z) based on the description of the problem. Commented Dec 17, 2018 at 15:06