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The parity-odd part of linear sky polarization is represented by the pseudoscalar potential $P_B$. At linear order an axisymmetric scalar cosmological perturbation gives no physical B mode. Tensor or vector sources and lensing conversion of E-mode polarization are outside that scalar-source statement.

The parity-even part of linear sky polarization can be represented by a scalar potential $P_E$ through $Q\pm iU={ð}_{\pm }^2(P_E\pm i{P}_B)$. It is naturally separated from B-mode polarization by the parity of its spherical multipoles. Linear scalar cosmological perturbations generate E modes; the potential has an unobservable low-multipole kernel.

With vectors and independent tensors neglected, finite scalar cosmological perturbations of the local spatial scale can be encoded by $h_{ij}=a_{\mathrm{bg}}^2{e}^{2\zeta }{\delta }_{ij}$. Scalar lapse and shift potentials complete its 3+1 decomposition of spacetime. The exact shift-square term is $h^{ij}{N}_i{N}_j$, independently of the parametrization of the scalar shift.

A scalar cosmological perturbation has two scalar gauge freedoms. In Newtonian gauge in cosmology, these remove the scalar shift and spatial shear, leaving the lapse potential $\Phi$ and spatial-curvature potential $\Psi$:A perfect fluid has no scalar anisotropic stress. The traceless spatial Einstein field equations therefore set $\Phi =\Psi =\varphi$ for nonzero spatial Fourier modes, producing the line element in the question. Homogeneous perturbations include background and gauge changes and require separate treatment. The single potential describes the physical scalar perturbation in this gauge.

A general scalar cosmological perturbation has four scalar metric functions: a lapse perturbation, a scalar shift, a spatial curvature perturbation and a scalar spatial shear. The two scalar functions in a cosmological gauge transformation can set the shift and shear to zero. For nondegenerate spatial modes this gives conformal Newtonian gauge, leaving the two potentials $\psi$ and $\varphi$. The scalar traceless spatial Einstein equation makes their difference proportional to scalar anisotropic stress. In general relativity, with no scalar anisotropic stress and the usual boundary conditions, $\psi =\varphi$. Pressure alone does not spoil equality; free-streaming relativistic species can do so through their anisotropic stress.

Write $\Psi =(\varphi +\psi )/2$, the Weyl lensing potential. A conformal rescaling leaves the unparameterized null geodesics unchanged, although the affine parameter changes. To first order, the conformal optical metric for scalar perturbations hasThe only nonzero connection coefficients at this order areHere spatial indices are raised with ${\delta }^{ij}$. An affine parameter $\lambda$ for this rescaled metric therefore obeysThe first line is also ${\eta }_{,\lambda \lambda }+2{\eta }_{,\lambda }d\Psi /d\lambda +2x_{,\lambda }^i{\eta }_{,\lambda }{\partial }_i\Psi =0$, since the derivative of the potential is along the ray.

Put $v^i=d{x}^i/d\eta$. The chain rule gives $x_{,\lambda \lambda }^i={\eta }_{,\lambda }^{2}d{v}^i/d\eta +v^i{\eta }_{,\lambda \lambda }$. Eliminating the affine acceleration givesThe extra spatial-gradient term inside the parentheses does not replace the spatial term already contained in the total derivative.

At reception, the null condition is $|v{|}^2=1+4{\Psi }_0$. With the specified unit propagation vector $e$, this impliesThe zero-order trajectory is $x^i=e^i(\eta -{\eta }_0)$. In every term already containing a first-order potential, we may substitute $v=e$. Defining the transverse projector $P_{\perp }^{ij}={\delta }^{ij}-e^i{e}^j$, the first-order geodesic equation becomesIntegrating with the reception position and velocity gives the first-order photon trajectory in scalar perturbations:Its first derivative at reception is $e^i+2e^i{\Psi }_0$, and differentiating twice gives the previous equation. For a first-order calculation, evaluating the integrals on the unperturbed ray is sufficient; shifting their evaluation points by the first-order displacement changes them only at second order. This is the Born approximation in cosmological lensing.

To fix the angular-deflection sign, define ${\chi }_s={\eta }_0-{\eta }_s>0$, $n_{\mathrm{obs}}=-e$, and $n_{\mathrm{src}}=x_s/|x_s|$. The latter is the unlensed coordinate direction to the source, while the former is its apparent direction measured by the observer. The longitudinal integral changes the radial distance, not the direction at this order. The transverse displacement isDividing by ${\chi }_s$ yieldsThis is the angular deflection kernel for cosmological lensing. It depends on the sum of the two metric potentials; setting them equal reduces it to the familiar twice-the-Newtonian-potential expression. With angular derivatives, the lensing potential is $2{\int }_0^{{\chi }_s}[({\chi }_s-\chi )/({\chi }_s\chi )]\Psi d\chi$, and its angular gradient is $\alpha$.

Gravitational lensing displaces apparent source positions and produces magnification and shear through angular derivatives of the deflection. Weak gravitational lensing of galaxy shapes probes the foreground mass distribution, including dark matter. Lensing also remaps the cosmic microwave background, smoothing its acoustic features and converting part of its E-mode polarization into B modes. Strong deflections can produce multiple images, though a single first-order ray calculation does not describe that regime completely.

A scalar cosmological perturbation at fixed wavevector carries no helicity or distinguished transverse direction. Its metric and scalar sources are unchanged by rotations around $\mathbf{k}$, and the scalar baryon peculiar velocity is longitudinal. With scalar initial conditions, rotational invariance of the linear evolution therefore makes $\Theta$ a function only of $\mu =\hat{\mathbf{k}}\cdot \mathbf{e}$. Expand it using the stated normalization, without an extra $(2\ell +1)$ factor:In the Thomson kernel, $1+v^2=4/3+(2/3)P_2(v)$. The spherical harmonic addition theorem and Orthogonality of Legendre polynomials giveOnly the monopole and quadrupole survive; the quadrupole's $(-i{)}^2=-1$ gives the angular scattering integralThis also explains why an isotropic distribution has zero net collision term.

The streaming term $ik\mu \Theta$ and the Legendre polynomial recurrence relation yield, after division by $(-i{)}^{\ell }$, $k[(\ell +1){\Theta }_{\ell +1}/(2\ell +3)-\ell {\Theta }_{\ell -1}/(2\ell -1)]$. With ${\mathbf{v}}_b=i\hat{\mathbf{k}}{v}_b$, the velocity collision source is $+{\tau }^{\prime }{v}_b$ in the dipole equation; the gravitational gradient source is $+k\psi$. Therefore the photon Boltzmann hierarchy isSet the absent ${\Theta }_{-1}$ contribution to zero. This is algebraically the printed hierarchy. In particular,Since ${\tau }^{\prime }<0$, these collision terms damp the nonzero multipoles and drive the dipole to ${\Theta }_1=-v_b$. The latter sign follows from the paper's velocity and $(-i{)}^{\ell }$ conventions, not a reversed physical photon velocity.

A cosmological gauge transformation changes the coordinates used to identify a perturbed spacetime with its homogeneous background. It can produce apparent metric perturbations even in a physically unperturbed FRW metric. For ${\tilde{x}}^i=x^i+{\xi }^i(\tau ,\mathbf{x})$ and unchanged time, the inverse differentials are, to first order,Inserting these into the background line element gives the transformed componentsThe spatial components can now depend on position, although no physical density inhomogeneity or spacetime curvature perturbation has been introduced. For a scalar displacement ${\xi }^i={\partial }^{i}L$, these are pure coordinate-generated scalar cosmological perturbations: $E=-L$, $C=-{\nabla }^{2}L/3$, $B=-L^{\prime }$ and $A=0$ in the transformed coordinates.

The gauge problem is that individual perturbation variables mix this coordinate freedom with physical fluctuations. A nonzero coordinate-dependent field therefore need not represent a physical effect. One must either choose a definite gauge, such as Newtonian gauge in cosmology, or construct invariant combinations, while ensuring that observables do not depend on that choice.

The same invariant interval can be written as $g_{\mu \nu }d{x}^{\mu }d{x}^{\nu }={\tilde{g}}_{\alpha \beta }d{\tilde{x}}^{\alpha }d{\tilde{x}}^{\beta }$. Applying the chain rule to the differentials proves the metric transformation lawFor first-order scalar cosmological perturbations and ${\xi }^{\mu }=(T,{\partial }^{i}L)$, compare perturbations at the same background coordinate label. Expanding the Jacobians and the shifted background gives the linear metric gauge-transformation lawFor the background ${\bar{g}}_{00}=-a^2$, ${\bar{g}}_{ij}=a^2{\delta }_{ij}$, the Lie derivative components areHere primes denote conformal time and $\mathcal{H}$ is the conformal Hubble parameter. These expressions establish all four alternatives, including any chosen pair.

The $00$ component is $\delta g_{00}=-2a^{2}A$, immediately giving the lapse function perturbation transformation. The $0i$ component is $a^2{\partial }_{i}B$, giving the shift vector scalar-potential transformation. The trace of the spatial perturbation is $6a^{2}C$, so tracing the spatial Lie derivative gives the transformation of $C$. Its trace-free scalar part is $2a^2({\partial }_i{\partial }_j-{\delta }_{ij}{\nabla }^2/3)E$, giving the transformation of $E$. ThusAs usual, identifying scalar potentials from their derivatives uses the standard boundary conditions, or nonzero Fourier modes, to remove homogeneous ambiguities.

At first order, tensor cosmological perturbations are spatial transverse-traceless tensors. The coordinate-generated spatial perturbation consists of a trace term and symmetrized derivatives of the displacement. In Fourier space, the latter terms carry a factor $k_i$ or $k_j$. The transverse-traceless projector removes these longitudinal terms and the trace, including the derivative of a transverse vector displacement. Therefore the tensor perturbation is gauge invariant at linear order around the homogeneous background. This is a first-order statement, not a claim of automatic invariance at arbitrary perturbative order.

Decompose the mixed extrinsic curvature of a spatial hypersurface into $K^i{}_j=\frac{1}{3}K{\delta }^i{}_j+{\tilde{K}}^i{}_j$. In the long-wavelength approximation in cosmology, the inflationary shear damping law makes ${\tilde{K}}^i{}_j$ fall as the inverse local volume. During cosmic inflation, many e-folds therefore suppress the anisotropic expansion, while $K$ still measures the local isotropic expansion through the local volume Hubble parameter.

Neglecting spatial gradients at leading order gives approximately independent expanding patches. Restrict also to scalar cosmological perturbations, and neglect the independent frozen tensor cosmological perturbation. The remaining spatial scale can be written as a background scale times a local fluctuation:This is the nonlinear scalar metric in a cosmological gradient expansion. The exponential keeps the metric positive and permits finite $\zeta$; no expansion in its amplitude is needed. Scalar lapse and shift perturbations describe the remaining freedom in time slicing and spatial threading. Choose $N=1+\Psi$ and $N_i=-a_{\mathrm{bg}}^2{\partial }_{i}B$. The negative shift vector convention then produces the positive mixed term $2a_{\mathrm{bg}}^2{\partial }_{i}Bdtd{x}^i$.

The exact completed-square form hasThe quadratic shift expression must be understood as this norm. If the raised derivative in the displayed ansatz is intended as a flat derivative, the corresponding factors of the spatial scale are needed. In either convention the shift-square term is of second gradient order, and the leading long-wavelength approximation in cosmology is unaffected.

Damping of shear motivates locally isotropic scalar expansion, not the disappearance of every tensor perturbation. A time-independent anisotropic shape can survive; the conformally flat scalar ansatz additionally neglects that tensor sector.

Use a dot for the coordinate-time derivative and let $\bar{N}$ be the homogeneous background lapse function. The passive linear transformation is $\delta {\tilde{g}}_{\mu \nu }=\delta g_{\mu \nu }-{\mathcal{L}}_{\xi }{\bar{g}}_{\mu \nu }$. Its time-time and time-space components yield the scalar gauge transformations with a background lapse:For example, ${\bar{g}}_{00}=-{\bar{N}}^2$ gives $-({\mathcal{L}}_{\xi }\bar{g}{)}_{00}=2\bar{N}\dot{\bar{N}}{\xi }^0+2{\bar{N}}^2{\dot{\xi }}^0$; ${\bar{g}}_{0i}=0$ gives $-({\mathcal{L}}_{\xi }\bar{g}{)}_{0i}={\bar{N}}^2{\partial }_i{\xi }^0-a^2{\partial }_i\dot{\lambda }$. These establish the signs directly.

To preserve synchronous gauge in cosmology, both transformed quantities must remain zero. HenceIntegrating gives the residual synchronous-gauge freedomThe integration lower limit can be absorbed into $D$. Spatially homogeneous additions to $\lambda$ generate no spatial displacement, so they are physically irrelevant to this scalar parametrization.

The spatial transformation also gives $\tilde{\Phi }=\Phi +(\dot{a}/a){\xi }^0$ and $\tilde{E}=E+\lambda$. Thus fixing the lapse and shift perturbations does not fix the entire coordinate system; the arbitrary functions $C,D$ can still change the remaining scalar cosmological perturbations.

The result above gives $K=-3H+\kappa$. Since the shear is trace-free and already first order, it cannot contribute linearly to the quadratic tensor contraction $K_{ij}{K}^{ij}=\mathrm{tr}(K^2)$. Thereforeup to second-order scalar cosmological perturbations. Combining these with the supplied spatial Ricci scalar gives the Hamiltonian constraintIts background part is the flat Friedmann equation $6H^2=16\pi G\bar{\rho }$. Subtracting that equation proves the linearized cosmological Hamiltonian constraint:Here $\rho$ is the energy density measured by the hypersurface normal, as required by the normal-normal projection of the Einstein field equations.

Write $h_{ij}={}^{(3)}{g}_{ij}$, and retain only first-order scalar cosmological perturbations. The inverse induced metric and inverse lapse function areThe flat background spatial connection vanishes, and $N_i=-a^2{B}_{,i}$ is already first order. ConsequentlyThe comma on $B_{,i}$ matters: this is the gradient of the scalar shift vector potential, not an unrelated vector field. Substituting the time derivative of $h_{ij}$ into the negative extrinsic curvature convention givesRaising the first index with the perturbed inverse induced metric is essential. Its $\Phi$ and $E$ terms cancel the corresponding terms multiplying the background Hubble parameter, leavingFor the scalar shear potential $\chi =a^2(\dot{E}-B)/\bar{N}$, raised spatial derivatives mean ${\partial }^i=a^{-2}{\delta }^{ij}{\partial }_j$ to the required order. Thus the final term is $-{\partial }^i{\partial }_j\chi$. Taking the trace identifies the scalar expansion perturbation:Separating the trace and trace-free parts now yieldsAll signs follow from the source's $d{x}^i-N^{i}dt$ shift vector convention and negative extrinsic curvature definition.

Let $\Gamma =-\dot{\tau }=a{\bar{n}}_e{\sigma }_T$ and let $D={\partial }_{\eta }+\mathbf{e}\cdot \nabla$ differentiate along the unperturbed photon trajectory. Use $1+(\mathbf{e}\cdot \hat{\mathbf{m}}{)}^2=4/3+(2/3)P_2(\mathbf{e}\cdot \hat{\mathbf{m}})$ to decompose the Thomson scattering gain term:The Orthogonality of Legendre polynomials identifies $Q$ as a pure photon quadrupole. Dropping $Q$ is an approximation, justified when tight coupling makes the scattering quadrupole small; the supplied unpolarized equation already neglects polarization corrections. The Cosmic microwave background line-of-sight solution requested here omits that quadrupole source.

Set $Y=\Theta +\psi$. Adding $D\psi =\dot{\psi }+\mathbf{e}\cdot \nabla \psi$ to the Photon Boltzmann equation with Thomson scattering yieldsAlong $\mathbf{x}(\eta )={\mathbf{x}}_0-({\eta }_0-\eta )\mathbf{e}$, the integrating factor is $e^{-\tau }$ because $d{e}^{-\tau }/d\eta =\Gamma e^{-\tau }$. ConsequentlyAn early optically thick boundary makes $e^{-\tau }Y$ vanish for finite initial perturbations, and $e^{-\tau ({\eta }_0)}=1$. ThereforeHere $\dot{\varphi }$ and $\dot{\psi }$ are partial time derivatives, not total derivatives along the ray. We work at first order in scalar cosmological perturbations, evaluate sources along the background radial null geodesic in FLRW spacetime, and ignore observer peculiar velocity. Perturbed paths and lensing affect this first-order source solution only at higher perturbative order. A moving observer adds its local Doppler dipole.

The original PDF, unlike the transcribed TeX, also asks for the instantaneous-last-scattering limit and its interpretation. For $g(\eta )={\delta }^{(D)}(\eta -{\eta }_{\ast })$ with no later scattering, and a matter-era growing mode with constant $\varphi$ and $\psi$, the Integrated Sachs-Wolfe effect vanishes. HenceThe four terms are the intrinsic emission temperature, the gravitational potential at emission, the observer's potential, and the Doppler CMB anisotropy. The two potentials describe the gravitational redshift; the observer term is independent of direction and is absorbed into the measured temperature monopole. The intrinsic temperature and emission potential form the Sachs-Wolfe combination. On superhorizon scales for adiabatic initial conditions in matter domination, ${\Theta }_0=-2\psi /3$, giving the ordinary Sachs-Wolfe effect ${\Theta }_0+\psi =\psi /3$. The Doppler sign is positive with the paper's propagation direction $\mathbf{e}$; the viewing direction is $-\mathbf{e}$.