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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.

In this article the rotational constant is a frequency,Its dimensions are ${\mathrm{s}}^{-1}$, since the Planck constant has dimensions $\mathrm{kg}{\mathrm{m}}^2{\mathrm{s}}^{-1}$ and a moment of inertia has dimensions $\mathrm{kg}{\mathrm{m}}^2$. Some authors instead call $hB={\hslash }^2/(2I)$ the energy-valued rotational constant; this is the convention in the earlier synthetic molecular beam electric resonance diagram. A third convention uses the spectroscopic wavenumber $\tilde{B}=B/c$, with $c$ the speed of light. Always check units before using $\nu =2B(J+1)$.

If $B$ is known and the two masses are specified, the rigid two-atom model gives the bond length $r=\sqrt{h/(8{\pi }^2\mu B)}$. This is why rotational spectroscopy can constrain molecular geometry. A real fitted constant can depend on molecular vibration, so this inference alone does not establish an equilibrium bond length.

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.

In Stokes Raman scattering, matter gains internal energy and the photon loses the same amount. For a quantum harmonic oscillator representing one molecular vibration, $E_v=h{\nu }_v(v+1/2)$ with $v=0,1,\dots$. The transition $v\to v+1$ gives ${\nu }_s={\nu }_0-{\nu }_v$, requiring ${\nu }_0>{\nu }_v$ for an outgoing photon. Even a molecule initially in its vibrational ground state can undergo this process. The shared name does not turn it into the absorption/emission-band Stokes shift used for fluorescence.

The Stokes shift compares the lowest absorption-band maximum with the fluorescence emission spectrum maximum. Relaxation often makes $\Delta \lambda >0$. This comparison differs from using an arbitrary excitation wavelength. A spectral-density maximum can move when changing from per unit wavelength to per unit spectroscopic wavenumber; specify the spectral convention.^[ref]

An original illustrative energy budget makes the distinction concrete. Excitation at $450\mathrm{nm}$ and emission at $520\mathrm{nm}$ correspond to about $2.76$ and $2.38\mathrm{eV}$ per photon, using $E=hc/\lambda$ with Planck constant $h$ and speed of light $c$ in vacuum. The difference, about $0.37\mathrm{eV}$, must go elsewhere in the complete energy balance. These chosen wavelengths are not measured band maxima for a particular fluorophore.

Anti-Stokes emission has shorter wavelength than excitation. Additional energy, for example from a thermally populated level of molecular vibration or multiple excitation photons, makes it possible. The one-photon, initially relaxed drawing does not cover these cases.^[ref]

An otherwise forbidden electronic transition gains intensity when accompanying molecular vibration supplies a symmetry factor that permits a nonzero electric dipole transition moment. A symmetry-allowed total vibronic state specifies a possible transition, while the coupling determines its actual strength.

A vibrational spectrum resolves transitions associated with molecular vibrations. In the harmonic approximation, a normal mode of angular frequency $\omega$ has levelsFor a diatomic molecule, $\omega \simeq \sqrt{k/\mu }$, where $k$ is the effective bond stiffness and $\mu$ the reduced mass. Replacing an atom by a heavier isotope therefore usually lowers the associated vibrational frequency without changing the basic electronic bonding.

Infrared spectroscopy detects a vibration when it couples to the electric dipole moment; Raman spectroscopy instead involves changing polarizability. Their selection rules can make complementary modes visible. Anharmonicity shifts frequencies and permits weaker transitions absent from the ideal harmonic picture.