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Use the exact scalar-metric normalization specified by the preceding brightness equation:The temperature source is therefore $-(h^{\prime }-h_s^{\prime })/6-{\mu }^2{h}_s^{\prime }/2$. Let $W=e^{ik\mu (\tau -{\tau }_0)}$, so $W^{\prime }=ik\mu W$ and $W^{\prime \prime }=-k^2{\mu }^{2}W$. The formula just derived becomesFor $k\ne 0$ the last term may be integrated twice by parts:No field equation beyond the supplied brightness relation has been used for this identity.

For scalar velocity the supplied photon continuity equation givesThe sign follows from division by $i$: with the Fourier streaming term $+ik\mu$, the velocity projection has the plus $h^{\prime }$ term shown here. Combining it with the lower endpoint from the integrations by parts gives the complete endpoint terms in the synchronous Sachs-Wolfe formula:The local observer bracket is an angular photon monopole plus photon dipole. Removing the local mean temperature and photon dipole, or restricting the result to measured multipoles $\ell \ge 2$, permits omitting that bracket. It must be retained for the literal full brightness amplitude. The $k=0$ limit should be taken in the original line-of-sight integral, rather than treating the separate inverse-$k$ endpoint terms as independently defined.

The final target formula printed in the PDF does not follow from its preceding equations. With the stated streaming and scalar conventions, the decoupling terms proportional to $h^{\prime }-h_s^{\prime }$ and $h_s^{\prime \prime }$ have plus signs as derived above, not the printed minus signs. Moreover the integral has no additional outside factor $1/2$ when its bracket is $(h^{\prime }-h_s^{\prime })/6-h_s^{\prime \prime \prime }/(2k^2)$. Equivalently one may retain an outside $-1/2$ only if that bracket is doubled to $(h^{\prime }-h_s^{\prime })/3-h_s^{\prime \prime \prime }/k^2$. Omitting the observer photon monopole and photon dipole does not fix these discrepancies.

For a direct counterexample to the integral coefficient, take $h_s=0$, $h^{\prime }=\epsilon$ constant, $\mu =0$, and ${\delta }_{\gamma \ast }={\delta }_{\gamma \ast }^{\prime }=0$. A scalar emission velocity with $i\mathbf{k}\cdot {\mathbf{v}}_{\ast }=-\epsilon /2$ satisfies the stated continuity equation; its projection along this transverse photon direction is zero. Keep $\epsilon ({\tau }_0-{\tau }_{\ast })\ll 1$ so perturbation theory is valid. The original brightness equation or the line-of-sight integral giveswhereas the printed final expression gives $-\epsilon ({\tau }_0-{\tau }_{\ast })/12$. All $h_s$ observer terms vanish in this example, so an observer-term convention cannot remove the contradiction. The boxed corrected identity supplies the requested integration-by-parts result with every boundary contribution and its observable higher-multipole version specified.

Before decoupling, rapid Thomson scattering makes the photon distribution nearly isotropic in the local photon-fluid rest frame. A local blackbody temperature perturbation changes its energy density by ${\delta }_{\gamma }=4\delta T/T$. A first-order boost by the fluid velocity adds a temperature photon dipole $\mathbf{n}\cdot \mathbf{v}$; it does not generate a photon quadrupole until second order in that boost. Thus the photon brightness perturbation at instantaneous decoupling isHigher angular moments are suppressed by the photon mean-free-path to perturbation-length ratio in the tight-coupling approximation. The idealized equilibrium assumption neglects those finite-mean-free-path corrections, polarization effects and finite thickness of the last-scattering surface. It is not an assertion that collisionless photons remain isotropic after decoupling. The distinction between a fluid treatment and the subsequent hierarchy is discussed in the primary synchronous and Newtonian perturbation treatment.

Let $\mu =\hat{\mathbf{k}}\cdot \mathbf{n}$ and $W(\tau )=e^{ik\mu (\tau -{\tau }_0)}$. Multiplication of the synchronous photon brightness equation by its integrating factor givesIntegration from ${\tau }_{\ast }={\tau }_{\mathrm{dec}}$ to ${\tau }_0$ and division by four yieldFor the real-space observation point ${\mathbf{x}}_0$, the corresponding background ray is $\mathbf{x}(\tau )={\mathbf{x}}_0-\mathbf{n}({\tau }_0-\tau )$. Fourier inversion gives the synchronous Sachs-Wolfe line-of-sight formulaEvery source is evaluated at its appropriate point on the ray, rather than at the observation position. Here $\mathbf{n}$ is photon propagation direction, opposite to the outward observer-to-source sky direction; changing that convention changes the written sign of the Doppler projection consistently.

The first term is the intrinsic photon temperature at last scattering. On wavelengths larger than the sound horizon, it retains primordial adiabatic information; inside that horizon the photon-baryon fluid has acoustic energy density oscillations, producing the acoustic temperature peaks. The second term is the Doppler effect from the fluid's bulk motion at emission. A regular long-wavelength velocity is gradient-suppressed; the Doppler contribution is important near and below the acoustic horizon, with its acoustic phase displaced from that of the energy density oscillation.

The integral is the gravitational energy shift produced by the changing spatial metric tensor along the photon path. In synchronous gauge it includes endpoint gravitational redshifts as well as evolving-potential effects. It must not be identified solely with the Integrated Sachs-Wolfe effect: in Newtonian gauge the same total separates into the ordinary last-scattering Sachs-Wolfe effect, local observer terms, and an integral of time-varying potentials. The ordinary contribution is important on scales outside the sound horizon at decoupling. A late integrated contribution is strongest on large angular scales when dark energy or curvature changes the potentials; evolution around matter-radiation equality also contributes near the first acoustic scales.

The relevant comoving sound scale is $r_s({\tau }_{\ast })={\int }^{{\tau }_{\ast }}{c}_s(\tau )d\tau$, while the projection distance is $D_{\ast }\simeq {\tau }_0-{\tau }_{\ast }$. Modes project roughly to photon temperature multipole $\ell \sim k{D}_{\ast }$, and the acoustic scale is $\ell \sim D_{\ast }/r_s$, with successive peak spacing of order $\pi D_{\ast }/r_s$. At much shorter wavelengths Silk damping and the finite last-scattering thickness suppress energy density and velocity anisotropies; the instantaneous-decoupling approximation alone does not model that cutoff. These scale distinctions explain why the three terms cannot be assigned one common range of important wavelengths.

The comoving observer measures $E=a{p}^0$, so $q=aE=a^2{p}^0$. Write $\mathcal{H}=a^{\prime }/a$. The null geodesic condition impliesThe time component of the geodesic equation, using $d\tau /d\lambda =p^0$, is thereforeDifferentiate $q=a^2{p}^0$. The homogeneous expansion terms cancel:At zeroth order $p^i/p^0=n^i$, with ${\delta }_{ij}{n}^i{n}^j=1$. The difference between coordinate velocity and physical unit direction is already first order and multiplies $h^{\prime }$, so it can be discarded in this expression. We obtain the synchronous photon momentum redshiftIn the unperturbed universe $q$ is constant and physical photon energy redshifts as $a^{-1}$; the displayed term is the additional anisotropic first-order redshift.

To check the direction, let $v^i=p^i/p^0$. Combining the spatial and temporal geodesic equations cancels their homogeneous Hubble terms and givesEvery remaining connection or explicit metric-derivative term is first order. A local orthonormal spatial frame defines the physical unit vector by $n^i=v^i+\frac{1}{2}{h}^i{}_j{v}^j+O(h^2)$. Its derivative adds only first-order terms, soConsequently direction deflection multiplied by an already first-order temperature anisotropy is second order. The first-order brightness calculation may use a straight unperturbed ray without neglecting the first-order gravitational frequency change.

Write $V=\dot{R}$, $a=2/(5-\beta )$ andHere $u$ is the outward radial velocity; $U,D,P$ are dimensionless. The relationsare sufficient for the reduction. In smooth post-shock gas the spherical continuity equation, Euler equations and adiabatic pressure equation areFor example, the mass density material derivative is $C{R}^{-\beta }(V/R)[(U-\xi )D^{\prime }-\beta D]$, and the velocity material derivative is $(V^2/R)[(U-\xi )U^{\prime }+qU]$. After cancellation of the dimensional factors the spherical blast wave in a power-law ambient density obeysPrimes here mean $d/d\xi$. In the pressure equation the scale derivative $2q-\beta =-3$ explains the constant term. Once the profiles are found, their finite energy integral fixes $A$ throughNo global constant in $p/{\rho }^{\gamma }$ is assumed: the shock wave generates different entropies in shells swept up at different times.

Assume $\gamma >1$. The upstream ideal gas is at rest in the laboratory frame and has mass density ${\rho }_1=C{R}^{-\beta }$. In the shock frame its incident speed is $V$. The strong-shock limit of the Rankine-Hugoniot conditions for a perfect gas gives compression $\chi =(\gamma +1)/(\gamma -1)$ and downstream shock-frame speed $V/\chi$. Transforming back to the laboratory frame gives $u_2=V(1-1/\chi )=2V/(\gamma +1)$.

The normal momentum jump yieldswhere the cold upstream pressure has been neglected. Thus the boundary values in the normalization of part (b) areThe velocity jump uses the relative speed $V-u_2$, not the laboratory speed $u_2$. This distinction is essential to all three shock wave values.

Let the equilibrium ideal gas be at rest, with uniform ${\rho }_0,p_0$ and ${\mathbf{B}}_0=B_0{\mathbf{e}}_z$. This local wave calculation neglects gravity, viscosity and magnetic diffusion. Define the adiabatic sound speed $c\equiv v_s=(\gamma p_0/{\rho }_0{)}^{1/2}$, the Alfvén speed $v_a=B_0/({\mu }_0{\rho }_0{)}^{1/2}$, and a wavevector $\mathbf{k}=k(\sin \theta ,0,\cos \theta )$, where $k=|\mathbf{k}|$ and $\theta$ is its angle to the equilibrium magnetic field. Perturbations have dependence $e^{i\mathbf{k}\cdot \mathbf{x}-i\omega t}$ and phase speed $v=\omega /k$.

The linearized ideal magnetohydrodynamic equations for mass conservation, pressure, induction and momentum giveThe relation between pressure and mass density follows from eliminating $\mathbf{k}\cdot {\mathbf{u}}^{\prime }$ between the two material evolution equations; it applies to the nonzero-frequency wave sector. The induction result automatically has $\mathbf{k}\cdot \mathbf{b}=0$.

The $y$ MHD wave polarization is perpendicular to both the field and wavevector. It has $\delta p=\delta \rho =0$ and obeysThus the transverse Alfvén wave hasIn the $x$-$z$ plane the two remaining velocity components instead obey the symmetric matrix eigenproblemTaking its determinant gives $v^4-(c^2+v_a^2)v^2+c^2{v}_a^2{\cos }^2\theta =0$. Solving this quadratic equation in $v^2$ derives both magnetoacoustic wave phase speeds,The plus sign is the fast magnetosonic wave, the minus sign the slow magnetosonic wave. The radicand is nonnegative, since it equals $[(c^2-v_a^2{)}^2+4c^2{v}_a^2{\sin }^2\theta ]/4$. Both squared speeds are nonnegative. Each nonzero mode has forward and backward propagation; an advected entropy disturbance is a separate zero-frequency mode in this static equilibrium.

Set $ϵ=v_a^2/c^2\ll 1$. Expanding the square root in part (a) givesHence, to first absolute order in $ϵ$,The high-pressure limit of magnetosonic waves can also be expressed as nonnegative speeds. Retaining the next squared-speed term gives the first relative correction for the slow mode:The exact Alfvén speed along the propagation direction remains $v_a|\cos \theta |$.

The fast magnetosonic wave is predominantly an acoustic wave, with velocity almost parallel to $\mathbf{k}$ and appreciable mass density and gas pressure variations. Its magnetic pressure correction is weak. The Alfvén wave is exactly incompressible, polarized perpendicular to the $\mathbf{k}$-${\mathbf{B}}_0$ plane and restored by magnetic tension.

The slow magnetosonic wave is polarized inside that plane and almost perpendicular to $\mathbf{k}$. Indeed, the leading sound speed part of the matrix in (a) annihilates a vector perpendicular to $\mathbf{k}$, so only the smaller magnetic restoring force acts on that MHD wave polarization. Its compression is smaller by order $ϵ$, with gas pressure changes almost cancelling magnetic pressure changes. It is often called the pseudo-Alfvén wave MHD wave polarization in this high pressure limit. Thus two distinct almost-incompressible magnetic MHD wave polarizations have the same leading speed, while the third is predominantly sound. For exactly perpendicular propagation the Alfvén wave and slow magnetosonic wave frequencies vanish; for parallel propagation the two transverse magnetic MHD wave polarizations are degenerate.

Take $B_0=B_z>0$, $v_a=B_0/\sqrt{{\mu }_0{\rho }_0}$ and $\zeta =k(z-v_{a}t)$. The transverse ideal magnetohydrodynamic induction equation and magnetohydrodynamic momentum equation are satisfied by $u_x=-B_x/\sqrt{{\mu }_0{\rho }_0}$, $u_y=u_z=0$, with constant mass density. However, the longitudinal magnetohydrodynamic momentum equation also requiresHere $B_x^2=a^2{B}_0^2{\cos }^2\zeta$ is not constant. Balancing its gradient would requireThe adiabatic pressure equation instead gives ${\partial }_{t}p=0$, since this velocity is divergence-free and has no component along the spatial variation. No choice of $p_{\ast }(t)$ removes the spatially varying time derivative. Thus a nontrivial finite-amplitude linearly polarized sinusoidal Alfvén wave is not an exact pure traveling wave in a compressible ideal fluid. The magnetic-pressure obstruction to a linearly polarized Alfvén wave is quadratic in $a$, so the wave is a valid linear approximation and drives a compressive response at higher order. In an incompressible model pressure is a constraint variable; it is not governed by the same compressible adiabatic evolution law.

Allowing longitudinal flow and varying mass density does not rescue a regular one-dimensional periodic traveling solution with precisely this transverse magnetic field. To see this, let $U=u_z-v_a$ and use the traveling coordinate. Integrated continuity, induction and transverse momentum givewhere $m,A,D$ are constants. Eliminating $u_x$ givesAt any zero of $B_x$, finite mass density and velocity require the right side to vanish. Wherever $B_x\ne 0$, the coefficient then vanishes and fixes $\rho ={\mu }_0{m}^2/B_0^2$, constant; continuity fixes $U$, and adiabatic entropy transport fixes $p$. The longitudinal magnetohydrodynamic momentum equation again contradicts the varying $B_x^2$. The alternative $m=0$ makes transverse momentum force $B_x$ constant. Trivial cases such as $a=0$ or $k=0$ are excluded from the nontrivial wave conclusion.

Use a steady axisymmetric flow governed by ideal magnetohydrodynamics and take the Newtonian gravitational potential to be axisymmetric. Work locally on connected regular magnetic flux surfaces with nonzero poloidal flow. As usual for a wind without an imposed toroidal loop voltage, $E_{\varphi }=0$; regularity on an included rotation axis guarantees this condition. It is needed to exclude steady cross-field flow in an annular domain.

Since $\nabla \varphi ={\mathbf{e}}_{\varphi }/R$, the poloidal magnetic flux function givesThe poloidal magnetic field is divergence-free. The zero toroidal electric field, with $\mathbf{E}=-\mathbf{u}\times \mathbf{B}$, gives ${\mathbf{u}}_p\parallel {\mathbf{B}}_p$. Define $k$ by $\rho {\mathbf{u}}_p=k{\mathbf{B}}_p$. Steady mass conservation then givesThus $k=k(\psi )$ is the magnetohydrodynamic mass loading, the mass flux per unit poloidal magnetic flux. Subtracting $(k/\rho )\mathbf{B}$ from the full velocity leaves only an azimuthal component. The cross product isThe steady ideal magnetohydrodynamic induction equation requires $0=\nabla \times (\omega \nabla \psi )=\nabla \omega \times \nabla \psi$. Consequently the field-line angular velocity is $\omega =\omega (\psi )$ and

Neither gas pressure nor axisymmetric gravity exerts an azimuthal torque. Multiplying the azimuthal momentum equation by $R$ combines the cylindrical curvature terms into angular-momentum transport:Insert $\rho {\mathbf{u}}_p=k{\mathbf{B}}_p$ and use ${\mathbf{B}}_p\cdot \nabla k=0$. It follows thatso the magnetohydrodynamic angular-momentum invariant isThe two terms account respectively for matter and Maxwell stress tensor transport of angular momentum.

Finally, subtracting $\gamma$ times the logarithmic mass density equation from the logarithmic pressure equation gives $D\ln (p/{\rho }^{\gamma })/Dt=0$. The specific entropy of a fixed-composition ideal gas is a function of $p/{\rho }^{\gamma }$. Steadiness and axisymmetry reduce its advection to ${\mathbf{u}}_p\cdot \nabla s=0$. HenceThese flux-label functions are local on connected regular surfaces; disconnected components with the same numerical label need not share constants without an additional matching condition.

On one poloidal flux surface, the field-line angular velocity $\omega (\psi )$ is constant. In a frame rotating at that rate the relative velocity is $\mathbf{v}=\mathbf{u}-R\omega {\mathbf{e}}_{\varphi }=k\mathbf{B}/\rho$, parallel to the magnetic field. The Lorentz force is perpendicular to this motion, and the Coriolis force also does no work. Centrifugal acceleration comes from the effective potential $-R^2{\omega }^2/2$. Since entropy is constant along the surface, pressure work integrates to specific enthalpy $w$, with $dw=dp/\rho$ along the flow.

Therefore kinetic energy relative to the rotating poloidal flux surface, enthalpy and effective gravitational-plus-centrifugal potential have a constant sum:This is a corotating energy integral, not the laboratory energy alone: Maxwell stress tensor stresses can transfer energy and angular momentum in the laboratory frame. In fact the expression equals $|\mathbf{u}{|}^2/2+w+\Phi -\omega R{u}_{\varphi }$, namely the total magnetohydrodynamic Bernoulli invariant minus $\omega \ell$. Differential rotation between surfaces does not invalidate the argument, because the chosen angular velocity is fixed along the particular surface being followed.

All invariants in this part refer to the selected magnetic field line. Its squared poloidal Alfvén number isAt the Alfvén surface crossing it equals unity, so ${\rho }_a={\mu }_0{k}^2$ and $M_{Ap}^2=1/y$. Subtract the two azimuthal integrals in part (a) to findFinite azimuthal fields and velocity at $y=1$ require the Alfvén-surface regularity condition for an axisymmetric wind $\ell =\omega R_a^2$. HenceHere $v_{\varphi }$ denotes the azimuthal velocity relative to field-line rotation. For gravity of a point mass at $z=0$, $\Phi =-GM/R=-GM/(R_{a}x)$.

Insert these quantities in the corotating energy integral with $w=0$, and divide by the gravitational energy scale $GM/R_a$. This gives the cold radial magnetohydrodynamic wind integral,withBoth are dimensionless. The apparent $0/0$ at $x=y=1$ must be evaluated on the smooth solution branch, not assigned an arbitrary value.

A constant-energy curve obeys $f_x+f_{y}dy/dx=0$. At a nondegenerate magnetosonic point, the vanishing coefficient of the mass density derivative requires simultaneous vanishing of its numerator:The curve must admit a real crossing slope. Expanding about such a point gives $f_{yy}{m}^2+2f_{xy}m+f_{xx}=0$, $m=dy/dx$, so a regular saddle point of a scalar function crossing has $f_{xy}^2-f_{xx}{f}_{yy}>0$. These are the form of the slow and fast regularity conditions in a warm wind, with the enthalpy contribution retained in $f$.

In the strictly cold approximation used to obtain the displayed $f$, the slow speed is zero. There is a cold-limit degeneracy of a slow magnetosonic point: a finite-speed interior slow crossing need not survive. Indeed,so this cold equation has no ordinary sub-Alfvénic slow critical point. A small retained enthalpy is needed to locate that point, which can approach the launching boundary as the sound speed tends to zero. The fast critical point still obeys the simultaneous conditions above wherever the reduced expression is nonsingular.

To distinguish Eulerian and Lagrangian fluid perturbations, use $\mathbf{\xi }=({\xi }_x,{\xi }_y,{\xi }_z)$ for the fluid displacement, $\varphi ={\Phi }^{\prime }$ for the Eulerian potential perturbation, and $\delta \rho$ for the Eulerian mass density perturbation. The equilibrium mass density $\rho (z)$ is the profile from part (a); subscript $z$ below denotes differentiation. Linearized mass conservation and the isothermal equation of state giveThe velocity perturbation is $-i\omega \mathbf{\xi }$. Linearized momentum initially has the formSince $\nabla {\Phi }_0=-c_s^2\nabla \ln \rho$, the pressure and perturbed mass density gravity terms combine into $-\rho \nabla (c_s^2\delta \rho /\rho )$. Define $W=c_s^2\delta \rho /\rho +\varphi$. The closed linearized system isThe $y$ component is a neutral transverse sector; it vanishes for nonzero-frequency modes.

To prove reality rather than assume it, multiply momentum by $\rho {\mathbf{\xi }}^{\ast }$ and integrate over $z$. With $I=\int \rho |\mathbf{\xi }{|}^{2}dz>0$, integration by parts and the conjugate continuity equation giveUsing the conjugate Poisson equation for Newtonian gravity in the last integral givesTake disturbances with finite $I$, finite pressure and gravitational integrals, and vanishing pressure work and gravitational boundary terms. For $k\ne 0$, require the potential to decay at both infinities and the fluid displacement to make $\rho {\xi }_z^{\ast }W\to 0$ as well. In a source-free outer region the decaying potential is proportional to $e^{-|k||z|}$. The isothermal displacement energy of a self-gravitating slab then givesSince $I$ is strictly positive for a nonzero displacement, ${\omega }^2$ is real. Positive values describe oscillations; negative values give pure growth and decay. The result is the energy identity for the pressure-gravity displacement operator, a self-adjoint operator in the mass density-weighted inner product. The identity excludes an oscillatory growing eigenmode under these boundary conditions. At $k=0$ the same identity applies when its boundary terms vanish and its integrals remain finite, with the gravitational integral involving only $|{\varphi }_z{|}^2$.

Let ${\rho }_i={\rho }_n=\rho$ and $\mu ={\mu }_{in}={\mu }_{ni}$. Use normal modes proportional to $e^{i\mathbf{k}\cdot \mathbf{x}-i\omega t}$, so $\mathrm{Im}\omega <0$ means damping. The linear displacements satisfyFor a nonzero-frequency perturbation, linearizing the ideal magnetohydrodynamic induction equation givesProject both momentum equations with $\mathsf{P}=I-\mathbf{k}{\mathbf{k}}^T/k^2$. This removes the pressure gradients. The magnetic tension is already transverse to $\mathbf{k}$, and, with $a=k_{\parallel }{v}_A$, the projected equations areHere $v_A=B_0/\sqrt{4\pi \rho }$ uses the ion density, as stipulated. For either of the two independent transverse polarizations, the equal-density ion-neutral Alfvén dispersion relation follows fromIts determinant is $\omega [{\omega }^3+2i\mu {\omega }^2-a^2\omega -i\mu a^2]$. Removing the static neutral-displacement factor givesThe removed factor is a time-independent neutral displacement with no velocity or magnetic perturbation, a relabeling of the homogeneous neutral fluid rather than a fourth physical wave. Direct elimination in the first-order system for $(u_i,u_n,\delta B)$ gives exactly the same cubic. This also handles the zero-frequency limits without introducing a displacement-label mode.

The neutral displacement equation gives, for each nonzero-frequency root,The vector displacements are parallel as complex polarization vectors, with this scalar amplitude and phase ratio.

For the rapidly damped root, $\mu -i{\omega }_d=-\mu +a^2/(4\mu )+O(a^4/{\mu }^3)$, and henceThe two fluids move almost equally and oppositely; collisions remove the counterflow at a rate close to $2\mu$.

For either weakly damped Alfvén wave, expand the ratio around unity:Equivalently, the exact slippage relation is ${\mathbf{\xi }}_i-{\mathbf{\xi }}_n=-(i\omega /\mu ){\mathbf{\xi }}_n$. Thus the relative slippage has size $|a|/(\sqrt{2}\mu )\ll 1$ and is primarily a small phase difference. The fluids almost share a common displacement, but magnetic tension acts directly on the ions and drag accelerates the neutrals. This imperfect locking causes ambipolar damping.

The sign of the damping is also checked by the perturbation energy balance. With periodic or vanishing-flux boundaries,It follows by dotting each linear momentum equation with its velocity and the induction equation with $\delta \mathbf{B}/(4\pi )$; magnetic-tension work cancels magnetic-energy change, pressure contributes only boundary flux, and the two drag terms combine into the negative square. Even though spatial diffusion was neglected, ion-neutral friction still dissipates the wave energy.