0
$\begingroup$

In terms of the temperature of the CMB, the universe is amazingly homogeneous. The temperature anisotropies are one part in $10^5$. However, instead of CMB fluctuations, there must be a way of quantifying the degree of homogeneity with respect to the mass distribution in the galaxies and clusters. Can we quantify the fluctuation in mass distribution in terms of a quantity like $\frac{\Delta M}{\bar{M}}$? Here $\bar{M}$ would mean, something like a mean mass averaged over a scale $R$ sufficiently large to include many many galaxies and $\Delta M$ being a measure of deviation from that $\bar{M}$.

$\endgroup$
1
  • $\begingroup$ Note that CMB measurements do not imply that the universe is homogeneous, since these measurements are always done from a vantage point (the solar system). CMB fluctuations imply that the universe is isotropic or, if you want to be rigorous, it implies that the universe was isotropic at z=1100. $\endgroup$
    – Thiago
    Commented Nov 30, 2020 at 16:38

1 Answer 1

0
$\begingroup$

In fact we use density instead of mass. Density fluctuation is quantified by $\delta = \frac{\rho-<\rho>}{<\rho>}$. Cosmological parameters associated with density fluctuations are the Growth factor D(t) such that $\delta(x,t)=\delta(x)D(t)$ and the Growth rate $f=\frac{d\ln D}{d \ln a}$ with a the scale parameter.

I hope this answers your question

$\endgroup$
2
  • $\begingroup$ I am asking how do you do it from the observational point of view, and what is the order of fluctuation of $\Delta M/\bar{M}$. Your answer does not tell what is the level of homogeneity or inhomogeneity from the galaxy surveys. $\endgroup$ Commented Nov 30, 2020 at 17:09
  • $\begingroup$ CMB inhomogeneity have grow and became galaxies and clusters of today. So, from an observational point of view we measure the correlation fonction between galaxy or the peculiar velocity field due to gravity force due to the matter density field. Estimation of inhomogeneity can be given by the cosmological parameter $f\sigma_8$. $\endgroup$
    – 2PiOmega
    Commented Nov 30, 2020 at 18:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.