Gauge Transformations in Cosmological Perturbation Theory

A cosmological perturbation depends on how points in the physical spacetime are identified with points in an ideal FLRW background. Changing that identification changes coordinate perturbations without changing the spacetime. This page fixes one passive convention, derives its first-order transformation rules, and uses them to separate gauge choices from invariant physics.

Helpful background. Einstein Equations introduce covariance, and relativistic matter supplies the stress–energy variables. Formulations of general relativity distinguishes coordinates from evolution; the cosmology overview locates the background geometry; and PDEs and Green functions explains the Fourier-mode split. The metric variables are defined in Scalar, Vector, and Tensor Decomposition.

A passive change of coordinates

Use the scalar metric convention of the preceding page,

$$ds^2=a^2\left[-(1+2\phi)d\eta^2 +2\partial_iB\,d\eta\,dx^i +\left((1-2\psi)\delta_{ij} +2\partial_i\partial_jE\right)dx^i dx^j\right].$$

Let the new coordinate labels be

$$\widetilde x^\mu=x^\mu+\xi^\mu, \qquad \xi^\mu=(T,\,\partial^iL+L_{\rm T}^i), \qquad \partial_iL_{\rm T}^i=0.$$

Write $\mathcal H=a'/a$. Our passive perturbation rule is

$$\boxed{ \widetilde{\delta Q}=\delta Q-\mathcal L_\xi\bar Q },$$

where the bar denotes the FLRW background. This sign is part of the convention. An active-map source often prints the opposite sign for the same named generator; changing only one line from such a source corrupts every invariant below.

For the metric,

$$(\mathcal L_\xi\bar g)_{\mu\nu} =\xi^\alpha\partial_\alpha\bar g_{\mu\nu} +\bar g_{\alpha\nu}\partial_\mu\xi^\alpha +\bar g_{\mu\alpha}\partial_\nu\xi^\alpha.$$

Applying this expression component by component gives

$$\begin{aligned} \widetilde\phi&=\phi-\mathcal HT-T', &\widetilde\psi&=\psi+\mathcal HT,\\ \widetilde B&=B+T-L', &\widetilde E&=E-L, \end{aligned}$$

The origin of each term is visible before any gauge is chosen. The background $00$ component changes both because $a(\eta)$ is evaluated on a shifted time slice and because $d\eta$ itself changes; these effects produce $-\mathcal HT$ and $-T'$ in $\widetilde\phi$. The spatial background metric changes under the time shift, producing $+\mathcal HT$ in $\widetilde\psi$. The scalar threading $L$ shifts $E$, while its time derivative enters $B$. This component-by-component derivation is a useful defense against mixing conventions that happen to use the same letters.

With the vector signs used on the preceding page,

$$\widetilde S_i=S_i-L_{{\rm T}i}', \qquad \widetilde F_i=F_i-L_{{\rm T}i}, \qquad \widetilde h_{ij}=h_{ij}.$$

The tensor perturbation is already invariant at first order. Malik and Wands derive the active/passive dictionary and the corresponding first-order metric rules in Malik and Wands 2009, §§6.1–6.4.1, printed pp. 17–25, eqs. (6.1)–(6.45) (Open PDF). Their displayed generator signs are translated here as one complete package. Explicitly, the dictionary to their variables is

$$T=-\alpha_{\rm MW},\qquad L=-\beta_{\rm MW},\qquad L_{\rm T}^i=-\gamma_{\rm MW}^i, \qquad S_i^{\rm here}=-S_i^{\rm MW}.$$

The last sign reflects the opposite sign of their vector shift in the metric; it must be translated together with the generator signs.

Matter perturbations

For any background scalar $f(\eta)$,

$$\widetilde{\delta f}=\delta f-f'T.$$

In particular,

$$\widetilde{\delta\rho}=\delta\rho-\rho'T, \qquad \widetilde{\delta p}=\delta p-p'T.$$

For a simple example, start with exactly homogeneous dust, $\delta\rho=0$ and $\rho'=-3\mathcal H\rho$. A spatially varying change of time label produces $\widetilde{\delta\rho}/\rho=3\mathcal HT$: the new constant-time surface samples different stages of the same homogeneous history. This apparent density fluctuation is a gauge mode. The simultaneous change $\widetilde\psi=\mathcal HT$ makes its uniform-density curvature $-\widetilde\psi-\mathcal H\widetilde{\delta\rho}/\rho'$ vanish, as it must.

Define the scalar momentum potential by $\delta T^0{}_i=\partial_i\delta q$. Taking the Lie derivative of the background mixed tensor $\bar T^\mu{}_\nu={\rm diag}(-\rho,p,p,p)$ gives

$$\widetilde{\delta q}=\delta q+(\rho+p)T.$$

Indeed, the only scalar contribution to $(\mathcal L_\xi\bar T)^0{}_i$ is $-(\rho+p)\partial_iT$. Subtracting the Lie derivative therefore adds $(\rho+p)T$ to the momentum potential. A separately conserved component $A$ obeys the same rule with $\rho_A+p_A$; if components exchange momentum, the transformation is unchanged but their evolution equations acquire transfer terms. Anisotropic stress has no homogeneous FLRW value, so its first-order scalar, vector, and tensor parts are already gauge invariant.

This sign follows from the index placement; it should not be guessed from a velocity variable defined in another convention. The transverse momentum is already invariant because the background supplies no transverse vector.

Invariant combinations

Introduce the scalar shear potential

$$\sigma\equiv E'-B, \qquad \widetilde\sigma=\sigma-T.$$

The time shift then cancels in the Bardeen potentials

$$\boxed{ \Phi=\phi-\mathcal H\sigma-\sigma', \qquad \Psi=\psi+\mathcal H\sigma }.$$

Matter invariants built from the same clock are

$$\delta\rho_{\rm gi}=\delta\rho-\rho'\sigma, \qquad \delta p_{\rm gi}=\delta p-p'\sigma, \qquad \delta q_{\rm gi}=\delta q+(\rho+p)\sigma.$$

For vectors, $F_i'-S_i$ is invariant, as is $h_{ij}$ in the tensor sector. Two especially useful invariant curvature variables are

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

The first requires $\rho'\ne0$ and the second $\rho+p\ne0$. Their dynamics and conservation conditions are the subject of the next page. Gauge invariance alone does not make a variable directly observable; it only removes the arbitrary first-order coordinate identification.

There is also a geometrical interpretation. A choice such as $\delta\rho=0$ selects hypersurfaces by a physical clock, while a condition on $E$ or $B$ selects how spatial coordinates are carried from one hypersurface to another. An invariant combination is the value a perturbation would have on a specified slicing and threading, expressed without actually changing coordinates. This interpretation explains why a denominator matters: if the background clock stops changing, that clock cannot identify a unique nearby time slice.

Gauge choices and residual freedom

For a nonzero Fourier mode, Newtonian gauge sets $\widetilde E=\widetilde B=0$. Choosing

$$L=E, \qquad T=E'-B=\sigma$$

does this, and the surviving metric variables are exactly $\widetilde\phi=\Phi$ and $\widetilde\psi=\Psi$. Uniform-density slicing uses $T=\delta\rho/\rho'$ when $\rho'\ne0$. Comoving slicing uses $T=-\delta q/(\rho+p)$ when $\rho+p\ne0$; on that slice $\mathcal R=-\widetilde\psi$.

A gauge condition can fail to fix all coordinates. Synchronous conditions $\widetilde\phi=\widetilde B=0$ determine differential equations for $T$ and $L$, leaving integration functions that generate familiar residual gauge modes. A temporal condition does not by itself fix the spatial threading. Moreover, the $k=0$ mode cannot be divided by $k^2$ to define the scalar longitudinal split and can overlap a redefinition of the background. Gauge conditions at a bounce, at $\rho'=0$, or for a vacuum-like component must be checked rather than imposed by a singular formula.

Two common misconceptions are worth separating. First, setting a perturbation to zero by a coordinate choice does not show that the associated physical effect vanishes; the effect reappears in invariant combinations. Second, dropping a term because it is small is a dynamical approximation, not gauge fixing. Newtonian gauge sets two coordinate variables exactly, whereas a subhorizon expansion discards terms only after comparing their scales in an already defined system.

Exercises

Verify directly that $\Phi$, $\Psi$, $\zeta$, and $\mathcal R$ are invariant under the transformations on this page.

Solution

For example,

$$\begin{aligned} \widetilde\Phi &=\phi-\mathcal HT-T' -\mathcal H(\sigma-T)-(\sigma'-T')=\Phi,\\ \widetilde\Psi &=\psi+\mathcal HT+\mathcal H(\sigma-T)=\Psi. \end{aligned}$$

Similarly,

$$-\widetilde\psi-\mathcal H \frac{\widetilde{\delta\rho}}{\rho'} =-\psi-\mathcal HT -\mathcal H\frac{\delta\rho-\rho'T}{\rho'}=\zeta,$$

and

$$-\widetilde\psi+\mathcal H \frac{\widetilde{\delta q}}{\rho+p} =-\psi-\mathcal HT +\mathcal H\frac{\delta q+(\rho+p)T}{\rho+p}=\mathcal R.$$

Each cancellation uses the full passive sign package. The denominators also show the stated domain restrictions. Since no Einstein equation entered, the result tests covariance and definitions rather than a particular matter model.

References

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