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For parabolic bands confined by an infinite well of width $L$, a same-index transition has edge $E_{g,j}+{\pi }^2{\hslash }^2{n}^2(1/m_e^{\ast }+1/m_{h,j}^{\ast })/(2L^2)$. Above it the in-plane pair kinetic energy is ${\hslash }^2{k}_{\parallel }^2/(2{\mu }_j)$ with reduced mass ${\mu }_j^{-1}=(m_e^{\ast }{)}^{-1}+(m_{h,j}^{\ast }{)}^{-1}$.

The reduced mass is the effective mass for the two-particle relative coordinate of two bodies with constant positive masses $m_1,m_2$:Both bodies can respond to their mutual force, so their relative acceleration is greater than if one were fixed. Consequently $0<\mu <\min (m_1,m_2)$. This explains why using either individual mass, or their sum, generally gives the wrong internal oscillation rate. The same reduced mass describes a two-body problem, a free two-mass spring, and the approximate molecular vibration of a diatomic molecule.^[ref]

For equal masses $m_1=m_2=m$, $\mu =m/2$. Two $1\mathrm{kg}$ bodies therefore have $\mu =0.5\mathrm{kg}$: in their stationary centre of mass frame a relative speed $2m/s$ means individual speeds $1m/s$ and total kinetic energy $1\mathrm{J}$. With $m_1=1\mathrm{kg}$ and $m_2=100\mathrm{kg}$, instead $\mu =100/101\mathrm{kg}\approx 0.990099\mathrm{kg}$. At the same relative speed, the individual speeds are $200/101$ and $2/101m/s$, and the total kinetic energy is $200/101\mathrm{J}$. These are calculations for stipulated masses and velocities.

For fixed $m_1$ and $q=m_2/m_1>0$, $\mu /m_1=q/(1+q)$. The heavy-partner limit $q\to \infty$ gives $\mu \to m_1$ from below; the relative error incurred by replacing $\mu$ with $m_1$ is $(m_1-\mu )/\mu =1/q$. At $q=100$ this is $1\%$. Treating the heavy body as fixed is thus a controlled approximation to internal motion, rather than an exact consequence of its being heavier.

For a rigid heteronuclear diatomic rotor, $E_J={\hslash }^{2}J(J+1)/(2I)$ and electric-dipole absorption has $\Delta J=+1$. Therefore the $J^{\prime }\to J^{\prime }-1$ line frequency is $h{J}^{\prime }/(4{\pi }^{2}I)$ and adjacent absorption lines are spaced by $h/(4{\pi }^{2}I)$. With reduced mass $\mu$, the bond-length estimate is $r=\sqrt{I/\mu }$. The conversion assumes frequency rather than angular frequency; centrifugal distortion makes real high-J spacings imperfectly uniform.

The inverse-length constant in the infinite-nuclear-mass hydrogenic spectrum, $R_{\infty }=m_e{e}^4/[8{ϵ}_0^2{h}^{3}c]$. A nucleus of finite mass replaces it by $R_M=R_{\infty }\mu /m_e$, where $\mu$ is the electron-nucleus reduced mass. Some spectroscopic fits call $Z^2{R}_M$ an effective Rydberg constant; the charge-inclusive and reduced-mass-only conventions must be distinguished.

Take an invented diatomic molecule with a Morse potential, setting the separated-atom energy to zero:Here $a$ is an inverse length, $r_e$ the equilibrium bond length, and ${\omega }_e=a\sqrt{2D_e/\mu }$, with reduced mass $\mu$. These stipulated parameters describe a simple model, not a measured chemical substance. For the one-dimensional Morse oscillator, energy eigenvalues above the minimum areConsequently,The harmonic estimate $A/2=0.1000\mathrm{eV}$ is slightly larger than this model's exact zero-point energy. The physical bound-level restriction matters: the quadratic expression is not an infinite ladder of vibrational quantum states.

For a separate thermal illustration using this same stipulated $D_0$, assume an ideal gas, fully active molecular rotation of the parent, negligible thermal molecular vibration, and negligible electronic excitation for all species. Each monatomic product contributes $\delta h=5RT/2$; the diatomic molecule contributes $7RT/2$. Thus the net increment is $3RT/2$, with molar gas constant $R$. At $T=300\mathrm{K}$,This last result follows from the stated thermal model, not a universal correction for every chemical bond. Frozen rotations, accessible excited states, different fragment composition, or a condensed phase change the thermal increments.

Choose positions in one inertial frame and let $M=m_1+m_2$. The centre of mass and two-particle relative coordinate areThe lighter body is farther from the centre of mass. Reconstructing both positions from these equations is part of solving the two-body problem; $\mathbf{r}$ alone describes their separation, not either body's laboratory position.^[ref]

Substitute the reconstructed velocities into the kinetic energy. The result is the exact two-particle instance of kinetic-energy decomposition about the centre of mass:The mixed dot product has coefficient $-m_1{m}_2/M+m_2{m}_1/M=0$; the coefficient of $|\dot{\mathbf{r}}{|}^2$ is $(m_1{m}_2^2+m_2{m}_1^2)/M^2=\mu$. This algebra needs constant positive masses but no particular force law. In the stationary centre of mass frame, the fraction of internal kinetic energy carried by body 1 is $m_2/M$, and by body 2 is $m_1/M$: the lighter body carries more.

For a closed pair with no external forces, write $\mathbf{F}$ for the force on body 2 and $-\mathbf{F}$ for that on body 1, as required by Newton's third law. Adding and subtracting their Newton's second law equations yieldsThus the centre of mass moves uniformly while the two-particle relative coordinate behaves like one body of reduced mass. Equal-opposite internal forces suffice for these equations. For an internal central force, $\mathbf{F}=f(r)\hat{\mathbf{r}}$ with $r=|\mathbf{r}|>0$, the internal torque vanishes and conservation of angular momentum gives constant ${\mathbf{L}}_{\mathrm{rel}}=\mu \mathbf{r}\times \dot{\mathbf{r}}$. If the central force comes from a time-independent potential energy $U(r)$, then $f(r)=-dU/dr$ and the internal mechanical energy $E_{\mathrm{rel}}=\mu |\dot{\mathbf{r}}{|}^2/2+U(r)$ is conserved. The kinetic energy split alone does not establish either conservation law.

Consider the exact two-body problem with masses $m_1,m_2$. In the inertial centre of mass frame, its internal kinetic energy is $K=m|\dot{\mathbf{r}}{|}^2/2$, where $\mathbf{r}$ is their relative position, $m=m_1{m}_2/(m_1+m_2)$ is the reduced mass, and $U=-\kappa /r$ with $\kappa =G_N{m}_1{m}_2$. Take a noncollision elliptic Kepler orbit with semi-major axis $a$ and orbital eccentricity $0<e<1$. Write $\eta$ for eccentric anomaly, to avoid confusing it with total energy $E$. With mean motion $n=\sqrt{\kappa /(m{a}^3)}$ and orbital period $P=2\pi /n$, Kepler's equation and the orbit geometry giveThe time origin is periapsis. Since $dt=(1-e\cos \eta )d\eta /n$, the time average of $1/r$ over one orbital period is exactly $1/a$. Therefore $⟨K{⟩}_P=\kappa /(2a)$ and $⟨U{⟩}_P=-\kappa /a$, independently of orbital eccentricity. Equal steps in eccentric anomaly are not equal time steps; unweighted averaging of a drawing of the ellipse gives the wrong answer.

The original dimensionless example in both diagrams has $m=a=\kappa =1$ and $e=0.6$. At periapsis, $r=0.4$, $K=2$, $U=-2.5$; at apoapsis, $r=1.6$, $K=0.125$, $U=-0.625$. Both have $E=-0.5$, but $2K+U$ is respectively $1.5$ and $-0.375$. Only the orbital period averages are $⟨K{⟩}_P=0.5$ and $⟨U{⟩}_P=-1$. The instantaneous zero of $2K+U$ occurs at $r=a$, not at every point of the orbit.

The endpoint term can also be evaluated exactly: $G=m\mathbf{r}\cdot \dot{\mathbf{r}}=mn{a}^{2}e\sin \eta$. Starting at periapsis, a window ending at eccentric anomaly $\eta$ hasAt $T=P/4$, Kepler's equation gives $\eta \simeq 2.09133$ and a positive residual $\simeq 0.3314\kappa /a$. At $T=P/2$ or $P$, $G$ returns to zero and the residual vanishes exactly. The lower panel shows the running residual and its endpoint expression coinciding; the shaded quarter-orbital period window explains why a short observation need not satisfy the infinite-time theorem.

A Wannier-Mott exciton has an internal wave function extending over many unit cells. Its simplest model assumes a three-dimensional bulk crystal, approximately parabolic valence band and conduction band, positive isotropic scalar effective masses $m_e^{\ast },m_h^{\ast }$, a constant scalar relative permittivity ${ϵ}_r$, and a dilute pair population. Replace microscopic crystal forces by those effective masses and dielectric screening, and neglect exchange corrections and coupling to phonons. These assumptions must hold on the spatial and energy scales of the particular exciton.

With relative coordinate $\mathbf{r}={\mathbf{r}}_e-{\mathbf{r}}_h$, the internal Hamiltonian operator is the three-dimensional hydrogen atom problem with a reduced mass and a screened Coulomb potential:Here $e$ is the elementary charge, ${ϵ}_0$ the vacuum permittivity, and $\hslash$ the reduced Planck constant. The internal energy eigenvalues, radius scale, and exciton binding energies areThe principal quantum number is $n=1,2,\dots$. This reduction and these scales are derived in Excitons in Semiconductor Quantum Dots by Matjaž Humar (§3, printed pp. 5–6, equations (13)–(26)).

To check the scales by dimensional analysis, write $C=e^2/(4\pi {ϵ}_0{ϵ}_r)$, which has units of energy times length. Then $a_B^{\ast }={\hslash }^2/(\mu C)$ is a length and $R^{\ast }=\mu C^2/(2{\hslash }^2)$ is an energy. Rescaling $r=a_B^{\ast }\rho$ makes $H_{\mathrm{rel}}/R^{\ast }=-{\nabla }_{\rho }^2-2/\rho$, the dimensionless hydrogen atom operator. The Bohr radius $a_B^{\ast }$ is a wave function scale, not the radius of a classical orbit; for this model's lowest internal bound state, the mean separation is $3a_B^{\ast }/2$.

For the diagrams' hypothetical inputs $m_e^{\ast }=0.10m_0$, $m_h^{\ast }=0.40m_0$, ${ϵ}_r=10$ and $E_g=1.500\mathrm{eV}$, where $m_0$ is the free electron mass, independent substitution gives $\mu =0.080m_0$, $a_B^{\ast }\simeq 6.615\mathrm{nm}$ and $R^{\ast }\simeq 10.885\mathrm{meV}$. Here electronvolts express energy. The exciton binding energies for $n=1,2,3$ are approximately $10.885$, $2.721$ and $1.209\mathrm{meV}$. These are model calculations, not measured material parameters. The large-radius approximation additionally requires $a_B^{\ast }$ to be much larger than the crystal's unit cell length scale. In a thin layer, with anisotropic effective masses, or with spatially varying dielectric screening, this three-dimensional formula is not generally valid.