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A diatomic molecule has one internal bond-stretching vibrational coordinate. Near its equilibrium separation, approximate the bond by a spring of stiffness $K$ and use the reduced mass $\mu =m_1{m}_2/(m_1+m_2)$. Its frequency is $\omega =\sqrt{K/\mu }$ and its quantum harmonic oscillator levels are ${ϵ}_r=(r+1/2)\hslash \omega$.

For independent molecules in thermal equilibrium, set $x=\hslash \omega /(k_{B}T)={\theta }_v/T$. The single-molecule canonical partition function is a geometric series:Using $U=-\partial \log (z_v^N)/\partial \beta$, where $\beta =1/(k_{B}T)$, givesPer mole replace $N{k}_B$ by $R$. The zero-point energy is present in $U_v$ but is temperature-independent and contributes nothing to the heat capacity.

For $T\ll {\theta }_v$, occupation of the first excited vibrational state is exponentially suppressed: $C_{V,v}\simeq N{k}_B{x}^2{e}^{-x}\to 0$. The vibrational degree of freedom is frozen out. For $T\gg {\theta }_v$, $C_{V,v}\to N{k}_B$. This recovers classical equipartition: the quadratic kinetic term and quadratic elastic term each contribute $k_B/2$ per molecule.

The vibrational heat capacity of a diatomic gas must be distinguished from the total heat capacity. In a dilute gas with rotations already active and vibration still frozen, translation contributes $3R/2$ and rotation $R$, so molar $C_V\simeq 5R/2$. Once vibration is active it adds $R$, giving $C_V\simeq 7R/2$. For an ideal gas of a fixed number of intact molecules, $C_P=C_V+R$, so the corresponding values are $7R/2$ and $9R/2$. Very low temperatures also freeze rotations, while sufficiently high temperatures bring anharmonicity, dissociation or electronic excitations; the harmonic plateau is restricted to the range where those effects are negligible.

Use the nonrelativistic Coulomb potential energy $V=-Z{e}^2/(4\pi {ϵ}_{0}r)$, with the nucleus treated as fixed. Since the wavefunction has zero orbital angular momentum, its Laplacian isSubstituting into the time-independent Schrödinger equation givesFor a constant energy eigenvalue, the coefficient of $1/r$ must vanish. Therefore $\boxed{\beta =Z/a_0}$, where $a_0$ is the Bohr radius. Wavefunction normalization then requiresChoosing the irrelevant overall phase so that $\alpha$ is real and positive gives $\boxed{{\alpha }^2={\beta }^3/\pi }$. The remaining constant term is the hydrogen ground-state energyReplacing $m_e$ by the electron-nucleus reduced mass gives the usual finite-nuclear-mass refinement.

For an attractive inverse-square potential, let the reduced mass be $\mu$, potential $-k/r$, and nonzero orbital angular momentum $\ell$. Conservation of angular momentum makes the orbit planar. The orbit equation isPositive relative energy gives $e>1$ and an unbound hyperbolic Kepler orbit, arriving from and returning to infinity. Its asymptotes satisfy $\cos {\phi }_{\infty }=-1/e$. At infinity $E=\mu v_{\infty }^2/2$ and $\ell =\mu b{v}_{\infty }$ for impact parameter $b$, so the deflection angle isThis describes gravitational encounters in the Newtonian two-body approximation. The body speeds up near periapsis while total energy remains positive. It is distinct from an elliptic bound orbit and from the zero-energy parabolic limit; repulsive Coulomb scattering also gives hyperbolae but uses a different orientation of the force and orbit.

For two bodies interacting through $U({\mathbf{r}}_1-{\mathbf{r}}_2)$, defineTheir kinetic energy separates exactly:Without external forces, the centre of mass moves uniformly while $\mu \ddot{\mathbf{r}}=-\nabla U$. The relative orbit is a one-body problem with reduced mass $\mu$, not generally either individual mass. For $m_2\gg m_1$, $\mu \simeq m_1$; for equal masses $m$, $\mu =m/2$. If the interaction is central, relative angular momentum is conserved and the relative motion is planar.

A central potential conserves angular momentum and confines a nonradial orbit to a plane. With reduced mass $\mu$, the radial motion obeysTurning points satisfy $E=U_{\mathrm{eff}}$. A radially bounded allowed interval gives bound motion, whereas an interval reaching infinity permits escape. The energy sign alone classifies these only with suitable assumptions about $U(\infty )$ and barriers. Circular orbits satisfy $U_{\mathrm{eff}}^{\prime }=0$ and are radially stable when $U_{\mathrm{eff}}^{\prime \prime }>0$. Bound orbits in a general central potential need not close: successive periapses can precess.

For attractive inverse-square gravity, $U=-\kappa /r$ andEllipses have $E<0$, $e<1$, and the attracting centre at a focus; parabolic and hyperbolic escape have $e=1$ and $e>1$. Conservation of angular momentum gives equal swept areas in equal times. The inverse-square law additionally yields $T^2=4{\pi }^2{a}^3/[G(m_1+m_2)]$. These are Kepler's laws; a harmonic central potential is another example with closed bound orbits, while generic central laws produce precession.

In a gravitational assist, a spacecraft's asymptotic speed relative to a planet is approximately unchanged but its direction turns. Transforming to the frame in which the planet has velocity $\mathbf{V}$ givesThe spacecraft can gain or lose orbital energy; the planet supplies the compensating energy and momentum, with a very small velocity change because of its large mass.

Repulsive Coulomb scattering instead has $U={\kappa }_C/r$. For incident relative kinetic energy $E$ and impact parameter $b$, the hyperbolic orbit givesThis Rutherford scattering formula follows by equating the incident annulus $2\pi bdb$ to the scattered solid-angle ring. It assumes classical elastic scattering by a point Coulomb centre; screening, finite nuclear size or quantum effects can change the applicable regime.

For the attractive inverse-square force, choose zero potential energy at infinity, so $U=-A/r$. At either radial turning point $r_p,r_a$, the kinetic energy is purely tangential, with conserved angular momentum $J$. ThusIts two roots are the periapsis and apoapsis distances. Their sum is $r_p+r_a=-A/E$, whereas the ellipse has $r_p+r_a=2a$. Therefore $a=-A/(2E)$, with $E<0$ for a nondegenerate bound ellipse. The circular case follows by the double-root limit.

For the two-body problem, define $\mathbf{r}={\mathbf{r}}_B-{\mathbf{r}}_A$, total mass $M=m_A+m_B$, reduced mass $\mu =m_A{m}_B/M$, and center of mass $\mathbf{R}$. The relative equation is $\mu \ddot{\mathbf{r}}=\mathbf{F}$, while the isolated center of mass has $\ddot{\mathbf{R}}=0$. Once the relative orbit is known,The relative energy uses $\mu$ in its kinetic energy and $A=G{m}_A{m}_B$ in the gravitational potential energy.

In the equal-mass case the initial relative speed is $u$, $\mu =m/2$, and the center-of-mass velocity is half B's initial velocity. ThusNegative relative energy requires $u^2<4Gm/a_1$. The escape threshold is $u_E=2\sqrt{Gm/a_1}$; equality is marginal parabolic escape, so a strictly bound pair has $u<u_E$.

For $u=u_E/2=\sqrt{Gm/a_1}$, $E_{\mathrm{rel}}=-3G{m}^2/(4a_1)$ and the relative semimajor axis is $a=2a_1/3$. The initial transverse speed is below the relative circular value $\sqrt{2Gm/a_1}$, so the initial turning point is apoapsis. HenceChoose the initial separation along $+x$ from A to B, with B's initial velocity $+u\hat{\mathbf{y}}$. Conservation of angular momentum gives relative periapsis speed $3u$. At the opposite end of the ellipse the direction is $-\hat{\mathbf{y}}$, soThus B has the same speed as initially but the opposite velocity. A has $2u\hat{\mathbf{y}}$ at that instant. A value $3u/2$ for B would be its center-of-mass-frame speed, not the requested initial laboratory frame result.

The relative orbital period follows from Kepler's third law with total gravitating mass $2m$:In the center-of-mass frame each star follows an ellipse of semimajor axis $a_1/3$ and eccentricity $1/2$, with the center of mass at a focus; they remain opposite on their simultaneous position vectors. In the laboratory frame add the uniform drift $ut/2$ along $y$. The laboratory paths of an equal-mass binary are therefore not closed ellipses, even though their relative motion repeats.

For two bodies interacting through a potential energy that depends only on their separation, introduce the centre of mass coordinate and the two-particle relative coordinate:Their kinetic energy separates exactly:The centre of mass moves uniformly if there is no external force, while the internal two-body problem reduces to one effective particle of reduced mass $\mu$:The original trajectories follow from ${\mathbf{r}}_1=\mathbf{R}+(m_2/M)\mathbf{r}$ and ${\mathbf{r}}_2=\mathbf{R}-(m_1/M)\mathbf{r}$.

For Newtonian gravity, $U=-G{m}_1{m}_2/r$, so the relative acceleration is $\ddot{\mathbf{r}}=-GM\mathbf{r}/r^3$. Conserved relative angular momentum and energy then give the usual conic sections. For a bound orbit of relative semimajor axis $a$,The two bodies share this orbital period and orbit their common centre of mass with semimajor axes in inverse proportion to their masses. If $m_2\gg m_1$, then $\mu \simeq m_1$, recovering the fixed-heavy-centre approximation. Using $M$ as the internal inertial mass would confuse centre-of-mass translation with relative motion.

Molecular transport across gradients. A gas can transport momentum, thermal energy and species abundance even when its average bulk velocity is zero. Molecules arriving at a point carry properties acquired near their preceding collisions. A small gradient makes the populations arriving from opposite sides slightly different, producing transport from larger to smaller values of the carried property. A continuum transport law requires a mean free path $\lambda$ small compared with the gradient length, and relaxation times short compared with the observation time.

For a dilute hard-sphere gas,where $d$ is molecular collision diameter. Both speed and collision distance matter: high speed alone does not determine a transport coefficient.

A simple kinetic estimate makes the common structure explicit. For a quantity $a(y)$ carried per particle, a molecule moving at angle $\theta$ to $+y$ brings approximately $a(y-\lambda \cos \theta )$. Expanding to first order and averaging an isotropic velocity distribution givessince $⟨\cos \theta ⟩=0$ and $⟨{\cos }^2\theta ⟩=1/3$. This mean-free-path argument estimates the coefficients; a detailed collision calculation changes their numerical prefactors and distinguishes different relaxation rates.

Momentum transport and viscosity. For a shear flow with mean velocity $U_x(y)$, the transported directed momentum per molecule is $m{U}_x$. The net flux of $x$ momentum in the $+y$ direction isHere $\eta$ is dynamic viscosity in $\mathrm{Pa}\mathrm{s}$. The associated shear stress on a material surface has the corresponding traction sign convention; the flux carries momentum from the faster-moving layer towards the slower-moving one. The kinematic viscosity is $\nu =\eta /(nm)$, with units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$. Viscosity smooths velocity gradients, and the lost ordered mechanical energy becomes heat.

Heat transport. Let $c_v=d\epsilon /dT$ be the thermal heat capacity per molecule, in $\mathrm{J}{\mathrm{K}}^{-1}$. Substituting $a=\epsilon (T)$ gives Fourier's law,Thermal conductivity has units $\mathrm{W}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$. For a monatomic gas, translational equipartition gives $c_v=3k_B/2$. Rotational or vibrational excitations in other gases can increase the energy carried per molecule, while collisions govern how those modes exchange energy. A temperature gradient can therefore be transported differently from directed momentum even when the dimensional estimates look similar.

Species transport. For a dilute tracer or a mixture with prescribed total density and negligible bulk flow, molecules diffuse down concentration gradients. The first-gradient number flux is Fick's first law,The diffusivity $D$ has units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$; multiplying number flux by the species mass gives mass flux. More generally one uses diffusion flux relative to the mixture's bulk velocity and specifies mass or mole fractions. The approximation is not a claim that an arbitrary pure-gas pressure gradient can remain at rest while only a Fick flux acts. In a mixture, collision cross-sections and relative molecular speeds, including the reduced mass, determine the species diffusivities.

Dependence on density and temperature. The kinetic estimates of gas transport coefficients predict, for fixed collision diameter and molecular heat capacity,Thus dynamic viscosity and thermal conductivity are approximately density independent in the dilute continuum regime. Reducing density reduces the number of carriers but increases their mean free path in the same proportion. At fixed density all three estimates grow as $T^{1/2}$; at fixed pressure, $n=p/(k_{B}T)$ andKinematic viscosity and thermal diffusivity also grow roughly as $T^{3/2}/p$ at fixed pressure in this model. Real interaction potentials make effective collision cross-sections temperature dependent, and internal excitations change $c_v$. Dense-gas correlations and boundaries invalidate the simple cancellation. In particular, when the mean free path approaches the apparatus scale, transport becomes free molecular; it cannot be described by extending the same local, pressure-independent Fourier and viscosity coefficients indefinitely.

Heat diffusion equation. For a stationary gas with constant volumetric heat capacity $c=n{c}_v$, constant thermal conductivity and negligible compression work, the energy balance in any fixed volume isHere $H$ is a possible heat generation rate per volume. The divergence theorem and Fourier's law yieldThermal diffusivity $\alpha$ has units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$ and is of order $\overline{v}\lambda /3$. In a moving or compressing gas, the material derivative, pressure-volume work and viscous heating must also enter the internal-energy equation. The displayed pure diffusion equation describes the approximation in which those effects are absent or negligible.

Representative solutions. With no source, a steady one-dimensional slab between temperatures $T_a,T_b$ at $x=0,L$ has $T=T_a+(T_b-T_a)x/L$ and uniform heat flux $-\kappa (T_b-T_a)/L$. Thus a constant spatial gradient transports heat without changing local temperature in time when the boundaries continually supply and remove it.

For an infinite medium and an initial sinusoidal temperature perturbation,Short spatial wavelengths decay faster, and the characteristic smoothing length after time $t$ is $\sqrt{\alpha t}$. A finite slab with both ends held at $T_b$ has a sine expansion with factors $e^{-\alpha (n\pi /L{)}^{2}t}$; its slowest relaxation time is $L^2/({\pi }^2\alpha )$.

An instantaneous heat input $Q_A$ per unit transverse area at $x=0$ in an otherwise uniform infinite medium gives the Gaussian solutionIts integral $\int c(T-T_b)dx=Q_A$ conserves the deposited heat. The peak falls as $t^{-1/2}$ while its width grows as $\sqrt{t}$; diffusion redistributes energy instead of transporting a sharp front at a single velocity.

Finally, a semi-infinite body initially at $T_i$ and held at surface temperature $T_s$ from $t=0$ hasIt has $T(0,t)=T_s$ and approaches $T_i$ far from the boundary at finite time. These solutions demonstrate steady conduction, mode decay, spreading of localized heat and boundary-driven diffusion. Their idealized behaviour at extremely short times or arbitrarily long distances must be interpreted within the continuum regime rather than as an assertion of instantaneous microscopic communication.

For two isolated point masses, introduce total mass $M_T=M+m$, reduced mass $\mu =Mm/M_T$ and relative separation $r$. The relative acceleration is $\ddot{\mathbf{r}}=-G{M}_T\mathbf{r}/r^3$. Conserved specific angular momentum $h=|\mathbf{r}\times \dot{\mathbf{r}}|$ fixes a plane and a constant areal velocity $h/2$. The specific orbital energy isFor $h\ne 0$, the trajectory is a conic:Negative energy gives an ellipse, including the circular limit $e=0$; zero energy gives a parabola; positive energy gives a hyperbola. Bound ellipses obey $\epsilon =-G{M}_T/(2a_o)$ andThus Kepler's laws of planetary motion follow from angular momentum conservation and the inverse-square force. For a circular orbit of radius $r$, $v_c=\sqrt{G{M}_T/r}$, while escape from that radius requires $v_{\mathrm{esc}}=\sqrt{2G{M}_T/r}$. Energy determines whether the orbit is bound; angular momentum also determines its shape. These ideal results neglect extended-body perturbations, drag and relativistic corrections.

Prefer molecular rotational spectroscopy or rovibrational spectroscopy, because nuclear masses enter their reduced mass directly. For an isotopologue with nuclear reduced mass ${\mu }_N$,Isotope substitution therefore produces appreciable shifts and distinct sets of lines, assuming the same electronic potential and approximately unchanged bond length and force constant. A suitable heteronuclear molecule can provide electric-dipole rotational/vibrational spectra; other molecules may require Raman spectroscopy.

Atomic electronic energies instead depend on the electron-nucleus reduced mass $m_{e}M/(m_e+M)$, whose fractional change is suppressed by $m_e/M$. High-resolution atomic spectroscopy can still detect isotope shifts, including nuclear-size and many-electron mass effects; it is not intrinsically incapable of identifying isotopes. Molecular spectra generally offer the stronger mass sensitivity for this purpose.

In the rigid-rotor approximation, $E_J={\hslash }^{2}J(J+1)/(2I)$ and electric-dipole absorption has $\Delta J=+1$. A line labelled $J^{\prime }$ therefore hasThe original graph gives adjacent spacing about $0.115\mathrm{THz}$; for example the $J^{\prime }=5$ line is near $0.575\mathrm{THz}$. Thus $I\simeq 1.46\times 10^{-46}\mathrm{kg}{\mathrm{m}}^2$. Using the reduced mass,Centrifugal distortion of a nonrigid molecule explains small departures from equal spacing at higher $J$.

Separation of variables in the central Coulomb potential gives ${\psi }_{nlm}=R_{nl}(r)Y_l^m(\theta ,\varphi )$. The radial wavefunction depends on $n$ and $l$, independently of $m$. For the hydrogen radial wavefunction in the first excited shell, put $\rho =r/a_0$. The Bohr radius is $a_0=4\pi {ϵ}_0{\hslash }^2/(\mu e^2)$, where $\mu$ is the electron–proton reduced mass. Choose the real phasesNear the origin, $R_{20}\to 1/(\sqrt{2}{a}_0^{3/2})$, whereas $R_{21}\sim r/(\sqrt{24}{a}_0^{5/2})$. More generally, regular hydrogen radial wavefunctions behave as $r^l$. At large $r$, both displayed functions have a polynomial factor proportional to $r$ multiplying $e^{-r/(2a_0)}$; $R_{20}$ approaches zero from below and $R_{21}$ from above with these phase choices.

The $2s$ radial wavefunction crosses zero once, at $r=2a_0$, and has its negative minimum at $4a_0$. The $2p$ radial wavefunction has no radial node at positive $r$, and reaches its maximum at $2a_0$. The count is $n-l-1$; neither the origin nor the limit at infinity is counted as a radial node. These are sketches of $R$, rather than of $rR$ or the radial probability density $r^2|R{|}^2$.

Separate the center-of-mass motion from the relative motion. Two equal quark masses give reduced mass $\mu =m_q/2$. Replacing the hydrogen Coulomb strength by $a>0$ gives the Coulomb binding energies and length scaleThe factor of two in the reduced mass matters: using $m_q$ directly would double the binding energy. These are relative binding energies measured from the free quark–antiquark threshold; the rest energy $2m_q{c}^2$ is excluded.

A classical orbiting electron accelerates and therefore radiates according to the Larmor formula. Its mechanical energy becomes more negative, its radius decreases, and the orbit spirals inward rather than remaining a stable atom. The Bohr model replaces this prediction with postulated nonradiating stationary states and quantized orbital angular momentum, together with photon emission or absorption only between states, $h\nu =|E_n-E_m|$.

For a heavy nucleus, $m_{e}vr=n\hslash$. Since the de Broglie wavelength is $h/(m_{e}v)$, this impliesThe electronic wave fits an integer number of wavelengths around the orbit. This is a standing-wave consistency interpretation of the quantization rule, not a classical mechanism that prevents an accelerating charge from radiating. For finite nuclear mass, the two-body relative coordinate uses reduced mass $\mu =m_{e}M/(m_e+M)$; the next derivation retains that correction. In the limit $M\gg m_e$, it reduces to the fixed-nucleus electron description above.

Let $r$ be the electron-nucleus separation and use the reduced mass $\mu =m_{e}M/(m_e+M)$. Coulomb attraction supplies the relative centripetal force, and Bohr quantization fixes the relative angular momentum:Eliminating $v$ givesThe total energy is $E=\mu v^2/2-Z{e}^2/(4\pi {ϵ}_{0}r)=-Z{e}^2/(8\pi {ϵ}_{0}r)$, henceHere $a_0$ is the ordinary Bohr radius and $R_{\infty }$ the infinite-nuclear-mass Rydberg constant. The electron's radius about the centre of mass is $M{r}_n/(M+m_e)$; for a heavy nucleus this distinction and the reduced-mass correction are small, but the supplied nuclear mass permits them to be stated precisely.

Separate center-of-mass motion and replace $m$ by the reduced mass $\mu =mM/(m+M)$ in the relative-coordinate formulas. Then $r_n$ is the proton-electron separation, energy levels and the Rydberg constant scale by $\mu /m$, and transition wavelengths increase by $m/\mu$. The electron's own orbital radius about the center of mass is $M{r}_n/(M+m)$; the proton radius is $m{r}_n/(M+m)$. The relative speed retains $\alpha c/n$, while the electron's individual speed acquires $M/(M+m)$. Replacing the mass in the radius alone but leaving energies unchanged would be inconsistent.

Inside a rectangular barrier with $E<V$, the stationary Schrödinger equation givesIts independent solutions are $e^{\pm \alpha x}$ withFor an opaque barrier the amplitude is suppressed by approximately $e^{-\alpha L}$ and the probability by $e^{-2\alpha L}$. Matching contributes a prefactor, so this exponential approximation is not an exact transmission coefficient for a transparent or very thin barrier. For two moving particles the relative-coordinate mass is their reduced mass.

At a separation of $1$ fm, the proton Coulomb energy is approximately $V=e^2/(4\pi {\epsilon }_{0}r)\simeq 1.44$ MeV, whereas $k_{B}T\simeq 0.862$ keV at the stated solar-core temperature. As a deliberately crude rectangular-barrier estimate, take a width of $1$ fm and use the two-proton reduced mass $\mu =m_p/2$. ThensoTreating one proton as fixed and using $m_p$ instead gives about $0.59$. Both estimates only say that this arbitrarily shortened barrier would be fairly transparent; the opaque-barrier exponential is itself uncontrolled for such a small exponent.

There is an important conditioning issue. If the protons have already reached a separation of one femtometre, most of the long-range Coulomb suppression has already been overcome. The rectangular estimate cannot represent the probability for thermal protons starting far apart. Their classical turning separation at this energy is about $1.7\times 10^3$ fm, not one femtometre.

Positronium is the electromagnetic bound state of an electron and a Positron. Their equal masses give reduced mass $m_e/2$, half the hydrogenic Rydberg binding scale and approximately twice the electron-proton separation of hydrogen. Its spin singlet and triplet differ in hyperfine energy; annihilation also affects their physical lifetimes and splitting.

A binary star of total mass $M$, reduced mass $\mu$ and separation vector $\mathbf{a}$ has far-field potential functionwhere $\theta$ is the angle between $\mathbf{x}$ and $\mathbf{a}$. The dipole term vanishes in centre of mass coordinates. Averaging a circular coplanar binary over its fast phase gives $\overline{\Phi }=-GM/x-G\mu a^2/(4x^3)$.