Slow-roll inflation - Cosmology Notes

Motivation¶

The $\Lambda$ CDM model is a perfectly satisfactory model of the universe for $t \gtrsim 10^{-3}$ s (scales at nuclear reactions), consistent with Big Bang nucleosynthesis, Hubble law, and growth of large-scale structure. However, the initial condition of the model (hot and dense in thermal equilibrium) is not explained.

Therefore, the model is not complete. $\Lambda$ CDM has a number of structure problems/unanswered questions:

Flatness problem: Why is the universe so flat today?
Horizon problem: Why is the CMB so homogeneous on scales that were never in causal contact?
“Density fluctuation” problem: What is the origin of the primordial seeds for structure formation?

These are initial condition problems — and these are what inflation addresses. (What is CDM or what is $\Lambda$ are questions inflation does not address.)

Inflation condition¶

If we take $\rho = \text{constant}$ , then $a \sim e^{\text{const} \times t}$ , so $\ddot{a} > 0$ . But such accelerated expansion never ends, so it cannot be followed by radiation and matter eras. We need a mechanism to realize a finite period of inflation. This leads us to postulate a scalar field.

<Figure size 800x400 with 1 Axes> — Figure 1:A schematic for $\dot a$ . We can see that $\dot a$ increases during inflation, then decreases rapidly in the radiation era, decreases more slowly in the matter era, and increases again in the dark energy era.

Are we simply pushing the problem of initial conditions back to $t_{inf}$ ?

The answer is no, because slow-roll inflation is an attractor solution.

Aspects of inflation¶

Homogeneous and isotropic cosmology with inflation¶

Why does inflation solve the flatness and horizon problems? What do you have to do to get a period of primordial inflation? Why is slow roll an attractor solution?

Basic equations¶

For a Friedmann-Lemaître-Robertson-Walker (FLRW) universe, we have:

\begin{cases} H^2 = \dfrac{\rho}{3 M_{pl}^2} - \dfrac{k}{a^2}, \qquad M_{pl}^2 = \dfrac{1}{8\pi G}, \qquad k = \begin{cases} 1 \quad \text{closed} \\ 0 \quad \text{flat} \\ -1 \quad \text{open} \end{cases} \\[6pt] \dfrac{\ddot{a}}{a} = -\dfrac{4\pi G}{3} (\rho + 3P) \\[6pt] \dot \rho = -3 H (\rho + P) \end{cases}

(5)

The comoving Hubble radius¶

Define the comoving Hubble radius as $(aH)^{-1}$ . If $a \sim t^p$ , then

(aH)^{-1} \sim \frac{1}{p} t^{1-p}.

(6)

Radiation/matter domination ( $p < 1$ ): $(aH)^{-1}$ increases with time.
Inflation ( $p > 1$ ): $(aH)^{-1}$ decreases with time.

The flatness problem¶

Define $\Omega(t) = \dfrac{\rho(t)}{\rho_c}$ , where $\rho_c = \dfrac{3 H^2}{8\pi G}$ . Then

\Omega(t) - 1 = \Omega_k = \dfrac{k}{(aH)^2}.

(7)

Radiation/matter domination: $\Omega_k$ increases with time. CMB observations constrain $|\Omega_k| \lesssim 0.0007$ today. This requires extreme fine-tuning of $\Omega_k$ at the beginning of the radiation era.
Inflation: $\Omega_k$ decreases during inflation, solving the fine-tuning problem. To satisfy the observational constraint, we need
$|\Omega_k(t_{pl})| \sim 10^{-60}.$
(8)

Number of e-folds¶

Define the number of e-folds as

N \equiv \ln\left( \frac{a_{end}}{a} \right).

(9)

If $a = a_{ini}$ , we have

N_\star \equiv \ln\left( \frac{a_{end}}{a_{ini}} \right).

(10)

To solve the flatness problem, we need $N_\star \gtrsim 50$ .

Quantum fluctuations around the background inflationary solution¶

How do these become the initial conditions for the $\Lambda$ CDM model?

<Figure size 640x480 with 1 Axes> — Figure 3:Hubble horizon evolution. Scales inside the horizon evolve; scales outside the horizon freeze.

Realizing inflation with a scalar field¶

Inflation requires $\ddot{a} > 0$ , which from the second Friedmann equation implies

\rho + 3P < 0 \quad \Rightarrow \quad P < -\frac{\rho}{3}.

(14)

This requires negative pressure. A cosmological constant gives accelerated expansion but never stops. We introduce a scalar field $\phi$ with a potential $V(\phi)$ to achieve a finite period of inflation.

If $\phi = \text{const}$ , this is equivalent to a cosmological constant since $\rho_\phi = \text{const}$ .

Slow-roll inflation: We consider a potential $V(\phi)$ such that $\phi$ moves very slowly down the potential.

Scalar field action and equations¶

The action for a scalar field is

S_\phi = \int d^4 x \sqrt{-g} \left( -\frac{1}{2} \partial_\mu \phi \partial_\nu \phi \, g^{\mu \nu} - V(\phi) \right).

(15)

The stress-energy tensor is

T_{\mu \nu} = -\frac{2}{\sqrt{-g}} \frac{\delta S_\phi}{\delta g^{\mu \nu}} = \partial_\mu \phi \partial_\nu \phi - g_{\mu \nu} \left( \frac{1}{2} \partial_\alpha \phi \partial^\alpha \phi + V(\phi) \right).

(16)

Varying the action with respect to $\phi$ gives the Klein-Gordon equation

\nabla_\mu \nabla^\mu \phi = \frac{\partial V}{\partial \phi}.

(17)

For a flat FLRW metric $g_{\mu\nu} = \text{diag}(-1, a^2(t)\delta_{ij})$ , this becomes

\boxed{\ddot{\phi} + 3H\dot{\phi} + \frac{dV}{d\phi} = 0}.

(18)

From $T^0_0$ and $T^i_j$ , we obtain the energy density and pressure:

\rho = \frac{1}{2} \dot{\phi}^2 + V(\phi), \qquad P = \frac{1}{2} \dot{\phi}^2 - V(\phi).

(19)

Slow-roll conditions¶

Inflation requires $P/\rho \ll -1/3$ , i.e., $\dot{\phi}^2 \ll V$ . This is the first slow-roll condition:

Under these conditions, the equations simplify dramatically:

H^2 \approx \frac{1}{3M_{pl}^2} V(\phi), \qquad 3H\dot{\phi} \approx -V'(\phi).

(20)

Slow-roll parameters¶

It is convenient to define dimensionless slow-roll parameters in terms of the potential.

Proof 1 (Relation to dynamics)

From the slow-roll equations, we have $\dot{\phi} = -V'/(3H)$ . Substituting into the definition of $\epsilon_V$ :

\epsilon_V = \frac{M_{pl}^2}{2} \left( \frac{V'}{V} \right)^2 = \frac{M_{pl}^2}{2} \left( \frac{3H\dot{\phi}}{V} \right)^2.

(22)

Using $H^2 \approx V/(3M_{pl}^2)$ , we get $V \approx 3M_{pl}^2 H^2$ , so

\epsilon_V \approx \frac{M_{pl}^2}{2} \left( \frac{3H\dot{\phi}}{3M_{pl}^2 H^2} \right)^2 = \frac{1}{2M_{pl}^2} \left( \frac{\dot{\phi}}{H} \right)^2.

(23)

This shows $\epsilon_V$ measures the kinetic energy relative to the potential.

Similarly, one can show that $\eta_V$ measures the acceleration term $\ddot{\phi}/(H\dot{\phi})$ .

One can also show that

\frac{\ddot{a}}{a} \approx H^2 (1 - \epsilon_V).

(24)

Inflation ends when either $\epsilon_V$ or $|\eta_V|$ becomes of order 1.

Number of e-folds in terms of the potential¶

The number of e-folds can be expressed as an integral over the field.

Proof 2 (Number of e-folds)

We have $dN = d\ln a = H dt$ . Using $dt = d\phi/\dot{\phi}$ and the slow-roll relation $\dot{\phi} \approx -V'/(3H)$ ,

dN = H \frac{d\phi}{\dot{\phi}} = H \left( -\frac{3H}{V'} \right) d\phi = -\frac{3H^2}{V'} d\phi.

(25)

Using $H^2 \approx V/(3M_{pl}^2)$ ,

dN = -\frac{V}{M_{pl}^2 V'} d\phi.

(26)

Thus the number of e-folds between a field value $\phi$ and the end of inflation $\phi_{end}$ is

\boxed{N(\phi) = \int_{\phi_{end}}^{\phi} \frac{V}{M_{pl}^2 V'} \, d\phi'}.

(27)

Example 3 (Quadratic potential)

Consider $V(\phi) = \frac{1}{2} m^2 \phi^2$ .

Slow-roll parameters:
$\epsilon_V = \frac{M_{pl}^2}{2} \left( \frac{2}{\phi} \right)^2 = 2 \left( \frac{M_{pl}}{\phi} \right)^2, \qquad \eta_V = 2 \left( \frac{M_{pl}}{\phi} \right)^2.$
(28)
Both are small when $\phi \gg M_{pl}$ .
Number of e-folds: Using $V/V' = \phi/2$ ,
$N = \int_{\phi_{end}}^{\phi} \frac{\phi}{2M_{pl}^2} d\phi = \frac{\phi^2 - \phi_{end}^2}{4M_{pl}^2}.$
(29)
Inflation ends when $\epsilon_V \approx 1$ , i.e., $\phi_{end} \approx \sqrt{2} M_{pl}$ . For $N \approx 60$ , we need $\phi \approx \sqrt{4N} M_{pl} \approx 15 M_{pl}$ .

Slow-roll inflation is an attractor solution¶

One of the key features of slow-roll inflation is that it is an attractor: independent of the initial conditions $(\phi_i, \dot{\phi}_i)$ , the solution $\phi(t)$ converges to the slow-roll trajectory.

We demonstrate this for the quadratic potential $V(\phi) = \frac{1}{2} m^2 \phi^2$ .

a. The slow-roll solution¶

From the slow-roll equation $3H\dot{\phi} = -V'$ , with $H^2 \approx V/(3M_{pl}^2)$ :

\dot{\phi}_{SR} = -\frac{V'}{3H} = -\frac{m^2 \phi}{3\sqrt{\frac{V}{3M_{pl}^2}}} = -\frac{m^2 \phi}{3\sqrt{\frac{m^2 \phi^2}{6M_{pl}^2}}} = -\frac{m M_{pl}}{\sqrt{3/2}} \frac{\phi}{|\phi|}.

(30)

For $\phi > 0$ , $\dot{\phi}_{SR} = -\sqrt{\frac{2}{3}} m M_{pl}$ — a constant (negative) velocity.

b. Convergence to the slow-roll solution¶

We want to show that any trajectory approaches this solution.

\frac{d\dot{\phi}}{d\phi} = -\frac{\sqrt{3}}{M_{pl}} \sqrt{\frac{1}{2}\dot{\phi}^2 + \frac{1}{2}m^2\phi^2} - \frac{m^2\phi}{\dot{\phi}}.

(31)

Proof 3 (Attractor behavior for quadratic potential)

Start from the full Friedmann and Klein-Gordon equations:

H^2 = \frac{1}{3M_{pl}^2} \left( \frac{1}{2} \dot{\phi}^2 + \frac{1}{2} m^2 \phi^2 \right), \qquad \ddot{\phi} + 3H\dot{\phi} + m^2 \phi = 0.

(32)

We can study the phase space $(\phi, \dot{\phi})$ . Define dimensionless variables:

x = \frac{\phi}{M_{pl}}, \qquad y = \frac{\dot{\phi}}{m M_{pl}}, \qquad \tau = m t.

(33)

Then the equations become:

\frac{dx}{d\tau} = y, \qquad \frac{dy}{d\tau} = -3\tilde{H} y - x,

(34)

where $\tilde{H} = H/m = \sqrt{\frac{1}{6}(y^2 + x^2)}$ .

The slow-roll solution corresponds to $y = -\sqrt{2/3}$ (for $x>0$ ). To study convergence, consider a small deviation: $y = -\sqrt{2/3} + \delta y$ , $x = x_{SR} + \delta x$ . Linearizing the equations around the slow-roll trajectory shows that perturbations decay exponentially in time, demonstrating that the slow-roll solution is an attractor.

Alternatively, we can derive a first-order equation for $\dot{\phi}(\phi)$ . From

\frac{d\dot{\phi}}{d\phi} = \frac{\ddot{\phi}}{\dot{\phi}} = -\frac{3H\dot{\phi} + m^2\phi}{\dot{\phi}} = -3H - \frac{m^2\phi}{\dot{\phi}}.

(35)

Substituting $H = \sqrt{\frac{1}{3M_{pl}^2}\left( \frac{1}{2}\dot{\phi}^2 + \frac{1}{2}m^2\phi^2 \right)}$ , we get

\frac{d\dot{\phi}}{d\phi} = -\frac{\sqrt{3}}{M_{pl}} \sqrt{\frac{1}{2}\dot{\phi}^2 + \frac{1}{2}m^2\phi^2} - \frac{m^2\phi}{\dot{\phi}}.

(36)

This is a first-order ODE for $\dot{\phi}(\phi)$ . The slow-roll solution is a particular solution. One can show numerically (or analytically by linearizing around it) that all trajectories in the phase plane flow toward this solution as $\phi$ decreases.

Thus, slow-roll inflation is indeed an attractor: regardless of the initial field velocity, the system quickly converges to the slow-roll trajectory, making the predictions of inflation insensitive to initial conditions. This is why inflation solves the fine-tuning problems of the standard Big Bang model.