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On a length scale $\ell$, pressure communicates support over a sound-crossing time $\ell /v_s$, whereas self-gravity acts over a time of order $(G{\rho }_0{)}^{-1/2}$. When the former is longer, gravity can amplify an overdensity before pressure smooths it. Thus Jeans instability occurs at scales of order $\ell \gtrsim v_s/\sqrt{G{\rho }_0}$; the precise numerical factor depends on whether scale means wavelength or inverse wavenumber.

Take a uniform static background, small perturbations, an inviscid barotropic fluid, and $\delta p=v_s^2\delta \rho$ with $v_s^2=(dp/d\rho {)}_0>0$. An infinite Newtonian homogeneous density cannot itself have zero background gravitational acceleration while satisfying the unmodified Poisson equation for Newtonian gravity. Specify the Jeans swindle: subtract the homogeneous field, or regard an external prescription as balancing it, and apply Poisson's equation only to perturbations. The linearized continuity equation and Euler equations for an inviscid fluid areTaking a time derivative of the first equation and substituting the second yields ${\partial }_t^2\delta \rho =v_s^2{\nabla }^2\delta \rho +4\pi G{\rho }_0\delta \rho$. A plane wave $e^{i(\mathbf{k}\cdot \mathbf{x}-\omega t)}$ therefore givesFor $0<k<k_J$ there is exponential growth at rate $\sqrt{4\pi G{\rho }_0-v_s^2{k}^2}$. At $k_J$ the density mode is neutral to this linear order, and above it there are acoustic oscillations. The critical wavelength is $2\pi /k_J=\sqrt{\pi }{v}_s/\sqrt{G{\rho }_0}$. Transverse velocity perturbations decouple as zero-frequency vortical modes in this idealization.

For the attractive Yukawa potential, assume a real screening parameter $\alpha \ge 0$. Its Green function obeys $({\nabla }^2-{\alpha }^2)(-Gm{e}^{-\alpha r}/r)=4\pi Gm{\delta }^{(3)}(\mathbf{r})$, including the delta source. Thus $\delta {\varphi }_{\mathbf{k}}=-4\pi G\delta {\rho }_{\mathbf{k}}/(k^2+{\alpha }^2)$. The same fluid elimination gives Jeans instability with Yukawa gravity:There exists a growing nonzero-wavenumber mode precisely whenEquality gives no exponentially growing nonzero mode. If screening exceeds this threshold, the finite-range gravitational response is too weak to overcome pressure at any wavelength. Otherwise all sufficiently long but nonconstant waves remain unstable, with a longer minimum unstable wavelength than in Newtonian gravity. For $\alpha >0$ the long-wave gravitational contribution also scales as $k^2$, so the growth rate tends to zero as $k\to 0$; screening does not produce a second positive lower cutoff in wavenumber. A homogeneous screened background can have the constant gravitational potential $-4\pi G{\rho }_0/{\alpha }^2$, with zero force, so no Newtonian background subtraction is needed for nonzero screening. The limits $k\to 0$ and $\alpha \to 0$ are nonuniform. An exact $k=0$ density change is not one of the growing finite-wavelength modes. In the pressureless edge case $v_s=0$, every finite $k>0$ is unstable for positive density and finite screening.

Use $G=c=1$ and the physical real part of the printed plane wave. Its null wavevector obeys $k^{\alpha }{k}_{\alpha }=0$, so each component satisfies $\square h_{\alpha \beta }=0$. Direct differentiation of the defined stress tensor givesThe trace is proportional to $k^{\nu }{k}_{\nu }$ for a single plane wave and hence vanishes. ThereforeThis is local conservation of energy and momentum conservation; spatial integration gives conserved total charges when the boundary flux vanishes. For a finite-amplitude complex phasor, one takes the real field before quadratic products and may time-average: $⟨{\tau }_{\mu \nu }⟩=k_{\mu }{k}_{\nu }{H}^{\alpha \beta }{\overline{H}}_{\alpha \beta }/(64\pi )$ in transverse-traceless gauge. Its energy density is nonnegative and its flux travels in the positive $z$ direction.

A collective coordinate describes a position, orientation, or another parameter of a family of static solitons. For a family $\Phi (\mathbf{x};q)$, promote $q$ to a slowly varying $q(t)$ and substitute into the field action. In a scalar theory with unit kinetic coefficient, this gives the collective-coordinate effective LagrangianFor a gauge-theory soliton, one also solves the Gauss law constraint in gauge theory constraint and projects out pure gauge transformations; arbitrary variations of gauge representatives do not define the physical metric. Tangent vectors to an exactly equal-energy family are zero modes in field theory. If $V=E_0$ is constant, the Euler-Lagrange equations of this moduli-space approximation arethe geodesic equations of its Riemannian metric. The approximation neglects radiation and deformation modes; it describes motion sufficiently slow that these omitted degrees of freedom remain unexcited to the required accuracy.

In collective-coordinate quantization, take the wavefunction measure $\sqrt{\det g}{d}^{n}q$ and the minimal scalar Hamiltonian operatorThe Laplace-Beltrami operator supplies coordinate-invariant kinetic energy. Global identifications and statistics must be imposed on the wavefunctions; the classical metric alone does not choose them. Curvature-ordering terms and loop corrections are additional quantum input.

For the phi-four kink, use the normalization and profile of Question 1. Substituting $\varphi (x,t)={\varphi }_K(x-X(t))$ givesThe metric is constant because of translation invariance. This proves the translational dynamics of a phi-four kink: classically the centre moves at constant velocity, and quantum mechanicallyPlane waves label the continuous translational momentum, with no position-dependent potential. Uniform-motion Lorentz invariance upgrades the dispersion to $\sqrt{M^2+p^2}$; the displayed collective Lagrangian is its small-velocity expansion. Small perturbations also include an internal shape mode and continuum radiation, which this single collective coordinate omits. The fluctuation operator of a phi-four kink in this normalization isthe translational eigenfunction has ${\omega }^2=0$, the shape mode has ${\omega }^2=3c^2$, and continuum modes in the spectrum start at ${\omega }^2=4c^2$. Thus the free-coordinate states describe the kink's low translational energies, not its full excitation spectrum or quantum mass correction.

For two Abelian Higgs vortices at critical coupling, the static energy is $2M_v$ and the Abelian Higgs vortex moduli space has four real dimensions. Let $z_1,z_2$ be their positions, $Z=(z_1+z_2)/2$, and $q=z_1-z_2$. The centre of mass decouples; the relative metric is rotationally symmetric and can be writtenAt large separation, $F(\rho )\to M_v/2$, recovering two free particles. Although there is no static separation potential, the nonconstant metric produces velocity-dependent interaction. Coincidence is smooth in the relative coordinate for two identical vortices $w=q^2$, not in the double-valued $q$. Smoothness gives $F(\rho )\sim C{\rho }^2$ for $C>0$ near zero. A head-on geodesic continues through $w=0$ to the opposite real ray, so $q$ changes its line by $\pi /2$: the vortices scatter through a right angle. This geometric argument does not require an explicit formula for $F$.

With ordinary bosonic exchange statistics, relative wavefunctions are single-valued in $w$ and smooth at coincidence. In the separated polar coordinate they obey $\Psi (\rho ,\theta +\pi )=\Psi (\rho ,\theta )$, with even integer angular labels. Their kinetic operator isThe apparent singularity at $\rho =0$ must be resolved with the smooth $w$ coordinate and regularity, rather than arbitrary boundary conditions on a punctured cone. The free centre-of-mass motion and the asymptotically free relative geometry give quantum scattering states; a flat static energy does not imply that the metric is flat or that scattering is absent. This is not a prediction of a discrete family of static two-vortex bound separations. The smooth collision geometry is developed in David Tong's arxiv.org/abs/hep-th/0509216.

For a Skyrmion of baryon number one, the Skyrmion hedgehog ansatz iswhere $\mathbf{\sigma }$ are the Pauli matrices. Include a centre $\mathbf{X}$ and an orientation $A\in SU(2)$ through $U=A{U}_0(\mathbf{x}-\mathbf{X})A^{-1}$. Hedgehog symmetry identifies spatial rotations with opposite internal rotations, so there are three independent orientation coordinates, not six. Since $A$ and $-A$ give the same field, the physical orientation space is $SO(3)$, with SU(2) group as its double cover. Write $A^{-1}\dot{A}=i\mathbf{\omega }\cdot \mathbf{\sigma }/2$. The leading collective Lagrangian has the formwhere $\Lambda >0$ is the rotational moment of inertia obtained by integrating the profile's field kinetic energy.

For the fermionic quantization appropriate to baryons, the Finkelstein-Rubinstein constraints on the double cover impose $\Psi (-A)=-\Psi (A)$. In SU(2) representations, the central element $-1$ acts by $(-1{)}^{2j}$, so $j$ must be half-integer. Left and right group actions supply isospin and spin angular momentum; hedgehog symmetry makes their magnitudes equal. The rotational quantization of a unit Skyrmion therefore givesThe $j=1/2$ level has four spin-isospin states and models the nucleon doublet, proton and neutron, each with two spin states. The $j=3/2$ level has sixteen states and models the Delta baryon quartet, each with four spin states. The rotor predicts a splitting $3{\hslash }^2/(2\Lambda )$. Without the fermionic sign, single-valued functions on $SO(3)$ would instead allow integer $j$, which is a different quantization. High rotor levels can couple to deformation and pion radiation; this semiclassical approximation does not establish that its entire formal tower consists of stable particles.

For the Fourier transform convention of equation 3, insert the given power spectrum as $S_F$. With $t=\nu L/2$,The Fourier transform of a Gaussian, or the order-zero Hankel transform, also givesThus the coherent attenuation in a white-noise random medium isFor a unit incident plane wave, the Fresnel propagator leaves the initial envelope unchanged, so the answer reduces to $\boxed{m=e^{-k_0^2{\mu }^{2}Lx/(2\pi )}}$. If instead the given numerical function is interpreted as the self-reciprocal Hankel transform spectrum suggested by the printed equations 4 and 5, then $Q(r)=2{\mu }^{2}L{e}^{-r^2/L^2}$ and the alternative isThe factor $2\pi$ ambiguity is present in the original PDF. No unique numerical attenuation coefficient follows until the power spectrum convention is fixed. The covariance strength $Q(0)$ has units of length in this longitudinal white noise model, making both attenuation exponents dimensionless.

Use the time-harmonic wave convention $\psi e^{-i\omega t}$ and write $\mathbf{z}=(y,z)$, $E(0,\mathbf{z})=E_0(\mathbf{z})$, with a deterministic incident wave envelope. The printed speed ratio is inconsistent with $k=k_{0}n$: the refractive index used below is $n=c_0/c$. Substituting $\psi =e^{i{k}_{0}x}E$ in the Helmholtz equation givesThe paraxial approximation discards ${\partial }_x^{2}E$. The usual weak-fluctuation model also linearizes $n^2-1=2\mu W+O({\mu }^2{W}^2)$, givingDropping the quadratic contrast is an additional weak-fluctuation assumption, not a consequence of small propagation angles. Although the linearized random potential generates attenuation of order ${\mu }^2$, it does not retain every effect of that order in the literal finite-correlation index: the discarded quadratic contrast can also produce a mean wave phase shift. This linearization must precede a Gaussian white noise limit: the square of ideal white noise has no ordinary pointwise meaning.

The split-step Fourier method alternates free-space diffraction, $U_h=e^{hD}$, with a random phase screen,For jointly Gaussian random fields, Gaussian phase averaging gives $\mathbb{E}{e}^{i{\varphi }_h}=e^{-\mathrm{Var}({\varphi }_h)/2}$. More generally, a product of $p$ fields and $q$ conjugate fields picks up $\exp (i{\sum }_a{ϵ}_a{\varphi }_h({\mathbf{z}}_a))$, with signs ${ϵ}_a=+1$ or $-1$. Its screen average is determined entirely by the wave phase covariance matrix. The deterministic diffraction step acts on each coordinate, with opposite signs on factors formed by complex conjugation. This is the basis of the field-moment equations.

There is an important closure qualification. A stationary Gaussian random field need not have independent longitudinal increments. For finite-correlation fluctuations, the unlinearized parabolic wave equation instead givesIn the weak-fluctuation model, the exact first-moment equation isThe last term cannot in general be replaced by a constant times $m$. An exact generic solution of the linearized model iswhere $\mathcal{T}$ denotes time ordering of the propagation operators. For the unlinearized model, add $i{k}_0{\mu }^2{W}^2/2$ inside the propagation generator. The covariance function of the medium is needed to evaluate this expression; One-point Gaussian distributions alone would not even determine the joint wave phase statistics.

For example, omit diffraction and take a longitudinal autocovariance function $R(s)=\mathbb{E}[W(s,\mathbf{z})W(0,\mathbf{z})]$. Direct Gaussian phase averaging givesEven here, unit-variance stationary Gaussian random fields with different $R$ give different answers. In the general problem the diffraction and multiplication operators do not commute, so this scalar attenuation cannot simply be multiplied by $U_x$ without an additional approximation.

The standard closed answer uses the Markov approximation for a random medium, made explicit in part ii. Let the longitudinally integrated covariance kernel beReplace the medium by longitudinal Gaussian white noise with this strength. A screen of thickness $h$ then has $\mathbb{E}[{\varphi }_h({\mathbf{z}}_a){\varphi }_h({\mathbf{z}}_b)]=k_0^2{\mu }^{2}hB({\mathbf{z}}_a-{\mathbf{z}}_b)$ and is independent of the incoming field. For the moment $M=\mathbb{E}{\prod }_a{E}_a^{({ϵ}_a)}$, expanding both steps to order $h$ givesIn particular, the coherent attenuation in a white-noise random medium and its Fresnel propagator solution areEquivalently, write $Wdx=d{\beta }_x$, where ${\beta }_x(\mathbf{z})$ is Brownian motion in $x$ with transverse covariance kernel $\mathbb{E}[d{\beta }_x(\mathbf{z})d{\beta }_x({\mathbf{z}}^{\prime })]=B(\mathbf{z}-{\mathbf{z}}^{\prime })dx$. The Stratonovich integral formulation is $dE=DEdx+i{k}_0\mu E\circ d{\beta }_x$. Its Itô integral form isThe mean of the Itô integral vanishes, independently confirming the attenuation drift. For $x>0$ in two transverse dimensions,Thus a unit plane wave has $m=e^{-\gamma x}$. For a general incident wave envelope, the Fresnel propagator supplies its spreading. The attenuation is redistribution between the coherent and diffuse wave fields, rather than wave absorption: for a field and its conjugate at the same point, the screen contribution in the second-moment equation cancels.

For a wavefunction $\Psi (x,t)$, the probability density and probability current areMultiply the Schrodinger equation by ${\Psi }^{\ast }$, multiply its complex conjugate by $\Psi$, and subtract. The real potential energy terms cancel, leavingwhich is the probability continuity equation ${\rho }_t+j_x=0$.

For a stationary state, ${\rho }_t=0$, so $j_x=0$ and $j$ is constant in space. A normalizable wavefunction has vanishing current at spatial infinity, forcing this constant to be zero. By contrast the nonnormalizable plane wavehas stationary density $|A{|}^2$ and nonzero currentwhen $k\ne 0$.

For unit plane wave incidence ${\psi }_i=e^{i{k}_0{\hat{\mathbf{r}}}_0\cdot \mathbf{r}}$, define the scattering wavevector difference $\mathbf{q}=k_0({\hat{\mathbf{r}}}_0-\hat{\mathbf{r}})$ andThe outgoing Green function for the three-dimensional Helmholtz equation givesThe Born far-field pattern is $A$, a Fourier transform sample of the potential. Expanding the far-field Rytov exponential in $1/r$ gives the same leading outgoing term, while the logarithmic representation preserves its accumulated phase and amplitude correction.

For a progressive plane wave, writeThe momentum equation gives ${\rho }_0\omega U=kP$, while the dispersion relation is $\omega =c_{0}k$. Therefore the pressure and longitudinal velocity amplitudes obeyThe constant $Z$ is the acoustic impedance.

Substitute the harmonic plane waveinto the elastic wave equation. The vector identitygivesIf $\mathbf{A}$ is parallel to $\mathbf{K}$, this becomesand gives a longitudinal P-wave. If $\mathbf{A}\cdot \mathbf{K}=0$, it becomesand gives either polarization of an S-wave.

For $\mathbf{K}=(k,0,m)$, convenient polarization vectors areThe SV-wave lies in the $xz$ propagation plane, whereas the SH-wave is perpendicular to that plane.

For a real short-range spherically symmetric potential, the stationary wavefunction must solve the Schrodinger equation, be regular at the origin whenever the potential is finite there, and satisfy the scattering boundary conditionHere $e^{ikz}$ is the incident plane wave, the second term is an outgoing spherical wave, and its coefficient $f(\theta )$ is the scattering amplitude.

Comparing the stated asymptotic expression with the partial-wave expansion of a scattering amplitude giveswhere $P_l$ is a Legendre polynomial. The outgoing-to-incoming coefficient in channel $l$ isPartial-wave unitarity requiresor equivalentlyThe real scattering phase shift ${\delta }_l$ is defined modulo $\pi$ by

For a one-dimensional wavefunction,The probability continuity equation isFor a stationary state $\psi (x,t)=\varphi (x)e^{-iEt/\hslash }$, the probability density is independent of time, so ${\partial }_{x}j=0$ and the probability current is constant in space.

For the plane waveone obtainsThis is a momentum eigenstate with momentum $p=\hslash k$ and, when it satisfies the free Schrodinger equation, energy $E={\hslash }^2{k}^2/(2m)$. Its constant density and current describe a spatially uniform beam carrying probability in the sign of $k$. It is not a normalizable wavefunction, so it represents an idealized state rather than a localized particle.

The basic state has velocity fieldand buoyancy $B=N^{2}z$. For disturbances independent of $y$, the linearized equations areSubstituting a plane wave proportional to $\exp [i(kx+mz-\omega t)]$ and eliminating $p$, $v$, and $b$ gives the dispersion relationThus the requested coefficients are

An instability exists precisely when some wavenumber pair makes ${\omega }^2<0$. Since the Brunt–Väisälä frequency satisfies $N^2>0$, this is possible exactly whenThe basic relative vorticity is $\Lambda {\mathbf{e}}_z$, so its absolute vorticity is $(f+\Lambda ){\mathbf{e}}_z$. Its Ertel potential vorticity is thereforeThe instability criterion can consequently be written as $fQ<0$: the vertical absolute vorticity has the opposite sign to the planetary vorticity. This is inertial instability, approached most directly by disturbances with $|m/k|\gg 1$.

Let the group velocity make an angle $\theta$ above the horizontal. A convenient right-handed choice of rotated unit vectors isThe first vector is parallel to the wavevector, the second is parallel to the group velocity, and their perpendicularity is the defining geometry of an internal-wave phase and group velocity. ThusWriting ${\Delta }_{\xi \zeta }={\partial }_{\xi }^2+{\partial }_{\zeta }^2$, the equation becomesFor an inviscid plane wave proportional to $e^{i(k\xi -\omega t)}$, its dispersion relation is $\omega =N\sin \theta$.

The Rytov approximation at first order writes the total profile as $f_R=f_i{e}^{{\chi }_1}$ and identifies the first logarithmic perturbation with the Born relative field:Hence, in the upper half-space,Expanding the exponential function to first order shows that its reflected component isThe exponentiated expression is the Rytov approximation to the total field; only its term linear in $\alpha$ is a single specular reflected plane wave.

Putand let $q_1=nk\cos {\theta }_T$ be the downward vertical wavenumber in the lower half-space. The given form of Snell's law saysWith the $e^{-i\omega t}$ convention, write the incident, reflected, and transmitted plane waves asThe two interface conditions at $z=0$ giveSolving this linear system gives the reflection and transmission coefficients at a scalar-wave interfaceThus the exact fields are

For the plane wave $A^{\mu }(x)={ϵ}^{\mu }(k)e^{-ik\cdot x}$, the equations from part a becomeIn the rest frame $k^{\mu }=(m,0)$, transversality sets ${ϵ}^0=0$ while the three spatial components remain independent. Hence a Proca field has three polarization states, as required for a massive spin-one particle. A covariant orthonormal choice obeys

Write the density as a hydrostatic reference profile $\hat{\rho }(z)$ plus a small perturbation ${\rho }^{\prime }$, and define buoyancy and buoyancy frequency byThe Boussinesq approximation to the Navier-Stokes equation, together with mass conservation, iswhere $D/Dt$ is the material derivative, $\nu$ is kinematic viscosity, and $\kappa$ is mass diffusivity. Linearizing about rest and eliminating pressure and buoyancy givesFor a plane wave proportional to $e^{i(kx+mz-\omega t)}$, with $K^2=k^2+m^2$, the viscous-diffusive dispersion relation isIn the inviscid limit this becomesFor weak diffusion the two oscillatory roots areso a freely evolving Fourier mode decays.

A single plane wave is also an exact solution of the nonlinear equations. Every field depends only on its phase $\Theta =\mathbf{k}\cdot \mathbf{x}-\omega t$, while incompressible flow gives $\mathbf{u}\cdot \mathbf{k}=0$. Therefore $\mathbf{u}\cdot \nabla$ annihilates both $\mathbf{u}(\Theta )$ and $b(\Theta )$, and all nonlinear advection terms vanish.

For a boundary-forced wave with real $\omega$, nonzero $\nu$ or $\kappa$ instead makes the bulk vertical wavenumber complex, attenuating the propagating beam. Because diffusion raises the spatial order of the equations, additional short vertical-wavenumber roots form viscous and scalar boundary layers; they allow a no-slip velocity condition and a scalar no-flux condition to accompany impermeability. These layers and bulk attenuation become essential near critical internal-wave reflection, where the inviscid reflected wavelength collapses.

Two sinusoidal signals are in phase quadrature when their phase difference is a quarter cycle, for example $\cos \varphi$ and $\sin \varphi$. Their time averages have zero correlation over a cycle. In a plane wave, this separates fields such as displacement and its time derivative, velocity.

Phase velocity is the velocity of a surface of constant phase. For a one-dimensional monochromatic wave it is $c_p=\omega /k$.