Bardeen Potentials and Curvature Perturbations

The Bardeen potentials describe scalar lapse and spatial-curvature perturbations without a coordinate artifact. Two related variables, $\zeta$ and $\mathcal R$, measure curvature on uniform-density and comoving slices. Their most important property is conditional: for a regular adiabatic mode, curvature is conserved at leading order in a long-wavelength expansion, but entropy, energy transfer, anisotropic stress, or a nonregular mode can invalidate that conclusion.

Helpful background. Einstein Equations and relativistic matter introduce the constraints and their sources. Formulations of general relativity explains the role of slicing, the cosmology overview locates the FLRW background, and PDEs and Green functions provides Fourier methods. The sign conventions are derived in Gauge Transformations in Cosmological Perturbation Theory.

Potentials on zero-shear slices

Work at first order on spatially flat FLRW, with conformal time $\eta$, $\mathcal H=a'/a$, and a prime denoting $d/d\eta$. In the passive transformation convention of the linked page, define

$$\sigma=E'-B, \qquad \Phi=\phi-\mathcal H\sigma-\sigma', \qquad \Psi=\psi+\mathcal H\sigma.$$

In Newtonian, or longitudinal, gauge $B=E=0$, so $\sigma=0$ and the scalar metric is

$$ds^2=a^2\left[-(1+2\Phi)d\eta^2 +(1-2\Psi)\delta_{ij}dx^i dx^j\right].$$

Here $\Phi$ is the lapse potential and $\Psi$ the spatial-curvature potential. Some sources exchange their names, so the metric—not the letter—is the reliable convention. These combinations and their interpretation as longitudinal-gauge values follow from Malik and Wands 2009, §7.1, printed pp. 30–32, eqs. (7.1)–(7.10) (Open PDF), after translating the gauge-generator sign as a whole.

For the scalar anisotropic-stress potential defined by

$$\left.\delta T^i{}_j\right|_{\rm TF} =\left(\partial^i\partial_j -\frac13\delta^i{}_j\nabla^2\right)\Pi,$$

the traceless spatial Einstein equation in this normalization gives

$$\Psi-\Phi=8\pi Ga^2\Pi.$$

Thus $\Phi=\Psi$ for a perfect fluid or any other source with vanishing scalar anisotropic stress. Equality is a field-equation statement, not part of the definition of either potential.

Curvature on physical slicings

Let $\delta T^0{}_i=\partial_i\delta q$. The two curvature perturbations used throughout cosmology are

$$\boxed{ \zeta=-\psi-\mathcal H\frac{\delta\rho}{\rho'}, \qquad \mathcal R=-\psi+\mathcal H\frac{\delta q}{\rho+p} }.$$

On a uniform-density slice, $\delta\rho=0$ and $\zeta=-\psi$; on a comoving slice, $\delta q=0$ and $\mathcal R=-\psi$. Consequently $\zeta$ requires $\rho'\ne0$, while $\mathcal R$ requires $\rho+p\ne0$. A pure cosmological constant has neither a useful density clock nor an independent comoving velocity.

The two curvatures are not identically equal at finite wavelength. Define the comoving density perturbation

$$\delta\rho_{\rm com} =\delta\rho+\rho'\frac{\delta q}{\rho+p} =\delta\rho-3\mathcal H\delta q,$$

where the second equality uses total background energy conservation. Directly from the definitions,

$$\zeta-\mathcal R =\frac{\delta\rho_{\rm com}}{3(\rho+p)}.$$

In general relativity on a flat background, the scalar constraints combine to give

$$\nabla^2\Psi=4\pi Ga^2\delta\rho_{\rm com}.$$

Equivalently,

$$\zeta-\mathcal R =\frac{\nabla^2\Psi}{3(\mathcal H^2-\mathcal H')}.$$

Thus a finite $\Psi$ makes the difference gradient-suppressed. It is $\mathcal O(k^2/\mathcal H^2)$ only when $\mathcal H^2-\mathcal H'=\mathcal O(\mathcal H^2)$; that estimate fails near a vacuum-like limit even though the exact expression remains the right one. This is the precise long-wavelength sense in which the two variables coincide in our sign convention. The constraint and curvature relations are derived in Malik and Wands 2009, §§7.4–7.6 and §8.2.1, printed pp. 39–42 and 46–48, eqs. (7.44)–(7.63) and (8.16)–(8.29) (Open PDF); their comoving-curvature symbol has the opposite overall sign from the $\mathcal R$ defined here.

The conservation equation

Define the adiabatic background sound speed and the nonadiabatic pressure by

$$c_a^2=\frac{p'}{\rho'}, \qquad \delta p_{\rm nad}=\delta p-c_a^2\delta\rho.$$

The combination $\delta p_{\rm nad}$ is gauge invariant whenever the background derivatives defining it are regular. Also define the invariant velocity–shear potential

$$\mathcal V=\frac{\delta q}{\rho+p}+\sigma.$$

The first-order energy-conservation equation alone gives the exact result

$$\boxed{ \zeta' =-\frac{\mathcal H}{\rho+p}\,\delta p_{\rm nad} -\frac13\nabla^2\mathcal V }.$$

No gravitational field equation is needed for this identity. With the declared $e^{i\mathbf k\cdot\mathbf x}$ convention, its gradient term is $+k^2\mathcal V/3$. Malik and Wands give the energy equation and this curvature evolution law in Malik and Wands 2009, §8.2.2, printed pp. 48–49, eqs. (8.32)–(8.37) (Open PDF).

For a controlled gradient expansion $k/\mathcal H\ll1$, $\zeta$ is conserved at leading order if $\delta p_{\rm nad}=0$ and $\mathcal V$ remains regular. “Superhorizon” in this statement means long compared with the instantaneous Hubble scale; it does not mean that two points are causally disconnected. The expansion is undefined at $\mathcal H=0$ and must then be reformulated.

The qualifications are physical. Relative entropy between components creates $\delta p_{\rm nad}$ even when each component is internally barotropic. Nonadiabatic energy transfer adds further sources to the component curvature equations. Anisotropic stress enters the metric and momentum constraints; a growing or $k^{-2}$ stress can prevent $\mathcal V$ from being regular, so its gradient term need not disappear. A decaying mode can fail the same regularity test. Conservation is therefore a result with hypotheses, not a definition of adiabaticity.

A constant-equation-of-state example

Consider one perfect fluid with constant $w=p/\rho\gt-1/3$ in its expanding, regular growing mode. On scales $k\ll\mathcal H$, the Einstein equations give $\Phi=\Psi$ and $\Psi'\simeq0$. The momentum constraint then reads

$$\mathcal H\Phi=-4\pi Ga^2\delta q, \qquad 4\pi Ga^2(\rho+p)=\frac32(1+w)\mathcal H^2.$$

Inserting these into the definition of $\mathcal R$ yields

$$\zeta\simeq\mathcal R \simeq-\frac{5+3w}{3(1+w)}\Phi.$$

During matter domination this gives $\Phi\simeq-3\mathcal R/5$; during radiation domination it gives $\Phi\simeq-2\mathcal R/3$. The latter assumes a perfect radiation fluid, so it excludes the free-streaming neutrino stress that separates the potentials. These are signed relations in this page’s convention: a positive primordial curvature perturbation corresponds to a negative potential well. The derivation also explains why a conserved $\mathcal R$ need not imply a constant potential through a change of equation of state.

Exercises

Starting from

$$\delta\rho'+3\mathcal H(\delta\rho+\delta p) -3(\rho+p)\psi' +(\rho+p)\nabla^2\mathcal V=0,$$

derive the evolution equation for $\zeta$.

Solution

Differentiate $\zeta=-\psi-\mathcal H\delta\rho/\rho'$ and use $\rho'=-3\mathcal H(\rho+p)$ to write

$$\zeta=-\psi+\frac{\delta\rho}{3(\rho+p)}, \qquad (\rho+p)'=-3\mathcal H(1+c_a^2)(\rho+p).$$

It follows that

$$\zeta'=-\psi' +\frac{\delta\rho'}{3(\rho+p)} +\frac{\mathcal H(1+c_a^2)\delta\rho}{\rho+p}.$$

Substitute the perturbed energy equation for $\delta\rho'$. The terms proportional to $\psi'$ cancel, while the pressure terms combine into $\delta p-c_a^2\delta\rho=\delta p_{\rm nad}$. The result is

$$\zeta' =-\frac{\mathcal H}{\rho+p}\delta p_{\rm nad} -\frac13\nabla^2\mathcal V.$$

This derivation also shows why the conservation statement is local energy conservation plus a gradient expansion; it is not a consequence of choosing Newtonian gauge.

References

• Malik, K. A., and D. Wands, “Cosmological Perturbations,” Physics Reports 475, 1–51 (2009), DOI, Open PDF.