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The astrophysical S factor separates the dominant Coulomb barrier suppression from a charged-particle nuclear reaction cross-section $\sigma (E)$:Here $E$ is relative kinetic energy, $\mu$ is reduced mass, $\alpha$ is the fine-structure constant, and $Z_1,Z_2$ are atomic numbers. The astrophysical S factor has units of energy times area. Factoring out a barrier does not remove all nuclear physics: nuclear levels and reaction mechanisms can still make $S(E)$ vary strongly.

For a dilute nonrelativistic Maxwell-Boltzmann velocity distribution of relative motion, multiplying the relative-energy density by speed and nuclear reaction cross-section cancels the explicit factors of $E$, givingThus the Gamow peak describes a rate-integral kernel, not a histogram of reacting stars or even the unweighted particle population. Plasma screening and departures from the assumed Maxwell-Boltzmann velocity distribution require additional physics.

In a local harmonic approximation, the vibrational wavenumber obeys $\tilde{\nu }=(2\pi c{)}^{-1}\sqrt{k/\mu }$. Comparisons at fixed reduced mass therefore use squared frequency ratios. A local stretching force constant is not identical to a bond-dissociation energy.

For a spherical square well of radius $a$, depth $V_0$ and reduced mass $\mu$, the regular zero-orbital angular momentum solution has reduced radial wavefunction $U=A\sin kr$ inside and $U=B{e}^{-qr}$ outside. Wavefunction matching at a finite potential step gives $k\cot (ka)=-q$, where $k^2=2\mu (V_0+E)/{\hslash }^2$ and $q^2=-2\mu E/{\hslash }^2$. The $n$th radial branch first reaches $E=0$ at $ka=(n+1/2)\pi$, for $n=0,1,\dots$. A normalizable negative-energy state on that branch requires $V_0>V_{s,n}$. Thus exactly one such state exists for $V_{s,0}<V_0\le V_{s,1}$; the new zero-energy $s$-wave solution at the upper equality is not normalizable.

For the first $\ell =1$ branch, the zero-energy exterior solution has $U\propto r^{-1}$. The regular interior solution is $U\propto r{j}_1(kr)$, where $j_1$ is a Spherical Bessel function. Matching logarithmic derivatives gives $1+z{j}_1^{\prime }(z)/j_1(z)=-1$, or $j_0(z)=0$, with $z=a\sqrt{2\mu V_0}/\hslash$. Its first threshold is $z=\pi$, so $V_{p,0}={\pi }^2{\hslash }^2/(2\mu a^2)$. Higher values of orbital angular momentum increase the positive centrifugal potential, so none can bind below this first $p$-wave threshold. Spin-dependent nuclear forces are absent from this central-well model.

A heteronuclear diatomic molecule consisting of carbon and sulfur. Its permanent electric dipole moment permits an electric-dipole rotational spectrum. The linear rigid rotor model relates its rotational constant to its reduced mass and bond length.

In the two-body problem, these coordinates separate the common translation from the internal separation. For total mass $M$ and reduced mass $\mu$, the kinetic energy is $M{\dot{R}}^2/2+\mu {\dot{r}}^2/2$. The conjugate momenta are $P=M\dot{R}$ and $p=\mu \dot{r}$ for the ordinary Newtonian kinetic term.

For a circular Kepler orbit, neglecting stellar spin, the orbital angular momentum of a binary star is $J=\mu a^2\Omega =\mu \sqrt{GMa}$, where $M$ is total mass, $\mu$ is the reduced mass, and $a$ is separation. Kepler's third law also gives $J=G^{2/3}{P}^{1/3}{M}_1{M}_2/[(2\pi {)}^{1/3}{M}^{1/3}]$. Distances from the center of mass are $a{M}_2/M$ and $a{M}_1/M$; adding their individual orbital angular momenta proves the formula.

For fixed component masses, total mass $M=M_1+M_2$ and reduced mass $\mu =M_1{M}_2/M$, in the weak-field slow-motion circular orbits, $E=-GM\mu /(2a)$ and $J=\mu \sqrt{GMa}$ obey $dE/dJ=\Omega$. The gravitational-wave energy and angular-momentum balance makes the secular loss tangent to this circular family. Substituting $v^2=GM/a$ into the instantaneous quadrupole luminosity of a Kepler binary and differentiating $E$ gives the displayed inspiral equation. Integration givesThe orbit is quasicircular, with small radial drift. The formal coalescence time is not a valid extension of the weak-field point-mass model into the final strong-field or stellar-contact phase.

For two identical fermions of mass $m$ in the same one-dimensional quantum harmonic oscillator, use center of mass $X=(x_1+x_2)/2$ and separation $r=x_1-x_2$. The corresponding oscillator masses are $2m$ and reduced mass $m/2$, with the same frequency $\omega$. If $N,n$ are their occupation numbers, the unperturbed energy is $\hslash \omega (N+n+1)$. The fermionic exchange symmetry requires even $n$ with a spin-one-half singlet state and odd $n$ with a spin-one-half triplet state.

For a delta potential $\lambda \delta (r)$ the first-order shift is $\lambda |{\phi }_n^{\mathrm{rel}}(0){|}^2$. Every odd relative wavefunction vanishes at contact, so these states remain exact unaffected eigenstates of the spatial problem. For the even relative ground state, the oscillator length is $\sqrt{2}\ell$, where $\ell =\sqrt{\hslash /(m\omega )}$, and the shift is $\lambda /(\sqrt{2\pi }\ell )$, independent of $N$. This separates interaction-sensitive singlets from contact-insensitive triplets.

Let $r$ be the instantaneous bond length, $r_e>0$ its equilibrium value, and $q=r-r_e$ the stretching displacement. Assume one fixed electronic quantum state, a smooth potential energy $V(r)$ with a nondegenerate local minimum at $r_e$, and omit molecular rotation initially. The Taylor series has no linear term because $V^{\prime }(r_e)=0$:Keeping only the quadratic term is the harmonic approximation near a potential minimum. The bond force constant $k$ has units $\mathrm{N}{\mathrm{m}}^{-1}$. It measures local stiffness; it is not the bond-dissociation energy, which concerns the depth and distant shape of the entire well.

The two-particle relative coordinate explains the reduced mass. For stretching along a fixed axis in a stationary centre of mass frame, let $u_1,u_2$ be the nuclear displacements from equilibrium. Then $u_2-u_1=q$ and $m_1{u}_1+m_2{u}_2=0$, soHere $T$ is the internal kinetic energy, and $m_1,m_2>0$ are the two nuclear masses. The lighter partner moves farther; the reduced mass is smaller than either individual mass. The restoring force $-dV/dq\simeq -kq$ gives simple harmonic motion:The angular frequency is $\omega$, the ordinary frequency is $\nu$, and ${\tilde{\nu }}_{\mathrm{harm}}$ is the corresponding spectroscopic wavenumber; $c$ is the speed of light. Use $c$ in $\mathrm{cm}{\mathrm{s}}^{-1}$ to obtain ${\tilde{\nu }}_{\mathrm{harm}}$ in $\mathrm{c}{\mathrm{m}}^{-1}$. The symbol $k$ here is a bond force constant, not an angular wavenumber.

For an original dimensionless example, choose $m_1=m_2=2$ and $k=4$ in consistent model units. Then $\mu =1$ and $\omega =2$. Releasing at $q(0)=0.1$ with $\dot{q}(0)=0$ gives $q(t)=0.1\cos (2t)$ and $u_1=-q/2$, $u_2=q/2$. Each atomic nucleus moves only half the stretching amplitude, while the conserved internal mechanical energy above $V_0$ is $\frac{1}{2}k(0.1{)}^2=0.02$. Fixing one partner instead would give $\omega =\sqrt{2}$, demonstrating why a one-body calculation changes the model.

For two particles of equal mass $m$ and opposite charges $\pm q$ separated by $s$, the centre of mass lies halfway between them. In a nonradiating classical model, each radius is $s/2$ and $k_e{q}^2/s^2=m{\omega }^{2}s/2$. Hence ${\omega }^2=2k_e{q}^2/(m{s}^3)$ and $T=2\pi \sqrt{m{s}^3/(2k_e{q}^2)}$. Equivalently, relative motion has reduced mass $m/2$. Radiation and quantum binding invalidate an indefinitely stable classical orbit for actual microscopic charges.

Take two point masses joined by a massless spring of spring constant $k>0$ and equilibrium length $\ell >0$. Assume one-dimensional motion along the spring, no external force, no damping, and an extension $s=x_2-x_1-\ell$ satisfying Hooke's law; keep $x_2>x_1$. For a real spring this is a model within its elastic range. Its potential energy is $U=k{s}^2/2$, soHere $\omega$ is angular frequency and $\nu$ is frequency. Choose $|A|<\ell$ so the model bodies do not cross. The internal normal mode has displacements $u_1=-m_{2}s/M$ and $u_2=m_{1}s/M$ about equilibrium offsets from the centre of mass; its amplitude ratio is $|u_1|/|u_2|=m_2/m_1$. The other normal mode is collective translation with zero frequency and no spring extension. This free pair has different boundary conditions from normal modes of two equal masses between three springs.

For $k=10N/m$ and two $1\mathrm{kg}$ masses, $\omega =\sqrt{20}{\mathrm{s}}^{-1}\approx 4.47214{\mathrm{s}}^{-1}$ and $\nu \approx 0.711763\mathrm{Hz}$. Replacing the second mass by $100\mathrm{kg}$ gives $\omega =\sqrt{10.1}{\mathrm{s}}^{-1}\approx 3.17805{\mathrm{s}}^{-1}$ and $\nu \approx 0.505802\mathrm{Hz}$. Fixing the second endpoint instead would give $\sqrt{10}{\mathrm{s}}^{-1}$: the free heavy partner still moves. The free pair's angular frequency exceeds that fixed-end limit by the factor $\sqrt{1+1/q}$, about $1.004988$ at $q=100$.

The Geiger-Nuttall law expresses the extreme sensitivity of an alpha decay half-life to its Q value. In a restricted range of heavy nuclei, a useful local form is ${\log }_{10}[T_{1/2}/T_{\ast }]\simeq a{Z}_d/\sqrt{Q_{\alpha }}+b$, where $Z_d$ is the daughter atomic number, and the constants depend on the chosen units and nuclear family. It is named for Hans Geiger and John Mitchell Nuttall.

The Coulomb barrier outside the daughter has $V(r)=2Z_d\alpha /r$ in natural units. Its outer turning point is $r_t=2Z_d\alpha /Q_{\alpha }$. The WKB approximation gives a penetration probabilitywhere $\mu$ is the daughter-alpha reduced mass. The leading large-barrier exponent is $2\pi (2Z_d\alpha )\sqrt{\mu /(2Q_{\alpha })}$, producing the approximate $Z_d/\sqrt{Q_{\alpha }}$ dependence. Finite nuclear size modifies the exponent and the slowly varying intercept. The nuclear decay constant also contains an assault frequency and an alpha-cluster formation probability. Extra orbital angular momentum raises the barrier further, and odd-nucleon structure can introduce substantial hindrance. Thus extrapolation from even-even decays gives an order-of-magnitude reference rather than a precision prediction for an odd-odd nucleus.

For a binary star viewed far from its centre of mass, let $\mathbf{r}$ be the separation and $\mu$ the reduced mass, $M=M_1+M_2$ the total mass, $r=|\mathbf{r}|$ and $x=|\mathbf{x}|$. The Taylor expansion of the point-mass Newtonian gravitational potential has zero dipole because ${\sum }_A{M}_A{\mathbf{x}}_A=0$. Its second moment is ${\sum }_A{M}_A{x}_{Ai}{x}_{Aj}=\mu r_i{r}_j$, soThis holds for an instantaneous separation in any orbit, not only a circular orbit. The mass quadrupole moment convention $Q_{ij}=\mu (r_i{r}_j-r^2{\delta }_{ij}/3)$ gives ${\Phi }_2=-3G{Q}_{ij}{x}_i{x}_j/(2x^5)$.

For a slowly evolving circular binary star with separation $a$, reduced mass $\mu$ and orbital angular frequency $\Omega$, the leading gravitational wave losses satisfy $\dot{E}=\Omega \dot{J}$. One direct check uses the standard quadrupole angular-momentum loss lawwhere $Q$ is the mass quadrupole moment and ${ϵ}_{ijk}$ is the Levi-Civita symbol. For $\mathbf{r}=a(\cos \Omega t,\sin \Omega t,0)$, put $A=\mu a^2$; then $Q_{xx}=A(1/6+\cos (2\Omega t)/2)$, $Q_{yy}=A(1/6-\cos (2\Omega t)/2)$, $Q_{xy}=A\sin (2\Omega t)/2$, and $Q_{zz}=-A/3$. Substitution gives ${\dot{J}}_z=-32G{A}^2{\Omega }^5/(5c^5)$ and the quadrupole formula gives $\dot{E}=-32G{A}^2{\Omega }^6/(5c^5)$. Their ratio proves the balance relation. The same relation follows from the rotating wave's angular harmonic number two and frequency $2\Omega$. The additional angular-momentum flux law, not the energy flux alone, is needed to justify preservation of secular circularity.

Gross atomic structure denotes the leading nonrelativistic Coulomb binding levels, before fine and hyperfine splittings. For a hydrogenic two-body system, the reduced mass controls the binding energy and characteristic separation. Fine structure is smaller by roughly ${\alpha }^2$; nuclear magnetic hyperfine effects have an additional electron-to-nuclear mass suppression.

Quantizing the stretching coordinate gives a quantum harmonic oscillator with reduced mass $\mu$:Here $v$ is the vibrational quantum number and $\hslash =h/(2\pi )$ the reduced Planck constant. Consecutive energy eigenvalues have gap $\hslash \omega$; the ground state has zero-point energy $\hslash \omega /2$ above the well minimum. The offset $V_0$ cancels from every transition energy.

A short proof uses creation and annihilation operators: with $\ell =\sqrt{\hslash /(\mu \omega )}$, write $q=\ell (a+a^{†})/\sqrt{2}$. The Quantum Hamiltonian becomes $\hat{H}-V_0=\hslash \omega (a^{†}a+1/2)$. Nonnegativity of $a^{†}a$ gives the lowest level; repeated application of $a^{†}$ raises its energy by $\hslash \omega$ each time. This also gives $⟨q{⟩}_v=0$ andThe ground state therefore has a nonzero spread of bond lengths, even though its mean stretching displacement vanishes. A stationary energy eigenstate has a time-independent position probability density; the classical oscillation curve is not a trajectory of an atomic nucleus in that state.

The parabolic model formally extends $q$ to the whole real line, whereas an actual bond length satisfies $r>0$. It is useful for low levels concentrated near $r_e$, with typical stretching displacements much smaller than $r_e$. It fails near breakup of the chemical bond and when neglected anharmonicity becomes substantial.

Keep the same bond force constant $k$ and approximate the nuclear masses of ${}^1\mathrm{H}$, ${}^2\mathrm{H}$, and ${}^{35}\mathrm{Cl}$ by $1u$, $2u$, and $35u$, respectively. Here $u$ is the unified atomic mass unit, and $\mathrm{D}$ denotes deuterium. These are chosen mass number approximations, not measured atomic masses or natural-abundance averages. With the same chlorine isotope in both molecules,This is the isotope scaling of diatomic vibration prediction. Doubling the light partner's mass nearly doubles the reduced mass, lowering the frequency by about $28.3\%$. The factor differs from $1/\sqrt{2}$ because the chlorine partner also moves.

For the same small initial stretch $A$ and zero velocity, both models start with the same internal mechanical energy $k{A}^2/2$, but the DCl oscillation period is longer by $\sqrt{72/37}\approx 1.394972$. In the centre of mass frame, $u_{\mathrm{H}}=-35q/36$ and $u_{\mathrm{Cl}}=q/36$ for HCl; $u_{\mathrm{D}}=-35q/37$ and $u_{\mathrm{Cl}}=2q/37$ for DCl. These ratios explain which partner moves and are separate from the change in frequency.

The common-$k$ assumption belongs to the leading Born-Oppenheimer approximation: fixed nuclear positions and electric charges determine the electronic potential energy, while nuclear masses govern motion on that curve. Quantitative comparison requires actual atomic masses, departures from that approximation, and separation of anharmonicity and molecular rotation from the measured band. Even with common $k$, the quantum harmonic oscillator predicts a different zero-point energy for each isotope.^[ref]

An actual infrared spectroscopy route is to compare gaseous hydrogen chloride containing ordinary hydrogen with its deuterium counterpart. A documented teaching experiment uses a stopcock-equipped gas cell with potassium bromide windows and a Nicolet Avatar instrument for Fourier transform spectroscopy. A nitrogen-purged flask and drying tube supply the mixture; the sample also contains sulfur dioxide. An empty sample-chamber background precedes the filled-cell scan. The absorbance trace and peak-picked spectroscopic wavenumbers are the readout. The analysis assigns and indexes ${}^1{\mathrm{H}}^{35}\mathrm{Cl}$ and ${}^2{\mathrm{H}}^{35}\mathrm{Cl}$ transitions separately and propagates transition uncertainties. The instructions distinguish resolvable chlorine isotope peaks from adequate precision for the intended parameter analysis. ^[ref]

A separate published teaching dataset reports a Nicolet Nexus 670 instrument, a $10\mathrm{cm}$ gas cell at $20\mathrm{Torr}$, and chlorine isotope peaks resolved to baseline. These apparatus details describe that dataset, not the preceding experiment. ^[ref]

The inference connects several models. Fit correctly assigned transitions to separate molecular rotation from molecular vibration, then compare the inferred vibrational scales using isotope scaling of diatomic vibration. Agreement supports the approximate separation between a nearly unchanged chemical bond stiffness and a changed nuclear mass. It does not make a real molecule an exact classical spring: anharmonicity shifts the fundamental transition away from the local harmonic oscillator scale, and coupling to molecular rotation shifts individual lines.

For an independent test, calculate the reduced masses from the specified isotopes and compare the inferred values $k=\mu (2\pi c{\tilde{\nu }}_{\mathrm{harm}}{)}^2$. Agreement within uncertainty tests the common-curvature assumption; choosing separate $k$ values just to fit each frequency would not test it. The documented background addresses instrument and ambient contributions, but does not by itself subtract window absorption, identify contaminating species, or calibrate the spectroscopic wavenumber scale. Those require separate controls. Finite spectral resolution, blended peaks, incorrect isotope assignments, and departure from the harmonic approximation near a potential minimum limit the comparison. No numerical measured line positions are assumed here.

A monatomic ion with exactly one electron. Its nonrelativistic Coulomb potential energy levels scale as $Z^2$ at fixed principal quantum number, apart from reduced mass corrections.

A family of hydrogen emission lines terminating at a fixed lower principal quantum number $n_f$. In the nonrelativistic Bohr model, their inverse wavelengths are $1/\lambda =R_H(1/n_f^2-1/n_i^2)$ for integers $n_i>n_f$. Holding $n_f$ fixed and increasing $n_i$ gives lines approaching the series limit $1/\lambda =R_H/n_f^2$. Reduced mass shifts the Rydberg constant, while additional interactions produce corrections to this gross spectral pattern.