GLOBAL KNOWLEDGE ARK

Cosmology · The cosmological constant and other dark-energy models

Dynamical-field proposals and evolving behaviour

Compare kinetic and potential energy in one exact rolling-field model, then check what its background solution does and does not establish.

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

  • Calculate pressure and energy density from a field’s kinetic and potential contributions.
  • Explain why a field and its expansion history must satisfy coupled equations.
  • Separate an exact homogeneous example from a tested dark-energy theory.
01

A rolling field has an energy ledger

A homogeneous canonical scalar field provides one way to model a changing stress-energy contribution. Canonical means its kinetic energy has the usual positive sign; the example here has no spatial gradients.

Write its kinetic energy density as K and its potential energy density as V. Then ε = K+V and P = K−V. All four quantities have energy-per-volume units.

If potential energy exceeds kinetic energy, pressure is negative. If K is positive, however, w = (K−V)/(K+V) is greater than −1: negative pressure need not be exactly the constant-Λ relation.

For nonnegative K and V with positive total density, −1 ≤ w ≤ 1. This algebraic range applies to the stated canonical ledger; it is not a classification of every proposed dark-energy theory.

02

Use one field, potential and expansion together

Choose an arbitrary positive inverse-time scale H_* and dimensionless time u = H_*t. On the finite interval 0.5 ≤ u ≤ 4, supply a spatially flat universe containing only this homogeneous scalar field.

Define E_* = c²H_*²/(8πG), an energy density. The exact example has K/E_* = u⁻² and V/E_* = 3+6/u+2/u², so both contributions are positive throughout the interval.

Its scale factor is a = u exp(u−1), with a = 1 at u = 1, and H/H_* = 1+1/u. These expressions belong to the same solution as the energy ledger; H_* is not a fitted present Hubble rate.

The density ε/E_* = 3(1+1/u)² obeys the scalar-only Friedmann equation 3H²/H_*² = ε/E_*. The normalized acceleration is (ä/a)/H_*² = 1+2/u, positive throughout the supplied interval.

One supplied exact canonical-scalar background; time and energy units are arbitrary reference scales
uaK/E_*V/E_*ε/E_*P/E_*w
0.50.30326542327−19−0.703704
1111112−10−0.833333
25.4365640.256.56.75−6.25−0.925926
480.3421480.06254.6254.6875−4.5625−0.973333
03

Near vacuum-like is different from constant

At u = 1, K/E_* = 1 and V/E_* = 11. The total density is 12E_*, while the pressure is −10E_*, giving w = −5/6 rather than −1.

At u = 2, K/E_* = 1/4 and V/E_* = 13/2. Therefore w = −25/27. The pressure-to-density ratio has changed even though the same field model is being used.

For this solution, w = −1+2/[3(u+1)²]. It stays above −1 at every finite positive u. A negative pressure close to vacuum-like behaviour can arise from a rolling canonical field.

The displayed finite evolution is a mathematical construction. Extending a limiting trend or comparing it with measured expansion requires additional assumptions and evidence; the table is not a forecast for our Universe.

One coupled canonical scalar: positive energy ledger and evolving wThis original exact background is a spatially flat expanding FLRW universe containing only one homogeneous, minimally coupled, positive canonical scalar field, with no spatial gradients. It omits ordinary matter, radiation, interactions and inhomogeneities. The arbitrary supplied rate H_* is positive in s⁻¹, u = H_*t is dimensionless and the displayed finite domain is 0.5 ≤ u ≤ 4. E_* = c²H_*²/(8πG) is an energy density in J/m³. Panel A uses linear u and a zero-based ordinate from 0 to 24 for energy density divided by E_*. Dashed blue square samples show K/E_* = u⁻², giving 4, 1, 1/4, 1/16 at u = 0.5, 1, 2, 4. Solid orange circles show V/E_* = 3+6/u+2/u², giving 23, 11, 13/2, 37/8. Energy density is ε = K+V and pressure is P = K−V. Panel B has a separate signed dimensionless ordinate from −1.05 to −0.65 for w = (K−V)/(K+V) = −1+2/[3(u+1)²]. Dotted green diamonds give w = −19/27, −5/6, −25/27 and −73/75 at those four u values. A dashed violet horizontal reference marks w = −1, which this rolling field does not reach at finite positive u. The samples and series have distinct shapes and dashes; direct formulas identify them without color. At u = 1, ε = 12E_* and P = −10E_*, with ε+P = 2E_* positive. The ledger is part of a coupled solution: a = u exp(u−1), H/H_* = 1+1/u and χ = √2 ln u. With the explicit SI normalization K = c²χ̇²/(16πG), V = E_*U(χ), U = 3+6 exp(−χ/√2)+2 exp(−√2χ), the field obeys χ̈+3Hχ̇+H_*²U_χ = 0, the scalar-only Friedmann equation, Ḣ = −χ̇²/2 and continuity. The algebraic first reading requires only the K,V ledger; optional displayed formulas explain its dynamic consistency. H_*, a = 1 at u = 1 and all quantities are supplied mathematical references, not fitted current parameters. An exact homogeneous solution does not establish perturbation or quantum stability, full cosmic history, observational viability or a microscopic explanation of our Universe.A rolling field can be near vacuum-likeOne positive canonical scalar only; homogeneous, no gradients.u = H_*t is dimensionless; H_* is arbitrary, not measured.Finite supplied interval 0.5 ≤ u ≤ 4; no ordinary matter.A · Kinetic and potential energy densities / E_*K = u⁻²V = 3+6/u+2/u²061218240.51234Dimensionless time u = H_*tB · Pressure / density: w = (K−V)/(K+V)w = −1 + 2/[3(u+1)²]-1-0.9-0.8-0.70.51234Dashed reference w = −1 · not reached at finite uAt u = 1: K = E_*, V = 11E_*, so ε = 12E_* and P = −10E_*.Thus w = −5/6; at u = 2 it is −25/27.Optional coupled check: χ = √2 ln u; a = u exp(u−1).E_* = c²H_*²/(8πG); K = c²χ̇²/(16πG).U(χ) = 3+6 exp(−χ/√2)+2 exp(−√2χ), with V = E_*U.χ̈ + 3Hχ̇ + H_*²U_χ = 0; H/H_* = 1+1/u.No perturbation, quantum-stability or viability claim follows.
This original exact background is a spatially flat expanding FLRW universe containing only one homogeneous, minimally coupled, positive canonical scalar field, with no spatial gradients. It omits ordinary matter, radiation, interactions and inhomogeneities. The arbitrary supplied rate H_* is positive in s⁻¹, u = H_*t is dimensionless and the displayed finite domain is 0.5 ≤ u ≤ 4. E_* = c²H_*²/(8πG) is an energy density in J/m³. Panel A uses linear u and a zero-based ordinate from 0 to 24 for energy density divided by E_*. Dashed blue square samples show K/E_* = u⁻², giving 4, 1, 1/4, 1/16 at u = 0.5, 1, 2, 4. Solid orange circles show V/E_* = 3+6/u+2/u², giving 23, 11, 13/2, 37/8. Energy density is ε = K+V and pressure is P = K−V. Panel B has a separate signed dimensionless ordinate from −1.05 to −0.65 for w = (K−V)/(K+V) = −1+2/[3(u+1)²]. Dotted green diamonds give w = −19/27, −5/6, −25/27 and −73/75 at those four u values. A dashed violet horizontal reference marks w = −1, which this rolling field does not reach at finite positive u. The samples and series have distinct shapes and dashes; direct formulas identify them without color. At u = 1, ε = 12E_* and P = −10E_*, with ε+P = 2E_* positive. The ledger is part of a coupled solution: a = u exp(u−1), H/H_* = 1+1/u and χ = √2 ln u. With the explicit SI normalization K = c²χ̇²/(16πG), V = E_*U(χ), U = 3+6 exp(−χ/√2)+2 exp(−√2χ), the field obeys χ̈+3Hχ̇+H_*²U_χ = 0, the scalar-only Friedmann equation, Ḣ = −χ̇²/2 and continuity. The algebraic first reading requires only the K,V ledger; optional displayed formulas explain its dynamic consistency. H_*, a = 1 at u = 1 and all quantities are supplied mathematical references, not fitted current parameters. An exact homogeneous solution does not establish perturbation or quantum stability, full cosmic history, observational viability or a microscopic explanation of our Universe. Original Ark teaching model.
04

Optional mathematical path: check the dynamics

Introduce a dimensionless field coordinate χ = √2 ln u. In explicit SI energy-density conventions, define K = c²χ̇²/(16πG) and V = E_*U(χ), where U = 3+6 exp(−χ/√2)+2 exp(−√2χ). This is a constant rescaling of a canonical field.

The field equation is χ̈+3Hχ̇+H_*² dU/dχ = 0. The expanding H must simultaneously satisfy Friedmann with ε = K+V; selecting an arbitrary expansion history independently would generally violate the coupled system.

Here χ̇ = √2H_*/u and Ḣ = −H_*²/u² = −χ̇²/2. Substitution also satisfies the field equation and local continuity ε̇+3H(ε+P) = 0.

These checks explain why the supplied ledger is dynamically consistent. The first-reading K/V calculation above requires only addition and division; it does not require solving the field equation.

05

An exact background is one stage of a physical proposal

This example omits ordinary matter, radiation, inhomogeneities and interactions with other fields. It demonstrates a self-consistent homogeneous solution, not an account of all cosmic epochs.

A fuller proposal must specify its perturbations, couplings and physical regime. Stability, initial-condition dependence, quantum corrections and compatibility with other observations require further analysis.

Measuring an expansion history close to w = −1 would not automatically select a constant term over every dynamical field. Different models can produce similar background behaviour within the available precision and assumptions.

The useful comparison is explicit: identify the proposed mechanism, derive its background and other observable consequences consistently, then compare those consequences with data and their uncertainty model.

Worked reasoning

Compute the pressure at u = 1

Does this field have exactly the pressure-density relation of a constant Λ at the reference time?

  1. The supplied kinetic and potential densities are E_* and 11E_*.
  2. Thus ε = 12E_* and P = −10E_*.
  3. Their ratio is w = −10/12 = −5/6, so ε+P = 2E_* remains positive.

No: it has negative pressure and is rolling, with w = −5/6 rather than −1.

Try the reasoning

  1. Find w at u = 2 from the table’s kinetic and potential densities.

    Show the answer and reasoning

    w = −25/27.

    In E_* units, ε = 27/4 and P = −25/4.

  2. May H(t), the field and its potential be chosen independently and assumed consistent?

    Show the answer and reasoning

    No.

    They must satisfy the field, conservation and gravitational equations together.

  3. Does an exact homogeneous solution establish observational viability and perturbation stability?

    Show the answer and reasoning

    No.

    Those are additional physical and observational tests beyond the background construction.

What to carry forward

  • A canonical field’s kinetic/potential balance determines its pressure and density.
  • A consistent dynamical example couples field evolution to expansion.
  • Similar background behaviour does not establish the same mechanism or complete a theory’s tests.

References

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

  1. Dynamics of dark energyEdmund J. Copeland; M. Sami; Shinji TsujikawaTeaching resourcearxiv.orgRead source