
#1
Jan1812, 04:56 AM

P: 10

Hi there. My interest in cosmology is purely amateurish, so bear with my ignorance. I have some questions regarding following Fridman's equation:
[tex]H^{2}=\frac{8\pi G\rho}{3}\frac{kc^{2}}{a^{2}}+\frac{\Lambda c^{2}}{3}[/tex] 1. Lambda is usually given in units of s^{2}, or in units of m^{2}. When given in s^{2} is it already multiplied with c^{2}, so the last term is λ/3, right or wrong? 2. What are the units of k  normalized spatial curvature? I can't find it anywhere. It is just given as 0,1, or 1. It can't be unitless or equation wouldn't end in 1/s, required units for Hubble parameter. It must also be in m^{2}, right or wrong? 3. What is exactly Hubble parameter? Same thing as Hubble constant, or to say, Hubble constant is current value of Hubble parameter? Any help would be appreciated, thanks. sm 



#2
Jan1812, 06:47 AM

Sci Advisor
P: 1,552





#3
Jan1912, 02:16 AM

Sci Advisor
P: 4,721

[tex]H^2 = H_0^2\left({\Omega_m \over a^3} + {\Omega_r \over a^4} + {\Omega_k \over a^2} + \Omega_\Lambda\right)[/tex] Here [itex]\Omega_x[/itex] is the current density fraction for component [itex]x[/itex]. In the equation above, I have included matter (normal matter + dark matter), radiation, curvature, and a cosmological constant. These make for a fraction in that the sum of all of the [itex]\Omega_x[/itex] values is equal to one, and they have no dimensions. Also useful in some situations is to get rid of the units of [itex]H_0[/itex] by defining: [tex]h = {H_0 \over 100km/s/Mpc}[/tex] This convention was made before we had an accurate measurement of the Hubble parameter, so a round number not too far away from the very poorlymeasured value was used. This is useful in many areas of cosmology where there is a measurement where the precise value of the measurement depends upon the Hubble constant, but doesn't actually measure it. But in general, if you want units, you should be able to read them right off of the equation. The units of the Hubble expansion rate are inverse time (or, equivalently, velocity over distance). So just equate the units and solve. 



#4
Jan1912, 04:06 AM

P: 10

Fridman's equation helpAlso, critical density equation is ρ_{crit}= 3H^{2}/8piG. Does the ρ here apply to all density contributors you mentioned in your post? Furthermore, how the curvature density parameter is defined, and is it 0 for flat universe? 



#5
Jan1912, 01:31 PM

Sci Advisor
P: 4,721

[tex] {k c^2 \over a^2} \to H_0^2 {\Omega_k \over a^2}[/tex] So: [tex]\Omega_k = {k c^2 \over H_0^2}[/tex] But perhaps an easier way to see it is from the definition that the current value of the Hubble parameter is [itex]H_0[/itex]: [tex]H(a=1) = H_0[/tex] It is really easy to see from the way I wrote the Hubble parameter earlier that: [tex]\Omega_m + \Omega_r + \Omega_\Lambda + \Omega_k = 1[/tex] So if we measure the matter density, the radiation density, and the dark energy density, the remainder is the curvature. (so yes, it's zero for a flat universe) 



#6
Jan1912, 04:31 PM

P: 10

In any case, thank you Chalnoth, you've been very helpful to me. 



#7
Jan1912, 08:37 PM

Sci Advisor
P: 4,721

For example, supernova observations are very good at measuring the ratio [itex]\Omega_m/\Omega_\Lambda[/itex], while CMB observations are good at measuring the total density, [itex]H_0^2(\Omega_m + \Omega_r + \Omega_\Lambda)[/itex] as well as providing a nearlyexact measurement of the radiation density [itex]H_0^2 \Omega_r[/itex]. So if you combine CMB observations, such as WMAP, with a nearby measurement of the Hubble expansion rate [itex]H_0[/itex], you get a really accurate measurement of how flat the universe is. 


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