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Take the two-dimensional electron gas in the $xy$ plane, the Hall bar along $x$, and $\mathbf{B}=B\hat{\mathbf{z}}$ with $B>0$ perpendicular to it. An electron has charge $-e$. The confining electron potential energy $U(y)$ is approximately flat inside a wide strip and rises at either boundary; smooth quenched disorder adds hills and valleys. For a slowly varying edge, each spin-resolved Landau level approximately followsHere $s=\pm 1$, and $Y$ is the guiding centre. For $\mathbf{A}=(-By,0,0)$, $Y={\ell }_B^2{k}_x$ and the edge group velocity isThus electrons move along $+x$ at the upper edge and $-x$ at the lower edge. The Lorentz force gives counterclockwise bulk cyclotron motion in this convention. At a hard boundary its interrupted circles become directed skipping orbits; a smooth boundary gives the corresponding guiding-centre drift. Smooth bulk quenched disorder generally gives closed drift contours around localized hills or valleys, with extended trajectories appearing near a broadened level's critical energy. Realistic electrostatic screening can produce alternating compressible quantum Hall edge strips and incompressible quantum Hall edge strips, so the unscreened bending curves are an effective sketch rather than an exact self-consistent edge profile.

In the original diagram, contacts are ordered around the electron propagation direction: upper-left $1\to 2\to 3$, then lower-right $4\to 5\to 6$. Let $I_1=I$, $I_4=-I$ and the other leads be ideal equilibrating voltage probes. Define $R_H=(V_2-V_6)/I$ and $R_{xx}=(V_2-V_3)/I$. These choices fix the voltage sign; reversing the field reverses chirality and the Hall polarity. For spin-resolved plateau index $N$,At fixed sheet carrier density and increasing field, $N$ decreases, giving successively higher Hall plateaus; longitudinal peaks occur between them. Unresolved spin branches can suppress odd plateaus. The sketch uses resistance magnitudes, and every integer step is not promised in an arbitrary sample.

For Fermi-energy oscillations in a two-dimensional electron gas, first distinguish fixed density from fixed reservoir chemical potential. At negligible temperature the former is determined by state counting. Each resolved Landau level contains $eB/h$ states per unit area, soIgnoring spin and broadening for the counting illustration, if $N<\nu <N+1$ then $N$ levels are full and level $N$ is partly full: $\mu =E_N$. At fixed $n_s$, increasing $B$ increases both level energy and degeneracy. The partly filled level rises in energy until the lower levels can contain all electrons; $\mu$ then drops to the next lower level. The resulting sawtooth is periodic in $1/B$, with $\Delta (1/B)=e/(n_{s}h)$ per resolved branch. At an exactly integer filling in an ideal gap, any chemical potential within the gap gives the same zero-temperature particle number; the plotted discontinuity denotes its limiting values.

At fixed $B$ while $n_s$ increases, the ideal chemical potential stays at a partly occupied level and steps upward after each additional density $eB/h$. It therefore has a staircase superposed on the rising zero-field trend, not a repeated downward sawtooth. Landau-level broadening and finite temperature smooth both ideal graphs. With an externally pinned chemical potential, density changes instead; the fixed-density sawtooth cannot simultaneously be imposed. In a realistic disordered sample, the Fermi energy sweeps localized states between transport-carrying level centres. The following schematic omits spin for clarity and does not attempt an interaction-corrected density of states.

For $N$ transmitted spin-resolved quantum Hall edge states, the Landauer-Büttiker formula reduces, in that contact order, toZero probe currents imply $V_2=V_3=V_1$ and $V_5=V_6=V_4$. Hence $I=(N{e}^2/h)(V_1-V_4)$ and the boxed resistance results follow. The spin-resolved conductance quantum arises because the one-dimensional density of states times velocity is $1/h$ per unit energy; a reservoir imbalance supplies $e^2\Delta V/h$ per channel. Its value does not depend on edge slope or sample width.

On a plateau, opposite-moving edge channels are separated by a wide bulk with a mobility gap. Ordinary elastic scattering on one edge has no counterpropagating channel into which to reflect; closed bulk orbits do not connect the contacts. Changing $B$ or $n_s$ can alter the occupancy of localized states without changing the number of transmitting edge channels. This accounts for the finite plateau width, which an entirely clean set of sharp levels would not explain. Accuracy requires negligible backscattering between opposite edges, low enough current to avoid breakdown, equilibrating contacts and negligible probe loading. The finite-width plateau and the precision of its quantized value are related but distinct observations.

An antidot is a depleted potential hill, not a potential well. Its closed edge circulates oppositely to the outer boundary. In a constriction, tunnelling through an antidot orbit can transfer an electron from a forward edge into the returning edge: resonant antidot backscattering produces conductance dips. For $N$ otherwise open spin-resolved channels and one active resonance,At symmetric coupling a single resonance can remove $e^2/h$; asymmetry or thermal averaging reduces its depth. If the background constriction is closed and the resonant route instead supplies forward transmission, peaks occur. The lead connectivity must therefore accompany the conductance sketch. In the fixed-area single-particle limit, successive alignments change enclosed flux by ${\Phi }_0=h/e$, giving the Aharonov-Bohm antidot periodThe effective orbit area, not merely the etched island area, enters. Charging, screening and spin can modify this simple sequence; two coupled spin branches can yield a near-$h/(2e)$ pattern without changing the elementary flux quantum.^[ref]

Use $e>0$, $B>0$ along $+z$, and $U(y)$ for the electron's confinement potential energy. The spin-resolved Landau level branches in the smooth-edge approximation areThe spin labels denote upper and lower energies; the sign of a material's effective $g$ factor is not inferred from the stated positive splitting. In units of $\hslash {\omega }_c$, the four occupied bulk energies are $1/3,2/3,4/3,5/3$. A chemical potential in $(5/3,7/3)$ therefore gives filling factor four. The branches rise at both confining edges and cross the common Fermi energy once at each edge. Each crossing is one spin-resolved quantum Hall edge state; there are four on either side. The two spin branches within an orbital level are separated by $\hslash {\omega }_c/3$, while the intervening gap to the next orbital pair is $2\hslash {\omega }_c/3$.

For the spin separation at a smooth quantum Hall edge, linearize the confinement energy at a crossing. Compensating a spin energy difference by a potential-energy change givesHere $V^{\prime }$ is interpreted as the specified slope of electron potential energy. If it denotes electrostatic voltage per length instead, the denominator is $e|V^{\prime }|$. The two edges of a neck and the two sides of an antidot have oppositely directed slopes; the formula is a positive local separation on any one edge. For local filling two, only the $n=0$ spin pair is occupied at the neck minimum: $E_{0,+}+U_{\min }<E_F<E_{1,-}+U_{\min }$. Raising both branches with the confinement produces two crossings on each side, as shown in the original neck sketch. This fixed-slope estimate precedes self-consistent electrostatic screening.

An antidot is a depleted potential hill. guiding centres circulate around its closed equipotentials, with chirality opposite to the outer sample edge. A phase-coherent round trip contributes magnetic phase $2\pi BS/{\Phi }_0$, with ${\Phi }_0=h/e$. A localized orbit resonant with the incident energy can transfer an electron between extended channels. If the outgoing channel returns toward the injecting reservoir, the process is resonant reflection and reduces electrical conductance. If coupling instead connects an otherwise blocked incident channel to the far reservoir, it is resonant transmission and adds electrical conductance. Transmission peaks require that forward connectivity; a closed orbit by itself does not guarantee them. Interchannel scattering can change which orbit participates.^[ref]

For the antidot resonance conductance bounds with one active spin-resolved resonance, writeHere ${\Gamma }_L,{\Gamma }_R$ are energy broadenings; the corresponding escape rates are ${\Gamma }_L/\hslash ,{\Gamma }_R/\hslash$. In a reflection-dominated configuration with two otherwise transmitted channels, $G=(e^2/h)(2-\mathcal{L})$: a $2e^2/h$ background can dip to $e^2/h$. In a transmission-dominated configuration with one spectator channel and the second otherwise blocked, $G=(e^2/h)(1+\mathcal{L})$: peaks rise from $e^2/h$ toward $2e^2/h$. These are the two ideal channel-count examples plotted below. More generally, $G=(e^2/h)(N_{\mathrm{open}}-R)$ or $(e^2/h)(N_{\mathrm{dir}}+T)$; a different spectator count changes the integer offset. The constriction filling alone does not specify every tunnel coupling or resonance height. Equal couplings permit unit resonant transfer in the isolated-level model; asymmetry, temperature and dephasing reduce the contrast. Holding the effective orbit area approximately fixed gives $\Delta B\simeq h/(eS)$ for one branch, not a universal charging-corrected period.

In quantum-dot versus antidot resonance topology, a quantum dot is a potential well holding a finite electron island. Its closed guiding-centre circulation is reversed relative to an antidot at the same field. With two barriers in series, current follows source $\to$ dot $\to$ drain; off resonance there need not be a direct conducting path. A gate shifts a level or an addition electrochemical potential into alignment with the reservoirs and produces a electrical conductance peak. For an isolated coherent spin-resolved level, the zero-temperature peak is at most $e^2/h$ and reaches it only for equal couplings; unresolved noninteracting spin degeneracy doubles this bound. In weak sequential tunnelling, peak height depends on rates and temperature rather than being quantized. Charging produces Coulomb blockade peaks with spacing set by addition energy and gate lever arm. An antidot also supports discrete orbit/charging resonances, but can be side-coupled to already open channels and produce dips. Gate voltage changes both orbit area and confinement, whereas a magnetic sweep directly changes enclosed flux.

Coulomb interactions require a self-consistent confinement potential. A gradual decrease in electron density cannot leave arbitrary uncompensated donor charge at no cost. Partly occupied Landau levels have a large charge response: electrostatic screening flattens their effective energy near the Fermi energy over compressible quantum Hall edge strips. Between them, integer-filled incompressible quantum Hall edge strips have a gap, weak density response and most of the potential variation. In an idealized cross-section the effective branch thus alternates between flat, partially occupied portions and rising portions crossing a gap; its carrier density varies oppositely. The bare levels still have Landau level and spin splittings locally, but the common electrostatic potential is no longer an imposed straight slope. exchange interaction can additionally renormalize spin gaps. Widths must follow electrostatics and the density gradient, so the unscreened $E_Z/|V^{\prime }|$ separation is not a quantitative screened-strip width.^[ref]

Ridge push is a horizontal pressure force integrated with depth, equivalently a difference in gravitational potential energy per horizontal area. It is not the weight of a slab multiplied by the surface slope at one chosen point.

For an explicit compensated-column derivation, let $y$ measure depth below the ridge seafloor; the old seafloor is at $y=h$. Set $b={\rho }_m\alpha \Delta T$. The hot column has greater pressure. Its pressure excess over the old column isThe second line follows by integrating $\delta \rho (s)=b(1-s/a)$ upwards from the compensation level. The two expressions agree at $y=h$ because $({\rho }_m-{\rho }_w)h=ba/2$. Integrating this positive pressure excess gives the ridge-push force in a plate model, directed away from the ridge:Consistently retaining only first order in thermal expansion givesIf the topographic term is retained within the same linear-density column model, it adds $1.52\times 10^{11}\mathrm{N}{\mathrm{m}}^{-1}$, giving $F\simeq 2.75\times 10^{12}\mathrm{N}{\mathrm{m}}^{-1}$. This is a useful small correction, not a claim that all other second-order material effects have been included. Both values use the intended hot initial plate, not the inconsistent printed zero-age profile.

A positive electric potential difference between gate and substrate attracts electrons towards the insulating interface and repels the mobile electron holes of the p-type semiconductor substrate. The exposed, negatively charged acceptor impurities form a depletion region; the thin mobile electron layer at its surface is the semiconductor inversion layer. The conduction band and valence band bend downwards towards the positively biased gate, because the electron potential energy is $-e\psi$.

To specify the conventional MOS inversion threshold, let $N_A$ be the ionized acceptor density, $n_i$ the intrinsic carrier concentration, andIn the neutral bulk the intrinsic level lies $e{\varphi }_F$ above the semiconductor chemical potential. A positive surface potential ${\psi }_s$ lowers both band edges by $e{\psi }_s$, giving, in the nondegenerate approximation,At ${\psi }_s={\varphi }_F$, the surface is intrinsic: $n_s=p_s=n_i$. The usual onset of strong inversion instead meansThus the surface electron density has grown to the bulk majority-carrier density. The chemical potential need not actually enter the conduction band. The phrase “just forms” can also refer to the earlier weak-inversion crossover; the diagram uses the conventional strong-inversion threshold and labels it explicitly.

Within the depletion approximation the negative space charge is approximately $-e{N}_A$ over a widthwhere ${ϵ}_s$ is the absolute semiconductor permittivity. Its outer edge is where the bands become flat and mobile electron holes recover their bulk density. The electrical insulator blocks steady carrier transport while sustaining an electric field; the metal gate supplies positive surface charge balancing depletion and inversion charge. Under gate bias, the gate and substrate electrochemical potentials are different, so a single equilibrium Fermi level should not be drawn through the whole biased stack.

Dielectric screening is the rearrangement of a medium's charges in response to an applied charge. In a metal, mobile electrons accumulate near a positive perturbation and reduce its long-range potential. Thomas-Fermi screening treats the electron gas locally as an equilibrium gas whose density responds to the local electrostatic potential, ignoring gradients of the density on the microscopic wavelength scale.

Take $e>0$, background permittivity ${\epsilon }_b$, and electrostatic potential $\varphi$. An electron's potential energy is $-e\varphi$, so the local kinetic Fermi energy becomes $\mu +e\varphi$. For a three-dimensional spin-degenerate parabolic band at zero temperature,where $g(E_F)=dn/d\mu$ is the spin-summed density of states per volume per energy. The induced charge is ${\rho }_{\mathrm{ind}}=-e\delta n=-e^{2}g(E_F)\varphi$. Combining it with Poisson equation givesFor a point external charge $Q$, spherical symmetry and decay at infinity selectThus the bare $1/r$ potential is cut off exponentially. Fourier transformation makes the same result transparent: $\varphi (\mathbf{q})={\rho }_{\mathrm{ext}}(\mathbf{q})/[{\epsilon }_b(q^2+k_{\mathrm{TF}}^2)]$, so the static longitudinal dielectric factor is $1+k_{\mathrm{TF}}^2/q^2$ relative to the background. For a free electron band, $g(E_F)=3n/(2E_F)$ and ${\lambda }_{\mathrm{TF}}^2=2{\epsilon }_b{E}_F/(3n{e}^2)$.

The local linear approximation requires a weak perturbation and slowly varying fields on the relevant microscopic scale. It omits nonlocal Friedel oscillations and atom-scale structure; a nonlinear local-density treatment is needed near a sufficiently strong charge. Finite temperature replaces $g(E_F)$ by the appropriate compressibility derivative $\partial n/\partial \mu$. The nondegenerate limit gives classical screening rather than the degenerate Thomas-Fermi length. A larger density of states means that a smaller potential can move more carriers, hence stronger screening.

Write $S=S_0+S_I$, where $S_0$ contains the quadratic kinetic term and mass term, and $S_I=-\lambda \int d^{D}x\varphi (x{)}^3/3!$. ThenLet $Z_0=\int D\varphi e^{i{S}_0}$ and let $⟨\cdot {⟩}_0$ denote the unit-normalized free Gaussian vacuum expectation value. The unnormalized numerator ${\mathcal{N}}_N$ is consequentlyThe Minkowski path integral is understood with the usual scalar Feynman propagator pole prescription, as a formal perturbation series. A real cubic potential energy is unbounded below; no nonperturbative stable vacuum is being assumed here.

Set $\Delta (x-z)=⟨T\varphi (x)\varphi (z){⟩}_0$, $K_i(x)=\Delta (y_i-x)$, and $t=\Delta (0)$, with a common regularization understood for coincident points. The zero-order one-point term has one field and vanishes. At first order, Wick's theorem gives three choices for the contraction with the external field, leaving a tadpole diagram:The second-order term contains seven fields, so it vanishes by the vanishing odd correlators of a free Gaussian field. HenceThe coefficient is $3/3!=1/2$, or symmetry factor $S=2$ when each diagram is weighted by $1/S$. No spacetime integral has been evaluated. The one-point tadpole diagram is shown in the first panel of the following figure; its other panels organize solution.

Write $h_0=-{\hslash }^2{\nabla }^2/(2m)+V(\mathbf{r})$ for the one-particle Hamiltonian operator. The second quantization of the kinetic energy, external potential energy and repulsive contact interaction isThe factor $1/2$ counts each unordered particle pair once. Normal ordering excludes self-interaction: the interaction acts on two distinct bosons. The fixed particle number sector has $\hat{N}=\int d^{3}r{\hat{\psi }}^{†}\hat{\psi }=N$. In three dimensions, this contact interaction is understood as a regulated low-energy interaction with a matched coupling, as in renormalization of a three-dimensional dilute-gas contact coupling; the finite-mode calculation below uses its projected matrix elements.

For a uniform continuum gas with pair interactions depending on relative positions, a Galilean boost changes every particle momentum by $p_0\hat{\mathbf{x}}$ and leaves the interaction energy unchanged. Write ${\hat{P}}_x={\sum }_j{\hat{p}}_{j,x}$. The transformed many-body Hamiltonian operator isThe resting ground state has $⟨{\hat{P}}_x⟩=0$, so the Galilean boost energy of an interacting gas isThe mass here is the bare particle mass, because this is an exact transformation of the microscopic kinetic energy. Substituting the Fermi-liquid effective mass $m^{\ast }$ would omit the change in interaction energy encoded by the Landau quasiparticle energy functional. This conclusion requires the Principle of Galilean relativity; a gas in a lattice or a fixed external potential energy landscape need not obey it.

Differentiate the Lagrangian with respect to the generalized velocities:The Legendre transform cancels the vector-potential energy term, giving $H=p\cdot \dot{x}-L=m|\dot{x}{|}^2/2+V$. Eliminating velocity with $m\dot{x}=p-eA$ therefore yieldsFor the given gauge, set ${\pi }_1=p_1+eB{x}_2/2$, ${\pi }_2=p_2-eB{x}_1/2$, ${\pi }_3=p_3$. Hamilton's equations give ${\dot{x}}_j={\pi }_j/m$, ${\dot{p}}_1=(eB/2){\dot{x}}_2$, ${\dot{p}}_2=-(eB/2){\dot{x}}_1$, ${\dot{p}}_3=0$. Hence the conserved quantities areThe two transverse combinations are components of magnetic pseudomomentum. The printed conserved combinations omit the electric charge factor $e$. With the stated vector potential and general electric charge they need the factors displayed above; as printed they hold only if $e=1$ or if electric charge is already absorbed into the field convention.

The initial data give $K_1=K_2=p_3=0$. Thus $p_1=eB{x}_2/2$, $p_2=-eB{x}_1/2$ throughout, andSolving with the initial point givesConsequently $x_1^2+x_2^2=a^2$: the guiding centre is the origin and the signed cyclotron frequency fixes the rotation direction. The energy is the constant $e^2{B}^2{a}^2/(2m)$.

The orthogonal coordinate change preserves the kinetic energy and takes the particle exchange operator to $(x_{+},x_{-})\mapsto (x_{+},-x_{-})$. Thus the spatial wave function is even in $x_{-}$ for bosons and odd for fermions.

The total potential energy and Hamiltonian separate asAssume $1+2\kappa >0$ so the relative oscillator is confining. Its spectrum is $E_{n_{+},n_{-}}=\hslash \omega (n_{+}+1/2)+\hslash \Omega (n_{-}+1/2)$, and its parity in the two-particle relative coordinate is $(-1{)}^{n_{-}}$.

The lowest boson state uses $n_{+}=n_{-}=0$, whereas the lowest fermion state uses $n_{+}=0$, $n_{-}=1$. ThereforeFor an oscillator of frequency $\Omega$, $⟨x_{-}^2{⟩}_n=(n+1/2)\hslash /(m\Omega )$. Since $(x_1-x_2{)}^2=2x_{-}^2$,The factor three is the extra relative-coordinate excitation required by antisymmetry. The supplied oscillator wave functions correspond to kinetic energy $P^2/(2m)$; the extra factor $1/2$ printed before that term in the hint is inconsistent with those functions and the stated energies. The calculation uses the conventional kinetic energy dictated by particle mass $m$.

Use the normalized Rayleigh quotient $R(\psi )=⟨\psi ,H\psi ⟩/⟨\psi ,\psi ⟩$. Its real directional derivative in an arbitrary complex direction $\eta$ isIt vanishes for every direction precisely when $(H-R)\psi =0$, since testing $\eta =(H-R)\psi$ gives its squared norm. Conversely an eigenvector makes every derivative zero. Stationarity over all states is stronger than stationarity in a restricted trial family.

Normalize the Gaussian as ${\psi }_{\alpha }=(\alpha /\pi {)}^{1/4}{e}^{-\alpha x^2/2}$, with $\alpha >0$. Its moments are $⟨X^2⟩=1/(2\alpha )$, $⟨X^4⟩=3/(4{\alpha }^2)$ and $⟨P^2⟩={\hslash }^2\alpha /2$. The printed potential energy consequently givesPut $a=\alpha \hslash /(m\omega )$. Then $\mathcal{E}/(\hslash \omega )=a/4+1/(4a)+15/(256a^2)$, whose second derivative is $1/(2a^3)+45/(128a^4)>0$. It diverges at both positive endpoints, so its sole stationary point is the unique global minimum. Its equation isThe variational principle therefore yields the optimized upper boundThe printed minimizer is inconsistent with the printed quartic coefficient. At $a=3/2$ the derivative is $5/48>0$, so this is not the stationary Gaussian for that potential energy. It is nevertheless a legitimate trial vector and gives the simple, weaker bound $\boxed{E_0\le 109\hslash \omega /192}$. If the quartic coefficient were instead $5m^2{\omega }^3/(16\hslash )$, four times the printed value, then $a=3/2$ would indeed minimize the family and its energy would be $31\hslash \omega /48$. This possible coefficient correction is identified explicitly rather than silently used. The reusable calculation is Gaussian trial for a quadratic-plus-quartic oscillator.

The Time-dependent Schrödinger equation, with the Hamiltonian for a particle of mass $m$ in a real potential energy $V$, isA stationary state is an energy eigenstate: its wavefunction evolves by a single global phase,Its probability density is independent of time. For a normalized wavefunction, the probability density and probability current areSubtracting the complex conjugate of the Time-dependent Schrödinger equation, multiplied by $\Psi$, from the original equation multiplied by ${\Psi }^{\ast }$ cancels the real potential energy terms and gives the continuity equation ${\partial }_t\rho +{\partial }_{x}j=0$. Thus $j$ measures signed probability flow in the positive $x$ direction.

At a finite jump in potential energy, the wavefunction and its first derivative must be continuous. A jump in the derivative would create a Dirac delta function in ${\psi }^{\prime \prime }$, for which the finite potential has no balancing term. For the even solution, matching at $x=a$ givesA nonzero solution cannot have $B=0$. It also cannot have $\cos \beta =0$: the first condition would give $D=0$, while the second would force $B\kappa \sin \beta =0$, a contradiction. Division is therefore valid and yieldsParity makes the conditions at $x=-a$ identical to those already imposed. Finally the definitions of the decay and oscillation constants giveThe circle relation fixes the well strength; the tangent relation selects its even bound states.

Twice applying the creation operator or twice applying the annihilation operator changes the energy eigenvalue by $2\hslash \omega$, unless they give zero. The same orthogonality argument as in part (i) therefore givesAlso the factorization in part (a) givesNow $P=(a_{+}+a_{-})/2$ and $X=(a_{+}-a_{-})/(2im\omega )$. Squaring in the correct operator order yieldsBy part (i), the means are zero, so these are the variances. Taking positive square roots gives the requested standard deviations:These harmonic-oscillator number-state second moments also show equal mean kinetic energy and potential energy, each $E/2$. Their uncertainty product is $E/\omega =(n+\frac{1}{2})\hslash$, so the ground state saturates the Heisenberg uncertainty relation.

In a sufficiently tight binary star, the main sequence companion can fill its Roche lobe. Roche-lobe overflow then transfers gas through the inner Lagrange point towards the neutron star. The gas carries angular momentum, so it generally forms an accretion disk rather than falling radially onto the surface. Internal stresses carry angular momentum outwards and permit accretion inwards. Capture of a stellar wind is another possible supply mechanism if the companion does not fill its Roche lobe.

Near the neutron star, the rapidly increasing magnetic dipole field, with $B\propto r^{-3}$, becomes dynamically important when its magnetic pressure $B^2/(2{\mu }_0)$ is comparable to the inflow stresses. The accretion disk is truncated around a magnetospheric radius. Ionized material attaches to magnetic field lines and streams along them in magnetically channelled accretion, forming an accretion column above one or both magnetic poles. The Lorentz force impedes motion across the magnetic field, but does not forbid motion parallel to it. Gas can therefore reach the polar surface, with its gravitational potential energy released in a hot shock wave and radiation.

This description presumes the neutron star rotates slowly enough for material to reach it: sufficiently rapid rotation can centrifugally expel the inflow instead. A strong magnetic field by itself does not guarantee surface accretion.

cosmic inflation is an expanding epoch with $\ddot{a}>0$, equivalently a shrinking comoving Hubble radius $(aH{)}^{-1}=1/\dot{a}$. In general relativity it requires $\rho +3P/c^2<0$ for the total effective fluid. A scalar field whose potential energy dominates its kinetic energy has $P\simeq -\rho c^2$ and can produce approximately exponential expansion. slow-roll inflation is one realization; rapid expansion alone without acceleration does not have the required effect.

The horizon problem is that widely separated regions of the cosmic microwave background have nearly the same temperature despite lying outside one another's causal domains in a purely decelerating hot Big Bang extrapolation. Sufficient inflation stretches a previously small causally connected region across our observable universe; relevant perturbation wavelengths can begin inside the Hubble radius, leave it during inflation and re-enter later. This does not require local signals to outrun light. Inflation supplies a possible earlier causal history, not a proof that every possible initial state becomes homogeneous.

For the flatness problem, the Friedmann equation gives ${\Omega }_{\mathrm{tot}}-1=k{c}^2/(aH{)}^2$ when $a$ is length-valued. Since $aH$ grows during inflation,Pre-existing unwanted massive relics such as magnetic monopoles are also diluted: conserved number density falls as $a^{-3}$. Enough inflation, followed by reheating that does not recreate the relic problem, can make their abundance negligible. These were central motivations of the original proposal, whose first phase-transition implementation also faced a graceful-exit problem.^[ref]

quantum fluctuations of the inflating field and metric supply seeds for linear cosmological density perturbations. Simple single-field slow-roll models predict a dominant adiabatic mode with nearly Gaussian, nearly scale-invariant scalar cosmological perturbations, consistent with the cosmic microwave background; tensor cosmological perturbations are another possible test. These predictions are model-dependent, and consistency is not unique proof of inflation. reheating must connect the accelerated phase to the hot expanding universe. Inflation addresses causal reach, small cosmological spatial curvature and unwanted relic abundance, while providing a mechanism for primordial structure.

Use the geodesic radial comoving coordinate, in which the Friedmann-Lemaître-Robertson-Walker metric has the formOn a slice of constant cosmic time, the radial proper-length element is $R(t)\mathrm{d}\chi$. ThereforeIf one starts from a curvature coordinate with radial element $R\mathrm{d}r/\sqrt{1-k{r}^2}$, then $\chi ={\int }_0^r\mathrm{d}{r}^{\prime }/\sqrt{1-k{r}^{\prime 2}}$; the curvature coordinate $r$ itself is not generally the $\chi$ used above.

For a comoving galaxy, $\dot{\chi }=0$, so $\dot{D}=\dot{R}\chi =HD$. The definitions areThis recession law concerns proper distances on common cosmic-time slices. It is not a local special-relativistic relative velocity measurement; a large recession rate is not a signal exceeding the local speed of light.

For a comoving volume $V=V_0{R}^3$, total fluid energy is $E=\rho c^{2}V$. Adiabatic energy conservation gives $\mathrm{d}E+P\mathrm{d}V=0$. Expanding,henceHere $\rho$ is mass-equivalent energy density. Radiation loses energy both by dilution and by pressure work; pressureless matter only dilutes in this description.

In the time-unit convention for the cosmological constant, it represents a uniform vacuum contribution with pressure equal to minus its energy density. There is a units convention in the printed field equations: their term is $\Lambda /3$, so the symbol $\Lambda$ there has units of inverse time squared. If the geometrical constant is instead given in inverse length squared, this symbol equals $c^2{\Lambda }_{\mathrm{geom}}$. The curvature parameter $k$ labels constant-curvature spatial slices; with a conventional scale-factor normalization it is $+1,0,-1$ for positively curved, flat and negatively curved slices. It fixes local curvature sign, not every possible global spatial topology.

For a flat fluid-only model without a cosmological constant and with $P=w\rho c^2$, the two field equations giveThe two printed cases and their scale-factor laws are worked under the literal inline labels below.

Inflation is accelerated expansion, $\ddot{R}>0$, equivalently a decreasing comoving Hubble radius $(RH{)}^{-1}$ on an expanding branch. An approximately constant positive vacuum-like energy can give nearly exponential expansion. Enough inflation can put regions now widely separated inside a common earlier causal patch, suppress $|k|/(RH{)}^2$ and dilute unwanted relics such as monopoles. These are the horizon, flatness and relic-abundance motivations. Reheating is needed to populate the subsequent hot universe. Quantum fluctuations can seed primordial perturbations, but inflation alone is not a universal solution to the initial singularity or a proof that all initial conditions inflate.

Now adopt the printed natural units $c=\hslash =1$. For the Homogeneous massive-scalar equation in an expanding universe,The continuity equation givesWhere $\dot{\varphi }\ne 0$, this implies the expanding-universe Klein-Gordon equationA regular canonical-field solution extends it through isolated turning points by continuity. Purely dividing the conservation equation does not determine a trajectory on an interval where $\dot{\varphi }$ is identically zero; canonical field dynamics, or variation of its action, supplies the equation there. In particular a constant nonzero massive field is not an exact freely evolving solution.

With $\Lambda =k=0$,ThereforeThis inflation condition for a homogeneous massive scalar means a potential-dominated field has sufficiently negative pressure to accelerate expansion. In quadratic chaotic inflation, a large initial field with $V=m^2{\varphi }^2/2$ can enter slow roll, with $3H\dot{\varphi }\simeq -m^2\varphi$ and $H^2\simeq 4\pi G{m}^2{\varphi }^2/3$. The potential slow-roll parameter is ${ϵ}_V=1/(4\pi G{\varphi }^2)$, so a large field relative to the gravitational scale can support many e-folds. The field eventually rolls far enough that this regime ends and oscillation and reheating become relevant. Large potential energy with arbitrary initial kinetic energy does not by itself guarantee an inflationary interval.

Differentiating the flat Friedmann equation and then using the scalar equation,For the expanding branch,This Nonincreasing Hubble rate for a canonical inflaton decreases strictly where the field rolls and can be momentarily stationary at a turning point; the statement “always decreasing” means nonincreasing. This does not conflict with acceleration, because $\ddot{R}/R=H^2+\dot{H}$ remains positive when ${ϵ}_H=-\dot{H}/H^2<1$.

At horizon exit, larger comoving wavelengths leave earlier, when $H$ is larger. For a canonical field in slow roll, with reduced Planck mass ${\overline{M}}_{\mathrm{Pl}}=(8\pi G{)}^{-1/2}$,The decreasing expansion rate gives a red tensor tilt at this order. The scalar curvature spectrum also contains the conversion from field fluctuations to curvature:Here $N=\ln R$ counts elapsed e-folds. Thus the Decreasing Hubble rate does not alone fix scalar tilt. Evolution of ${ϵ}_H$ matters. For the quadratic slow-roll model, ${\eta }_V={ϵ}_V$ and $n_s-1\simeq -6{ϵ}_V+2{\eta }_V=-4{ϵ}_V<0$. If $N_{\mathrm{rem}}$ e-folds remain until the end, leading estimates give $n_s\simeq 1-2/N_{\mathrm{rem}}$ and tensor-to-scalar ratio $r_{\mathrm{TS}}\simeq 8/N_{\mathrm{rem}}$, with small end-of-inflation corrections.

The relevant contemporary comparison is a nearly scale-invariant but red scalar spectrum. The February2015 Planck analysis reported $n_s=0.968\pm 0.006$ for its temperature, low-polarization and lensing combination, and an upper limit $r_{0.002}<0.11$ at95percent confidence. The scalar tilt is qualitatively consistent with a slowly evolving inflaton, but that result does not establish the quadratic potential; its relatively large tensor prediction was already disfavored against lower-tensor alternatives in that analysis.^[ref]

Adopt the approximate mass-luminosity relation $L/L_{\odot }\simeq (M/M_{\odot }{)}^3$ and equal available fuel fractions. The main-sequence lifetime then scales as $M/L$, givingUsing an exponent $3.5$ instead gives about $5.4\times 10^7$ and $3.1\times 10^8$ years; these are model-dependent estimates. At $r=3r_s=6GM/c^2$ the Newtonian gravitational potential energy released from infinity is $GMm/r=m{c}^2/6$. Half remains as kinetic energy in a circular orbit, so the radiatable binding energy is $GMm/(2r)$:Using the whole potential energy loss would give $1/6$, which does not subtract the final kinetic energy. A fully relativistic calculation for a nonrotating black hole gives $1-\sqrt{8/9}\simeq 5.7\%$. When the surviving star exhausts hydrogen burning, it expands off the main sequence and can fill its Roche lobe. Binary mass transfer to the black hole can form an accretion disk and an X-ray binary; unstable binary mass transfer may instead produce a common envelope. This is conditional on the binary star remaining bound after the first remnant forms. The supplied black hole endpoint is an assumption, not a universal outcome for an isolated $8M_{\odot }$ star.

For a particle on a circle, the tangential speed is $v=r\dot{\varphi }$, the moment of inertia is $I=m{r}^2$, and the angular momentum is $L=I\dot{\varphi }=mrv$. Hence its kinetic energy isThere is no potential energy variation along the circle in the stated model.