# General Relativity, Friedmann Equations, and Accelerating Expansion: A Quick Overview

• FRW cosmology and accelerating expansion

Our discussion here will be focused on how FRW (Friedmann-Robertson-Walker) cosmology can be used to understand the accelerated expansion of the universe, but it is important to note that, taken alone, FRW cosmology does not necessarily imply such an accelerated expansion. One could also have a static universe. Indeed, Einstein originally proposed a static universe, where gravitational attraction was balanced by the gravitationally repulsive effects of a new cosmological constant term he introduced into his equations to achieve a static model. In this case, distant stellar systems would have no Doppler shift relative to us. This was a static universe with the geometry of a three-sphere, and it was uniform in density.

However, in 1922, Friedmann published a dynamical model that solved Einstein’s original field equations without a cosmological constant. This model was homogeneous, had the geometry of a three-sphere, and started at zero size with a Big Bang. This universe expands, reaches a maximum size, and then re-contracts to a point, ending in a Big Crunch. We point this out simply as historical context for what follows. We turn now to a derivation of the equations Friedmann would have used to produce this model, and that more generally constrain the evolution of any model universe given certain initial inputs.

• Derivation of the Friedmann equations from general relativity

Before we discuss general relativity, a bit of formalism is necessary. In our discussion of general relativity, we will be using the word “tensor” frequently. A “tensor” is just like a multi-dimensional matrix, and a matrix is just like a table of numbers. There are some differences — for instance, tensors must transform in a certain special way when one changes coordinates. However, since we will only be using what are called “rank two” tensors, it is sufficient to think of them as tables of numbers, where one index designates the row and the other the column. Note that Greek indices, such as $\alpha,\;\beta$, etc., take on values $0$ through $3$, whereas Latin indices such as $i,\;j$ etc. take on values $1$ through $3.$ Finally, we will use units where the speed of light $c=1$, and we employ the Einstein summation convention throughout (just sum over repeated indices where one is upstairs and the other downstairs).

The basic ideas in general relativity are simple: one needs only two elements. The popular adage is that “mass tells space how to curve, and space tells mass how to move,” which succinctly encapsulates both of these elements. Mass tells space how to curve via the Einstein equation,
$G_{\alpha\beta}\equiv R_{\alpha\beta}-\frac{1}{2}g_{\alpha\beta}R=8\pi G T_{\alpha\beta},$
where $G_{\alpha\beta}$ is the Einstein tensor, $R_{\alpha\beta}$ is the Ricci tensor, $R$ is the Ricci scalar, and $T_{\alpha\beta}$ is the energy-momentum tensor. The Ricci tensor and scalar describe the curvature of space, and the energy-momentum tensor describes the presence and flow of matter and energy. $g_{\alpha\beta}$ is the metric, which we discuss next. Finally, $G$ is Newton’s constant.

So how does mass move under the Einstein equation? Space tells mass how to move via the metric. Free particles follow geodesics, the straightest paths between points. If one considers the Earth, one quickly realizes that the straightest distance between two points depends on the shape of the space (or surface) in or on which one travels. For example, planes flying straight ahead without turning end up taking great circles because these are the geodesics in a spherical geometry such as the Earth’s surface.

Distance is measured via the metric, which describes the infinitesimal distance between two points. In a flat, Euclidean geometry, the metric is simple (recall the Pythagorean theorem):
$ds^2=dx^2+dy^2+dz^2,$
where $ds$ is the infinitesimal distance between two points separated by $dx$, $dy$, and $dz$ in the three spatial directions respectively.
However, there are two additional elements in the metric we will be using here: space is not necessarily static but may be expanding or contracting, and space may be intrinsically curved. If the universe is homogeneous and isotropic, as Hubble’s observations combined with the Copernican principle that we are not in a “special” location imply, the curvature at a given time must be the same everywhere in the universe. It can be either positive, negative, or zero. To account for the possibility of curvature, we write the metric using a sort of generalized spherical coordinates. To account for expansion or contraction, we multiply the spatial elements by a scale factor $a(t)$. Finally, as a consequence of special relativity, objects’ or events’ separation in time also contributes to the “distance” between them.
With all of these comments in mind, the metric is
$ds^2=-dt^2+a^2(t)\left[\frac{dr^2}{1-kr^2}+r^2\left(d\theta^2+\sin^2\theta d\phi^2\right)\right].$

$k=1$ is a positively curved universe (meaning it must be finite in extent). $k=-1$ is an open universe, which is curved like a horse’s saddle. $k=0$ is a flat universe, which in the usual three spatial dimensions is just a Euclidean geometry.

• The Curvature

To gain a bit more insight into what the curvature means, we can rewrite the metric in terms of the differential radius. Note that in our equation for the metric, to find the radius at a particular time we would integrate not $dr$ but rather $dr/\sqrt{1-kr^2}$, so it is really this second quantity that acts as a differential radius. It is this observation that motivates us to define the differential radius to be $d\chi=dr/\sqrt{1-kr^2}$. Using this in the equation for the metric, we have
$ds^2=-dt^2+a^2(t)\left[d\chi^2+f^2(\chi)\left(d\theta^2+\sin^2\theta d\phi^2\right)\right],$
where $f(\chi)=\sin\chi$ for $k=1$, $\chi$ for $k=0$, and $\sinh \chi$ for $k=-1$.

If one computes the integrals for the radius and circumference, one can observe that the ratio of the circumference to the radius is smaller than that of a circle for a $k=1$ universe and larger than that of a circle for a $k=-1$ universe. This Escher drawing illustrates what a $k=-1$ universe would look like: notice there are many more than $2\pi\approx 6$ devils around the circumference of a large circle in the drawing for every one devil along the radius.

We should note here that “the metric” can refer to two slightly different objects: one is an equation like the one we gave above, which gives the infinitesimal arc length. Since the metric is a quadratic form, it can also be written as a tensor and row and column vectors, as below:
$ds^2=\vec{dx}g_{\alpha\beta}\vec{dx}^{\tau},$
where $\vec{dx}\equiv(dt,dr,d\theta,d\phi)$, $g_{\alpha\beta}=diag(-1,a^2/(1-kr^2),a^2 r^2 ,a^2 r^2 \sin^2\theta)$, and superscript $\tau$ means “transpose.” In this context it is $g_{\alpha\beta}$ that is called the metric.

So we now have one of the two elements of general relativity, the metric. What about the other, the Einstein equation? To understand what it means, we need the Ricci tensor, the Ricci scalar, and the energy-momentum tensor.

The Ricci tensor, also called the Ricci curvature, is
$R_{\alpha\beta}=\frac{\partial \Gamma_{\alpha\beta}^{\gamma}}{\partial x^{\gamma}}-\frac{\partial \Gamma_{\alpha\gamma}^{\gamma}}{\partial x^{\beta}}+\Gamma_{\alpha\beta}^{\gamma}\Gamma_{\gamma\delta}^{\delta}-\Gamma_{\alpha\delta}^{\gamma}\Gamma_{\beta\gamma}^{\delta},$

where the $\Gamma$‘s are called the Christoffel symbols. The Christoffel symbols are complicated to calculate, but their conceptual role is simple: they just appear as coefficients in the geodesic equation in an arbitrarily curved spacetime. This equation simply describes the motion of a free particle. It is
$\frac{d^2x^{\alpha}}{d\tau^2}=-\Gamma_{\beta\gamma}^{\alpha}\frac{dx^{\beta}}{d\tau}\frac{dx^{\gamma}}{d\tau}.$

If we recall the problem of extremizing the action in classical mechanics, this equation will allow us to obtain an explicit formula for the Christoffel symbols. A geodesic must extremize the arc length $S=\int ds$ between its two endpoints. The metric gives us an expression for $ds$, and the Euler-Lagrange equations, used to extremize the action $S=\int L$, where $L$ is the Lagrangian in classical mechanics, give us a differential equation governing the path. Comparing this to the geodesic equation above, which also describes the path, yields an explicit formula for the Christoffel symbols. It is
$g_{\alpha\delta}\Gamma_{\beta\gamma}^{\delta}=\frac{1}{2}\left(\frac{\partial g_{\alpha\beta}}{\partial x^{\gamma}}+\frac{\partial g_{\alpha\gamma}}{\partial x^{\beta}}-\frac{\partial g_{\beta\gamma}}{\partial x^{\alpha}}\right),$
where we recall that $g_{\alpha\delta}$ is simply the metric. With the Christoffel symbols now in hand, the earlier equation for the Ricci tensor gives us  $R_{\alpha\beta}$. The Ricci tensor in turn gives us the Ricci scalar $R$, which is essentially just the trace of the Ricci tensor weighted by the components of the inverse metric $g^{\alpha\beta}$. In other words,
$R\equiv g^{\alpha\beta}R_{\alpha\beta}.$

• The Energy-Momentum Tensor

We now have the left hand side of the Einstein equation in hand. What about the right hand side, the energy-momentum tensor $T_{\alpha\beta}$? We will briefly try to develop some qualitative intuition for what the energy-momentum tensor, also called the stress-energy tensor, means. It is basically a table describing what is present in a region of space and also what is entering or leaving the region and what stresses are being exerted within the region (imagine their being exerted on a small test particle in the region, for instance). Note that stress is simply a force per unit area, so it has the same units as pressure. It is most intuitive to think of a force being exerted perpendicular to this unit area, but in fact it need not be, and can have components parallel to the unit area on which it is exerted as well.

The upper left-most entry of the stress-energy tensor is the energy density $E/V$, $V$ the volume, and all of the entries beneath describe the momentum density $p/V$. The entries along the top, excluding the energy density, represent the energy flux (so how much energy per unit volume is entering or leaving the region). The rest of the entries (i.e. those not in the first row or column of the tensor) represent the mechanical stress.

In general, the stress-energy tensor $T_{\alpha\beta}$ could be complicated, but we will focus on the simple case of an ideal perfect fluid, which has two advantages. First, it is not complicated. Second, it seems to describe the background universe (assumed to be nearly homogeneous and isotropic on large scales) quite well.

For an ideal perfect fluid,
$T_{\alpha\beta}=diag\left(-\rho,p,p,p\right),$
where $\rho$ is the energy density of the fluid and $p$ is the pressure.

So we now have everything we need to describe the background universe in the framework of general relativity. Using the Einstein equation with all of the above, we can find two independent equations governing the evolution of the background universe.

We first write down the elements of the Ricci tensor, as well as the Ricci scalar $R$:
$R_{00}=\frac{3\ddot{a}}{a},$

$R_{ij}=\left(\frac{\ddot{a}}{a}+\frac{2\dot{a}^2}{a^2}+\frac{2k}{a^2} \right)\delta_{ij},$
$R=6\left(\frac{\ddot{a}}{a}+\frac{\dot{a}}{a^2}+\frac{k}{a^2}\right),$
where we computed $R$ using the first two lines above via our earlier formula for the Ricci scalar. Note that $\delta_{ij}$ is the Kronecker delta, which is defined to be $1$ for $i=j$ and zero otherwise.

• The Friedmann Equations

Using the Ricci tensor and scalar in the left hand side of the Einstein equation and the stress-energy tensor on the right hand side, and demanding component by component equality (because the Einstein equation is in fact a set of equations governing each component of the curvature), we have two independent equations for $a$, one from $R_{00}$ and one from $R_{ij}$, since all of the $R_{ij}$‘s are the same.

These are
$H^2\equiv\left(\frac{\dot{a}}{a}\right)^2=\frac{8\pi G\rho}{3} -\frac{k}{a^2},$

and
$\dot{H}=-4\pi G(p+\rho)+\frac{k}{a^2}.$

These are the familiar Friedmann equations.

Explicitly differentiating the first of these to obtain $\dot{H}$ and setting it equal to the second equation above yields a third equation called the continuity equation:
$\dot{\rho}+3H(\rho+p)=0.$
See Copeland, Sami, & Tsujikawa (2006). Note that this equation can also be written as a statement of conservation of energy. With $dQ=0=dE+pdV$, (i.e. heat in is zero, so energy change is negative of work done by system) we have
$dE=-pdV\rightarrow\frac{d}{dt}\left(\rho a^3\right)=-p\frac{d}{dt}\left(a^3\right),$
where we have used $dE=d(\rho a^3)$ and $dV=d(a^3)$ and divided by $dt$.
Explicitly computing the derivatives in the above leads to the continuity equation.
See Misner, Thorne, & Wheeler (1973) for further detail.

• Accelerated Expansion?

When does accelerated expansion occur in this cosmology? We will have accelerated expansion when $\ddot{a}/a>0$. Using the fact that
$\dot{H}=-\left(\frac{\dot{a}}{a}\right)^2+\left(\frac{\ddot{a}}{a}\right)$
and adding the first Friedmann equation to the second to yield $\ddot{a}/{a}$, we find that
$\frac{\ddot{a}}{a}=-\frac{4\pi G}{3}\left(\rho+3p\right).$
Thus for accelerated expansion, we require that
$p < -\rho/3$

My treatment here is original, but the equations are checked with those in Hartle (2003) and Copeland, Sami, & Tsujikawa (2006) where appropriate.

• Comparison with Newtonian result

If we did not have general relativity, it would be extremely difficult to understand the accelerated expansion of the universe we now observe.  This is because in Newtonian gravity, the only types of expansion that are possible are constant or decelerated expansions.  Consider a homogeneous sphere with mass (equivalent to energy in units where $c=1$) density $\rho$ and radius $a$.  The acceleration at the surface of the sphere is simply
$\ddot{a}=-\frac{4\pi Ga^3\rho}{3a^2},$
where we have used Newton’s law of gravitation with $M=(4/3)\pi a^3\rho$.  This leads to
$\frac{\ddot{a}}{a}=-\frac{4\pi G\rho}{3}.$

Since $\rho\geq0$, $\ddot{a}/{a}$ will always be less than or equal to zero, and the expansion will always be constant or decelerated.  What this means is that  we require the framework of general relativity to understand the accelerated expansion we observe. If astronomers had discovered this acceleration before 1921, they would have checked their telescopes very carefully, because Newtonian physics would have told them to disbelieve their eyes.  They would have had to assume that the overall mass density in the universe was negative — certainly a conclusion significantly out of line with the physics we experience on Earth.  The reason general relativity makes accelerated expansion possible is because to the $\rho$ in the above equation it adds a pressure term $p$.

—-

Gravity: An Introduction to Einstein’s General Relativity, J.B. Hartle, 2003, Addison-Wesley.

Dynamics of Dark Energy, E.J. Copeland, M. Sami, and S. Tsujikawa, 2006, http://arxiv.org/abs/hep-th/0603057.

Gravitation, C. Misner, K. Thorne, and J. Wheeler, 1973, W.H. Freeman and Company.

I’m a 2nd year grad student in Astronomy at Harvard, working with Daniel Eisenstein on the effect of relative velocities between regular and dark matter on the baryon acoustic oscillations. I did my undergrad at Princeton, where I worked with Rich Gott on dark energy, Jeremy Goodman on dark matter, and Roman Rafikov on planetesimals. I also spent a year at Oxford getting a master’s in philosophy of physics, which remains an interest.

1. Congratulations for your clear deduction of the Friedmann equations.