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research28 min studyLesson 865 in Physics

Normalize primordial spectra and compare their scale dependence

Keep Fourier volume, polarization counting, slow-roll order, and observational transfer distinct.

Before you begin
On this page

The question for today

What you’ll understand

  • Distinguish dimensionful Fourier power from dimensionless variance per logarithmic interval.
  • Compute leading canonical single-field scalar and total tensor spectra.
  • Apply explicitly defined Hubble slow-roll parameters without mixing conventions.

A mode amplitude becomes a statistical prediction only after the Fourier covariance and its normalization are specified. Scalar and tensor powers are often quoted as dimensionless functions of wave number, while a Fourier-space correlator also carries a volume dimension. Tensor predictions additionally depend on how polarization tensors and their summed power are normalized. Within a canonical single-field slow-roll model these conventions lead to a simple tensor-to-scalar ratio, but that relation is a conditional theoretical prediction rather than a definition valid for every primordial mechanism.

Define which power is being quoted

For a statistically homogeneous real curvature field, the two-point Fourier expectation contains a delta function of the sum of momenta. Rotational invariance reduces its remaining dependence to k. With the Fourier convention already fixed, Pℛ has length-cubed units, whereas the dimensionless spectrum is k³Pℛ/(2π²). Integrating that spectrum over logarithmic wave number gives the variance, subject to smoothing and ultraviolet/infrared limits. A spectrum constant in the dimensionless convention therefore has P proportional to k⁻³, not a constant Fourier power.

Gaussianity is not required to define this two-point quantity. It is required if one wants all higher correlations to follow from it. The primordial field should also be specified: here it is the positive-spatial-curvature ℛ defined in the preceding lesson, agreeing with uniform-density ζ in the regular adiabatic large-scale regime. Reversing that field’s sign leaves its power unchanged but changes sign conventions for odd correlations, which matters later for a bispectrum.

⟨RkRk′⟩=(2π)3δ3(k+k′)PR(k),ΔR2(k)=k32π2PR(k),⟨R2⟩=∫dlog⁡k  ΔR2(k)\langle\mathcal R_{\mathbf k}\mathcal R_{\mathbf k'}\rangle=(2\pi)^3\delta^3(\mathbf k+\mathbf k')P_{\mathcal R}(k),\qquad \Delta_{\mathcal R}^2(k)=\frac{k^3}{2\pi^2}P_{\mathcal R}(k),\qquad \langle\mathcal R^2\rangle=\int d\log k\;\Delta_{\mathcal R}^2(k)
The variance integral requires an appropriate range or window when it does not converge. The Fourier power has volume units and the dimensionless spectrum is variance per logarithmic interval.↔ Scroll wide equations horizontally.

Count the tensor polarizations consistently

Retain general relativity, one canonical scalar with sound speed 1, an adiabatic positive-frequency initial state, slow background variation, and a regular attractor. At leading slow-roll order the late-time scalar power is H²/(8π²ε₁Mpl²), evaluated using background quantities near exit k=aH. This is the late-time approximation derived from the mode solution, not the exact finite-time amplitude at crossing. Extra fields, a modified sound speed, or an excited initial state can change the relation.

Write hij as a sum of two transverse-traceless polarization tensors times amplitudes hs. Choose unit contraction eijˢeijˢ′=δss′. The canonical tensor field for each polarization is aMplhs/2. Define total dimensionless tensor power as the sum of the two amplitude powers with this normalization. It becomes 2H²/(π²Mpl²), yielding r=16ε₁. Authors using polarization tensors normalized to 2 redistribute factors between amplitudes and the definition of total power; their convention must be translated before comparing coefficients.

ΔR2≃H∗28π2ϵ1∗MPl2,Δt2≃2H∗2π2MPl2,r≡Δt2ΔR2≃16ϵ1∗\Delta_{\mathcal R}^2\simeq\frac{H_*^2}{8\pi^2\epsilon_{1*}M_{\rm Pl}^2},\qquad \Delta_t^2\simeq\frac{2H_*^2}{\pi^2M_{\rm Pl}^2},\qquad r\equiv\frac{\Delta_t^2}{\Delta_{\mathcal R}^2}\simeq16\epsilon_{1*}
The star denotes evaluation near k=aH in the leading slow-roll approximation. Δt² sums two unit-normalized tensor polarization amplitudes; it is not the power of one polarization.↔ Scroll wide equations horizontally.

Define slow-roll derivatives before taking a tilt

Use increasing background e-fold number N=ln a, ε₁=−dlnH/dN and ε₂=dlnε₁/dN. To first order, scalar tilt is ns−1=−2ε₁−ε₂ and tensor tilt is nt=−2ε₁. A potential slow-roll parameter or an alternative η notation is not interchangeable with ε₂. The exact exit relation gives dlnk/dN=1−ε₁; replacing its reciprocal by one is part of the stated first-order approximation. Leading r=−8nt follows only within the same assumptions and order.

The function lab illustrates quadratic power scaling. Reset and set coefficient 1 and position 1; predict value 1. Reset and set coefficient 1 and position 2; predict value 4. Doubling a real amplitude multiplies its square by four. This fixed parabola does not contain random modes, gravitational polarizations, scale tilt, or inflationary parameters. Its coefficient control is not the cosmological scale factor, and the two parameter settings are fresh comparisons rather than a simulated expansion history.

ϵ1=−dlog⁡HdN,ϵ2=dlog⁡ϵ1dN,ns−1≃−2ϵ1−ϵ2,nt≃−2ϵ1,r≃−8nt\epsilon_1=-\frac{d\log H}{dN},\qquad\epsilon_2=\frac{d\log\epsilon_1}{dN},\qquad n_s-1\simeq-2\epsilon_1-\epsilon_2,\qquad n_t\simeq-2\epsilon_1,\qquad r\simeq-8n_t
Both ε parameters are dimensionless, assumed small over the relevant exit interval. Tilt is a derivative with respect to log k of the corresponding dimensionless spectrum; the consistency relation is model dependent.↔ Scroll wide equations horizontally.

Work it through

Compute two amplitudes and their ratio

At one pivot exit take H/Mpl=10⁻⁵ and ε₁=0.005 in the stated canonical slow-roll model. Find the scalar power, total tensor power, and r at leading order.

  1. The scalar denominator is 8π²×0.005=0.04π². Thus Δℛ²=10⁻¹⁰/(0.04π²)≈2.533×10⁻¹⁰; this is dimensionless variance per logarithmic interval.

  2. The summed tensor spectrum is 2×10⁻¹⁰/π²≈2.026×10⁻¹¹. One unit-normalized polarization contributes half of this total under the declared definition.

  3. Dividing gives r=0.08, agreeing with 16ε₁=0.08. These stipulated model numbers are an arithmetic example, not an inferred observational amplitude or a claim about a measured tensor signal.

The result

The leading scalar and total tensor powers are about 2.533×10⁻¹⁰ and 2.026×10⁻¹¹, with r=0.08.

The living notebook

Explore the model

Interactive lab

Interactive lab · function

Square a stipulated amplitude

Reset and set coefficient a=1,position x=1; predict y=1. Reset and set a=1,x=2; predict y=4.

Square a stipulated amplitudeThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe fixed parabola illustrates amplitude-squared scaling; it does not calculate a primordial spectrum, tensor ratio, or spectral tilt.

Pause & reason

Check your understanding

What does a scale-invariant dimensionless curvature spectrum imply for the Fourier power Pℛ in this convention?

Your turn

Try it yourself

At a pivot suppose ε₁=0.010 and ε₂=−0.030, both in the stated Hubble-flow convention. Find the first-order scalar tilt ns, tensor tilt nt and leading r. Can the scalar spectrum be slightly blue while H decreases?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use ns−1=−2ε₁−ε₂ and distinguish it from nt.

Compare with the solution

The scalar tilt is ns−1=−0.020+0.030=0.010, so ns=1.010. The tensor tilt is −0.020 and r=0.160, satisfying r=−8nt at this order. A decreasing ε₁ can more than compensate the decrease of H in the scalar amplitude, producing a slightly blue scalar spectrum locally. The values specify a model regime, not its viability against observations or its duration.

Keep the idea

Take it with you

  • Dimensionless power and Fourier power differ by k³/(2π²).
  • State both tensor-polarization and total-power conventions.
  • Slow-roll tilts and consistency relations have explicit parameter definitions and model assumptions.

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