Equation 1 (page 5) in “Expanding Confusion: common misconceptions of cosmological horizons and the superluminal expansion of the universe” by Davis and Lineweaver gives the general relativistic relation between velocity and cosmological redshift:$$v_{rec}\left(t,z\right)=\frac{c}{R_{0}}\dot{R}\left(t\right)\int_{0}^{z}\frac{dz^{\prime}}{H\left(z^{\prime}\right)}.$$In Figure 1 (page 7) they use this equation to show the relationship between velocity and redshift for a range of Fridemann-Robertson-Walker models (using different values of $\Omega_{m}$ and $\Omega_{\Lambda}$). They assume $\dot{R}\left(t\right)=\dot{R}_{0}$. enter image description here

How do they do that? In other words, what is the relationship between redshift and the density parameters $\Omega_{m}$, $\Omega_{\Lambda}$ and $\Omega_{r}$ (which they appear to ignore)?


2 Answers 2


The thing you are missing is perhaps $$H(z) = H_0\left((1-\Omega_\Lambda -\Omega_m)(1+z)^2 + (\Omega_\Lambda + \Omega_m)\frac{\rho(z)}{\rho_0}\right)^{1/2}$$ Where the densities $\rho$ dependend on the matter content, so they must be split into different epochs (of matter, radiation, etc. domination) to do the full integral but they just. This equation can be derived from the Friedmann equations. Once you plug in the values for the quantities measured today, $0$, you can compute the integral and obtain the velocity. You can find more details on the following link (https://ned.ipac.caltech.edu/level5/Peacock/Peacock3_2.html) or in any standard book on cosmology such as Mukhanov's "Physical Foundations of Cosmology" (2005), Chapter 2.

EDIT: The formula above holds for universes that are not spatially flat, $k\neq 0$.

Starting from the Friedmann equation: $$H^2(t) = \frac{8\pi G}{3}\rho(t)- \frac{k}{a^2} + \frac{\Lambda}{3}$$

It turns out that for the "Expanding Confusion" document, they take spatial curvature to be cero (flat universe) so that the density parameters add up to 1 (so you can always eliminate one of them, in your case to eliminate $\Omega_r$) and the universe to be composed only of matter, radiation and a cosmological constant so that with the usual definitions $$\rho_{crit} = \frac{3H_0^2}{8\pi G}\qquad \textrm{and}\qquad \Omega_X = \frac{\rho_X}{\rho_{crit}},$$ with $0$ denoting the values today and knowing how the different components of the universe behave with respect to the scale factor allows us to rewrite the Friedmann equation in terms of the scale factor. Recall for matter $\rho_m\propto a^{-3}$, for radiation $\rho_r\propto a^{-4}$ and for dark energy we assume consant in this case, then $$H(a) = H_0\sqrt{\Omega_\Lambda + \Omega_m a^{-3} + \Omega_r a^{-4}}$$ with $a_0 =1$. Now using the relation between scale factor and redshift $$\frac{a_0}{a(t)}=1+z$$ in the previous formula, this is were the problems arise, using $\Omega_r = 1- \Omega_m - \Omega_\Lambda$ and a scaling of radiation as $a^{-2}$ you get their result, however using the correct scaling for radiation you get: $$H(z) = H_0 (1+z)\left( 1 + \Omega_m z + \Omega_\Lambda\left(\frac{1}{(1+z)^2}-1\right) + \color{red}{2z\Omega_r + z^2\Omega_r} \right)^{1/2}$$

I hope this helps, but now I am also curios about how they get that formula...

  • $\begingroup$ I've just spotted equation 25 (page 20) in Davis and Lineweaver: $$H\left(z\right)=H_{0}\left(1+z\right)\left[1+\varOmega_{m}z+\varOmega_{\varLambda}\left(\frac{1}{\left(1+z\right)^{2}}-1\right)\right]^{1/2}.$$ Is this a variant of your equation? $\endgroup$
    – Peter4075
    May 15, 2018 at 17:35
  • $\begingroup$ Yes it is, a particular form for $\rho$ has already been used after assuming a specific matter content. $\endgroup$
    – ohneVal
    May 16, 2018 at 8:24
  • $\begingroup$ Right, I've had a look at Mukhanov's book and think one source of my confusion is that they (Davis and Lineweaver) are using $z$ instead of time to parameterize the Hubble parameter. That threw me. Any chance of a hint as to how I get from Davis and Lineweaver's equation to your/Mukhanov's equation? I feel I'm almost there but can't quite see it. Thank you. $\endgroup$
    – Peter4075
    May 17, 2018 at 17:44
  • $\begingroup$ Many thanks. I haven't had time to work through your calculations yet, but aren't D&L simply assuming $\varOmega_{r}=0$? Doesn't that give the correct answer? $\endgroup$
    – Peter4075
    May 17, 2018 at 20:05
  • $\begingroup$ I've given my derivation of D&L's equation 25 as an answer to my question. Hope it makes sense. Thanks. $\endgroup$
    – Peter4075
    May 18, 2018 at 18:19

I eventually realised that in order to answer my question I needed to understand the derivation of D&L's Equation 25: $$H\left(z\right)=H_{0}\left(1+z\right)\left[1+\varOmega_{m}z+\varOmega_{\varLambda}\left(\frac{1}{\left(1+z\right)^{2}}-1\right)\right]^{1/2}.$$

This is the Hubble parameter parameterised by redshift not time. The critical density at present is$$\rho_{c}=\frac{3H_{0}^{2}}{8\pi G}.$$

Therefore $$H_{0}^{2}=\frac{8\pi G\rho_{c}}{3}$$

and $$\varOmega=\frac{\rho}{\rho_{c}}=\frac{8\pi G\rho}{3H_{0}^{2}}.$$

The Friedmann equation is

$$H^{2}\left(t\right)=\frac{8\pi G}{3}\rho-\frac{kc^{2}}{R^{2}}.$$

Divide by $H_{0}^{2}$ $$\frac{H^{2}}{H_{0}^{2}}=\frac{8\pi G}{3H_{0}^{2}}\rho-\frac{kc^{2}}{R^{2}H_{0}^{2}}$$

$$\frac{H^{2}}{H_{0}^{2}}=\varOmega-\frac{kc^{2}}{R^{2}H_{0}^{2}}.$$Where $\varOmega=\varOmega_{m}+\varOmega_{\varLambda}+\varOmega_{r}$. To find $-kc^{2}$ set$$-kc^{2}=H_{0}^{2}R_{0}^{2}-\frac{R_{0}^{2}8\pi G\rho}{3}$$ and get$$\frac{H^{2}}{H_{0}^{2}}=\varOmega+\frac{H_{0}^{2}R_{0}^{2}}{H_{0}^{2}R^{2}}-\frac{R_{0}^{2}8\pi G\rho}{3R^{2}H_{0}^{2}}$$

$$\frac{H^{2}}{H_{0}^{2}}=\varOmega+\frac{R_{0}^{2}}{R^{2}}\left(1-\frac{8\pi G\rho}{3H_{0}^{2}}\right)$$ $$\frac{H^{2}}{H_{0}^{2}}=\varOmega+\frac{R_{0}^{2}}{R^{2}}\left(1-\varOmega\right).$$

Now write as a function of redshift $z$.

First, $$\frac{R_{0}^{2}}{R^{2}}=\left(1+z\right)^{2}$$giving$$H^{2}\left(z\right)=H_{0}^{2}\left[\varOmega+\left(1+z\right)^{2}\left(1-\varOmega\right)\right].$$


giving$$H^{2}\left(z\right)=H_{0}^{2}\left[\varOmega_{m}\left(1+z\right)^{3}+\varOmega_{\varLambda}+\varOmega_{r}\left(1+z\right)^{4}+\left(1+z\right)^{2}\left(1-\varOmega\right)\right].$$To simplify, let $\varOmega_{r}=0$ (as do Davis and Lineweaver)$$H^{2}\left(z\right)=H_{0}^{2}\left[\varOmega_{m}\left(1+z\right)^{3}+\varOmega_{\varLambda}+\left(1+z\right)^{2}\left(1-\varOmega_{m}-\varOmega_{\varLambda}\right)\right]$$ $$H^{2}\left(z\right)=H_{0}^{2}\left(1+z\right)^{2}\left[\varOmega_{m}\left(1+z\right)+\frac{\varOmega_{\varLambda}}{\left(1+z\right)^{2}}+1-\varOmega_{m}-\varOmega_{\varLambda}\right]$$ $$H^{2}\left(z\right)=H_{0}^{2}\left(1+z\right)^{2}\left[1+\varOmega_{m}z+\varOmega_{\varLambda}\left(\frac{1}{\left(1+z\right)^{2}}-1\right)\right]$$



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.