3 Structure Formation

The second is the Euler equation, which can be viewed as Newton’s “F=ma” for a continuous system,

$mn\left({\displaystyle \frac{\partial }{\partial t}}+𝐮\cdot \nabla \right)𝐮=-\nabla P$

(3.139)

The left-hand side interpreted as mass $\times$ acceleration, while the pressure $-\nabla P$ on the right-hand side provides the force.

The last of our equations is the equation of state which, for now, we leave general as

$P=P(n,T)$

(3.140)

For much of this section, we will use ideal gas equation, $P=n{k}_{B}T$, which is the appropriate equation of state for a non-relativistic fluid. In time, we will also apply these ideas to other fluids.

The simplest solution to these equations describes a static fluid with $𝐮=0$ and constant density and pressure

$n=\overline{n}\mathit{\hspace{1em}\hspace{1em} }\mathrm{and}\mathit{\hspace{1em}\hspace{1em} }P=\overline{P}$

This is a homogeneous and isotropic fluid. We take this to be our background and look at small perturbations. We take $𝐮$ to be small, and write

$n(𝐱,t)=\overline{n}+\delta n(𝐱,t)\mathit{\hspace{1em}\hspace{1em} }\mathrm{and}\mathit{\hspace{1em}\hspace{1em} }P(𝐱,t)=\overline{P}+\delta P(𝐱,t)$

The equations (3.138) and (3.139) are linearised to give

${\displaystyle \frac{\partial (\delta n)}{\partial t}}=-\nabla \cdot (\overline{n}𝐮)\mathit{\hspace{1em}\hspace{1em} }\mathrm{and}\mathit{\hspace{1em}\hspace{1em} }m\overline{n}{\displaystyle \frac{\partial 𝐮}{\partial t}}=-\nabla \delta P$

We can combine these to find,

${\displaystyle \frac{{\partial }^2(\delta n)}{\partial t^2}}=-\overline{n}\nabla \cdot {\displaystyle \frac{\partial 𝐮}{\partial t}}={\displaystyle \frac{1}{m}}{\nabla }^2\delta P$

At this point, we need to invoke the equation of state, relating $P$ to $n$. It will be useful to give a new name to the quantity $\partial P/\partial n$: we write it as

${\displaystyle \frac{\partial P}{\partial n}}=m{c}_s^2$

(3.141)

We can then relate $\delta P=m{c}_s^2\delta n$ to find that the density perturbations obey

$\left({\displaystyle \frac{\partial {}^2}{\partial t^2}}-c_s^2{\nabla }^2\right)\delta n=0$

(3.142)

This is the wave equation. As its name suggests, its solutions are waves of the form

$\delta n(𝐱,t)=A(𝐤)\cos \left(\omega t-𝐤\cdot 𝐱\right)+B(𝐤)\sin \left(\omega t-𝐤\cdot 𝐱\right)$

(3.143)

We call these sound waves. The solution key property of the solution is the wavevector $𝐤$ which determines the direction of travel of the wave and the wavelength $\lambda =2\pi /|𝐤|$. The frequency $\omega$ of the wave is given by

$\omega =c_s|𝐤|$

The proportionality constant $c_s$, defined in (3.141), has the interpretation of the speed of sound. (We’ll compute examples below.) Finally, $A(𝐤)$ and $B(𝐤)$ are two, arbitrary integration constants. Because the wave equation is linear, we can add together as many solutions of the form (3.143) as we like, with different integration constants $A(𝐤)$ and $B(𝐤)$. In this way, we can build up wavepackets with different profiles.

In what follows, we will often write the solution (3.143) in complex form,

$\delta n=C(𝐤)\exp \left(i(\omega t-𝐤\cdot 𝐱)\right)$

for some complex $C$. This is standard, albeit inaccurate notation. Obviously the number density $\delta n$ should be real. But because the wave equation is linear, we can always just take the real part of the right-hand-side to get a solution. This form of the solution is more useful simply because it’s quicker to write exponentials rather than cos and sin.

The equation for the sound speed (3.141) can alternatively be written in terms of the energy density $\rho =mn{c}^2$, as

${\displaystyle \frac{\partial P}{\partial \rho }}={\displaystyle \frac{c_s^2}{c^2}}$

(3.144)

Using $P=0$ suggests that $c_s=0$ for a non-relativistic fluid. But what this is really telling is simply that

$c_s\ll c$

(3.145)

This makes sense. The sound speed is related to the speed of the constituent particles in the fluid. In a non-relativistic fluid, this is necessarily much less than the speed of light.

In fact, we can do better and compute the sound speed as a function of temperature (which, itself, is related to the speed of the constituent particles). For the ideal gas, the equation of state is

$P=n{k}_{B}T$

It’s tempting to simply differentiate $\partial P/\partial n$, with $T$ fixed, to determine the speed of sound. But that’s a little too hasty: as $P$ and $n$ vary, it is quite possibly that $T$ varies as well.

To understand how this works, we need a little physical input. The energy of the ideal gas with some fixed total number of atoms $N=nV$ is (2.96)

$E={\displaystyle \frac{3}{2}}N{k}_{B}T$

If the volume changes, then the energy should change by the work done

	$dE=-PdV$	$\Rightarrow$	$\mathrm{ }\mathit{\hspace{1em}}{\displaystyle \frac{3}{2}}N{k}_{B}dT=-PdV$
		$\Rightarrow$	$\mathrm{ }\mathit{\hspace{1em}}{\displaystyle \frac{3}{2}}{\displaystyle \frac{dT}{T}}=-{\displaystyle \frac{dV}{V}}$

where, in the second line, we’ve used the equation of state. Integrating this expression tells us that $T^{3/2}V$ is constant. Alternatively, using $n=N/V$, we learn that $T{n}^{-2/3}$ is constant. Such changes are referred to as adiabatic. (Underlying this is the statement that entropy is conserved for adiabatic changes; you can learn more about this in the lectures on Statistical Physics.) This means that

${{\displaystyle \frac{\partial T}{\partial n}}|}_{\mathrm{adiabatic}}={\displaystyle \frac{2}{3}}{\displaystyle \frac{T}{n}}$

The speed of sound (3.141) should be computed under the assumption of such an adiabatic change. We then have

${{\displaystyle \frac{\partial P}{\partial n}}|}_{\mathrm{adiabatic}}={\displaystyle \frac{5}{3}}{k}_{B}T$

From this, we compute the speed of sound in an ideal gas to be

$c_s=\sqrt{{\displaystyle \frac{5k_{B}T}{3m}}}$

Before we proceed, I will briefly mention another, more rigorous, approach to get this result. We could treat $T(𝐱,t)$ as a new dynamical field with its own equation of motion. This equation of motion turns out to be

$\left({\displaystyle \frac{\partial }{\partial t}}+𝐮\cdot \nabla \right)T+{\displaystyle \frac{2T}{3}}\nabla \cdot 𝐮=0$

A full derivation of this needs the Boltzmann equation, and can be found in the lectures on Kinetic Theory. It’s straightforward to check that this equation combines with the continuity equation (3.138) to ensure that $T{n}^{-2/3}$ is indeed constant along flow lines.

Much of the discussion above goes through for general fluids if the equations are phrased in terms of the energy density $\rho$ instead of the number density $n$. In particular, the sound speed can be computed using (3.144). For a relativistic fluid, with $P=\rho /3$, we have

$c_s^2={\displaystyle \frac{1}{3}}{c}^2$

(3.146)

This time we see that the speed of sound is tied to the speed of light. Again, this is to be expected: in a relativistic fluid, any constituent particles are flying around at close to the speed of light. The difference in the speed of sound between a non-relativistic fluid (3.145) and a relativistic fluid (3.146) will prove to be one of the important ingredients in the story of structure formation.

We now perturb the constant background. We require that the perturbed gravitational potential $\mathrm{\Phi }+\delta \mathrm{\Phi }$ obeys

${\nabla }^2\delta \mathrm{\Phi }=4\pi Gm\delta n$

The same linearisation that we saw previously now shows that the wave equation (3.142) is deformed to

$\left({\displaystyle \frac{\partial {}^2}{\partial t^2}}-c_s^2{\nabla }^2-4\pi Gm\overline{n}\right)\delta n=0$

This is again solved by the ansatz

$\delta n=C(𝐤)\exp \left(i(\omega t-𝐤\cdot 𝐱)\right)$

but now with the frequency and wavevector related by

	${\omega }^2$	$=$	$c_s^2{k}^2-4\pi Gm\overline{n}$
		$=$	$c_s^2(k^2-k_J^2)$

Equations of this type, which relate the frequency to the wavenumber, are referred to as dispersion relations. In the second line we have defined the Jeans wavenumber $k_J$,

$k_J=\sqrt{{\displaystyle \frac{4\pi Gm\overline{n}}{c_s^2}}}$

The qualitative properties of the solution now depend on the wavenumber $𝐤$. For small wavelengths, or large wavenumbers $k>k_J$, the solutions oscillate as before. These are sound waves. However, when the wavelengths are large, then the gravitational background becomes important. Here the frequency is imaginary, which has the interpretation that perturbations $\delta n\sim e^{i\omega t}$ grow or decay exponentially. For $k\ll k_J$ we have

$\delta n\sim e^{\pm t/\tau }\mathit{\hspace{1em}\hspace{1em} }\mathrm{with}\mathit{\hspace{1em}\hspace{1em} }\tau \approx \sqrt{{\displaystyle \frac{1}{4\pi Gm\overline{n}}}}$

(3.149)

We learn that long wavelength perturbations no longer oscillate as sound waves. Instead, any perturbation that has a size larger than the Jeans length,

${\lambda }_J={\displaystyle \frac{2\pi }{k_J}}=c_s\sqrt{{\displaystyle \frac{\pi }{Gm\overline{n}}}}$

(3.150)

will typically grow exponentially quickly due to the effect of gravity. This is known as the Jeans’ instability.

The derivation above also gives us a clue as to the physical mechanism for the Jeans instability. It comes from attempting to balance the pressure and gravitational terms in the Euler equation (3.147). Consider an over-dense spherical region of radius $R$. In the absence of any pressure, this region would collapse with a time-scale $\tau$ given in (3.149). In a fluid, this collapse is opposed by the pressure. But the build-up of pressure is not instantaneous; it takes time given roughly by

$t_{\mathrm{pressure}}\sim {\displaystyle \frac{R}{c_s}}$

When $R$ is small, and the build-up of pressure stops the collapse and we get oscillating motion that we interpret as sound waves. In contrast, if $R$ is large we have , there is no time for the pressure to build. In this case, the system suffers the Jeans instability and is susceptible to gravitational collapse.

We need to revisit our equations once more. Consider a particle tracing out a trajectory $𝐱(t)$ in co-moving coordinates. The physical coordinates $𝐫(t)$ (called ${𝐱}_{\mathrm{phys}}$ in (1.14)) is given by

$𝐫(t)=a(t)𝐱(t)$

The physical velocity of a particle is

$𝐮=\dot{𝐫}=H𝐫+𝐯$

(3.151)

with $𝐯=a\dot{𝐱}$. In what follows, we will need to jump between physical and co-moving coordinates. The spatial derivatives are related simply by

${\nabla }_{𝐫}={\displaystyle \frac{1}{a}}{\nabla }_{𝐱}$

(3.152)

The temporal derivatives are a little more subtle since they differ depending on whether we keep $𝐫(t)$ fixed or $𝐱(t)$ fixed. In particular

	${{\displaystyle \frac{\partial }{\partial t}}|}_{𝐫}$	$=$	${{\displaystyle \frac{\partial }{\partial t}}|}_{𝐱}+{{\displaystyle \frac{\partial 𝐱}{\partial t}}|}_{𝐫}\cdot {\nabla }_{𝐱}$		(3.153)
		$=$	${{\displaystyle \frac{\partial }{\partial t}}|}_{𝐱}+{{\displaystyle \frac{\partial (a^{-1}𝐫)}{\partial t}}|}_{𝐫}\cdot {\nabla }_{𝐱}$
		$=$	${{\displaystyle \frac{\partial }{\partial t}}|}_{𝐱}-H𝐱\cdot {\nabla }_{𝐱}$

Now we come to the equations describing the fluid. The equations that we dealt with previously should be viewed as given in terms of physical coordinates $𝐫$. However, it will turn out that subsequent calculations are somewhat easier if done in co-moving coordinates. We just have to translate from one to the other.

We can make contact with the story of Section 1. Following (3.151), we write the velocity of the fluid as

$𝐮(𝐱,t)=Ha𝐱(t)+𝐯(𝐱,t)$

(3.154)

and the continuity equation becomes

${\displaystyle \frac{\partial n}{\partial t}}+3Hn+{\displaystyle \frac{1}{a}}\nabla \cdot (n𝐯)=0$

(3.155)

where we’ve used $\nabla \cdot 𝐱=3$. This form makes it clear that if we restrict to solutions in which $𝐯=0$, so the velocity of the fluid simply follows the expansion of spacetime, then we recover our earlier continuity equation (1.39), specialised to the case of non-relativistic matter,

${\displaystyle \frac{\partial n}{\partial t}}=-3Hn$

(Recall that the energy density is given by $\rho =m\overline{n}{c}^2$.) This has the familiar solution

$n(t)\sim {\displaystyle \frac{1}{a^3}}$

(3.156)

which simply tells us that the number density dilutes as the universe expands.

Now we perturb the fluid,

	$n(𝐱,t)$	$=$	$\overline{n}(t)+\delta n(𝐱,t)$
		$=$	$\overline{n}(t)\left[1+\delta (𝐱,t)\right]$

where $\overline{n}(t)$ is a spatially homogeneous density evolving as (3.156) and, in the second line, we’ve defined

$\delta ={\displaystyle \frac{\delta n}{\overline{n}}}={\displaystyle \frac{\delta \rho }{\overline{\rho }}}$

The perturbation $\delta$ is referred to as the density contrast.

Let’s now see what conditions the continuity equation (3.155) imposes on these perturbations. It reads

	${\displaystyle \frac{\partial }{\partial t}}(\overline{n}\delta )+3H\overline{n}\delta$	$=$	$-{\displaystyle \frac{1}{a}}\nabla \left[\overline{n}(1+\delta )𝐯\right]$
		$=$	$-{\displaystyle \frac{\overline{n}}{a}}\nabla \cdot 𝐯+𝒪(𝐯\delta )$

We drop the second term on the grounds that it is non-linear in the small quantities $\delta$ and $𝐯$. Using the fact that the background density $\overline{n}$ evolves as (3.156), this equation reduces to the simple requirement

$\dot{\delta }=-{\displaystyle \frac{1}{a}}\nabla \cdot 𝐯$

(3.157)

This is the first of our perturbed equations.

For our immediate purposes, the thing that should make us most happy is that we no longer have to worry about the Jeans’ swindle; our spatially homogeneous background satisfies the equations of motion as it should. At heart, the Jeans’s swindle was telling us that a spatially homogeneous fluid is inconsistent in Newtonian gravity. But it is perfectly consistent if we allow for an expanding universe, with the gravitational potential $\overline{\mathrm{\Phi }}$ for a homogeneous fluid driving the expansion of space, just as we learned in Section 1.

Next, we perturb around the background. We write $P=\overline{P}+\delta P$ and $\mathrm{\Phi }=\overline{\mathrm{\Phi }}+\delta \mathrm{\Phi }$ and, as before, $𝐮=Ha𝐱+𝐯$. The linearised Euler equation reads

$m\overline{n}a\left(\dot{𝐯}+H𝐯\right)=-\nabla \delta P-m\overline{n}\nabla \delta \mathrm{\Phi }$

(3.161)

where we’ve used the fact that $(𝐯\cdot \nabla )𝐱=𝐯$. If we drop the pressure and gravitational perturbation on the right-hand side, this equation tells us that $\dot{𝐯}=-H𝐯$, so the peculiar velocities redshift as $𝐯\sim 1/a$. This can be viewed as a consequence of Hubble friction, which slows the peculiar velocities as the universe expands.

Finally, the linearised Poisson equation is

${\nabla }^2\delta \mathrm{\Phi }=4\pi Gm\overline{n}{a}^2\delta$

(3.162)

Now we combine our three linearised equations (3.157), (3.161) and (3.162). Take the time derivative of (3.157) to get

$\ddot{\delta }={\displaystyle \frac{H}{a}}\nabla \cdot 𝐯-{\displaystyle \frac{1}{a}}\nabla \cdot \dot{𝐯}=-H\dot{\delta }-{\displaystyle \frac{1}{a}}\nabla \dot{𝐯}$

Take the gradient of (3.161) to get

	$m\overline{n}a\left(\nabla \cdot \dot{𝐯}-Ha\dot{\delta }\right)$	$=$	$-{\nabla }^2\delta P-m\overline{n}{\nabla }^2\delta \mathrm{\Phi }$
		$=$	$-m\overline{n}\left(c_s^2{\nabla }^2\delta +4\pi Gm\overline{n}{a}^2\delta \right)$

where in the second line we’ve used $\delta P=m{c}_s^2\delta n=m{c}_s^2\overline{n}\delta$ and the Poisson equation (3.162). We now combine these two results to get a single equation telling us how the density perturbation $\delta$ evolves in an expanding spacetime

$\ddot{\delta }+2H\dot{\delta }-c_s^2\left({\displaystyle \frac{1}{a^2}}{\nabla }^2+k_J^2\right)\delta =0$

(3.163)

where $k_J$ is the physical Jeans wavenumber given, as before, by $k_J^2=4\pi Gm\overline{n}/c_s^2$.

The most important addition from the expanding space time is the friction-like term proportional $2H\dot{\delta }$. This is referred to as Hubble friction or Hubble drag. (We saw an analogous term when discussing inflation in Section 1.5.)

To solve (3.163), it is simplest to find by working in Fourier space. We define

$\delta (𝐤,t)={\displaystyle \int d^{3}x{e}^{i𝐤\cdot 𝐱}\delta (𝐱,t)}$

where we are adopting the annoying but standard convention that the function $\delta (𝐱)$ and its Fourier transform $\delta (𝐤)$ are distinguished only by their argument. Since $𝐱$ is the co-moving coordinate, $𝐤$ is the co-moving wavevector.

The advantage of working in Fourier space is that the equation (3.163) decomposes into a separate equation for each value of $𝐤$,

$\ddot{\delta }(𝐤,t)+2H\dot{\delta }(𝐤,t)+c_s^2\left({\displaystyle \frac{k^2}{a^2}}-k_J^2\right)\delta (𝐤,t)=0$

(3.164)

The slightly unusual factor of $a$ in the final term arises because $k$ is the co-moving wavenumber and so $k/a$ is the physical wavenumber, but $k_J$ refers to the physical Jeans wavenumber. Our challenge now is to solve this equation.

Small wavelength modes have $k/a\gg k_J$. Here, the equation (3.164) is essentially that of a damped harmonic oscillator, with the expanding universe providing the friction term $H\dot{\delta }$. The solutions are oscillating sound-waves with the Hubble friction leading to an ever-decreasing amplitude.

If structure is to ultimately form in the universe, we need to find solutions that grow over time. These are supplied by the long-wavelength modes, with , which suffer from the Jeans’ instability. However, as we shall now see, the details of the Jeans’ instability are altered in an expanding universe.

In what follows, we will see that there are two length scales at play for the growing modes. One is the Jeans’ length scale ${\lambda }_J=2\pi /k_J$,

${\lambda }_J=c_s\sqrt{{\displaystyle \frac{\pi }{Gm\overline{n}}}}=c_{s}c\sqrt{{\displaystyle \frac{\pi }{G\overline{\rho }}}}$

(3.165)

Only modes with $\lambda >{\lambda }_J$ will grow. The other relevant physical length scale is that set by the expansion of the universe,

$d_H\approx c{H}^{-1}=c^2\sqrt{{\displaystyle \frac{3}{8\pi G\overline{\rho }}}}$

(3.166)

This is called the apparent horizon. In the standard FRW cosmology (with ordinary matter or radiation) the apparent horizon coincides with the particle horizon, defined in (1.24). In such a situation, it would make little sense to talk about perturbations with wavelength $\lambda >d_H$. This is because the Fourier mode of a perturbation is a coherent wave and causality would appear to prohibit the formation of such perturbations on distances greater than $d_H$ since there has been no time for light, or anything else, to cross this distance since the Big Bang.

This, however, is exactly the problem that is resolved by a period of inflation in the very early universe. The whole point of inflation is to stretch the particle horizon so that it sits way outside the apparent horizon. Indeed, we will see that perturbation modes with wavelengths $\lambda >d_H$ play an important role in the story of structure formation in our universe, strongly implying that a period of inflation is needed. In what follows, we will refer to the apparent horizon (3.166) simply as the “horizon”.

Af the wavelength of the mode is sufficiently long, so $k/a\ll k_J$, then we can approximate (3.164) as

$\ddot{\delta }(𝐤)+2H\dot{\delta }(𝐤)-{\displaystyle \frac{4\pi G\overline{\rho }}{c^2}}\delta (𝐤)=0$

(3.167)

Here we’ve left the $t$ argument in $\delta (𝐤,t)$ implicit, but kept the $𝐤$ argument because it tells us that the is the mode is in Fourier space rather than real space.

In a matter dominated universe, $a\sim t^{2/3}$ so $H=2/3t$. The third term is also related to the Hubble parameter through Friedmann equation $H^2=8\pi G\overline{\rho }/3c^2$. We then have

$\ddot{\delta }(𝐤)+{\displaystyle \frac{4}{3t}}\dot{\delta }(𝐤)-{\displaystyle \frac{2}{3t^2}}\delta (𝐤)=0$

Substituting in the power-law ansatz $\delta (𝐤,t)\sim t^n$, we find two solutions, one decaying and one growing

(3.168)

We see that the expansion of the universe slows down the rate at which objects undergo gravitational collapse, with the Hubble damping turning the exponential growth of the Jeans instability (3.149) into a power-law, one that scales linearly with the size of the universe.

We will work with the energy density $\rho (𝐱,t)$, rather than the number density $n(𝐱,t)$; for a non-relativistic fluid, they are related by $\rho =mn{c}^2$. We need to go through each of our original equations – continuity, Euler, and Poisson – and ask how they change for a general fluid. We will motivate each of these changes, but not derive them.

First, the continuity equation (3.155): this gets replaced by

${\displaystyle \frac{\partial \rho }{\partial t}}+3(1+w)H\rho +{\displaystyle \frac{1}{a}}(1+w)\nabla \cdot (\rho 𝐯)=0$

The equation of state parameter $w$ appears twice. The first of these is unsurprising, since it guarantees that this equation reduces to our previous continuity equation (1.39) when $𝐯=0$. We won’t derive the presence of the $(1+w)$ factor in the final term, but it arises in a similar way to the first.

The Euler equation (3.158) remains unchanged. Somewhat more subtle is the relativistic generalisation of the gravitational potential. In general relativity, both energy density and pressure gravitate. It turns out that the Poisson equation (3.159) should be replaced by

${\nabla }^2\mathrm{\Phi }={\displaystyle \frac{4\pi G}{c^2}}(1+3w)\rho a^2$

(3.169)

There is, in fact, a clue in the discussion above that strongly hints at this form. Recall that we avoided the Jeans swindle in an expanding spacetime by relating the gravitational potential to the acceleration in (3.160). The Poisson equation then became equivalent to the acceleration equation (1.52) which, in general, reads

${\displaystyle \frac{\ddot{a}}{a}}=-{\displaystyle \frac{4\pi G}{3c^2}}(1+3w)\rho$

We see the same distinctive factor of $(1+3w)$ appearing here.

Repeating the same steps as previously, the perturbation equation (3.164) is replaced by

$\ddot{\delta }(𝐤)+2H\dot{\delta }(𝐤)+c_s^2(1+w)\left({\displaystyle \frac{k^2}{a^2}}-(1+3w)k_J^2\right)\delta (𝐤)=0$

(3.170)

with $k_J=2\pi /{\lambda }_J$ is the physical wavenumber, defined as in (3.165). It differs from the non-relativistic Jeans’ length only by the expression for the speed of sound $c_s$.

Let’s now restrict to radiation with $w=1/3$. We know from (3.146) that the speed of sound for a relativistic fluid is

$c_s^2={\displaystyle \frac{1}{3}}{c}^2$

This means that there is no parametric separation between the Jeans length (3.165) and the horizon (3.166). Instead, we have ${\lambda }_J\approx c{H}^{-1}$. Any perturbation that lies inside the horizon does not grow. Instead, the pressure of the radiation causes the perturbation to oscillate as a sound wave.

Outside the horizon it’s a different story. In a radiation dominated universe, $a\sim t^{1/2}$ so $H=1/2t$. For wavenumbers $k/a\ll k_J$, the equation (3.170) governing perturbations becomes

${\ddot{\delta }}_r(𝐤)+{\displaystyle \frac{1}{t}}{\dot{\delta }}_r(𝐤)-{\displaystyle \frac{1}{t^2}}{\delta }_r(𝐤)=0$

Once again, substituting in the power-law ansatz $\delta (𝐤,t)\sim t^n$, we find two solutions, one decaying and one growing

(3.171)

We lean that perturbations in the density of radiation grow outside the horizon. Indeed, they grow faster than the linear growth (3.168) seen in the matter dominated era.

In contrast, the matter perturbations with wavelength larger than the horizon obey the same equation as the radiation perturbations in the radiation dominated era. (The term ${\overline{\rho }}_r{\delta }_r$ in (3.172) cannot now be neglected.) This means that those modes outside the horizon grow as $\delta \sim a^2$ as seen in (3.171).

We should be able to trust our equations on small length scales, for the simple reason that general relativity reduces to Newtonian gravity in this regime. However, when we get to perturbations whose length is comparable to the horizon $d_H$, we should be more nervous, since it seems plausible that the perturbations feel the curvature of spacetime in a way that our Newtonian approximation misses.

The only way to know if the Newtonian perturbation equation (3.164) is valid is to roll up our sleeves and perform the correct, general relativistic perturbation theory. This is a somewhat painful exercise that you will be given the opportunity to embrace in next year’s Part III cosmology course. There you will learn that the question “is our equation (3.164) valid?” has a short answer and a long answer.

The short answer is: yes.

The long answer is substantially more subtle. It turns out that the matter perturbation in general relativity is not diffeomorphism invariant, which means that the answer you get depends on the coordinates you use. This is bad. Indeed, one of the main philosophical lessons of general relativity is that the coordinates you use should not matter one iota. Moreover, this issue is particularly problematic for super-horizon perturbations with $\lambda \gg d_H$, and an important part of the relativistic approach is to understand the right, diffeomorphism invariant quantity to focus on. For most choices of coordinates (so called “gauges”) it turns out that the Poisson equation (3.162) is not valid on super-horizon scales. There is, however, a choice of coordinates -- conformal Newtonian gauge -- where the Poisson equation holds even on super-horizon scales and this is the one we are implicitly choosing⁹⁹ 9 More details can be found in the paper by Chisari and Zaldarriaga, “Connection between Newtonian simulations and general relativity, arXiv:1101.3555.. All of this is to say that you can trust the physics that we’ve derived here, but if you should be careful when comparing to analogous results derived in a general relativistic setting where the answers may look different because, although the symbols are the same, they refer to subtly different objects.

We are interested in the perturbations of the matter, since this is what we’re ultimately made of. If the density perturbations remain sufficiently small, so that the linearised analysis developed above holds then the linear analysis of this section remains valid which, in particular, means that each $\delta (𝐤)$ evolves independently, as seen in, for example, (3.164). The evolution of a perturbation of a given wavevector $𝐤$ from an initial time $t_i$ to the present can be distilled into a transfer function $T(k)$, defined as

$\delta (𝐤,t_0)=T(k)\delta (𝐤,t_i)$

(3.174)

The initial time $t_i$ is usually taken to be early, typically just after the end of inflation.

Key to understanding the physics is the question of when perturbations enter the horizon. Recall that, in physical coordinates, the apparent horizon is $d_H\approx c/H$ as in (3.166). It is simplest, however, to work in co-moving coordinates, where the apparent horizon is

${\chi }_H={\displaystyle \frac{c}{aH}}$

In the radiation dominated era, $a\sim t^{1/2}$ and so $H\sim 1/a^2$. In the matter dominated era, $a\sim t^{2/3}$ and so $H\sim 1/a^{3/2}$. In both cases, the co-moving horizon increases over time

(3.175)

The intuition behind this is that, as the universe expands, there is more that one can see and, correspondingly, the co-moving horizon grows.

The co-moving wavevector $𝐤$ remains unchanged over time. (This is the main advantage of working with the co-moving wavevector in the previous section. In contrast, the physical wavevector is ${𝐤}_{\mathrm{phys}}=𝐤/a$ shrinks over time as the physical wavelength ${\lambda }_{\mathrm{phys}}=2\pi /k_{\mathrm{phys}}=2\pi a/k$ is stretched by the expansion of the universe.) This means that, for each $𝐤$, there will be a time when the corresponding perturbation enters the horizon. It matters whether the time of entry occurs during the radiation or matter dominated eras.

At the time of matter-radiation equality (which occurred around $z=3400$), modes with wavenumber $k_{\mathrm{eq}}$ have just entered the horizon, where

$k_{\mathrm{eq}}={\displaystyle \frac{2\pi }{c}}{(aH)}_{\mathrm{eq}}$

Modes larger than this (i.e. with ) will then enter the horizon in the matter dominated era. Modes smaller than this (i.e. with $k>k_{\mathrm{eq}}$) will have entered the horizon during the radiation dominated era. Let’s look at each of these in turn.

First the long wavelength modes with . These were outside the horizon during the radiation era where $\delta$ grew as $a^2$, as seen in (3.171). As the universe entered the matter dominated era, the growth slows to $\delta \sim a$, as seen in (3.168). This means that, starting from an initial time $t_i$ evolve to their present day value

$\delta (𝐤,t_0)$

$=$

(3.176)

We learn that each mode grows by an amount independent of $𝐤$.

Things are more interesting for the short wavelength modes with $k>k_{\mathrm{eq}}$ which enter the horizon during the radiation era. Before entering the horizon, such modes grow as $\delta \sim a^2$ as in (3.171). However, when they enter the horizon, their growth slows to the logarithmic growth seen in (3.173). For our purposes, this is effectively constant. The growth only resumes when the universe enters the matter dominated era. This means that

	$\delta (𝐤,t_0)$	$=$	${\left({\displaystyle \frac{a_{\mathrm{enter}}}{a_i}}\right)}^2{\displaystyle \frac{a_0}{a_{\mathrm{eq}}}}\delta (𝐤,t_i)$		(3.177)
		$=$	${\left({\displaystyle \frac{a_{\mathrm{enter}}}{a_{\mathrm{eq}}}}\right)}^2\times \left[{\left({\displaystyle \frac{a_{\mathrm{eq}}}{a_i}}\right)}^2{\displaystyle \frac{a_0}{a_{\mathrm{eq}}}}\right]\delta (𝐤,t_i)\mathit{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em} }\mathrm{for}k>k_{\mathrm{eq}}$

The factor in square brackets is the same, constant amount (3.176) that the long wavelength modes grew by. However, the amplitude is suppressed by the factor of $a_{\mathrm{enter}}^2/a_{\mathrm{eq}}^2$, reflecting the fact that growth stalled during the radiation dominated era. For a given mode $𝐤$, the scale factor at horizon entry is given by

$k={\displaystyle \frac{2\pi }{c}}{(aH)}_{\mathrm{enter}}$

Using $a\sim t^{1/2}$ in the radiation era, we have $H=1/2t\sim 1/a^2$ so a given scale $k$ enters the horizon at $k\sim {(aH)}_{\mathrm{enter}}\sim 1/a_{\mathrm{enter}}$. We can then write ${(a_{\mathrm{enter}}/a_{\mathrm{eq}})}^2=k_{\mathrm{eq}}^2/k^2$.

Finally, all of this can be packaged into the transfer function (3.174). Assuming that the perturbations remain sufficiently small, so the linearised analysis is valid, the transfer function can be found in (3.176) and (3.177), we have

(3.178)

We will make use of this shortly.

There are two basic questions that we need to address:

• •

  What is the initial probability distribution?

• •

  How did this probability distribution subsequently evolve?

If we understand both of these well enough, we should be able to compare our results to the distribution of galaxies observed in the sky. We start by describing the initial probability distribution. We then see how it evolves in Section 3.2.3.

It may seem daunting to guess the form of the initial perturbations. However, the universe is kind to us and the observational evidence suggests that these perturbations take the simplest form possible. (We will offer an explanation for this in Section 3.5.)

First, the perturbations of each fluid component are correlated. In particular, the perturbation in any non-relativistic matter, such as baryons and cold dark matter, is the same: ${\delta }_B={\delta }_{CDM}$. Furthermore, perturbations in matter and perturbations in radiation are related by

${\delta }_m={\displaystyle \frac{3}{4}}{\delta }_r$

(3.183)

Perturbations of this kind are called adiabatic.

It may seem like a minor miracle that the perturbations in all fluids are correlated in this way. What’s really happening is that there is an initial perturbation in the gravitational potential (or, in the language of general relativity, in the metric) which, in turn, imprints itself on each of the fluids in the same way.

Logically, we could also have initial perturbations of the form $\delta {\rho }_m=-\delta {\rho }_r$. These are referred to as isocurvature perturbations because the net perturbation $\delta \rho =\delta {\rho }_m+\delta {\rho }_r=0$ gives no change to the local curvature of spacetime. There is no hint of these isocurvature perturbations in our universe.

Since we have adiabatic perturbations, we need only specify a probability distribution for a single component, which we take to be $\delta \equiv {\delta }_m$. We take this distribution to be a simple Gaussian

$\mathrm{Prob}\left[\delta (𝐤)\right]={\displaystyle \frac{1}{\sqrt{2\pi P(k)}}}\exp \left(-{\displaystyle \frac{\delta {(𝐤)}^2}{2P(k)}}\right)$

(3.184)

This expression holds for each $𝐤$ independently. This means that there is no correlation between perturbations with different wavelengths. This is an assumption, and one that can be tested since it means that, at least initially, all higher point correlation functions are determined purely in terms of the one-point and two-point functions. For example, $⟨\delta \delta \delta ⟩=0$.

Note that the power spectrum $P(k)$ arises in this distribution in the guise of the variance. This ensures that the two-point function is indeed given by

$⟨\delta (𝐤,t_i)\delta ({𝐤}^{\prime },t_i)⟩={(2\pi )}^3{\delta }_D^3(𝐤+{𝐤}^{\prime })P(k)$

(3.185)

It remains only to specify the form of the power spectrum $P(k)$ for these initial perturbations. These are usually taken to have the power-law form

$P(k)=A{k}^n$

(3.186)

for constants $A$ and $n$. The exponent $n$ is called the spectral index.

A power-law $P\sim k^n$ gives rise to a real space correlation function $\xi (r)\sim 1/r^{n+3}$. (Actually, one must work a little harder to make sense of the inverse Fourier transform (3.182) at high $k$, or small $r$.) The choice $n=0$ is what we would get if we sprinkle points at random in space; it is sometimes referred to as white noise. (We’ll build more intuition for this in Section 3.2.2 below.) Meanwhile, any means that $\xi (r)\to \mathrm{\infty }$ as $r\to \mathrm{\infty }$, so the universe gets more inhomogeneous at large scales, in contradiction to the cosmological principle. We’d like to ask: what choice of spectral index $n$ describes our universe?

To explain this mathematical property, we need some new definitions. We start by some simple dimensional analysis. The original perturbation $\delta (𝐱)=\delta \rho /\rho$ was dimensionless, so after a Fourier transform (3.180) the perturbation $\delta (𝐤)$ has dimension ${[\mathrm{length}]}^3$. The delta-function ${\delta }_D^3(𝐤)$ also has dimension ${[k]}^{-3}={[\mathrm{length}]}^3$ which means that the power spectrum $P(k)$ also has dimension ${[\mathrm{length}]}^3$. It is often useful to define the dimensionless power spectrum

$\mathrm{\Delta }(k)={\displaystyle \frac{4\pi k^{3}P(k)}{{(2\pi )}^3}}$

(3.187)

The factors of $2$ and $\pi$ are conventional. Because $\mathrm{\Delta }(k)$ is dimensionless, it makes sense to say that, for example, $\mathrm{\Delta }(k)$ is a constant. Unfortunately, as you can see, this does not give rise to the Harrison-Zel’dovich spectrum.

However, we can also look at fluctuations in other quantities. In particular, rather than talk about perturbations in the density, $\rho$, we could instead talk about perturbations in the gravitational potential: $\mathrm{\Phi }(𝐱)=\overline{\mathrm{\Phi }}(𝐱)+\delta \mathrm{\Phi }(𝐱)$. The two are related by the Poisson equation (3.169)

${\nabla }^2\delta \mathrm{\Phi }={\displaystyle \frac{4\pi G}{c^2}}(1+3w)\overline{\rho }{a}^2\delta \mathit{\hspace{1em}\hspace{1em} }\Rightarrow \mathit{\hspace{1em}\hspace{1em} }-k^2\delta \mathrm{\Phi }(𝐤)={\displaystyle \frac{4\pi G}{c^2}}(1+3w)\overline{\rho }{a}^2\delta (𝐤)$

(3.188)

We can then construct the power spectrum of gravitational perturbations

$⟨\delta \mathrm{\Phi }(𝐤)\delta \mathrm{\Phi }({𝐤}^{\prime })⟩={(2\pi )}^3{\delta }_D^3(𝐤+{𝐤}^{\prime })P_{\mathrm{\Phi }}(k)$

(3.189)

and the corresponding dimensionless gravitational power spectrum

${\mathrm{\Delta }}_{\mathrm{\Phi }}={\displaystyle \frac{4\pi k^3{P}_{\mathrm{\Phi }}(k)}{{(2\pi )}^3}}$

The Poisson equation (3.188) tells us that there’s a simple relationship between $P_{\mathrm{\Phi }}(k)$ and $P(k)$, namely

$P_{\mathrm{\Phi }}(k)\propto k^{-4}P(k)$

(3.190)

where the proportionality factor hides the various constants arising from the Poisson equation. We can write this as

$P(k)\propto k^4{P}_{\mathrm{\Phi }}(k)\propto k{\mathrm{\Delta }}_{\mathrm{\Phi }}$

We see that the Harrison-Zel’dovich spectrum arises if the initial gravitational perturbations are independent of the wavelength, in the sense that ${\mathrm{\Delta }}_{\mathrm{\Phi }}=\mathrm{constant}$. Such fluctuations are said to be scale invariant. We will see that such scale invariant perturbations in the gravitational potential are a good description of our universe, and hold an important clue to what was happening at the very earliest times. We will see what this clue is telling us in Section 3.5. First, however, it will be useful to pause to build some intuition for these different probability distributions.