Lecture 2 Equations of the Expanding Universe
2.1 Metric of space-time
Consider a flat piece of paper where points can be specified by coordinates \(x_1\) and \(x_2\). The distance between two points is given by \[\begin{equation} \Delta s^2 = \Delta x_{1}^{2} + \Delta x_{2}^{2} \tag{2.1} \end{equation}\]where \(\Delta x_{1}^{2}\) and \(\Delta x_{2}^{2}\) are the separations in the \(x_1\) and \(x_2\) coordinates.
What about if we replaced the paper with a rubber sheet that can expand? The coordinate system \(x_1\)–\(x_2\) will expand with the sheet, and the physical distance between the points will change over time.
Assuming that the expansion is uniform over the whole sheet, we now have \[\begin{equation} \Delta s^2 = a^2(t)\left[\Delta x_{1}^{2} + \Delta x_{2}^{2}\right] \tag{2.2} \end{equation}\]where \(a(t)\) measures the rate of expansion. \(x_1\) and \(x_2\) are comoving coordinates.
Our current definition of \(\Delta s\) only depends on the spatial distance between the points. However, in general relativity, we have to consider 4-dimensional space time, which may be curved.
In this case, the distance becomes \[\begin{equation} ds^2 = \sum_{\mu, \nu} g_{\mu\nu} dx^{\mu} dx^{\nu} \tag{2.3} \end{equation}\]where \(g_{\mu\nu}\) is the metric and \(\mu\) and \(\nu\) are indices taking the values 0, 1, 2, 3. \(x^{0}\) is the time coordinate and \(x^{1}, x^{2}, x^{3}\) are the spatial coordinates.
The spatial component of Equation (2.3) can be simplified by applying the Cosmological Principle (see Sec. 1.1) - i.e. the Universe should not have any preferred locations. This means that the spatial component must have constant curvature at all positions.
Changing to spherical polar coordinates, we now get \[\begin{equation} ds^2_3 = \dfrac{dr^2}{1 - kr^2} + r^2\left(d\theta^2 + \sin^2\theta d\phi^2\right) \tag{2.4} \end{equation}\]where \(ds_3^2\) considers only the spatial dimensions, and \(k\) is a constant which measures the curvature of space.
Incorporating this into the space-time metric gives \[\begin{equation} ds^2 = -c^2 dt^2 + a^2(t) \left[\dfrac{dr^2}{1 - kr^2} + r^2\left(d\theta^2 + \sin^2\theta d\phi^2\right)\right] \tag{2.5} \end{equation}\]Which is the Robertson-Walker metric. Here \(a(t)\) is the scale factor of the Universe - i.e. \(a(t) = r/r_0,\) where \(r_0\) is \(r\) at time \(t_0\).
2.2 The Einstein Equations
The Robertson-Walker metric evolves according to Einstein’s equation: \[\begin{equation} R_{\mu\nu} - \dfrac{1}{2}g_{\mu\nu}R = \dfrac{8\pi G}{c^4} T_{\mu\nu} \tag{2.6} \end{equation}\]where \({T_{\mu\nu}}\) is the energy-momentum tensor of matter present, \({R_{\mu\nu}}\) is the Ricci tensor, and \({R}\) is the Ricci scalar. The Ricci components give the curvature of space-time.
The energy-momentum tensor is assumed to be symmetric, meaning that there are potentially 10 Einstein equations. If \(T_{\mu\nu}\) has additional symmetries, then the number of equations is reduced.
2.3 Fluid equation
We consider that any possible components of the Universe (e.g. matter, radiation) can be described as perfect fluids. Perfect fluids have no viscosity or heat flow, and their equations of state can be described using only density, \(\rho\) and pressure, \(P\).
Perfect fluids have a well known relation between energy, \(E\), pressure, \(P\), and temperature, \(T\): \[\begin{equation} dE + PdV = TdS \tag{2.7} \end{equation}\]where \(dV\) is the change in volume, and \(dS\) is the change in entropy.
If we consider a sphere of radius \(a\), using \[\begin{equation} E = mc^2 \tag{2.8} \end{equation}\] we can relate the total energy contained within the sphere to its density via \[\begin{equation} E = \dfrac{4\pi}{3}a^3\rho c^2 \tag{2.9} \end{equation}\]Assuming that the sphere is expanding adiabatically, i.e. there is no change in entropy (\(dS = 0\)), differentiating Equation (2.7) with respect to \(t\) gives
\[\begin{equation} \dot{\rho} + 3\dfrac{\dot{a}}{a}\left(\rho + \dfrac{P}{c^2}\right) = 0 \tag{2.10} \end{equation}\]where \(\dot{x} = \dfrac{dx}{dt}, \ddot{x} = \dfrac{d^2x}{dt^2}\), etc. Equation (2.10) is known as the fluid equation.
As the energy of the perfect fluid depends only on \(P\) and \(\rho\), the energy-momentum tensor for a perfect fluid is therefore \[\begin{equation} T_{\mu\nu} = \text{diag}\left(-\rho c^2, P, P, P\right) \tag{2.11} \end{equation}\]This results in two independent Einstein equations, the time-time solution and the space-space solution.
2.4 Ricci tensor and Ricci scalar
Deriving the full Ricci tensor is beyond the scope of this course. The following is provided for completeness. The components of the Ricci tensor that are not equal to zero are those where \(\mu = \nu\), such that
\(R_{tt} = -3\dfrac{\ddot{a}}{a}\)
\(R_{rr} = \dfrac{a\ddot{a}}{1 - kr^2} + \dfrac{2\dot{a}^2}{1 - kr^2} + \dfrac{2k}{1-kr^2}\)
\(R_{\theta\theta} = r^2a\ddot{a} + 2r^2\dot{a}^2 + 2kr^2\)
\(R_{\phi\phi} = r^2 a\ddot{a}\sin^2\theta + 2r^2\dot{a}^2\sin^2\theta + 2kr^2 \sin^2\theta\)
2.5 Friedman Equation
Using the time-time solution of the Einstein equation gives \[\begin{equation} \left(\dfrac{\dot{a}}{a}\right)^2 + \dfrac{kc^2}{a^2} = \dfrac{8\pi G}{3}\rho \tag{2.15} \end{equation}\]which is the Friedman equation.
The space-space solution follows in a similar way:
\[\begin{equation} 2\dfrac{\ddot{a}}{a} + \left(\dfrac{\dot{a}}{a}\right)^2 + \dfrac{kc^2}{a^2} = -8\pi G \dfrac{P}{c^2} \tag{2.16} \end{equation}\] Which looks odd right now, but by subtracting the Friedman equation (Eqn. (2.15)) we find: \[\begin{equation} \dfrac{\ddot{a}}{a} = -\dfrac{4\pi G}{3}\left(\rho + \dfrac{3P}{c^2}\right) \tag{2.17} \end{equation}\]which is the known as the acceleration equation.
2.6 Natural units a.k.a. where have all the \(c^2\)’s gone?
From this point forward, you will see that the factor \(\left(k c^2/a^2\right)\) in the Friedman equation will magically become \(\left(k/a^2\right)\). This is because we often like to use natural units in Cosmology. Mass density \(\left(\rho\right)\) and energy density \(\left(\epsilon\right)\) are often used interchangeably in Cosmology, and are related by \(\epsilon = \rho c^2\). By converting to natural units, i.e. setting \(c=1\), mass density and energy density are equivalent. As you will see later, many important cosmological parameters are expressed in dimensionless units (e.g. the density parameters, discussed in sections XXXX. So don’t worry about our missing \(c^2\) factor, it’s just another maths trick.
2.7 An alternative derivation of the Friedman equation
The evolution of the scale factor, \(a\), can be derived using mostly Newtonian mechanics, in addition to two results from General Relativity:
Birkhoff’s theorem – for a spherically symmetric system, the force due to gravity at radius \(r\) is determined only by the mass interior to \(r\).
Energy contributes to the gravitating mass density, such that \(E = \rho_m + \dfrac{u}{c^2}\), where \(\rho_m\) is the density of matter, and \(u\) is the energy density of radiation and relativistic particles.
If \(\rho_0\) is non-zero, then \(\ddot{r}\) must also be non-zero. The Universe must either be expanding or contracting.
To get Eqn. (2.19) into a useful form, we use some clever maths magic. First, multiply both sides by \(\dot{r}\) to get: \[\begin{equation} \dot{r}\ddot{r} + \dfrac{4\pi}{3}\dfrac{G\rho_{0}}{r^2}\dot{r} = 0 \tag{2.20} \end{equation}\] We can use our mad differentiation skills to recognise that \[\begin{equation} \dfrac{d(\dot{r}^2)}{dt} = 2\dot{r}\ddot{r} \tag{2.21} \end{equation}\] Which leads to \[\begin{equation} \dfrac{1}{2}\dfrac{d(\dot{r}^2)}{dt} + \dfrac{4\pi G \rho_0}{3} \dfrac{1}{r^2}\dfrac{dr}{dt} = 0 \tag{2.22} \end{equation}\] Once again, we can use our handy \(1^{st}\) year maths skills to remind ourselves that \[\begin{equation} \dfrac{1}{r^2}\dfrac{dr}{dt} = -\dfrac{d(1/r)}{dt} \tag{2.22} \end{equation}\] so we can simplify it further to \[\begin{equation} \dfrac{d}{dt}\left[\dot{r}^2 - \dfrac{(8\pi G \rho_0 / 3)}{r}\right] = 0 \tag{2.23} \end{equation}\] As Eqn (2.23) is equal to zero, the expression contained in the brackets must be constant: \[\begin{equation} \dot{r}^2 - \dfrac{(8\pi G \rho_0 / 3)}{r} = k \tag{2.24} \end{equation}\] Our final step is to replace \(\rho_0\) with \(\rho\), and divide by \(r^2\): \[\begin{equation} \left(\dfrac{\dot{r}}{r}\right)^2 - \dfrac{8}{3}\pi G \rho = -\dfrac{k}{r^2} \tag{2.25} \end{equation}\] or \[\begin{equation} \left(\dfrac{\dot{r}}{r}\right)^2 + \dfrac{k}{r^2} = \dfrac{8\pi G}{3}\rho \tag{2.26} \end{equation}\]which is the same as the Friedman equation (Equation (2.15)), when \(c=1\).
2.8 The meaning of \(k\)
In the previous sections, \(k\) has been used to describe the curvature of the Universe. \(k\) has three possible values:
\(k = 0\): no curvature. If \(k=0\), then \(\dot{a}\) will always be positive. The expansion will continue but will be slowing down, as \(\rho\) will be decreasing. This is known as a flat Universe, or a critical Universe.
\(k = 1\): positive curvature. If \(k = 1\), then \(\dot{a}\) will initially be positive, but will reach a point where it changes sign. The Universe will expand to a certain point then collapse back in on itself (aka the big crunch). This is known as a closed Universe.
\(k = -1\): negative curvature. If \(k=-1\) then \(\dot{a}\) will always be positive and the expansion will continue forever (aka the big rip). This is known as an open Universe.
:::fyi In some derivations \(k\) can have different values from those listed here. However, they will scale such that \(k = -1, 0, 1\). :::
2.9 Exercises
Confirm for yourself how the behaviour of \(k\) affects the evolution of the scale factor (\(a\)) over time. For \(k=1\), at what point does the Universe fall back in on itself?
Sketch how the scale factor evolves with time for the three values of \(k\).
How does the inclusion of a cosmological constant, \(\Lambda\), change the Friedman equation?