research28 min studyLesson 866 in Physics

Track relative composition as an independent perturbation

Separate adiabatic time shifts, component entropy, and super-Hubble conversion.

On this page

What you’ll understand

  • Compute relative matter–radiation entropy with a gauge-invariant definition.
  • Weight component curvatures by enthalpy rather than density alone.
  • Explain how additional fields and conversion histories qualify single-clock predictions.

An adiabatic disturbance changes the local history along one background trajectory: every component is displaced by the same small clock shift. An isocurvature disturbance changes relative composition and cannot be removed by that one shift. It can coexist with curvature, correlate with it, or later generate it. Multiple inflationary fields provide one mechanism, but the existence of several fields alone does not determine which entropy modes survive reheating. The physical species, their interactions, and the conversion history must be specified before connecting an early fluctuation to an observable radiation pattern.

Compare component clocks, not equal density contrasts

For separately conserved fluids define ζi=−ψ−ℋδρi/ρi′ when each background derivative is nonzero. A common time shift cancels in each component combination. Adiabaticity means the ratios δρi/ρi′ agree, so the component curvatures agree. Equal fractional density contrasts are generally not the adiabatic condition because different equations of state dilute at different rates. A matter component and radiation provide the simplest concrete example.

Let c denote pressureless matter and γ photons, with no energy exchange in this illustration. Their relative entropy is Scγ=3(ζc−ζγ)=δc−3δγ/4. The ψ terms cancel, leaving a gauge-invariant composition disturbance. A compensated perturbation can have zero total density disturbance in a chosen flat slice while retaining nonzero relative entropy. That cancellation at one time does not imply the relative composition is unphysical or that total curvature will remain zero through later evolution.

ζi=−ψ−Hδρiρˉi′,Scγ=3(ζc−ζγ)=δc−34δγ\zeta_i=-\psi-\mathcal H\frac{\delta\rho_i}{\bar\rho_i'},\qquad S_{c\gamma}=3(\zeta_c-\zeta_\gamma)=\delta_c-\frac34\delta_\gamma
The matter–photon contrast formula assumes separate background conservation with wc=0 and wγ=1/3. Energy exchange changes the individual background derivatives and must be included before using this shortcut.↔ Scroll wide equations horizontally.

Weight the total curvature and follow conversion

The total uniform-density curvature is a weighted sum of component curvatures with weights proportional to their background density derivatives. For separately conserved components these are enthalpy weights, proportional to ρi+pi. Weighting simply by density would be wrong when radiation pressure matters. Even if the individual barotropic ζi remain constant on regular large scales, their evolving weights can change total ζ when they are unequal. This is one explicit way composition sources curvature.

During two-field inflation, perturbations can be resolved along the background field trajectory and perpendicular to it. A turn can couple the perpendicular entropy displacement to curvature outside the Hubble scale. Whether the entropy mode is light, decays, or survives into relative species abundances depends on the potential, kinetic geometry, and reheating couplings. A heavy field can sometimes be integrated out, but only within a controlled effective regime; the mere field count neither guarantees observable isocurvature nor justifies the canonical single-clock tensor ratio.

ζ=∑iρˉi′ρˉ′ζi=∑iρˉi+pˉiρˉ+pˉζifor separately conserved components\zeta=\sum_i\frac{\bar\rho_i'}{\bar\rho'}\zeta_i=\sum_i\frac{\bar\rho_i+\bar p_i}{\bar\rho+\bar p}\zeta_i\quad\text{for separately conserved components}
The first identity follows from the definitions when all component derivatives exist; the enthalpy form uses separate continuity equations. Total curvature need not be constant when unequal component curvatures have changing weights.↔ Scroll wide equations horizontally.

Describe the transfer only within its regime

A schematic large-scale transfer matrix can summarize linear evolution after defining an entropy variable and its normalization. Its curvature-source coefficient measures conversion; it is not fixed universally by the horizon-exit amplitude. Correlated initial modes require their cross power when computing final curvature power. Observational species entropy can also vanish after conversion even though entropy affected the final curvature. Thus absence of a surviving isocurvature component would not by itself establish that the primordial evolution involved only one dynamical field.

The matrix lab illustrates coupling without an area change. Reset and set diagonal scale 1,shear 0; predict final-matrix determinant 1 and singular values (1,1). Reset and set scale 1,shear 1; predict determinant 1 and singular values ((√5+1)/2,(√5−1)/2). At reset the displayed geometry is identity; Play or Step interpolates it, while these readouts always describe selected final A. This positive triangular 2D family is not a multifield perturbation solver or a physical entropy-transfer calculation.

(RS)f=(1TRS0TSS)(RS)∗\begin{pmatrix}\mathcal R\\\mathcal S\end{pmatrix}_{f}=\begin{pmatrix}1&T_{\mathcal RS}\\0&T_{\mathcal SS}\end{pmatrix}\begin{pmatrix}\mathcal R\\\mathcal S\end{pmatrix}_{*}
This schematic linear super-Hubble transfer uses explicitly normalized curvature and entropy amplitudes for the chosen model. Its triangular form and coefficients require the corresponding dynamical assumptions; fluid Scγ need not have the same normalization as field-space S.↔ Scroll wide equations horizontally.

Weight a matter–radiation composition disturbance

In a flat slice ψ=0 take equal background densities ρc=ργ, matter contrast δc=0.06 and photon contrast δγ=0.04. Find ζc,ζγ,Scγ and total ζ for separately conserved components.

  1. For matter ζc=δc/3=0.020, while radiation has ζγ=δγ/4=0.010. Equal background densities do not give equal curvature denominators.

  2. The relative entropy is 3(0.020−0.010)=0.030, matching 0.06−(3/4)0.04. The disturbance is not adiabatic.

  3. The enthalpy weights are 1 and 4/3, or 3/7 and 4/7 after normalization. Therefore ζ=(3×0.020+4×0.010)/7=0.100/7≈0.014286, rather than the simple average 0.015.

The result

The component curvatures are 0.020 and 0.010, entropy is 0.030, and total ζ≈0.014286 under the declared background conditions.

Explore the model

Interactive lab

Interactive lab · matrix

Observe shear without area change

Reset and set a=1,b=0; final-A readouts give det 1 and singular values (1,1). Reset and set a=1,b=1; det stays 1 and singular values become ((√5+1)/2,(√5−1)/2).

Observe shear without area changeThe selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Current readings: Area scale · det A: 1.4; Singular values · σmax, σmin: 1.597, 0.876; Condition number · κ₂(A): 1.823. All parameters are adjustable with the labeled controls below.Matrix A with first row 1.4, 0.6 and second row zero, one; display interpolates between identity and A while final determinant remains 1.4.−4−2024−2−1012x · equal geometric scaleyA = [ 1.4 0.6 ; 0 1 ]Display blend: 0% · final values below
Interpolated geometryOriginal unit squareSelected final transform
Area scale · det A
1.4
Singular values · σmax, σmin
1.597, 0.876
Condition number · κ₂(A)
1.823
t = 0 s

The selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. 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 forAt Reset geometry is identity; Play/Step interpolates identity↔A, while readouts describe final A. This matrix family does not evolve cosmological fields or determine entropy conversion.

Check your understanding

Which condition characterizes an adiabatic matter–photon disturbance in the separately conserved example?

Try it yourself

Use the stipulated linear transfer with ℛ*=0.010,S*=0.020,TRS=0.5 and TSS=0.25. Find the final amplitudes. If instead S*=0 with the same transfer, what happens?

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

Give me a hint

Apply the two matrix rows; the normalization of S is part of this supplied model.

Compare with the solution

The final curvature is 0.010+0.5×0.020=0.020 and final entropy is 0.25×0.020=0.005. With zero initial entropy, the supplied triangular transfer gives unchanged curvature 0.010 and zero entropy. These numbers demonstrate conversion algebra, not a universal prediction for every multifield model or an identification of field-space S with a particular late-time species entropy.

Take it with you

  • Adiabaticity compares component clock shifts rather than equal contrasts.
  • Total curvature uses derivative or enthalpy weights.
  • Entropy conversion and surviving isocurvature depend on dynamics and reheating.

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