Derivation of Friedman-Robertson-Walker (FRW) Equation
We shall use Newton’s theory of gravity, one of his theorem’s from his Principia, and the conservation of energy to derive the FRW equation which describes how changes with time. (Later on we’ll derive the FRW equation by substituting the metric \(g_μν\) into the EFE’s.)
In Newtonian theory, what determines the motions of the galaxies is gravity. What is the effect on the motion of the mass \\(m=m_j\) (Alice’s galaxy say) with some initial outward velocity \(V=H_0D\) due to the expansion of space. Note that \(V\) is a variable depending on \(D\). Very large values of \(D\) (corresponding to galaxies very far away) have enormous velocities \(V\) whereas for small values of \(D\) (corresponding to nearby galaxies) have very small velocities \(V\). Clearly \(V\) is different for each galaxy of mass \(m\) due to \(D\) being different. We assume that the motion of \(m\) is only effected by the gravitational field due to \(M\) (in other words, the gravitational force \(F_{\text{M,m}}\) is the only influence acting on \(m\)).
One of Newton’s theorem's states that the gravity due to all the masses surrounding \(m\) can be regarded as the gravity due to a point-mass \(M\) centered in the sphere of radius \(D\) where \(M\) is the sum of all the masses enclosed by that sphere.
The total energy \(E\) for each galaxy of mass \(m\) is constant and is given by \(E=KE+PE=1/2mv^2-MmG/D\). From this point onwards the steps we take might seem tedious, but really all we’ll be doing is a lot of algebra and making a lot of substitutions. First let’s multiply both sides of the energy-conservation equation by \(2/m\) to get
$$V^2-2\frac{MG}{D}=\frac{2E}{m}=KE.$$
Next let’s substitute \(V=dD/dt=(∆r)da/dt\) and \(D=a∆r\):
$$∆r^2\biggl(\frac{da}{dt}\biggr)-2\frac{MG}{a∆r}=\frac{2E}{m}=KE.$$
Next we can multiply both sides by \(1/a^2\) to get
$$∆r^2\biggl(\frac{da}{dt}\frac{1}{a^2}\biggr)-2\frac{MG}{a^3∆r}=\frac{2E}{a^2m}=\frac{KE}{a^2}.$$
We will also use Newton’s theorem to show that only the mass \(M\) inside the sphere of radius \(D\) contributes to the gravitational effect of a galaxy at a distance \(D\) away from the center. The mass is given by \(M=ρ(\frac{4}{3}πD^3)=ρ(\frac{4}{3}πa^3∆r^3\). It follows that
$$∆r^2\biggl(\frac{da}{dt}\frac{1}{a^2}\biggr)-2\frac{[ρ(\frac{4}{3}πa^3∆r^3)]G}{a^3∆r}=∆r^2\biggl(\frac{d^2a}{dt^2}\frac{1}{a^2}-\frac{8}{3}πρG\biggr)=\frac{K}{a^2}.$$
At some particular time \(t=constant\), the scaling factor is constant and does not vary with position (because we assumed the Universe is isotropic and homogeneous over large distances) and \(Δr^2K=K\). I will now write the left- and right-hand side of the equation in a way where you can clearly see both sides are proportional to \(Δr^2\) which must be the case since they are equal:
$$Δr^2\biggl(\frac{d^2a/dt^2}{a^2}-\frac{8}{3}πρG\biggr)=\frac{2}{ma^2}\biggl(\frac{1}{2}mV^2-\frac{MmG}{aΔr}\biggr).$$
Using the fact that \(V=Δrda/dt\) and \(M=ρ(\frac{4}{3}πa^3Δr^3)\), you can see that a \(Δr^2\) term can be factored out of the right-hand side of the equation and thus, by dividing both sides by \(Δr^2\), we get
$$\frac{(da/dt)^2}{a^2}-\frac{8}{3}πGρ=\frac{K}{a^2}.$$
Next, we'll let \(K=-κ\) and substitute it into the equation above; then we'll add \(\frac{8}{3}πGρ\) to both sides to get the FRW equation which is given by
$$H^2(t)=\frac{(da/dt)^2}{a^2}=8πGρ/3.\tag{6}$$
Some comments on the qualitative character of FRW Equation
The FRW equation is a differential equation whose solution is the scaling factor \(a(t)\). The scaling factor \(a(t)\) that we obtain from this equation will be different based on what \(ρ\) and \(κ\) are. The value of \(κ\) is related to the curvature and geometry of the space of the universe. If \(κ=1\), then the space of the Universe is a closed and bounded 3-sphere; if \(κ=-1\), the space is an open, 3-dimensional hyperboloid; if \(κ=0\), then space is flat (non-curved) and open. We know, empirically, from observations of the Cosmic Microwave Background (CMB) that \(κ≈0\) to within an accuracy of about 1%. We will assume that \(κ=0\) which simplifies the FRW equation to
$$H^2(t)=(da/dt)^2/a^2=\frac{8}{3}πGρ.\tag{7}$$
If we live in a Universe with a flat space with no curvature where \(κ=0\), since the left-hand side of the equation will always be positive, the left-hand side of the equation will always be positive. This means that Hubble’s parameter \(H(t)\) will remain constant with time which means that the Universe must always be expanding (it cannot be contracting).
For now we’ll just worry about studying the FRW equation when \(κ=0\) and (for now) we won’t worry about how \(a(t)\) changes based on the value of \(κ\). Given \((da/dt)^2/a^2=8πGρ/3\), \(a(t)\) will be different depending on what \(ρ\) is. For example, from the birth of the Universe until the Universe was about 10,000 years old, the energy content of the Universe was dominated by radiation and we can call the energy density in this primordial Universe \(ρ_r\). From when the Universe was 10,000 years old to a few billion years old, the distribution of energy was mostly in the form of masses (galaxies) at rest in the coordinates \(x^i)\). We can call the energy density of the universe during this time period \(ρ_M\). \(ρ_M\) and \(ρ_r\) are different; as a consequence of this, the right-hand side of the FRW equation will be different for \(ρ_M\) and \(ρ_r\). Therefore, \(a(t)\) will be different for both cases.
In a Universe where there is only radiation, all of the energy is in the photons present. The energy of each photon is given by \(E_γ=hf=hc/λ=k/λ\). The energy \(E_γ\) of each photon depends on the size of the box. If the box expands, for example, as space expands, the wave associated with each photon will expand with it. As its wavelength \(λ\) gets stretched (which is to say redshifted), the energy \(E_γhc/λ\) of each photon will decrease. For photons traveling along the x-axis in the box, let the distance between points A and B (which can be the two points on crests or troughs separated by one wavelength \(λ\) or, more generally, two points with the same phase \(Φ\) separated by \(λ\)) be \(D=a∆x\); then, \(D=a∆x=λ\). As the box grows, \(∆x\) stays the same while the scaling factor increases. Since \(∆x=constant\), we have \(a(constant)=λ\) and thus \(a∝λ\). We can substitute this into \(E_γ\) to obtain \(E_γ=hc/λ=hc/a∆x=k/a\). Thus, \(E_γ∝λ\).
We can choose any point in space to be the origin of a coordinate system \(x^i)\) which “follows” the expansion/contraction of space. Suppose we draw a box whose edges are separated by 1 coordinate unit. As space either expands or contracts, the box will expand or contract with it. \(∆x^1\), \(∆x^2\), and \(∆x^3\) do not change with this expansion/contraction. The actual physical distances between the edges of the box are given \(a∆x^1\), \(a∆x^2\), and \(a∆x^3\) which, in general, will change with time. The volume of the box is given by \(V=(a∆x^1)(a∆x^2)(a∆x^3)\) since \(∆x^1=∆x^2=∆x^3=1\). All of the photons will be traveling at the speed of light where some are going into the box and some are going out of the box, but we will assume that on average the total number of photons in the box is constant. The energy density in the box due to the radiation present is \(ρ_r=\frac{\text{total energy in box}}{\text{volume of box}}=E_{total}/V\). The energy of each photon is \(E_γ\) and if there are \(N\) (which we assume is constant) photons in the box then the total energy in the box is \(E_{total}=NE_γ\) (this is just the sum of the energies of each photon in the box). The energy density in the box is
$$ρ_r=E_{total}/V=NE_γ/a^3=N(k/a)/a^3=k/a^4.\tag{8}$$
where \(ρ_r∝1/a^4\). In a Universe where space is flat (but can be expanding/contracting) with only radiation in it, the FRW equation simplifies to
$$H^2(t)=(da/dt)^2/a^2=8πGρ_r/3=k/a^4.\tag{9}$$
This article is licensed under a CC BY-NC-SA 4.0 license.
References
1. Leonard Susskind. "Matter and radiation dominated universes". theoreticalminimum.com.