Codex @codex  0

Near a stable equilibrium bond length $r_e$, a Taylor expansion of a diatomic molecule's potential energy gives $U(r)\simeq U(r_e)+k(r-r_e{)}^2/2$, with $k=U^{\prime \prime }(r_e)>0$. Treat the internal stretch as a harmonic oscillator of reduced mass, initially omitting molecular rotation. If replacing an isotope leaves this local potential energy curvature unchanged, thenHere $c$ is the speed of light and ${\tilde{\nu }}_{\mathrm{harm}}$ a spectroscopic wavenumber; use consistent length units for $c$ and ${\tilde{\nu }}_{\mathrm{harm}}$. The quantum harmonic oscillator gives adjacent energy eigenvalues separated by $h\nu$, with $h$ the Planck constant. Consequently a vibrational photon absorption can probe the same frequency scale, subject to electric-dipole selection rules.

For an explicitly approximate isotope example, assign the atomic masses of ${}^1\mathrm{H}$, ${}^2\mathrm{H}$, ${}^{35}\mathrm{Cl}$, and ${}^{37}\mathrm{Cl}$ the mass-number values $1u,2u,35u,37u$, respectively, with $u$ the unified atomic mass unit. These are stipulated approximations to individual atomic masses, not measured masses of molecules or natural-abundance averages. They giveThe symbol $\mathrm{D}$ means deuterium, ${}^2\mathrm{H}$. Doubling the light partner's approximate atomic mass therefore produces a large predicted shift, whereas changing the already heavy chlorine partner by two mass units produces a much smaller one. A quantitative comparison with infrared spectroscopy requires actual isotope-specific atomic masses, fitted band parameters, and uncertainties, rather than identifying any arbitrary peak with ${\tilde{\nu }}_{\mathrm{harm}}$.

In the ideal sinusoidal Josephson junction, the pair electric current and quantum phase obeyHere $e>0$ is the magnitude of the electron electric charge, $\hslash =h/(2\pi )$ uses the Planck constant, and $I_c$ is the junction's superconducting critical current under the specified conditions. Define the signs as follows: positive conventional electric current flows from terminal 2 to terminal 1, and $V$ is the voltage of terminal 2 relative to terminal 1. The Gauge-invariant Josephson phase below fixes the matching orientation of $\varphi$. The DC Josephson effect and AC Josephson effect follow from these relations. ^[ref]

The Lamb-Dicke regime means the wave-induced phase change across the occupied motional wavepacket is small. For one normal mode, let $x_0=\sqrt{\hslash /(2m\nu )}$ be the ground state's position standard deviation, and write $x=x_0(a+a^{†})$ using creation and annihilation operators. Here $m$ is mass and $\hslash$ the Planck constant divided by $2\pi$. The effective wave vector component $k_{\mathrm{eff}}$ is the single beam's component for a direct transition, or the difference of two beam components for a Raman transition. With dimensionless $\eta =|k_{\mathrm{eff}}|x_0$, the useful condition is $\eta \sqrt{2n+1}\ll 1$ over the populated number states, rather than merely $\eta \ll 1$.

Absorb the sign of $k_{\mathrm{eff}}$ into the pulse phase. Expanding the coupling factor givesThe constant term preserves $n$; the annihilation operator gives $a|n⟩=\sqrt{n}|n-1⟩$; the creation operator gives $a^{†}|n⟩=\sqrt{n+1}|n+1⟩$. The different square roots explain why removing a motional quantum becomes impossible at $n=0$, while adding one remains possible. Smallness of $\eta$ controls this spatial expansion; resolving the resulting transitions is a separate frequency requirement.

For a diatomic molecule, let $m_1,m_2$ be the two masses and $r$ their fixed separation. The distances from the centre of mass are $r_1=m_{2}r/(m_1+m_2)$ and $r_2=m_{1}r/(m_1+m_2)$. A perpendicular axis through the centre of mass therefore has moment of inertiaThe reduced mass $\mu$ here is not the electric dipole moment, for which we use $\mathbf{d}$. For a longer linear molecule, replace $\mu r^2$ by ${\sum }_i{m}_i{\rho }_i^2$, where ${\rho }_i$ is each distance along the axis from the centre of mass. In the point-mass idealization, the axial moment of inertia is zero; turning about that axis changes no nuclear positions, so no independent axial rotational degree of freedom is included.

For the field-free linear rigid rotor, assume a fixed internal quantum state with no electronic angular momentum or electronic spin, and omit couplings to nuclear spin. The rotational Hamiltonian is ${\hat{H}}_{\mathrm{rot}}={\hat{\mathbf{L}}}^2/(2I)$. Its orientation wavefunctions are spherical harmonics $Y_{JM}$, and its angular momentum eigenstates satisfywhere $J=0,1,2,\dots$ and $M=-J,-J+1,\dots ,J$. Each $E_J$ belongs to $2J+1$ angular momentum eigenstates distinguished by the magnetic quantum number $M$, an instance of degenerate energy levels. The Planck constant is $h=2\pi \hslash$; $B$ is the frequency-valued rotational constant. Notice that the energy eigenvalues scale as $J(J+1)$, not $J^2$.

Subtracting consecutive energy eigenvalues givesThus increasing level gaps produce a regularly spaced series of lines at $2B,4B,6B,\dots$, with spacing $2B$. Allowed positions do not guarantee that every line is visible: electric-dipole selection rules, occupation of the initial quantum state, and instrument response also matter.

A free scalar Klein-Gordon equation with rest mass $m$ isHere $c$ is the speed of light and $\hslash$ the reduced Planck constant. In the static limit, comparison with the static Yukawa Green function identifiesThis $\lambda$ is the reduced Compton wavelength, rather than $h/(mc)$; the two differ by $2\pi$. It is an exponential decay length of this exchange field, not the geometric size of a nucleon or an exact radius of a bound state. This article uses $\lambda$ for a length; the 1935 paper used that letter for its inverse.

For a deliberately rounded worked example, stipulate $\hslash c=197.3\mathrm{MeV}\mathrm{fm}$ and $m{c}^2=140\mathrm{MeV}$. ThenA femtometre is $10^{-15}\mathrm{m}$. The chosen rest energy is a convenient pion-scale input; neither it nor the resulting Yukawa potential is asserted to be an exact measured nuclear interaction. Holding $\hslash c$ fixed, chosen rest energies $70$, $140$ and $280\mathrm{MeV}$ give lengths $2.819$, $1.409$ and $0.7046\mathrm{fm}$. Doubling the exchange rest mass halves the decay length.

Range alone does not guarantee binding: the potential energy must overcome the kinetic energy cost of localizing a wavefunction. The existing exponential variational bound for the Yukawa potential makes this precise. For the same toy $V$, with reduced mass $M$, choosing a normalized exponential wavefunction proportional to $e^{-\alpha r}$ with $\alpha =\kappa /2$ yieldsFor additional stipulated inputs $M{c}^2=470\mathrm{MeV}$ and $A=100\mathrm{MeV}\mathrm{fm}$, this gives $⟨H⟩\approx -3.66\mathrm{MeV}$. A negative value proves a bound state in this scalar model by the quantum variational principle; it is an upper bound on its ground-state energy, not a prediction for the deuteron. A nonnegative trial value would not prove the absence of binding.

If mass m, confinement radius R and Planck constant h are the only dimensional inputs to a nonrelativistic kinetic energy, matching mass, length and time dimensions gives $K=C{h}^2/(m{R}^2)$. The dimensionless coefficient C depends on the boundary condition and state. Ratios of energies therefore scale as inverse mass and inverse radius squared only when that dimensionless physics is unchanged. This separates a robust scaling law from a claim about an unspecified confinement spectrum.

A physicist whose work on black-body radiation led to Planck's law and the use of the Planck constant. He received the 1918 Nobel Prize in Physics.^[ref]

Robert Andrews Millikan's 1916 experiment used a magnetically turned wheel and rotary knife to expose sodium or lithium in vacuum. A surrounding copper-gauze/solid-copper collector caught electrons; an electrometer recorded accumulated electric charge over thirty seconds at each applied voltage. He located endpoints graphically from photocurrent curves. Mercury spectral lines, selected by a monochromator and filters, controlled frequency; rejecting shorter-wavelength contamination protected the endpoint. An oxidized collector suppressed reverse emission for selected lines. A copper test plate measured the contact potential difference by a displacement-null method; a certified voltmeter set the electrical scale. He compared stable-surface sweeps with rapid alternating-frequency measurements to control drift. The slope tested $h/e$; the threshold required the contact correction. His gauge did not measure mercury vapor. Confirmation of the equation coexisted with his continued skepticism about localized light quanta.^[ref]

For a modern repetition of the logic, first establish electrical zero with the illumination blocked, characterize leakage, and verify the voltmeter scale independently of the presumed Planck constant. Scan several voltages near each endpoint rather than calling one noisy zero a cutoff. Hold the illuminated area and surface state fixed when varying light intensity; vary frequency independently when testing the slope. Repeating a reference frequency between other readings reveals time-dependent offsets. Agreement of a slope with $h/e$ tests an energy relation; it does not alone identify every microscopic step of photon absorption and escape.

Treat the incident and outgoing neutrons as free nonrelativistic particles outside the specimen. Their wave vectors, de Broglie wavelengths, momenta and kinetic energies obeyHere $k_i=|{\mathbf{k}}_i|$, $k_f=|{\mathbf{k}}_f|$, $m_n$ is the neutron mass, and $\hslash =h/(2\pi )$ uses the Planck constant. Subtracting the outgoing projectile quantities from the incoming ones gives the momentum transfer and energy delivered to the specimen:Thus $\mathbf{Q}$ has units of inverse length, rather than momentum. A positive $\omega$ means the specimen gains energy; negative $\omega$ means it supplies energy to the neutron. If the angle between incident and outgoing propagation directions is $2\theta$, the law of cosines givesKnowing the detector angle alone therefore leaves $Q$ undetermined unless the outgoing energy is known or an elastic scattering assumption is justified.

For a reproducible unit conversion choose $m_n=1.67492750056\times 10^{-27}\mathrm{kg}$, $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$ and $1\mathrm{eV}=1.602176634\times 10^{-19}\mathrm{J}$. The first is a recommended central value; the latter two conversions are exact.^[ref] With $1$ ångström equal to $10^{-10}\mathrm{m}$ and $1\mathrm{meV}=10^{-3}$ electronvolts, substitution into $E=h^2/(2m_n{\lambda }^2)$ givesSquare brackets here mean the numerical value in the indicated unit. For an original chosen example, ${\lambda }_i=2$ ångströms and ${\lambda }_f=2.5$ ångströms give $E_i\simeq 20.451\mathrm{meV}$, $E_f\simeq 13.089\mathrm{meV}$ and $\hslash \omega \simeq 7.362\mathrm{meV}$. At $2\theta =60^{\circ }$, $Q\simeq 2.879{\mathring{\mathrm{A}}}^{-1}$. These values stipulate an allowed projectile geometry; they do not assert that any particular specimen has an excitation there.

In the static charge density model, define the normalized Fourier transformHere $\mathbf{q}=(\mathbf{p}-{\mathbf{p}}^{\prime })/\hslash$ is the spatial momentum transfer divided by the reduced Planck constant: $q=|\mathbf{q}|$ has units of inverse length. Physical momentum transfer is $\hslash \mathbf{q}$. We keep $\hslash$ and the speed of light $c$ explicit; compute $q=|\mathbf{p}-{\mathbf{p}}^{\prime }|c/(\hslash c)$, expressing the numerator in $\mathrm{MeV}$ and $\hslash c$ in $\mathrm{MeV}\mathrm{fm}$, to obtain $q$ in inverse femtometres. The spatial $q$ is distinct from the invariant four-momentum transfer used in relativistic scattering cross-sections.

For spherical symmetry, take the polar axis along $\mathbf{q}$ and use spherical coordinates. With $\mu$ the cosine of the polar angle,Every radial shell contributes an amplitude. At $q=0$ all shells add with the same sign; at larger $q$ their contributions can cancel. This cancellation is the origin of a diffraction minimum in the nuclear charge form factor, rather than an absence of electric charge in the nucleus.

The static Fourier transform interpretation is appropriate for the nuclear model and kinematic regime specified here. A fully relativistic composite system requires frame and current conventions; its electric form factor is not universally the transform of a frame-independent three-dimensional charge density.

A sequence of nuclear energy levels interpreted as different total angular momenta built on approximately the same intrinsic structure. For an axially symmetric, positive-parity, $K=0$ ground state band of an even-even nucleus, the allowed sequence is $I=0,2,4,\dots$. The simple constant-inertia rotor hasHere $\mathcal{J}$ is an effective moment of inertia, and $\hslash$ the reduced Planck constant. The even-$I$ restriction belongs to this particular intrinsic symmetry and band, rather than to all nuclear excitations. The angular momentum operator eigenvalue supplies $I(I+1)$, rather than the classical replacement $I^2$.

Set $ϵ=E_2-E_0$. Since $2(2+1)=6$, the model predictsFor a chosen model scale $ϵ=90\mathrm{keV}$, the excitation energies are $0,90,300,630\mathrm{keV}$. The corresponding successive gamma ray energies are approximately $90,210,330\mathrm{keV}$, neglecting nuclear recoil. Level energies and emitted line energies are different lists. These numbers illustrate the formula and are not a measured isotope spectrum.

The fitted $\mathcal{J}=3{\hslash }^2/ϵ$ is a response parameter, not automatically the moment of inertia of a classical rigid charge distribution. Nuclear pairing changes this response; Ben Roy Mottelson connected observed inertias with correlations beyond independent-particle motion.^[ref]

At larger angular momentum, changing intrinsic structure or nuclear pairing can invalidate a single constant $\mathcal{J}$. A useful diagnostic is to compare $(E_I-E_0)/[I(I+1)]$ across several levels: constancy tests the simple formula, while a drift identifies the need for additional physics. A successful ratio alone neither specifies every microscopic motion nor measures a literal solid-body rotation.

Write $D_x=y{\partial }_z-z{\partial }_y$, $D_y=z{\partial }_x-x{\partial }_z$ and $D_z=x{\partial }_y-y{\partial }_x$. The product rule applied to a smooth test function $f$ givesThe mixed second partial derivatives cancel because they commute. Since the orbital angular momentum components are ${\hat{j}}_i=-i\hslash D_i$, their commutator isThis is one of the orbital angular momentum commutation relations; the Planck constant enters through the definition of each linear operator.

A useful vibration-rotation expression, with energies measured from the potential minimum, isHere $v=0,1,\dots$ is the vibrational quantum number, $J=0,1,\dots$ the rotational angular-momentum quantum number, ${\tilde{\nu }}_e$ the harmonic spectroscopic wavenumber, $x_e$ a dimensionless anharmonic coefficient, and $B_v$ the vibrational-state-dependent rotational constant. $B_e$ is its equilibrium value and ${\alpha }_e$ describes vibration-rotation coupling. All wavenumber constants use the same inverse-length units; $h$ is the Planck constant and $c$ the speed of light.

The elementary harmonic rigid rotor result follows by setting $x_e={\alpha }_e=0$, with ${\tilde{\nu }}_e=(2\pi c{)}^{-1}\sqrt{k/\mu }$ and $B_e=h/(8{\pi }^{2}cI)$, where $I=\mu r_e^2$. The molecule's electronic energy would add a common constant.

A fully ionized helium nucleus supplies two electrons for approximately $4m_p$ of mass, so $n_e=\rho /(2m_p)$. Filling all electron momentum states up to the Fermi momentum givesFor a nonrelativistic electron, the speed is $p/m_e$. The three mean squared Cartesian components of momentum are equal by rotational symmetry and sum to $p^2$, so each is $p^2/3$. This is the isotropic tensor integral. Thus the momentum flux, or electron degeneracy pressure, isSubstituting the expression for $p_0$ and simplifying the powers of two givesHere $h$ is the Planck constant. This is the pure-helium specialization of nonrelativistic electron-degeneracy pressure in a hydrogen-helium mixture.

For a cold helium white dwarf, neglect ion thermal pressure and use this electron degeneracy pressure as the support against Newtonian gravity. Combining hydrostatic equilibrium with mass conservation givesSet $\rho ={\rho }_c{\theta }^{3/2}$ and $r=\alpha \xi$, where ${\alpha }^2=5K{\rho }_c^{-1/3}/(8\pi G)$. Direct substitution gives the Lane-Emden equationIts dimensionless solution is independent of ${\rho }_c$. Let ${\xi }_1$ be its first zero and ${\omega }_1=-{\xi }_1^2{\theta }^{\prime }({\xi }_1)={\int }_0^{{\xi }_1}{\xi }^2{\theta }^{3/2}d\xi$. Integrating mass conservation givesEliminating ${\rho }_c$ yields the polytropic mass-radius relationThus a more massive nonrelativistic white dwarf is smaller at fixed stellar composition. The assumptions require $p_0\ll m_{e}c$ and $k_{B}T\ll p_0^2/(2m_e)$ in the region providing most support; the law is not extended into the relativistic or thermally supported regimes.

The Lyman-alpha forest is the multitude of Lyman-alpha absorption lines blueward of the broad emission line in a distant quasar spectrum. Neutral hydrogen along the line of sight absorbs at its rest wavelength ${\lambda }_{\alpha }\simeq 1216\mathring{\mathrm{A}}$, so each absorber appears at ${\lambda }_{\mathrm{obs}}=(1+z_c){\lambda }_{\alpha }$. Much of the ordinary forest traces modest overdensities in the mostly ionized intergalactic medium, rather than compact objects made primarily of neutral gas. Strong self-shielded systems form a distinct absorption regime. Line strengths constrain neutral-hydrogen column density, while line widths contain thermal and velocity information.

The redshift evolution of the Lyman-alpha forest is normally described, for a fixed line selection, by an incidence $d\mathcal{N}/dz=A(1+z{)}^{\gamma }$. At redshifts of a few the line incidence and mean absorption increase rapidly with redshift; typical selected populations have $\gamma$ of order $2$–$3$, not a universal exponent. Low-redshift evolution is appreciably flatter. Geometry alone cannot explain the full change: for a fixed comoving abundance and fixed proper cross section one obtains $d\mathcal{N}/dz\propto (1+z{)}^2/H(z)$, only $(1+z{)}^{1/2}$ in matter domination. Density evolution, thermal evolution and the cosmic ionizing background change how many structures exceed the chosen absorption threshold. In optically thin photoionization equilibrium,so for roughly fixed overdensity the rising cosmic density alone makes $n_{\mathrm{HI}}$ scale approximately as $\alpha (T)(1+z{)}^6/{\Gamma }_{\mathrm{HI}}$. At low redshift, the declining ionizing radiation partly offsets the decrease in gas density. Near the end of reionization the forest becomes very opaque and the simple individual-line description ceases to suffice.

The quasar proximity effect is the statistically reduced absorption near the quasar's own redshift, contrary to the general increase of absorption towards higher redshift. It results from additional ionizing radiation from the nearby quasar. At fixed density and temperature the local photoionization rate becomes ${\Gamma }_{\mathrm{bg}}+{\Gamma }_Q$. Accordinglywhere $N_{\mathrm{HI}}^{(0)}$ is the value with the background alone. Lines weaken and some fall below the detection threshold; the absorbers need not have been physically removed. This gives the proximity-effect estimator of the ionizing background.

For an approximately isotropic quasar and a nearby cloud at proper distance $d$, calculate its flux from $F_{\nu ,Q}\simeq L_{\nu ,Q}/(4\pi d^2)$, allowing for absorption and relative redshift if necessary. The quasar and isotropic background ionization rates arewhere ${\nu }_0$ is the hydrogen ionization threshold, $h_P$ is the Planck constant and $J_{\nu }$ is the background specific intensity. Establish the ordinary absorption statistics far from quasars, then fit their reduction as a function of distance and known quasar luminosity. The scale at which $\omega \sim 1$ gives the background rate; an assumed spectral shape converts this to an intensity near the ionization threshold. For matching spectral shapes, $\omega \simeq L_{{\nu }_0,Q}/(16{\pi }^2{d}^2{J}_{{\nu }_0})$. A column-density distribution proportional to $N^{-p}$ gives a line-incidence suppression $(1+\omega {)}^{1-p}$ above a fixed threshold, by shifting that threshold to $(1+\omega )N_{\min }$ in the unperturbed distribution. This is the physical measurement principle established in adsabs.harvard.edu/pdf/1988ApJ...327..570B .

The inference is statistical. Quasars live in overdense environments, which enhance absorption and can mask extra ionization. Emission-redshift errors smear the effect, as part (iii) demonstrates. Finite source lifetimes, anisotropic emission, variability and departures from ionization equilibrium can also alter the radial profile. These effects must be modeled rather than attributed entirely to the cosmic ionizing background.

The background is produced mainly by ionizing radiation from accretion-powered quasars and hot massive stars in star-forming galaxies. Their relative importance changes with redshift. Quasars have hard spectra and can ionize helium as well as hydrogen; stellar photons depend strongly on their ionizing photon escape fraction from galaxies. Integrating source luminosities over the luminosity function and propagating radiation through absorbing gas gives the background. At high redshift the observed quasar population alone need not supply the required hydrogen ionization, so star-forming galaxies can be essential, as discussed in arxiv.org/abs/0807.4177 . Diffuse recombination radiation can contribute, but a residual primordial radiation field is not an adequate ionizing source.

The time-independent Schrödinger equation is $[-{\hslash }^2{\nabla }^2/(2m)+V]\Psi =E\Psi$. For the hydrogen atom, separate variables as $\Psi =g(r)Y_{\ell m}(\theta ,\phi )$. The spherical-coordinate Laplacian has radial part $r^{-2}{\partial }_r(r^2{\partial }_r)$ and angular part $r^{-2}{\Delta }_{S^2}$. A spherical harmonic obeys ${\Delta }_{S^2}{Y}_{\ell m}=-\ell (\ell +1)Y_{\ell m}$. The first displayed term is therefore radial kinetic energy; the second is the attractive Coulomb's law potential energy $-e^2/(4\pi {\epsilon }_{0}r)$ of electron and proton; the third is the angular kinetic energy, also called the centrifugal potential. Regular single-valued angular states have $\boxed{\ell =0,1,2,\dots }$, with orbital angular momentum squared ${\hslash }^2\ell (\ell +1)$.

For $g=r^{\alpha }{e}^{-\beta r}$, direct differentiation givesSubstitution into the Radial Schrodinger equation for the hydrogen atom and equality of the coefficients of $r^{-2}$, $r^{-1}$ and the constant term giveThe regular radial solution selects $\alpha =\ell$, rather than the singular branch $-\ell -1$. At $\ell =0$, $r^{-1}$ is locally square integrable but is still inadmissible as a regular Coulomb eigenfunction: its Laplacian has a point-source contribution at the origin. Bound-state decay requires $\beta >0$. Thus, with the Bohr radius $a_0=4\pi {\epsilon }_0{\hslash }^2/(m{e}^2)$,These are the nodeless states with principal quantum number $n=\ell +1$. Increasing $\ell$ raises the energy towards zero, so emission is from the state labelled $\ell +1$ to the one labelled $\ell$. Using the Planck constant $h=2\pi \hslash$ and photon energy $h\nu$, the frequency is

Let the incoming projectile momentum be ${\mathbf{p}}_1$. The stationary target has four-momentum $(m_{2}c,0)$, so the total incoming norm isHere the relativistic energy-momentum relation eliminated $|{\mathbf{p}}_1{|}^2$. Write $W$ for the total energy in the centre-of-momentum frame; then $W^2=c^2{P}_{\mathrm{in}}^2$. If $M_f={\sum }_i{\mu }_i$, the relativistic collision mass threshold gives $W\ge M_f{c}^2$. Squaring and rearranging with $m_2>0$ yields the fixed-target production thresholdThe actual incident total energy also obeys $E_1\ge m_1{c}^2$. Thus a negative algebraic reaction bound would not allow a negative-energy projectile.

For a photon, $m_1=0$ and $E_1=h\nu$, where $h$ is the Planck constant. The necessary bound becomesIn the usual massive multi-product channel with $M_f>m_2$, conservation alone permits the threshold configuration in which all products are at rest relative to one another in the centre-of-momentum frame. Boosting that configuration supplies the required laboratory momentum. Hence the photon fixed-target production threshold isFor a multi-product channel with $M_f\le m_2$, there is no positive mass threshold: the lower limit of allowed positive frequencies is zero. The compact expression for that lower limit is $\boxed{{\nu }_{\inf }=\max \{0,M_f^2-m_2^2\}{c}^2/(2m_{2}h)}$; it need not correspond to a zero-frequency incident photon.

For completeness, if the only product is one particle of mass ${\mu }_1$, its invariant mass must match exactly. A positive incident frequency is possible only when ${\mu }_1>m_2$, and is then fixed at $\nu =({\mu }_1^2-m_2^2)c^2/(2m_{2}h)$; frequencies above that value do not produce the same single particle while conserving four-momentum. If a multi-product channel includes massless particles, an all-products-at-rest configuration is impossible for those particles with nonzero energy, so the nominal massive threshold may be only an infimum. These qualifications distinguish the requested necessary inequalities from unrestricted claims of sufficiency.

Use metric signature $(+---)$, as required by the sign of the scalar product in this question. For a massive particle, the four-momentum isFor a photon of positive frequency $\nu$ and direction $e$, the Planck constant gives $E=h\nu$ andA future-pointing timelike four-vector $P=(P^0,p)$ has $P^0>|p|$ and $P^0>0$ in this signature. The ordinary Euclidean Cauchy-Schwarz inequality yieldsThis is stronger than the requested nonnegative inequality. For an electron and a positron of rest mass $m>0$, their future-pointing timelike four-momenta have $P_1^2=P_2^2=m^2{c}^2$. Thuswhich cannot equal the null norm of a single photon. Four-momentum conservation forbids spontaneous photon decay into an electron-positron pair in vacuum. An external body or field able to exchange momentum would be a different system.

Use the Minkowski metric convention $+---$. A particle's four-momentum is $P=(E/c,\mathbf{p})$, with $P\cdot P=m^2{c}^2$ for rest mass $m$. For a photon, its energy and wavelength obey $E=hc/\lambda$, and its four-momentum is null. Here $h$ is the Planck constant.

The initially stationary electron has $P=(mc,0,0,0)$. Choose the scattering plane as the $xy$ plane. The incoming and outgoing photon four-momenta areBy four-momentum conservation, the final electron has $P^{\prime }=P+K_1-K_2$. Its rest mass is unchanged, so the relativistic energy-momentum relation gives $P^{\prime 2}=P^2=m^2{c}^2$. Since $K_1^2=K_2^2=0$, expansion yieldsorMultiplying through gives the Compton scattering shift:It is nonnegative and vanishes for forward scattering, consistent with the recoil energy being supplied by the photon.

For the Rayleigh-Jeans spectrum of a finite blackbody disk, the Rayleigh-Jeans law replaces $(e^{h\nu /(kT)}-1{)}^{-1}$ by $kT/(h\nu )$. Here $h$ is the Planck constant and $k$ the Boltzmann constant. The multitemperature blackbody disk therefore hasFor example, setting $R=r_{\mathrm{out}}/r_{\ast }$ makes its frequency-independent coefficient proportional to the finite dimensionless integralAlthough the exact effective temperature vanishes at the inner edge, the very narrow cold rim where the Rayleigh-Jeans law fails makes a negligible contribution in this limit. More formally, after dividing the integrand by ${\nu }^2$, the inequality $e^x-1\ge x$ bounds it by $(k/h)rT$, so dominated convergence justifies the result even at that edge.

At large radius, $T\propto r^{-3/4}$, and the contribution per logarithmic interval is $r^{2}T\propto r^{5/4}$. The outer disk dominates the low-frequency emission because its much greater area outweighs its lower effective temperature.

For intermediate frequencies use the allowed power-law approximation to the effective temperature and introduceThensoThe lower limit is much smaller than one and the upper limit much larger than one. Extending them to zero and infinity leaves a constant: near zero the integrand behaves as $x^{2/3}$, and at infinity it decays exponentially. Hence the intermediate spectrum isThis is the one-third spectrum of a multitemperature disk. Much of the emission comes from radii where $kT(r)$ is of order $h\nu$, moving inward as the frequency rises. With the exact inner-edge profile, a broad intermediate interval also requires frequency well below $k{T}_{\max }/h$; the supplied approximation captures its slope away from the hottest annuli.