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Work away from $x=0$, where the inverse-square potential is singular. Variation of the phase-space action givesThe last equation is the constraint imposed by the Lagrange multiplier $e$. The canonical momentum conjugate to $t$ is $-E$, not $E$. Consequently the nonzero canonical Poisson brackets areIn particular,With the convention $\delta z=\alpha \{z,Q\}$, the Noether charge generating $\delta t=\alpha$ is $Q=-E$. Equivalently, the conserved physical energy is $E$; a convention that calls the energy the time-translation charge absorbs this minus sign into its parameter.

For the dilation charge,on the constraint surface. Its infinitesimal canonical transformations areFor constant $\alpha$, extending this by ${\delta }_{D}e=-\alpha e$ preserves the action: the kinetic terms are invariant and $E-H$ scales oppositely to $e$. Thus the dilation acts on time and position with their nonrelativistic relative scaling. Its Poisson bracket with the energy is

Put $C=E-H$. Direct variation of the Hamiltonian givessoThe supplied transformations are generated by the special conformal charge of inverse-square mechanicsChooseThen $\delta (eC)=0$ even for a time-dependent parameter. Expanding the kinetic-term variation, keeping the terms in $\dot{\beta }$, givesHenceFor a constant parameter only the boundary term remains, identifying $K$ as the Noether charge. Independently, the equations of motion giveUsing the same canonical Poisson bracket convention,Together, $\{D,E\}=-E$, $\{D,K\}=K$ and $\{E,K\}=2D$ form the sl2R Lie algebra of conformal mechanics. These are time translation, dilation and special conformal transformation. A useful normalization check is the Casimir elementon the constraint surface.

For the Monge-gauge Hamiltonian of a three-dimensional string, in Monge gauge, $X^{\prime m}=(0,1,{\Phi }^{\prime })$ and $P_m=(P_0,P_1,\Pi )$. The momentum constraint gives $P_1=-{\Phi }^{\prime }\Pi$. The other Virasoro constraint then readsChoosing the positive-energy branch,Substituting into the phase-space action leaves $I=\int dtd\sigma (\Pi \dot{\Phi }-\mathcal{H})$, soFor an infinite string this is an energy-density expression; even the undeformed string has infinite total energy. One may regulate the length or subtract its constant reference energy when a finite total energy is needed. The positive branch is unambiguous at the density level.

The canonical Hamiltonian field equations areConsequently $\Pi ={\Phi }^{\prime }$ impliesThe second equation also becomes $\dot{\Pi }={\Phi }^{\prime \prime }$, consistent with differentiating the first. These are one-direction traveling profiles along the string.

For the linear profile, the spatial string is the straight line $X^2=k{X}^1+kt$. It is tilted, and its coordinate intercept moves with speed $k$. This is also a uniformly moving straight string: the physical transverse velocity of a tilted string is the velocity normal to the line. With unit tangent $(1,k)/\sqrt{1+k^2}$ and coordinate velocity $(0,k)$, the normal speed isThus $k>1$ does not imply superluminal physical motion. Fixed-$\sigma$ labels have spacelike trajectories when $k>1$, but labels on a string can slide tangentially and are not identifiable material particles. Choosing $\dot{\sigma }=-k^2/(1+k^2)$ removes the tangential velocity and gives the subluminal velocity $(-k^2,k)/(1+k^2)$.

The induced worldsheet metric supplies a coordinate-invariant check:The string worldsheet remains Lorentzian for every finite $k$, including $k=1$ and $k>1$. Its constant energy density is $1+k^2$. This linear profile is a tilted boosted straight string, rather than a localized finite-energy wave.

In a point-particle description of quantum gravity, interactions of a massless spin-two field occur at localized vertices of Feynman diagrams. The gravitational coupling has negative mass dimension for spacetime dimension greater than two, so short-distance loop integrations generate higher-derivative counterterms that cannot all be absorbed into the Einstein action. A string worldsheet instead joins incoming and outgoing strings by a smooth surface. In particular, a closed string splits or joins through a smooth pair-of-pants surface. There is no invariantly distinguished point at which every part of an extended string interacts. The choice of a time slicing can draw a joining point, but it is not a physical pointlike vertex. This removes the point-interaction geometry responsible for arbitrarily localized interactions and introduces the string length ${\ell }_s=\sqrt{{\alpha }^{\prime }}$.

This geometric explanation is supported by the organization of the Polyakov path integral. Introducing a worldsheet metric makes the Nambu–Goto action amenable to gauge fixing. In the Euclidean integral, metrics on a fixed oriented surface are divided by worldsheet diffeomorphisms and Weyl transformations. What remains is a Riemann surface with a conformal structure, together with finitely many worldsheet moduli and the positions of external string vertex operators. The determinants from fixing the metric are represented by Faddeev-Popov ghosts. Schematically, for $n$ closed-string external states,Here ${\mathcal{M}}_{g,n}$ is the moduli space of genus-$g$ punctured surfaces. The dilaton weights a surface by its Euler characteristic $\chi =2-2g$, giving $g_s^{-\chi }$ before the normalization of external states. Every additional handle multiplies the contribution by $g_s^2$. The string genus expansion therefore plays the role of the ordinary loop expansion, with the sphere giving tree level and the torus giving one loop. The critical dimension cancels the worldsheet Weyl anomaly, making this reduction to conformal geometry consistent.

The massless symmetric traceless closed-string state is a graviton, and its long-wavelength interactions reproduce gravity. At distances comparable with ${\ell }_s$, however, an infinite tower of excited states contributes. This is more than a single particle form factor: the same amplitude contains graviton exchange, massive higher-spin exchange and string-scale softness.

The dimensionless variables in the given Virasoro–Shapiro amplitude correspond to $s={\alpha }^{\prime }S/4$, $t={\alpha }^{\prime }T/4$, $u={\alpha }^{\prime }U/4$, where capital letters denote the usual dimensionful Mandelstam invariants. Its constraint $s+t+u=-4$ identifies the external particles as the closed bosonic tachyons, not four gravitons. Nevertheless its intermediate massless pole directly exhibits graviton exchange. The overall normalization will be suppressed below; no additional copy of the amplitude formula is needed.

First, the amplitude is symmetric under permutations of $s,t,u$, expressing crossing symmetry. The poles in any channel areFor generic $t$, the residue can be calculated with the Gamma function pole at $-n$ and its recurrence:The product is $1$ when $n=0$. Indeed, the pole factor contributes $(-1{)}^{n+1}/n!$, while the two remaining Gamma ratios contribute $(-1{)}^n(t+2{)}_n^2/n!$. A residue polynomial of degree $2n$ shows exchange of spins up to $2n$, with leading Regge trajectory $J=2+2s$. At $n=1$, the massless residue is $-(t+2{)}^2$, containing the spin-two exchange. By crossing, the massless $t$-channel pole isfor generic $s$. Thus at large $s$ the pole behaves as $s^2/t$, the characteristic energy dependence of graviton exchange. Its massless propagator produces the long-range gravitational interaction. At low energies the massive poles can instead be expanded into local higher-derivative corrections to the effective gravitational action.

At the string scale one must retain the whole tower. The Gamma reflection formula rewrites the amplitude exactly asAt large $s$ and fixed $t$, away from the resonance poles or with a specified complex continuation, the Gamma ratio scales as $s^{t+1}$. Hence the Regge limit scales as $s^{2+2t}$, up to the signature factor and the displayed $t$-dependent coefficient. In particular, a fixed nonzero momentum transfer changes the point-graviton power law by a factor $s^{2t}$: the graviton belongs to an entire trajectory rather than remaining an isolated spin-two exchange at arbitrarily high energy.

There is stronger suppression when the scattering angle is fixed. Put $t\simeq -as$, $u\simeq -(1-a)s$ with $0<a<1$. Applying Stirling formula to the Gamma functions, after the same treatment of real-axis poles, gives the leading smooth envelopeThe leading $s\log s$ terms cancel and the remaining entropy-like combination is negative. In physical units the exponent is $-{\alpha }^{\prime }Sf(a)/2$. This fixed-angle softness of the closed-string amplitude is absent for an elementary pointlike gravitational vertex. Equivalently, short-distance gravitational scattering probes additional string degrees of freedom. In a four-dimensional long-distance effective setting, a massive exchanged mode contributes a Yukawa-type correction proportional to $e^{-M_{n}r}/r$, with couplings and tensor factors depending on the external states. Such terms are suppressed for $r\gg {\ell }_s$ and become relevant for $r\sim {\ell }_s$. The amplitude establishes the spectrum and short-distance change; it does not by itself give a universal static potential for arbitrary sources. Its bosonic tachyon also signals an unstable vacuum, so it must not be treated as a complete stable model of a gravitational force.

Finally, the modular group prevents overcounting equivalent surfaces and reorganizes the loop ultraviolet region. A torus is $\mathbb{C}/(\mathbb{Z}+\tau \mathbb{Z})$ with $\mathrm{Im}\tau >0$. A change of lattice basis identifiesThus the one-loop integral uses a fundamental domain for $\mathrm{PSL}(2,\mathbb{Z})$, for exampleThe measure $d^2\tau /(\mathrm{Im}\tau {)}^2$ is invariant, and the physical integrand, including matter, ghosts and external insertions, must respect torus modular invariance. In this domain $\mathrm{Im}\tau \ge \sqrt{3}/2$. The region of arbitrarily small proper time that causes ultraviolet divergences in a particle loop is therefore not an independent part of the string integral; a modular transformation relates it to another description already represented in the domain.

This does not remove every possible divergence. The cusp $\mathrm{Im}\tau \to \infty$ is a long tube, and factorization through it can produce infrared divergences, massless tadpoles or the bosonic tachyon divergence. At higher genus, degenerating cycles similarly factorize into propagating intermediate string states. The essential improvement is the relation between worldsheet geometry, the full spectrum and the quotient by large diffeomorphisms; infrared and unstable-background problems remain separate questions. A short-distance expansion that keeps only the graviton discards precisely this structure.

The PDF's displayed zero-mode term has no derivative. Read literally, $d_0\cdot d_0$ vanishes for classical Grassmann variables and supplies no zero-mode symplectic structure. The subsequent canonical-algebra requests therefore require the standard kinetic term $\frac{i}{2}{d}_0\cdot {\dot{d}}_0$. We use that intended correction explicitly; the rest of the displayed action fixes the nonzero-mode normalization.

The Ramond level operator, with vacuum-annihilating normal ordering, isThe canonical oscillator relations, for transverse indices $I,J$, areThe hermiticity convention is ${\alpha }_k^{†}={\alpha }_{-k}$ and $d_k^{†}=d_{-k}$. The commuting bosonic zero mode is supplied by the center-of-mass momentum. For $k>0$, define $a_k^I={\alpha }_k^I/\sqrt{k}$ and $f_k^I=d_k^I$. Their bosonic occupation numbers are $0,1,2,\dots$ and their fermionic occupation numbers are $0,1$. Therefore, on the Fock space generated from an oscillator vacuum,Equivalently, creation operators raise the level by $k$, since $[N_R,{\alpha }_{-k}^I]=k{\alpha }_{-k}^I$ and $[N_R,d_{-k}^I]=k{d}_{-k}^I$. The zero modes commute with $N_R$ and do not change the level. The multiplier imposesso the $N=0$ states are massless. In the Ramond sector the bosonic and fermionic oscillator zero-point contributions cancel, consistently with the stated zero intercept. The massless ground states are spacetime spinors, as the Ramond zero-mode Clifford algebra now shows.

Normalize ${\gamma }^I=\sqrt{2}{\hat{d}}_0^I$. ThenLet $v=|0⟩\ne 0$. Each positive-frequency bosonic annihilator commutes with ${\gamma }^I$, while each fermionic annihilator anticommutes with it. Applying either annihilator to ${\gamma }^{I}v$ therefore gives zero. Thus all eight ${\gamma }^{I}v$ are oscillator vacua. For real $a_I$, the hermitian operator $A={\sum }_I{a}_I{\gamma }^I$ satisfies $A^2=({\sum }_I{a}_I^2)1$, soThis proves the real independence of Clifford-generated vectors, and hence their linear independence over $\mathbb{R}$.

The same argument applies to the nonzero vacuum ${\gamma }^{1}v$, because $({\gamma }^1{)}^2=1$. It gives eight real-linearly independent oscillator vacua ${\gamma }^I{\gamma }^{1}v$. They include $v$ itself, at $I=1$. Products of two zero modes preserve vacuum annihilation just as products of one do.

For the chirality matrix ${\gamma }_9={\gamma }^1{\gamma }^2\cdots {\gamma }^8$, reversing eight anticommuting factors introduces $(-1{)}^{8\cdot 7/2}=1$. HenceMoving any ${\gamma }^I$ through the other seven factors also gives ${\gamma }_9{\gamma }^I=-{\gamma }^I{\gamma }_9$. If ${\gamma }_{9}v=v$, thenThe first collection has negative chirality; the second has positive chirality. If $u_{+}$ and $u_{-}$ have these respective chiralities, hermiticity gives $⟨u_{+},u_{-}⟩=⟨{\gamma }_9{u}_{+},u_{-}⟩=⟨u_{+},{\gamma }_9{u}_{-}⟩=-⟨u_{+},u_{-}⟩$, so they are orthogonal. Combining the two real-independent collections therefore gives at least sixteen real-linearly independent oscillator vacua, eight in each chirality.

The real qualification in the question matters: the particular eight vectors generated from an arbitrary complex $v$ need not be independent over $\mathbb{C}$. Nevertheless the dimension bound from paired Clifford involutions also follows from the full Clifford algebra. Define four commuting hermitian involutions $J_j=i{\gamma }^{2j-1}{\gamma }^{2j}$, $j=1,\dots ,4$. Their joint spectral projections preserve the vacuum space, so it contains a nonzero common eigenvector $w$. Multiplication by ${\gamma }^{2j-1}$ flips the eigenvalue of $J_j$ and leaves the other three eigenvalues unchanged. The sixteen products obtained by independently choosing whether to apply these four odd-indexed Gamma matrices to $w$ consequently have distinct joint eigenvalue quadruples. They are nonzero, mutually orthogonal oscillator vacua. Thus the unprojected vacuum space also has complex dimension at least sixteen, with eight states of each chirality in the minimal representation. A further chiral projection is an additional physical restriction, not part of the oscillator-vacuum conditions here.

The first-order fermionic kinetic term produces second-class momentum constraints. After eliminating them, the canonical graded Dirac brackets arewith other mixed brackets zero. The bracket of two odd variables is symmetric, consistently with the graded Poisson bracket. The constraint algebra isIt is the worldline supersymmetry algebra: the square of the fermionic constraint generates the Hamiltonian constraint.

With $\hslash =1$, quantization converts these toRepresent the fermions by Gamma matrices, ${\hat{\psi }}^m={\gamma }^m/\sqrt{2}$, where $\{{\gamma }^m,{\gamma }^n{\}}_{+}=2{\eta }^{mn}$, and ${\hat{p}}_m=-i{\partial }_m$. The wavefunction is a Dirac spinor. Its physical-state constraint iswhich is exactly the massless Dirac equation. Moreover ${\hat{\mathcal{Q}}}^2=\hat{\mathcal{H}}=-{\square }_D/2$, so this equation implies the remaining mass-shell constraint. Conversely, a solution of the massless Dirac equation satisfies both constraints. The Klein-Gordon constraint alone would not impose the spinorial first-order equation.

Take the primitive state vector $\mathbf{U}=(\rho ,p,u_x,u_y,u_z,B_x,B_y,B_z{)}^T$, with $\rho >0$. In the absence of gravity, the ideal magnetohydrodynamic equations are first order: expanding the material derivatives, pressure gradient and Lorentz force makes every term linear in a spatial derivative of $\mathbf{U}$, with coefficients depending on $\mathbf{U}$. Dividing momentum balance by $\rho$ therefore gives a quasilinear partial differential equation system ${\partial }_t\mathbf{U}+A_i(\mathbf{U}){\partial }_i\mathbf{U}=0$ with eight-by-eight matrices. The solenoidal magnetic-field constraint is an additional constraint on initial data; the ideal magnetohydrodynamic induction equation preserves it because the divergence of a curl vanishes.

For a one-dimensional simple wave in magnetohydrodynamics, write $\mathbf{U}=\mathbf{U}(q(x,t))$ with a nonzero state-space tangent. Substitution gives ${\mathbf{U}}_q{q}_t+A_x{\mathbf{U}}_q{q}_x=0$. A nonconstant profile requires the tangent to be a right eigenvector of $A_x$:The chain rule then gives the wave-speed equation:This is the Inviscid Burgers equation, including the special case of constant $v$. Its characteristic curves are $x=a+v_0(a)t$, with $v=v_0(a)$. As long as the mapping remains invertible,Thus a region with $v_0^{\prime }<0$ steepens: faster characteristics catch slower ones, and the first gradient catastrophe occurs at $t_{\ast }=-1/{\min }_a{v}_0^{\prime }(a)$ when the minimum is negative. A genuinely nonlinear compressive simple wave therefore forms a shock wave. Rarefactive profiles can spread instead; steepening is not inevitable for every initial profile.

To continue past this time, use weak solutions of the conservative mass, momentum, fluid total-energy equation and ideal magnetohydrodynamic induction equation. The Rankine-Hugoniot conditions fix the jumps and shock velocity, while physical entropy production selects admissible magnetohydrodynamic shocks. Diffusive shock wave layers are replaced by moving discontinuities, so their microscopic structure need not be explicitly resolved. In particular, the smooth adiabatic pressure equation must be replaced by total-energy conservation when imposing the ideal magnetohydrodynamic shock conditions.

For the Alfvén waves, impose the one-dimensional solenoidal magnetic-field constraint, so $B_x$ is constant and $\delta B_x=0$. Put ${\mathbf{B}}_{\perp }=(B_y,B_z)$ and $c=v-u_x$. The transverse components of the linearized ideal magnetohydrodynamic equations giveThe longitudinal momentum equation additionally gives $\delta p+{\mathbf{B}}_{\perp }\cdot \delta {\mathbf{B}}_{\perp }/{\mu }_0=0$. The Alfvén eigenvectors have $\delta \rho =\delta p=\delta u_x=0$, hence their transverse polarization is perpendicular to ${\mathbf{B}}_{\perp }$. For $s=\pm 1$, their speed and explicit right Alfvén characteristic eigenvectors areFor ${\mathbf{B}}_{\perp }\ne 0$ one may choose $(h_y,h_z)=(B_z,-B_y)$, up to a nonzero scalar factor. If ${\mathbf{B}}_{\perp }=0$, either transverse polarization is allowed and this characteristic speed is degenerate. These expressions describe the two propagating Alfvén branches for $B_x\ne 0$; if $B_x=0$, they coalesce with advected, nonpropagating transverse disturbances.

Integrating along an Alfvén simple wave leaves $\rho ,p,u_x,B_x$ unchanged and gives $d{\mathbf{u}}_{\perp }=-sd{\mathbf{B}}_{\perp }/\sqrt{{\mu }_0\rho }$ and ${\mathbf{B}}_{\perp }\cdot d{\mathbf{B}}_{\perp }=0$. Consequently the finite-amplitude nonlinear Alfvén wave relations areThe constant-magnitude condition is essential: otherwise a varying magnetic pressure would drive longitudinal compression. An arbitrary smooth phase profile gives an explicit family, $f_3=B_{\perp }\cos \Theta (x-v_{s}t)$ and $f_4=B_{\perp }\sin \Theta (x-v_{s}t)$, with $f_1,f_2$ as above. Direct substitution in the transverse momentum and ideal magnetohydrodynamic induction equation gives $(u_x-v_s){\mathbf{u}}_{\perp }^{\prime }=B_x{\mathbf{B}}_{\perp }^{\prime }/({\mu }_0\rho )$ and $(u_x-v_s){\mathbf{B}}_{\perp }^{\prime }=B_x{\mathbf{u}}_{\perp }^{\prime }$, confirming the solution without relying on the eigenvector argument. Longitudinal momentum holds because $p+B_{\perp }^2/(2{\mu }_0)$ is constant. The speed $v_s$ is independent of amplitude along this family: these linearly degenerate characteristic fields translate without steepening.

Let $e$ be specific internal energy, $Q$ the heat supplied per unit volume per unit time, and $E=\rho (e+|\mathbf{u}{|}^2/2)$. For an inviscid perfect gas in a prescribed Newtonian gravitational potential $\Phi$, the fluid total-energy equation isThere is no gravitational term if no external body is present. For a time-independent $\Phi$, adding potential energy gives the equivalent conservative form ${\partial }_t(E+\rho \Phi )+\nabla \cdot [(E+p+\rho \Phi )\mathbf{u}]=Q$. Subtracting the kinetic-energy equation and using mass conservation givesFor a perfect gas with constant specific-heat ratio $\gamma$, $e=p/[(\gamma -1)\rho ]$. The isothermal equation of state therefore makes $e=c_s^2/(\gamma -1)$ constant. Hence the required heat supply and heat loss areCompression requires cooling; expansion requires heating. The isothermal sound speed $c_s$ differs from the adiabatic sound speed $\sqrt{\gamma }{c}_s$.

Define the isothermal Mach numbers by ${\mathcal{M}}_i=u_i/c_s$. Orient the normal to a stationary isothermal shock along the flow, so the mass flux $j={\rho }_1{u}_1={\rho }_2{u}_2>0$. Tangential velocity is continuous for this planar hydrodynamic shock. Normal momentum conservation givesFor a genuine discontinuity $u_1\ne u_2$, factorization gives $u_1{u}_2=c_s^2$. Thus the isothermal shock jump relations areAcross a thin shock, gravitational potential and tangential kinetic energy are unchanged. The specific enthalpy $h=\gamma c_s^2/(\gamma -1)$ is the same on both sides, so the energy removed per unit area per unit time isPositive net cooling requires $u_1>u_2$, equivalently ${\rho }_2>{\rho }_1$ and ${\mathcal{M}}_1>1>{\mathcal{M}}_2$. Cooling across an isothermal shock permits only compression shocks; the formal expansion discontinuity requires energy supplied by the surroundings. The continuous state $u_1=u_2$ is not a shock.

For steady radial flow, mass conservation gives $r^2\rho u_r=\text{constant}$. Differentiating it and eliminating the mass density gradient from radial momentum balance givesA regular sonic point has $|u_r|=c_s$, so the right-hand side must vanish there as well. ThereforeThis is the Bondi sonic point for accretion; the same critical radius appears in an isothermal Parker wind. A general steady solution need not pass through a sonic point, but a smooth transonic one must obey both critical conditions.

Put $x=r/r_s$ and $\mathcal{M}=|u_r|/c_s$ on a branch whose flow direction is fixed. The radial equation becomesIntegrating yields the integrated isothermal Bondi flow relation:For a transonic branch, $x=\mathcal{M}=1$ gives $C=-3/2$. Expansion about the critical point gives $(\mathcal{M}-1{)}^2=(x-1{)}^2+O((x-1{)}^3)$, with slope $-1$ for inflow supplied by gas at rest at infinity and slope $+1$ for a transonic outflow.

For Isothermal Bondi accretion, integrate radial momentum balance once more, using $u_r\to 0$ and $\rho \to {\rho }_{\infty }$ at infinity:At the Bondi sonic point, $GM/r_s=2c_s^2$ and $u_r^2=c_s^2$, so ${\rho }_s=e^{3/2}{\rho }_{\infty }$. The positive inward isothermal Bondi accretion rate is consequentlyThe critical mass density and rate are fixed by the regular transonic solution and the specified reservoir at infinity; an arbitrary subsonic solution does not share this accretion rate.

For the self-gravitating incompressible slab, let the fluid occupy $-H<z<H$, with mass density ${\rho }_0=\Sigma /(2H)$ and a vacuum exterior. Poisson equation and hydrostatic equilibrium giveA constant ambient pressure could be added without changing the result. Choose perturbations proportional to $e^{ikx-i\omega t}$, using rotational symmetry in the horizontal plane to put the wavevector along $x$.

For the fluid displacement, the linearized momentum equation inside the uniform slab is $-{\omega }^2\mathbf{\xi }=-\nabla (p^{\prime }/{\rho }_0+{\varphi }^{\prime })$. For $\omega \ne 0$ it follows that $\mathbf{\xi }=\nabla \chi$, with $p^{\prime }/{\rho }_0+{\varphi }^{\prime }={\omega }^2\chi$. Incompressibility gives $\nabla \cdot \mathbf{\xi }=0$, henceNeutral limits of these surface modes follow by continuity. Stationary vorticity perturbations form a separate zero-frequency sector; the nonzero-frequency surface modes are irrotational by the momentum equation itself, without an extra irrotational-flow assumption.

The Eulerian mass density perturbation vanishes in the bulk because the equilibrium mass density is uniform and the displacement is divergence free. If the two vertical surface displacements are ${\eta }_{+}={\xi }_z(H)$ and ${\eta }_{-}={\xi }_z(-H)$, the surface density perturbation of a displaced uniform interface givesThus Poisson equation reduces to Laplace equation for ${\varphi }^{\prime }$ inside and outside. The potential is continuous, while integrating through each surface givesThe perturbing potential must decay at vertical infinity. The kinematic conditions are ${\eta }_{\pm }={\partial }_z\chi (\pm H)$. The free surface pressure condition is the vanishing Lagrangian pressure perturbation, $p^{\prime }+{\eta }_{\pm }{p}_0^{\prime }=0$; at the top this gives $p^{\prime }(H)/{\rho }_0=g_s{\eta }_{+}$.

Put $q=kH>0$. For even scalar-potential symmetry, chooseThen ${\eta }_{-}=-{\eta }_{+}$: this is a thickness-changing mode, not a vertical bending mode. Write $\eta ={\eta }_{+}$; the top derivative jump gives $-kC(\cosh q+\sinh q)=4\pi G{\rho }_0\eta$. Since $\cosh q+\sinh q=e^q$,Substitute these values into ${\omega }^2\chi (H)=g_s\eta +{\varphi }^{\prime }(H)$. Using $4\pi G{\rho }_0=g_s/H$ gives the even-mode dispersion relation for the surface modes of a self-gravitating incompressible slab:The positive term comes from the restoring hydrostatic free-surface pressure in the background gravitational field. The negative term is the attraction due to the perturbed gravitational potential of the displaced surfaces, which tends to reinforce the disturbance. Both effects ultimately involve the slab's gravity, but enter the boundary condition through different terms.

For odd scalar-potential symmetry, instead takeNow ${\eta }_{-}={\eta }_{+}$, so the surfaces bend in the same direction. The jump again gives $C=-4\pi G{\rho }_0\eta /(k{e}^q)$, but $\chi (H)=\eta \tanh q/k$ and ${\varphi }^{\prime }(H)=-4\pi G{\rho }_0\eta /[k(1+\coth q)]$. The same pressure condition yieldsThe bottom kinematic, jump and pressure conditions follow with the indicated parity and give the same dispersion relation, so no boundary condition has been discarded.

To determine the gravitational instability of an incompressible slab, factor the two dimensionless expressions asFor the even mode the bracket is strictly increasing, with derivative $1+e^{-2q}>0$, is negative at zero and tends to positive infinity. It therefore has a unique positive zeroThe even mode is unstable for $0<kH<q_c$, neutral at $q_c$, and oscillatory for $kH>q_c$. Negative ${\omega }_e^2$ gives a growing mode with $\omega =i\sqrt{-{\omega }_e^2}$ under the stated time convention. For the odd mode, $1-e^{-2q}<2q$ for every $q>0$, so the bracket is always positive: all odd modes with positive wavenumber are stable.

At long wavelengths $F_e=-q+2q^2+O(q^3)$ and $F_o=q-2q^2/3+O(q^3)$: a thickness or column-density disturbance has a gravitational instability, whereas a bending disturbance is restored. At short wavelengths both approach $q-1/2$, so both surface branches are stable. The slab equilibrium is therefore linearly unstable overall, because it admits the long-wavelength even modes despite the stability of its odd sector.

Use momentum conservation and conservation of energy in the laboratory inertial frame. The photon has momentum magnitude $E/c$, while the outgoing electron satisfies the relativistic energy-momentum relation. Its total energy and momentum obeySubstitute these into $E_e^2=m^2{c}^4+c^2|{\mathbf{p}}_e{|}^2$. Cancelling $m^2{c}^4$ and $(E_1-E_2{)}^2$ gives $m{c}^2(E_1-E_2)=E_1{E}_2(1-\cos \theta )$. Both photon energies are positive, so division yieldsThis is Compton scattering. The energies include the electron's rest energy; using only its kinetic energy in the mass-shell equation would give the wrong result. The outgoing photon has unchanged energy at $\theta =0$ and smaller energy for a nonzero scattering angle.

The wavefunction is a plane wave and a simultaneous formal eigenfunction of the momentum operator and the time-translation energy operator:Its momentum and energy are therefore definite, while its probability density is spatially uniform:For a particle of mass $m$, the probability current is $j=(\hslash /m)\mathrm{Im}(\overline{\varphi }{\partial }_x\varphi )=\hslash k{A}^2/m$, directed according to the sign of $k$.

On the whole real line, a nonzero constant-amplitude plane wave is not square-integrable and cannot be normalised to total probability one. It represents an idealised momentum eigenstate, interpreted through generalized eigenfunctions or as a limit of wave packets. Also, arbitrary $k,E$ do not automatically solve the Schrödinger equation for a specified potential energy: for a free particle, $E={\hslash }^2{k}^2/(2m)$, or this value plus the constant potential energy for a constant-potential region. If $A=0$, the zero function is not a physical quantum state.