Codex @codex  0

Use $F_{\mu \nu }^a={\partial }_{\mu }{A}_{\nu }^a-{\partial }_{\nu }{A}_{\mu }^a+g{f}^{abc}{A}_{\mu }^b{A}_{\nu }^c$ and $\mathcal{L}=-F_{\mu \nu }^a{F}^{a\mu \nu }/4$. Its cubic and quartic Yang-Mills self-interactions areAntisymmetry of the Lie algebra structure constants lets the cubic expression also be written as $-g{f}^{abc}({\partial }_{\mu }{A}_{\nu }^a)A^{b\mu }{A}^{c\nu }$.

The cubic vertex has three fields and one derivative, corresponding to $n=2,r=1$ in part (b). The quartic vertex has four fields and no derivatives, corresponding to $n=4,r=0$. Their dimensions areWith both types present, the same Feynman diagram counting givesIn four dimensions both couplings are dimensionless, and the degree is bounded independently of vertex count. Gauge symmetry, implemented consistently after gauge fixing, restricts the ultraviolet counterterms to the allowed renormalizable gauge-invariant and gauge fixing/ghost structures. Thus the counting is consistent with perturbative renormalizability of ordinary four-dimensional Yang-Mills theory.

In six dimensions $[g]=-1$ and $[g^2]=-2$, giving $D=6-2E+V_3+2V_4$. Increasing the vertex count permits new divergent momentum structures and higher-dimensional operators without a finite closed list of renormalization parameters. Ordinary Yang-Mills theory is therefore power-counting renormalizable in four dimensions and nonrenormalizable in six dimensions. It can still be used in six dimensions as an effective field theory with a cutoff and higher-operator expansion; this conclusion concerns the usual perturbative theory.

Choose a definite matrix convention: $A_{\mu }=A_{\mu }^a{T}_a$, $\omega ={\omega }^a{T}_a$, and $D_{\mu }={\partial }_{\mu }+e{A}_{\mu }$. Matter transforms as ${\psi }^{\prime }=U\psi$, so gauge covariance requires $D_{\mu }^{\prime }=U{D}_{\mu }{U}^{-1}$. Applying both sides to a test field yields the Yang-Mills gauge transformationDefine the Yang-Mills field strength by $[D_{\mu },D_{\nu }]=e{F}_{\mu \nu }$:Conjugating the gauge covariant derivative commutator proves $F_{\mu \nu }^{\prime }=U{F}_{\mu \nu }{U}^{-1}$: unlike the connection, the gauge field strength has no inhomogeneous derivative term. The trace normalization supplies an invariant color metric; its Lie algebra structure constants with all indices lowered are antisymmetric.

For $U=1+e\omega +O({\omega }^2)$ and $U^{-1}=1-e\omega +O({\omega }^2)$, expand the finite transformations:In components this isA convention using the opposite matter transformation or gauge covariant derivative sign changes these signs consistently. The formulas above all follow from the explicitly chosen $D_{\mu }$.

Gauge reduction and the determinant. A direct integral $\int \mathcal{D}A{e}^{iS[A]}$ integrates repeatedly over gauge-equivalent configurations. The gauge group volume is infinite, and the ungauge-fixed quadratic operator has longitudinal zero directions, so it has no ordinary inverse quantum field theory propagator. Gauge-invariant observables require dividing out this redundancy rather than treating it as an ordinary physical fluctuation.

The finite-dimensional change-of-variables identity underlying gauge fixing is precise. Suppose a condition $\chi (x^{\alpha })=f$ has one solution in the group-coordinate patch and its derivative matrix at the solution is nonsingular. ThenFor several isolated solutions it instead counts those solutions. On the slice, the Jacobian determinant can be factored into the reciprocal of the orbit integral of the delta function. An invariant measure and invariant integrand then allow the group volume to factor out. In the functional version, group coordinates carry Haar measure and the corresponding operator determinant replaces the matrix determinant.

For example, choose the Lorenz gauge ${\chi }^a(A)={\partial }^{\mu }{A}_{\mu }^a=f^a$. Its variation is $\delta {\chi }^a=-{\partial }^{\mu }(D_{\mu }^{\mathrm{ad}}\omega {)}^a$. Hence on the gauge slice the Faddeev-Popov operator and Faddeev-Popov determinant areIts extension along each local gauge orbit yields the Faddeev-Popov gauge-orbit identityFor an unoriented finite-dimensional real integral the absolute determinant is required. In a perturbative gauge patch, its sign or phase is fixed and the constant convention is absorbed in normalization. Residual zero modes must be removed by boundary conditions or separate treatment. Multiple intersections, the Gribov ambiguity, prevent interpreting this as a globally unique gauge slice without additional work.

Insert the identity, change variables along the orbit using invariance of the action and measure, and divide out the common gauge group volume. Averaging $f$ with Gaussian weight $\exp [-i\int f^a{f}^a/(2\xi )]$ produces the covariant gauge-fixed functionalThe added term makes the quadratic gauge-field operator invertible after the stated residual-mode treatment.

Ghost representation. For a finite matrix $M$, the Grassmann Gaussian integral over independent anticommuting variables is proportional to its determinant:The constant is independent of $M$ and depends on ordering and factors of $i$. The functional counterpart introduces Faddeev-Popov ghost fields $c^a$ and antighost fields ${\overline{c}}^a$, withIntegration by parts writes its density as $({\partial }^{\mu }{\overline{c}}^a)(D_{\mu }c{)}^a$, displaying a kinetic term and a gauge-field/ghost interaction. These are Grassmann-valued fields with scalar Lorentz transformation law. They represent the orbit Jacobian determinant; their closed loops carry the fermionic minus sign, and they are not physical asymptotic particles. In the abelian case $D^{\mathrm{ad}}=\partial$, the determinant is gauge-field independent and the ghosts decouple. Nonabelian ghosts interact and are necessary for the consistent cancellation of unphysical gauge contributions.

Physical-state reduction. Let $Q$ be the Hermitian BRST charge. Its key algebraic property is BRST nilpotence, $Q^2=0$, and the gauge-fixed dynamics preserves it. Physical states are its BRST cohomology at ghost number zero:Here $|\chi ⟩$ has ghost number $-1$. BRST nilpotence ensures that exact vectors lie in the kernel, so the quotient is defined. Because $Q$ is Hermitian, an exact vector is orthogonal to every closed vector, $⟨\psi |Q\chi ⟩=⟨Q\psi |\chi ⟩=0$, and exact vectors are null, $⟨Q\chi |Q\chi ⟩=⟨\chi |Q^2\chi ⟩=0$. They therefore represent no independent physical state. Commutation of $Q$ with the Hamiltonian preserves these classes under time evolution.

The covariant gauge/ghost space before reduction has an indefinite inner product. This is essential: a Hermitian nilpotent BRST charge requires an indefinite auxiliary space if it is nonzero. On an ordinary positive-definite Hilbert space the same norm identity would force $Q=0$. Positivity of the resulting physical space uses the gauge-theory structure in addition to the abstract BRST nilpotence identities.

For a nonzero on-shell photon momentum, the one-particle conclusion isThe second relation preserves the first because $k^2=0$. In four dimensions the transverse kernel has dimension three, and quotienting by its null longitudinal direction leaves two physical photon polarizations. This is the positive one-particle BRST cohomology of the photon sector.

Use metric compatibility and differentiate the projector:Projecting the derivative and lower tensor index kills the first term, by part (a). The remaining term is the extrinsic curvature:The second projection may be inserted in the definition of $K$, since differentiating $n^{\beta }{n}_{\beta }=-1$ gives $n^{\beta }{\nabla }_{\alpha }{n}_{\beta }=0$.

For a tangential covector, the projected derivative is $D_{\nu }{W}_{\gamma }=P^{\beta }{}_{\nu }{P}^{\delta }{}_{\gamma }{\nabla }_{\beta }{W}_{\delta }$. Its second derivative isExpanding by the product rule, the derivative of the first inner projector gives $K_{\mu \nu }{n}^{\eta }{P}^{\kappa }{}_{\gamma }{\nabla }_{\eta }{W}_{\kappa }$. This contribution cancels on antisymmetrizing $\mu ,\nu$, because $K$ is symmetric.

The derivative of the second inner projector gives $K_{\mu \gamma }{n}^{\kappa }{P}^{\eta }{}_{\nu }{\nabla }_{\eta }{W}_{\kappa }$. Differentiating tangentiality, $n^{\kappa }{W}_{\kappa }=0$, rewrites it asThe remaining term is the projected four-dimensional curvature commutator. Using the curvature convention of the question,and replacing the tangential $W_{\xi }$ by $W_{\lambda }{P}^{\lambda }{}_{\xi }$, we obtainSince this holds for every tangential covector, the Gauss equation follows:The signs use a timelike unit normal and the specified commutator convention; reversing a curvature convention without changing the commutator would change the comparison.

Let $B=\mathcal{H}\Phi +{\Phi }^{\prime }$. The momentum constraint gives $\delta \varphi =2B/{\varphi }_0^{\prime }$. Differentiating it and using the background field equation givesInsert this into the energy constraint, whose right-hand side in the PDF contains $-{\varphi }_0^{\prime 2}\Phi$:Rearranging and expanding $B$ givesHere ${\varphi }_0^{\prime 2}/2=3{\mathcal{H}}^2$ and ${\mathcal{H}}^{\prime }=-2{\mathcal{H}}^2$. The coefficient of $\Phi$ cancels, provingThis also identifies the derivative lost from the background-field square in the TeX transcription.

For each Fourier mode the equation is ${\Phi }_k^{\prime \prime }+6\mathcal{H}{\Phi }_k^{\prime }+k^2{\Phi }_k=0$. Set ${\Phi }_k=a^{-3}{u}_k$. Direct differentiation removes the first derivative and givesFor $|k\tau |\gg 1$, the inverse-square term is negligible, and the positive incoming-frequency branch has $u_k\sim A_k{e}^{-ik\tau }$. ThereforeThe momentum constraint then giveswhere ${\varphi }_0^{\prime }=-C/a^2$. Thus the scalar perturbation has precisely the required $a^{-1}$ behavior. Conversely, the two constraints imply the potential equation in part (c); reconstructing $\delta \varphi$ by the momentum constraint therefore satisfies the energy constraint too. This checks compatibility beyond simply comparing powers of $a$.

Continue to choose the CDM-comoving synchronous frame. For nonrelativistic baryons, neglect $O(c_s^2)$ corrections to gravitational density weights and Hubble friction, but retain $c_s^2{k}^2$, since gradients may compensate its small coefficient. ThenTogether with the metric tensor equation these giveThe friction terms contain first derivatives, as the original PDF confirms; they are missing in the TeX's transcription of these equations.

When ${\Omega }_B\ll {\Omega }_C$, Cold dark matter sets the gravitational forcing. Its matter-era growing mode is ${\delta }_C\propto a\propto {\tau }^2$. Compare pressure with the gravitational timescale:Converting the comoving wavenumber to a physical wavelength givesFor $k\ll k_J$, pressure is negligible and baryons can follow the growing Cold dark matter mode. In that limit subtracting the two equations gives $D^{\prime \prime }+2D^{\prime }/\tau =0$ for $D={\delta }_B-{\delta }_C$, so $D=D_0+D_1/\tau$. Relative to ${\delta }_C\propto {\tau }^2$, this offset vanishes: baryons catch up.

The phrase “only grow” needs this gravitational-growth interpretation. For a small baryon fraction the baryon equation is forced, not an autonomous Jeans-instability equation. At $k\gg k_J$, its slowly varying response isplus acoustic transients. It is suppressed rather than identically absent. For example, with constant sound speed, $\mathcal{H}=2/\tau$ and ${\delta }_C=A{\tau }^2$, the equation is ${\delta }_B^{\prime \prime }+2{\delta }_B^{\prime }/\tau +c_s^2{k}^2{\delta }_B=6A$. Its particular solution $6A/(c_s^2{k}^2)$ is constant, and its oscillatory homogeneous part decays as $1/\tau$. This demonstrates Jeans support versus forced baryon response; the scale above is not a theorem forbidding every forced or transient sub-Jeans change.

Before decoupling, tight photon coupling makes the effective sound speed large, about $1/\sqrt{3}$. The small-$c_s$ approximation just derived is not quantitative in that regime; the coupled photon-baryon fluid must be used. Its large pressure supports acoustic oscillations, and baryon clustering below the corresponding large Jeans/sound scale is inhibited while Cold dark matter wells grow. After decoupling the supplied sound speed is tiny and decreases further as the gas cools. The Jeans length drops sharply, so many previously supported modes fall into Cold dark matter wells and approach its growing density contrast. Modes still below the instantaneous pressure-support scale remain suppressed. The subsequent detailed thermal scaling depends on the gas's temperature evolution; the given sound-speed history suffices for this qualitative conclusion.

Let $p^0=d\tau /d\lambda$. A comoving observer measures photon energy $p=a{p}^0$, so $q=ap=a^2{p}^0$. The null condition is $({\delta }_{ij}+h_{ij})p^i{p}^j=(p^0{)}^2$. Using the stated Christoffel symbols in the time geodesic equation givesDivide by $p^0$ to differentiate with respect to conformal time. The background redshift term cancels when differentiating $a^2{p}^0$:In the perturbation term use the background relation $p^i/p^0={\hat{n}}^i$, givingWithout metric tensor perturbations, both $q$ and the propagation direction are constant along an unperturbed ray. Thus $q^{\prime }$ and $d{\hat{n}}^i/d\tau$ are first order. Multiplying them by derivatives of $f_1$, which is also first order, gives second-order terms. The term $q^{\prime }d{f}_0/dq$ is first order and must be retained. At linear order the Collisionless Boltzmann equation therefore readsTaking the Fourier transform in the spatial coordinates provesA first-order correction to the propagation velocity multiplying a spatial derivative of $f_1$ is likewise second order.

Define $\Delta t=t-t^{\prime }$, $\mathbf{r}=\mathbf{x}-{\mathbf{x}}^{\prime }$ in the three noncompact spatial directions, and $\Delta w=w-w^{\prime }$. The compact source must be the periodic Dirac deltaFor the wave operator as explicitly defined in the question, a retarded circle-compactified massless propagator isThe poles $\omega =\pm \sqrt{{\mathbf{k}}^2+k_n^2}-i0$ are both below the real axis, making the integral vanish for $\Delta t<0$. Performing just the frequency integral gives an equivalent expressionThe first time derivative of $\Theta (t)\sin (\Omega t)/\Omega$ jumps by one at zero, so applying ${\partial }_t^2+{\Omega }^2$ yields $\delta (t)$. Together with the completeness relations for the Fourier series and Fourier transform this verifies the specified source and normalization. The symbol for the wave operator follows the question's explicit sign convention; it is the negative of $g^{AB}{\nabla }_A{\nabla }_B$ for its mostly-plus metric tensor.

Take $u<0$ for inward radial velocity, and assume a constant specific-heat ratio $\gamma >1$. The isentropic flow has $p=K{\rho }^{\gamma }$ and adiabatic sound speed $c^2=\gamma K{\rho }^{\gamma -1}$. Steady spherical mass conservation and the radial Euler equation for fluid motion giveEliminating the mass density derivative gives $(u-c^2/u)u^{\prime }=2c^2/r-GM/r^2$, orDifferentiating $c^2\propto {\rho }^{\gamma -1}$ gives the corresponding sound speed equation:At a smooth sonic point, $u_s^2=c_s^2$. A finite derivative requires the numerator of the flow equation to vanish too. ThusThis regularity requirement concerns a smooth transonic branch, rather than an arbitrary singular solution reaching unit Mach number.

Integrating the momentum equation using the specific enthalpy $c^2/(\gamma -1)$ yields the Bernoulli equationFor a finite positive mass accretion rate, its large-radius solution hasThese expressions satisfy both the flow equations asymptotically and show $u\to 0$ and $\rho \to {\rho }_{\infty }$. In particular $u^{\prime }/u\sim -2/r$ agrees with the first boxed equation. Combining the Bernoulli equation with sonic regularity additionally gives $c_s^2=2c_{\infty }^2/(5-3\gamma )$, so a finite positive sonic point occurs in the range identified below.

For the small-radius inward free fall branch, neglecting enthalpy against $GM/r$ in the Bernoulli equation predicts $u^2\sim 2GM/r$. The continuity equation then fixes $\rho \propto r^{-3/2}$, and consequently $c^2\propto r^{-3(\gamma -1)/2}$. The neglected-to-leading ratio isIt tends to zero precisely whenThis makes the assumed supersonic flow and leading free fall self-consistent. At equality, the ratio is of constant order rather than vanishing; above it, the proposed asymptotic balance fails. The isothermal limit $\gamma =1$ also permits free fall, with logarithmic rather than power-law enthalpy.

For the stellar funnel, write the magnetic scalar potential as $\psi =\mu \cos \theta /r^2$, where $R=r\sin \theta$ and $z=r\cos \theta$. Its magnetic dipole field has $B_r=-2\mu \cos \theta /r^3$ and $B_{\theta }=-\mu \sin \theta /r^3$, so a magnetic-field-line equation isThe edge of the polar flux tube meets the stellar surface at ${\theta }_{\ast }\simeq a/R_{\ast }$. Thus, near the pole, ${\theta }^2\simeq {\theta }_{\ast }^{2}r/R_{\ast }$, and its transverse radius is $R\simeq r\theta$. This proves the dipolar flux-tube area lawEquivalently magnetic flux conservation gives $A|\mathbf{B}|=\mathrm{constant}$ and $|\mathbf{B}|\propto r^{-3}$. The stipulated small cross section keeps the tube close to the polar axis, where its direction is approximately radial; the approximation is not valid all the way to arbitrarily large $r$.

Outside the star the prescribed field is a gradient, hence $\nabla \times \mathbf{B}=0$ and $\mathbf{j}=0$. Its Lorentz force density is therefore $\mathbf{j}\times \mathbf{B}=0$. The magnetic-pressure and magnetic-tension parts cancel in this exact current-free dipole. Field-aligned velocity also has $\mathbf{u}\times \mathbf{B}=0$, consistent with a steady prescribed field. The following calculation uses the narrow polar funnel approximation.

Now steady mass conservation reads $\rho uA=\mathrm{constant}$. With $A\propto r^3$, replace the spherical factor 2 by 3:The Bernoulli equation is unchanged because the magnetic field supplies no work or force. On a gravity-dominated free fall branch, $u^2\sim 2GM/r$, but $\rho \propto r^{-5/2}$ and $c^2\propto r^{-5(\gamma -1)/2}$. HenceThe same bound follows from regular transonic accretion in a power-law tube: $u_s^2=c_s^2=GM/(3r_s)$ and the reservoir-matched Bernoulli equation gives $c_s^2=2c_{\infty }^2/(7-5\gamma )$. The stellar surface cuts off the mathematical small-$r$ limit; the criterion refers to a funnel that becomes gravity dominated and crosses its sonic point before reaching that surface.

Let $p_1(z)$ denote the Eulerian perturbation of a fluid variable that is the pressure amplitude, and keep $\rho$ constant. The linearized Euler equations and incompressible flow constraint areThe last equation gives $u=i{w}^{\prime }/k$. Substituting it into the first yieldsDifferentiating this formula and using the vertical momentum equation produces the Rayleigh equation for inviscid shear flow:For a growing normal mode, $c$ is nonreal, so $U-c$ never vanishes on the real $z$ axis. In each open region the base velocity is constant or linear and $U^{\prime \prime }=0$; therefore $w^{\prime \prime }-k^{2}w=0$. Decay away from the layer selects $e^{-kz}$ above and $e^{kz}$ below, while both exponentials are retained inside. Conveniently normalize these asThese are the spatially localized discrete modes relevant to instability, rather than all possible generalized neutral disturbances at a critical layer in a shear flow.

At each corner $U$ is continuous and $U^{\prime }$ jumps. The two sides have the same normal velocity and normal stress: there is no membrane, density jump or singular external force supplying a pressure jump. Thus $w$ and $p_1$ must be continuous. The pressure formula proves the required continuity of $(c-U)w^{\prime }+U^{\prime }w$. This is pressure matching at a piecewise-linear shear interface; it does not require $w^{\prime }$ to be continuous. Equivalently integrating the Rayleigh equation for inviscid shear flow through a corner gives $(U-c)[w^{\prime }]=[U^{\prime }]w$.

Write $\alpha =kd$, $y=c/U_0$, and $E=e^{2\alpha }$. Continuity of normal velocity givesAt $z=d$, the outside derivatives are $w_{+}^{\prime }=-kA$, $U_{+}^{\prime }=0$, while the inside derivatives are $w_{-}^{\prime }=-kB+kEC$, $U_{-}^{\prime }=U_0/d$. Hence pressure matching reduces toSimilarly at $z=-d$, the outside derivative is $kD$ and the inside derivative is $-kEB+kC$. Matching givesThe two equations can be written asA nonzero mode therefore requiresThis is the dispersion relation for the unbounded piecewise-linear shear layer.

With the phase convention $e^{ik(x-ct)}$, positive $\mathrm{Im}c$ gives exponential temporal growth. Since the displayed $c^2$ is real, a growing/decaying pair occurs precisely whenFor $0<\alpha \le 1/2$, $e^{-2\alpha }>1-2\alpha$ by the strict exponential tangent inequality, so every such mode is unstable. For $\alpha \ge 1/2$, define $f(\alpha )=2\alpha -1-e^{-2\alpha }$. Its derivative $2+2e^{-2\alpha }$ is positive, while $f(1/2)<0$ and $f(1)>0$. There is consequently exactly one root in that interval, andAt the cutoff the discrete roots merge at $c=0$; above it the roots are real and have no exponential growth. The physical wavelengths in the unstable band satisfy ${\lambda }_x=2\pi /k>2\pi d/{\alpha }_s\simeq 9.83d$.

This is a finite-thickness Kelvin-Helmholtz instability: long disturbances roll up the layer and draw kinetic energy from its shear. The off-diagonal $e^{-2kd}$ terms describe coupling of the two vorticity discontinuities. Short disturbances decay across the layer before strongly coupling its edges and are neutral. Indeed at $kd\to \infty$ the two phase speeds approach the velocities $\pm U_0$ of the separated edges. At $kd\to 0$, expansion of the numerator gives $c^2/U_0^2\to -1$, recovering the vortex sheet result $c=\pm i{U}_0$; the growth rate then behaves as $k|U_0|$.