The definition of the linear growth factor $D(z)$ of perturbations in a cosmological setting is usually normalized to unity at redshift $z=0$. So,

$$\delta(z) = D(z)\times \delta(z=0) \tag{1}$$

where $\delta$ is the density contrast of a perturbation (I am looking at adiabatic perturbations of cold dark matter).

In the cosmology text book by Dodelson and Schmidt, there is an expression given to calculate the growth factor $D$ as a function of the cosmological scale factor $a$. When a cosmological constant and curvature is present, it is:

$$D_{+}(a)=\frac{5 \Omega_{\mathrm{m}}}{2} \frac{H(a)}{H_0} \int\limits_0^a \frac{d a^{\prime}}{\left(a^{\prime} H\left(a^{\prime}\right) / H_0\right)^3} \tag{2}$$

From some online reading, I found out that: $$\frac{H(a)}{H_0} = \sqrt{\Omega_{R, 0} a^{-4} + \Omega_{M, 0} a^{-3}+\Omega_{K, 0} a^{-2}+\Omega_{\Lambda, 0}} \tag{3}$$

In eqn (3), the omega parameters are the density ratio parameters (relative to the critical density) today. I used the following values:

$\Omega_{R, 0} = 9.23640\times 10^{-5}$, $\Omega_{M, 0} = 0.3106$, $\Omega_{\Lambda, 0} = 0.6894$ and $$\Omega_{K, 0} = 1 - \Omega_{R, 0} - \Omega_{M, 0} - \Omega_{\Lambda, 0}$$

Performing the integration in eqn (2) numerically in Python and setting $a = 1$, I get: $$D_{+}(a=1)= 0.7848$$

But I would expect the growth factor $D$ today ($z=0$ or $a=1$) to be equal to 1. I am not sure what exactly I am doing wrong or if I have incorrectly interpreted the the expression for growth factor. Can someone help?


1 Answer 1


That expression for the growth factor is not normalized to 1 today. Instead it is normalized such that


during the matter epoch (i.e. before dark energy becomes important). This is another common choice.

The factor 0.78 precisely says how much perturbations today are suppressed due to dark energy.


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