Physics Asked by VeritasK on April 30, 2021
I’m not really knowledgeable on physics but was curious about this and couldn’t find any good answers related to it.
Yes.
"The curvature of the universe" is an imprecise term, and describing the curvature of a general four-dimensional spacetime takes 20 numbers at every point. But I'll assume that your phrase should mean the Ricci scalar curvature $R$, which is a single number at each point that is a kind of average curvature of spacetime (where the averaging is over various "directions" in which one can slice spacetime). In our universe, this number is uniform throughout space (on the largest scales) but decreasing with time. It is asymptotically approaching a constant, nonzero value.
On the largest scales our universe seems to be highly homogeneous, isotropic, and spatially flat (i.e., it has no spatial curvature, but it does have spacetime curvature). Such a universe is described by a Friedmann metric with $k=0$.
The Ricci scalar curvature of such a universe given by
$$R(t)=6left(frac{ddot{a}(t)}{a(t)}+frac{dot{a}(t)^2}{a(t)^2}right)tag1$$
where $a(t)$ is the dimensionless Friedmann scale factor, which was $0$ at the Big Bang and is $1$ today. The dots mean derivatives with respect to cosmological time $t$, and the units are such that $c=1$.
Beginning about 10 million years after the Big Bang, radiation stopped being important to its expansion rate. A solution to the Friedmann equations that takes into account dark energy, dark matter, and ordinary matter (but not radiation, because it is no longer important) is
$$a(t)=left(frac{1-Omega_Lambda}{Omega_Lambda}right)^{1/3}sinh^{2/3}left(frac32Omega_Lambda^{1/2}H_0 tright)tag2$$
where $H_0$ is the current Hubble constant and $Omega_Lambda$ is the current fraction of the energy density that is dark energy.
Substituting (2) into (1) gives the Ricci scalar curvature as a function of cosmological time,
$$R(t)=3Omega_Lambda H_0^2left[3+coth^2left(frac32Omega_Lambda^{1/2}H_0 tright)right]tag3.$$
This is a monotonically decreasing function. As $ttoinfty$, $Rto 12Omega_Lambda H_0^2$.
The relevant parameter values, derived from fitting observational data of the cosmic microwave background to the Lambda-CDM model, are $H_0=0.069text{ Gyr}^{-1}$ and $Omega_Lambda=0.69$. (The similarity of these values is a numerical coincidence.)
One finds that the current value of $R$ (at $t=13.7text{ Gyr}$) is $0.044text{ Gly}^{-2}$, and it is decreasing to the asymptotic value $0.039text{ Gly}^{-2}$.
Here is a graph, where the horizontal axis is cosmological time in billions of years and the vertical axis is the Ricci scalar curvature in inverse billions of lightyears squared:
The reason that $R$ is approaching a constant value is that the increasingly exponential expansion due to dark energy is causing our universe to approach a four-dimensional deSitter geometry, which has constant spacetime curvature. The spatial curvature remains zero, but the spacetime curvature decreases to a constant.
Correct answer by G. Smith on April 30, 2021
No
The curvature parameter $k$ of the universe remains constant throughout its evolution. If the universe is open ($k < 0$) , it will stay open, and if it is closed ($k > 0$), it will stay closed. That's because the amount of matter-energy of the universe is conserved, so if the density is greater than critical now it will forever be greater than critical.
Answered by Allure on April 30, 2021
Get help from others!
Recent Answers
Recent Questions
© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP