7 Multi-Component Universes

7.1 The Friedmann Equation

Let’s start from the first Friedmann equation \[ \left(\frac{\dot{a}}{a}\right)^{2} = \frac{8 \pi G}{3} \rho(t) - \frac{k}{R_0^{2} a^{2}} + \frac{\Lambda}{3} \] where I have now also included the cosmological constant term into the mix. It is conceptually useful to include the last two terms on the right as density terms. Note that this is a convenience, especially in the case of curvature, which isn’t a energy density. However, let us define : \[ \begin{aligned} \rho_{k}(a) &= -\frac{3k}{R_{0}^{2}a^{2} 8 \pi G} \\ \rho_{\Lambda}(a) &= \frac{\Lambda}{8\pi G} \end{aligned} \] where we have kept the time/scale factor dependence explicit. The Friedmann equation can now be written as \[ \left(\frac{\dot{a}}{a}\right)^{2} = \frac{8 \pi G}{3} (\rho_{m}(a) + \rho_{r}(a) + \rho_{k}(a) + \rho_{\Lambda}(a)) \] where we have now explicitly broken out the matter and radiation components explicitly. We could also generalize the cosmological constant to be a general dark energy term with $w\neq 1$ and get \[ \left(\frac{\dot{a}}{a}\right)^{2} = \frac{8 \pi G}{3} (\rho_{m}(a) + \rho_{r}(a) + \rho_{k}(a) + \rho_{DE}(a)) \] Multiplying and dividing by $H_{0}^{2}$ on the right, and now explicitly putting in the scale factor dependence into the equation, we find \[ H(a)^{2} = \frac{H_{0}^{2}}{\rho _\text{crit,0}}(\rho_{m,0} a^{-3} + \rho_{r,0} a^{-4} + \rho_{k,0} a^{-2} + \rho_{DE,0} a^{-3(1+w)}) \] A few comments are necessary here:

The subscript $0$ refers to the present day value of the quantity.
For dark energy, this expression is only correct for a constant equation of state. For a time-varying equation of state, you would need to go back to the continuity equation to work out the variation with scale factor. The case of the cosmological constant is $w=-1$.

If we now define $\Omega$ for each individual component $X$ \[ \Omega_{X}(a) \equiv \frac{\rho_{X}(a)}{\rho _\text{crit}(a)} \] we have \[ H(a)^{2} = H_{0}^{2} \left[\Omega_{m,0} a^{-3} + \Omega_{r,0} a^{-4} + \Omega_{k,0} a^{-2} + \Omega_{DE,0} a^{-3(1+w)} \right] \] or \[ H(z)^{2} = H_{0}^{2}\left[ \Omega_{m,0} (1+z)^{3} + \Omega_{r,0} (1+z)^{4} + \Omega_{k,0} (1+z)^{2} + \Omega_{DE,0} (1+z)^{3(1+w)} \right] \] These forms of the Friedmann equations are the forms that get used the most in cosmology - well worth committing to memory!

If, instead of dividing by $H_{0}^{2}$, we divided by $H(a)^{2}$, then we recover a very useful relationship \[ \sum_{i} \Omega_{i}(a) = 1 \] We note that this includes $\Omega_{k}$ as well, unlike our discussion earlier. In what follows, we’ll continue to treat curvature as having its corresponding $\Omega$, instead of treating it separately.

7.2 Distances and Ages

We can now write out the integrals to compute distances and ages. We have \[ \chi = c \int_{0}^{z} \frac{dz}{H(z)} = \frac{c}{H_{0}} \int_{0}^{z} \frac{1}{E(z)} \, dz \] where \[ E(z) \equiv \left[ \Omega_{m,0} (1+z)^{3} + \Omega_{r,0} (1+z)^{4} + \Omega_{k,0} (1+z)^{2} + \Omega_{DE,0} (1+z)^{3(1+w)} \right]^{1/2} \] Similarly, the time integral yields \[ t = \frac{1}{H_{0}} \int_{0}^{a} \frac{1}{a E(a)} \, da = \frac{1}{H_{0}} \int_{-\infty}^{\ln a} \frac{1}{E(a)} \, d\ln a \] which gives us the age of the Universe \[ t_{0} = \frac{1}{H_{0}} \int_{0}^{1} \frac{1}{a E(a)} \, da = \frac{1}{H_{0}} \int_{-\infty}^{0} \frac{1}{E(a)} \, d\ln a \]

A note on units here. As discussed earlier, the uncertainty in the Hubble constant is often parametrized as $H_{0} = 100 h \text{\,km/s/Mpc}$, which results in an $h^{-1}$ factor in both distances and times.

In general, these integrals must be done numerically, but there are some useful analytic results in the case of two-component Universes. Baumann discusses these in his textbook, and I’d encourage you to explore these there.

7.3 $\Omega$ variation with time

If component $X$ has a constant equation of state $w$, we can write out how it varies with time \[ \Omega_{X}(z) = \frac{\Omega_{X,0} (1+z)^{3(1+w)}}{E(z)^{2}} \]

7.4 Our Fiducial Model

The table presents our fiducial, standard model of cosmology. The results here are very similar to the legacy results from the Planck satellite, rounded up. These were derived from here, which is a large table exploring many different options. I am using the values from the base_plikHM_TT_lowl_lowE_post_BAO chain (rounded up), which sets the curvature contribution to zero. You may explore other variants (this file should be posted to Canvas). All of these are values today.

Parameter	Value
$\Omega_m$	0.31
$\Omega_\Lambda$	0.69
$\Omega_b h^2$	0.02225
$H_0$	70 km/s/Mpc**
$T_\text{CMB}$	2.7255 K
$\Omega_\gamma h^2$	$2.473 \times 10^{-5}$
$\Omega_\nu h^2$	$1.68 \times 10^{-5}$
$\Omega_r h^2$	$4.15 \times 10^{-5}$
$\Omega_{k}$	0

One caveat : I’ve rounded the Hubble constant here, but there is tension between high and low redshift estimates of the Hubble constant as of this writing.

7.5 The Flatness Problem

The above table fixed the curvature to be flat (i.e. $k=0$). If one attempts to constrain it, we find $\Omega_{k,0}=0.001 \pm 0.002$. We can now ask how this value changes as we go to the very early Universe. We find \[ \Omega_{k}(z) \to \frac{\Omega_{k,0}}{\Omega_{r,0}} \frac{(1+z)^{2}}{(1+z)^{4}} \text{\,as\,}z \to \infty \] If we assume the best-fit non-zero value, then we find that $\Omega_{k}$ must have been vanishingly small but non-zero in the very early Universe. The challenge is to find the physics that would force the curvature to almost perfectly vanish in the early Universe - this is the flatness problem that gets solved by inflation.

# Multi-Component Universes ## The Friedmann Equation Let's start from the first Friedmann equation $$ \left(\frac{\dot{a}}{a}\right)^{2} = \frac{8 \pi G}{3} \rho(t) - \frac{k}{R_0^{2} a^{2}} + \frac{\Lambda}{3} $$ where I have now also included the cosmological constant term into the mix. It is conceptually useful to include the last two terms on the right as density terms. Note that this is a convenience, especially in the case of curvature, which isn't a energy density. However, let us define : $$ \begin{aligned} \rho_{k}(a) &= -\frac{3k}{R_{0}^{2}a^{2} 8 \pi G} \\ \rho_{\Lambda}(a) &= \frac{\Lambda}{8\pi G} \end{aligned} $$ where we have kept the time/scale factor dependence explicit. The Friedmann equation can now be written as $$ \left(\frac{\dot{a}}{a}\right)^{2} = \frac{8 \pi G}{3} (\rho_{m}(a) + \rho_{r}(a) + \rho_{k}(a) + \rho_{\Lambda}(a)) $$ where we have now explicitly broken out the matter and radiation components explicitly. We could also generalize the cosmological constant to be a general dark energy term with $w\neq 1$ and get $$ \left(\frac{\dot{a}}{a}\right)^{2} = \frac{8 \pi G}{3} (\rho_{m}(a) + \rho_{r}(a) + \rho_{k}(a) + \rho_{DE}(a)) $$ Multiplying and dividing by $H_{0}^{2}$ on the right, and now explicitly putting in the scale factor dependence into the equation, we find $$ H(a)^{2} = \frac{H_{0}^{2}}{\rho _\text{crit,0}}(\rho_{m,0} a^{-3} + \rho_{r,0} a^{-4} + \rho_{k,0} a^{-2} + \rho_{DE,0} a^{-3(1+w)}) $$ A few comments are necessary here: - The subscript $0$ refers to the present day value of the quantity. - For dark energy, this expression is only correct for a constant equation of state. For a time-varying equation of state, you would need to go back to the continuity equation to work out the variation with scale factor. The case of the cosmological constant is $w=-1$. If we now define $\Omega$ for each individual component $X$ $$ \Omega_{X}(a) \equiv \frac{\rho_{X}(a)}{\rho _\text{crit}(a)} $$ we have $$ H(a)^{2} = H_{0}^{2} \left[\Omega_{m,0} a^{-3} + \Omega_{r,0} a^{-4} + \Omega_{k,0} a^{-2} + \Omega_{DE,0} a^{-3(1+w)} \right] $$ or $$ H(z)^{2} = H_{0}^{2}\left[ \Omega_{m,0} (1+z)^{3} + \Omega_{r,0} (1+z)^{4} + \Omega_{k,0} (1+z)^{2} + \Omega_{DE,0} (1+z)^{3(1+w)} \right] $$ These forms of the Friedmann equations are the forms that get used the most in cosmology - well worth committing to memory! If, instead of dividing by $H_{0}^{2}$, we divided by $H(a)^{2}$, then we recover a very useful relationship $$ \sum_{i} \Omega_{i}(a) = 1 $$ We note that this includes $\Omega_{k}$ as well, unlike our discussion earlier. In what follows, we'll continue to treat curvature as having its corresponding $\Omega$, instead of treating it separately. ## Distances and Ages We can now write out the integrals to compute distances and ages. We have $$ \chi = c \int_{0}^{z} \frac{dz}{H(z)} = \frac{c}{H_{0}} \int_{0}^{z} \frac{1}{E(z)} \, dz $$ where $$ E(z) \equiv \left[ \Omega_{m,0} (1+z)^{3} + \Omega_{r,0} (1+z)^{4} + \Omega_{k,0} (1+z)^{2} + \Omega_{DE,0} (1+z)^{3(1+w)} \right]^{1/2} $$ Similarly, the time integral yields $$ t = \frac{1}{H_{0}} \int_{0}^{a} \frac{1}{a E(a)} \, da = \frac{1}{H_{0}} \int_{-\infty}^{\ln a} \frac{1}{E(a)} \, d\ln a $$ which gives us the age of the Universe $$ t_{0} = \frac{1}{H_{0}} \int_{0}^{1} \frac{1}{a E(a)} \, da = \frac{1}{H_{0}} \int_{-\infty}^{0} \frac{1}{E(a)} \, d\ln a $$ A note on units here. As discussed earlier, the uncertainty in the Hubble constant is often parametrized as $H_{0} = 100 h \text{\,km/s/Mpc}$, which results in an $h^{-1}$ factor in both distances and times. In general, these integrals must be done numerically, but there are some useful analytic results in the case of two-component Universes. Baumann discusses these in his textbook, and I'd encourage you to explore these there. ## $\Omega$ variation with time If component $X$ has a constant equation of state $w$, we can write out how it varies with time $$ \Omega_{X}(z) = \frac{\Omega_{X,0} (1+z)^{3(1+w)}}{E(z)^{2}} $$ ## Our Fiducial Model The table presents our fiducial, standard model of cosmology. The results here are very similar to the legacy results from the Planck satellite, rounded up. These were derived from [here](https://wiki.cosmos.esa.int/planck-legacy-archive/images/4/43/Baseline_params_table_2018_68pc_v2.pdf), which is a large table exploring many different options. I am using the values from the `base_plikHM_TT_lowl_lowE_post_BAO` chain (rounded up), which sets the curvature contribution to zero. You may explore other variants (this file should be posted to Canvas). All of these are values today. | Parameter | Value | | ------------------- | ---------------------- | | $\Omega_m$ | 0.31 | | $\Omega_\Lambda$ | 0.69 | | $\Omega_b h^2$ | 0.02225 | | $H_0$ | 70 km/s/Mpc** | | $T_\text{CMB}$ | 2.7255 K | | $\Omega_\gamma h^2$ | $2.473 \times 10^{-5}$ | | $\Omega_\nu h^2$ | $1.68 \times 10^{-5}$ | | $\Omega_r h^2$ | $4.15 \times 10^{-5}$ | | $\Omega_{k}$ | 0 | One caveat : I've rounded the Hubble constant here, but there is tension between high and low redshift estimates of the Hubble constant as of this writing. ## The Flatness Problem The above table fixed the curvature to be flat (i.e. $k=0$). If one attempts to constrain it, we find $\Omega_{k,0}=0.001 \pm 0.002$. We can now ask how this value changes as we go to the very early Universe. We find $$ \Omega_{k}(z) \to \frac{\Omega_{k,0}}{\Omega_{r,0}} \frac{(1+z)^{2}}{(1+z)^{4}} \text{\,as\,}z \to \infty $$ If we assume the best-fit non-zero value, then we find that $\Omega_{k}$ must have been vanishingly small but non-zero in the very early Universe. The challenge is to find the physics that would force the curvature to almost perfectly vanish in the early Universe - this is the **flatness problem** that gets solved by inflation.

Parameter	Value
\(\Omega_m\)	0.31
\(\Omega_\Lambda\)	0.69
\(\Omega_b h^2\)	0.02225
\(H_0\)	70 km/s/Mpc**
\(T_\text{CMB}\)	2.7255 K
\(\Omega_\gamma h^2\)	\(2.473 \times 10^{-5}\)
\(\Omega_\nu h^2\)	\(1.68 \times 10^{-5}\)
\(\Omega_r h^2\)	\(4.15 \times 10^{-5}\)
\(\Omega_{k}\)	0

7.1 The Friedmann Equation

7.2 Distances and Ages

7.3 \(\Omega\) variation with time

7.4 Our Fiducial Model

7.5 The Flatness Problem