GLOBAL KNOWLEDGE ARK

Cosmology · Inflation and primordial perturbations

An early accelerated phase as a scalar-field proposal

Use an exact canonical scalar background to calculate negative pressure, decreasing expansion rate and an accelerating scale factor without assuming an observational fit.

Preparation: ratios, powers of ten and ordinary arithmetic. Physical quantities and notation are introduced as they are needed.

Back to the chapter

After reading, you can

  • Relate a homogeneous canonical scalar’s kinetic and potential energy to its pressure.
  • Verify acceleration in an exact supplied scalar-only background.
  • Identify what a consistent background solution leaves unestablished about the early Universe.
01

A proposal must explain its expansion rate

A prescribed expansion function is useful for comparing geometry, but a dynamical proposal also specifies what produces that expansion. One widely studied possibility uses a homogeneous canonical scalar field with kinetic energy and a potential.

This lesson gives one exact solution in spatially flat scalar-only general relativity. It is an original teaching construction using the canonical equations. No observation is used to choose its field value, energy scale or potential.

Use natural units c = ℏ = 1 and the reduced Planck mass M defined by M² = (8πG)⁻¹ in those units. The field φ and M have energy units, H has inverse natural-time units equivalent to energy, and energy density has energy to the fourth power.

For a homogeneous field, write K_φ = φ̇²/2, ρ_φ = K_φ + V and p_φ = K_φ − V. In these natural units pressure and energy density have the same units. The symbol K_φ here denotes kinetic energy density, rather than spatial curvature.

02

Choose H as a function of the field

Choose positive reference values H_i and M and prescribe H(φ) = H_i exp[−(φ − φ_i)/(2M)]. On the monotonic field branch, the flat canonical Hamilton–Jacobi equations give φ̇ = −2M² dH/dφ and V = 3M²H² − 2M⁴(dH/dφ)².

The derivative is dH/dφ = −H/(2M). It follows that φ̇ = MH, K_φ = M²H²/2 and V = 5M²H²/2. Equivalently, V(φ) = (5/2)M²H_i² exp[−(φ − φ_i)/M].

The total energy density is ρ_φ = 3M²H², satisfying the flat scalar-only Friedmann equation. The canonical field equation φ̈ + 3Hφ̇ + dV/dφ = 0 is also satisfied by these expressions. This is a dynamically consistent solution rather than a freely assigned pressure curve.

Since dφ/dN = φ̇/H = M, the displacement (φ − φ_i)/M equals N = ln(a/a_i). Over the chosen finite interval 0 ≤ N ≤ 4, H/H_i = exp(−N/2). The expansion rate falls even though the scale factor will accelerate.

03

Potential energy can outweigh kinetic pressure

The kinetic fraction K_φ/ρ_φ is 1/6 and the potential fraction V/ρ_φ is 5/6 throughout this solution. Their difference gives w = p_φ/ρ_φ = 1/6 − 5/6 = −2/3.

The pressure is negative because the potential contributes with opposite signs to pressure and energy density. This statement concerns the field stress-energy in the gravitational equations, not an ordinary external force acting on the wall of a cosmic container.

At N = 2, H/H_i ≈ 0.367879. In units M²H_i², kinetic energy density is about 0.0676676, potential energy density 0.338338 and their sum 0.406006. The fractions remain the same while the densities fall.

The table separates the changing dimensional quantities, normalized to a common energy-density scale, from the fixed dimensionless fractions. Positive total energy density and negative pressure coexist in this supplied solution.

Exact flat canonical scalar example; all energy densities in the table use the common scale M²H_i²
NH/H_iK_φ/(M²H_i²)V/(M²H_i²)ρ_φ/(M²H_i²)p_φ/(M²H_i²)
010.52.53−2
10.6065310.1839400.9196991.103638−0.735759
20.3678790.06766760.3383380.406006−0.270671
40.1353350.009157820.04578910.0549469−0.0366313
04

Acceleration does not require constant H

Differentiate H along the field trajectory to obtain Ḣ = (dH/dφ)φ̇ = −H²/2. Thus ε_H = −Ḣ/H² = 1/2, and ä/(aH²) = 1 − ε_H = 1/2 is positive.

Integrating in time gives a/a_i = [1 + H_i(t − t_i)/2]² on the expanding branch. The scale factor has positive second derivative, while H = H_i/[1 + H_i(t − t_i)/2] decreases. These two properties are compatible.

At N = 4, the dimensionless elapsed time is H_i(t − t_i) = 2[exp(2) − 1] ≈ 12.7781. This interval is not assigned a number of seconds: H_i was left as an arbitrary positive reference.

The example is exact and has ε_H = 0.5. It should not be called a calculation in which ε_H is much smaller than one, and no slow-roll approximation is needed to obtain these background values. Its perturbation spectrum has not been calculated here.

A decreasing Hubble rate and fixed scalar energy fractionsAn exact supplied flat canonical scalar-only general-relativistic background uses c = ℏ = 1, reduced Planck mass M² = (8πG)^−1, positive H_i, and H(φ) = H_i exp[−(φ−φ_i)/(2M)]. The monotonic branch has φdot = MH and (φ−φ_i)/M = N = ln(a/a_i). The finite plot spans N from 0 to 4 and has a linear H/H_i axis starting at zero. Solid blue circles at N = 0, 1, 2, 4 give 1, 0.606531, 0.367879, 0.135335. A separate common-denominator fraction bar shows K_φ/ρ_φ = 1/6 and V/ρ_φ = 5/6; it is not a bar of changing absolute energy. Here K_φ = M²H²/2, V = 5M²H²/2, total ρ_φ = 3M²H² and pressure p_φ = −2M²H². The accompanying table uses the common energy-density unit M²H_i²: at N = 0 kinetic, potential and pressure are 0.5, 2.5, −2; at N = 2 they are 0.0676676, 0.338338, −0.270671; at N = 4 they are 0.00915782, 0.0457891, −0.0366313. Fixed fractions give w = −2/3. Exact Hdot = −H²/2 gives ε_H = 1/2 and ä/(aH²) = 1/2, so acceleration coexists with falling H. In natural units φ, M and H have energy units and density/pressure have energy-to-the-fourth units. K_φ denotes kinetic density, not spatial curvature. The solution satisfies the Friedmann and canonical field equations, but this finite display is not an observational fit, a small-ε approximation, an exit, particle coupling, reheating mechanism, thermal state or perturbation-spectrum calculation. Its energy scale is arbitrary; stopping at N = 4 does not supply missing dynamics.An exact scalar background can accelerate as H fallsFlat scalar-only GR · natural units c = ℏ = 1.M² = (8πG)⁻¹; (φ − φ_i)/M = N from 0 to 4.H/H_i = exp(−N/2); this is a supplied exact proposal.Expansion rate H/H_i · linear scale, starts at zero00.250.50.75101234E-folds N · dimensionlessFixed fractions of ρ_φ = K_φ + VKinetic: 1/6Potential: 5/6w = (K_φ − V)/ρ_φ = −2/3.ε_H = 1/2; ä/(aH²) = 1/2 > 0 while H decreases.The energy fractions stay fixed while densities fall.All densities below use the same unit M²H_i².NK_φVp_φ00.52.5-220.0676680.338338-0.27067140.0091580.045789-0.036631Pressure has the same energy-density units here.The potential is exponential; no exit or reheating is calculated.This exact ε_H = 0.5 example uses no small-ε slow-roll limit.No measured field, energy scale or perturbation spectrum is shown.
An exact supplied flat canonical scalar-only general-relativistic background uses c = ℏ = 1, reduced Planck mass M² = (8πG)^−1, positive H_i, and H(φ) = H_i exp[−(φ−φ_i)/(2M)]. The monotonic branch has φdot = MH and (φ−φ_i)/M = N = ln(a/a_i). The finite plot spans N from 0 to 4 and has a linear H/H_i axis starting at zero. Solid blue circles at N = 0, 1, 2, 4 give 1, 0.606531, 0.367879, 0.135335. A separate common-denominator fraction bar shows K_φ/ρ_φ = 1/6 and V/ρ_φ = 5/6; it is not a bar of changing absolute energy. Here K_φ = M²H²/2, V = 5M²H²/2, total ρ_φ = 3M²H² and pressure p_φ = −2M²H². The accompanying table uses the common energy-density unit M²H_i²: at N = 0 kinetic, potential and pressure are 0.5, 2.5, −2; at N = 2 they are 0.0676676, 0.338338, −0.270671; at N = 4 they are 0.00915782, 0.0457891, −0.0366313. Fixed fractions give w = −2/3. Exact Hdot = −H²/2 gives ε_H = 1/2 and ä/(aH²) = 1/2, so acceleration coexists with falling H. In natural units φ, M and H have energy units and density/pressure have energy-to-the-fourth units. K_φ denotes kinetic density, not spatial curvature. The solution satisfies the Friedmann and canonical field equations, but this finite display is not an observational fit, a small-ε approximation, an exit, particle coupling, reheating mechanism, thermal state or perturbation-spectrum calculation. Its energy scale is arbitrary; stopping at N = 4 does not supply missing dynamics. Original Ark teaching model.
05

A consistent solution still needs an exit and evidence

A mathematically consistent scalar background demonstrates a possible mechanism for acceleration. It does not show that this field or this particular exponential potential existed in the early Universe.

The stated example includes neither other fluids nor spatial curvature. It also gives no transition that ends the accelerating phase and no particle couplings that transfer the field’s energy to a thermal population. Restricting the displayed interval does not create those missing physical mechanisms.

A larger proposal must connect an early phase to the later thermal history, explain the relevant fluctuations and compare predictions with observations. The chosen energy fractions alone do not determine a measured primordial spectrum, tensor amplitude or reheating temperature.

A useful reading sequence separates these tasks: first check the background equations, then study the transition and energy transfer, and then ask which statistical predictions can be tested. Each connection introduces assumptions that should remain visible.

Worked reasoning

Calculate the scalar equation of state

What w follows when the kinetic and potential fractions are 1/6 and 5/6?

  1. Use p_φ = K_φ − V and ρ_φ = K_φ + V.
  2. Divide both terms by ρ_φ, giving w = K_φ/ρ_φ − V/ρ_φ.
  3. Substitute 1/6 − 5/6 = −2/3, which is below the acceleration threshold −1/3 in this scalar-only flat model.

w = −2/3. The corresponding exact solution has ä/(aH²) = 1/2 > 0.

Try the reasoning

  1. Does accelerated expansion require exactly constant H?

    Show the answer and reasoning

    No.

    Here H decreases as exp(−N/2), yet the scale factor has positive second derivative.

  2. At N = 2, has the potential fraction fallen below 5/6 because V has decreased?

    Show the answer and reasoning

    No.

    Both kinetic and potential energy densities decrease with H², so their fixed fractions remain 1/6 and 5/6.

  3. Does stopping the table at N = 4 supply a reheating mechanism?

    Show the answer and reasoning

    No.

    A display endpoint is not a dynamical exit or an interaction that transfers field energy to other populations.

What to carry forward

  • A canonical scalar’s potential contributes positively to energy density and negatively to pressure.
  • An exact background can accelerate while H decreases.
  • Background consistency, an exit, thermalization and observational evidence are separate parts of an early-Universe proposal.

References

Sources supporting the explanations, physical distinctions and examples in this lesson.

  1. TASI Lectures on InflationDaniel BaumannCourse notesarxiv.orgRead source