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Two objects orbiting their common center of mass under their mutual gravity.

A breathing mode is a collective normal mode in which a structure expands and contracts radially while preserving its overall symmetry. For a uniform scaling pattern, the displacement of point $i$ is proportional to its position relative to the center of mass, as displayed above. In a central spring framework this pattern is a normal mode only when applying the stiffness matrix returns a multiple of the same pattern with the appropriate mass matrix; radial motion alone does not establish an eigenvector.

Every point in the convex hull of a subset of ${\mathbb{R}}^d$ is a convex combination of at most $d+1$ points of that subset. Thus in the plane three ingredients always suffice, however many samples were supplied. The representation of $p=(2,1)$ in convex hull illustrates this bound. Two ingredients need not suffice: the centroid of a triangle with only its three vertices as input is on none of their three pairwise line segments.

A short proof outline removes redundant ingredients. With more than $d+1$ points, the augmented vectors $(s_i,1)$ are linearly dependent. A dependence lets the weights vary while preserving both their sum and the represented point. Move along it until one positive weight first becomes zero; all weights stay nonnegative. Repeating leaves at most $d+1$ positive weights.

In his 1911 paper, Constantin Carathéodory used convex geometry to investigate the possible Fourier coefficients of positive harmonic functions. Its introduction interprets convex mixtures through the center of mass, connecting finite-dimensional geometry to constraints on a function's coefficients.^[ref]

For a closed system with fixed total mass, summing the constituent Newton's second law equations makes pairwise internal forces cancel when Newton's third law applies. The remaining resultant external force is the derivative of total momentum, equal to $M{\ddot{\mathbf{r}}}_{\mathrm{CM}}$. This governs translation of the center of mass; external torque additionally governs changes of angular momentum.

For fixed classical particle masses, differentiating the center of mass position gives the displayed mass-weighted velocity. The total momentum is $M{\mathbf{V}}_{\mathrm{CM}}$. The relativistic zero-momentum frame instead has velocity $c^2\mathbf{P}/E$; treating rest mass as the only weight there generally fails.

The point through which the resultant gravitational force acts when that force has a single point of application. In a uniform gravitational field, it coincides with the center of mass; a nonuniform field need not give the same point.

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 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.

Contact slip is the tangential relative velocity between the two material points instantaneously in contact. For a nonrotating box on a belt it is $w=v-U$; for a rigid body, use the velocity of its contact point, including the contribution from angular velocity. Zero contact slip defines rolling without slipping at a rolling contact, even when the center of mass moves.

A point mass $m$ strikes a free rod of equal mass, initially at rest, perpendicularly at distance $h$ from its center of mass. Assume a smooth elastic collision and rod moment of inertia $I$ about its centre. The final translational speeds are equal precisely when $h^2=2I/m$. Indeed, momentum conservation then fixes both final velocities to $v_0/2$; angular impulse gives $I\omega =m{v}_{0}h/2$, and conservation of kinetic energy gives $I{\omega }^2=m{v}_0^2/2$. Eliminating $\omega$ proves the formula. A uniform rod has $I=m{l}^2/12$, hence $|h|=l/\sqrt{6}$, which lies within the rod.

An exciton is a bound state of an electron and an electron hole. Their opposite electric charges attract through the Coulomb potential, producing a neutral quasiparticle with internal relative motion and motion of its center of mass. In a semiconductor, promoting an electron from the valence band into the conduction band leaves an electron hole; the pair can occupy either an exciton bound state or an unbound scattering state. Merely creating the two charge carriers does not establish binding. A neutral exciton can transport excitation energy without transporting net electric charge. Coupling this material excitation strongly to a photon can form a polariton.

The striking optical consequence is that an exciton optical transition can absorb a photon with less energy than the threshold for creating unbound electrons and electron holes. The difference is the exciton binding energy, not an extra one-electron energy band inside the band gap. The Wannier-Mott exciton supplies a useful large-radius model; the Frenkel exciton describes the opposite, tightly localized limit.^[ref]

For the ideal Wannier-Mott exciton in energy bands whose edges occur at the same crystal momentum, the pair excitation energy, measured from the unexcited crystal, isHere $\hslash \mathbf{K}$ is the pair's total crystal momentum; the last term is center of mass kinetic energy. For a permitted photon absorption transition with negligible recoil energy, $\hslash \omega \simeq E_g-E_{B,n}$. An electric-dipole selection rule and a nonzero transition matrix element are still needed: a bound state need not produce an observable spectral line. If the relevant energy band edges have different values of crystal momentum, a phonon can also be needed to conserve crystal momentum; subtracting a binding scale from a band gap alone does not specify the optical threshold.

For example, the lowest internal bound state in this hypothetical model costs $1.489115\mathrm{eV}$ to create optically, yet needs only $0.010885\mathrm{eV}$ more to dissociate. The large creation energy and the small exciton binding energy answer different physical questions. An electron-hole recombination event can instead release excitation energy by photon emission or to other degrees of freedom; an emitted photon need not reproduce the original absorption energy if relaxation intervened.

An expected value is the probability-weighted arithmetic mean of a real-valued random variable. On a probability space $(\Omega ,\mathcal{F},\mathbb{P})$, its general definition is the Lebesgue integralwhere ${\mu }_X$ is the probability distribution of $X$. For an integrable random variable, this is a finite real number. With a probability mass function $p_X$, use ${\sum }_{x}x{p}_X(x)$; with a probability density function $f_X$, use ${\int }_{\mathbb{R}}x{f}_X(x)dx$. These are two ways of computing the same weighted average, rather than different definitions for different experiments.

For example, suppose $X$ takes the values $-1,2,8$ with probabilities $1/2,3/10,1/5$. ThenThe most likely outcome is $-1$, and $1.7$ is not a possible outcome at all. The expected value summarizes all outcomes and their weights; it need not predict the result of one trial. Geometrically, placing masses equal to the probability masses at the outcome positions gives a center of mass at $1.7$, since ${\sum }_x{p}_X(x)(x-1.7)=0$. This balance-point interpretation also explains the connection to a weighted arithmetic mean. ^[ref].

For the dimensionless random variable $Y$ with probability density function $f_Y(y)=2y$ on $[0,1]$, the expected value is $\mathbb{E}[Y]={\int }_0^{1}y(2y)dy=2/3$. Integrating the probability density function itself gives one; to compute the expected value, multiply by the outcome first. ^[ref].

Why is this useful? Linearity of expectation lets us compute an average total by adding average contributions, even for dependent random variables. And for independent and identically distributed integrable random variables, the strong law of large numbers says their sample mean approaches their common expected value almost surely. The theorem supplies the link to repeated observations; the weighted-average definition itself does not promise that any particular finite run will be close to the expected value.

There is a useful historical connection to valuing uncertain rewards. In his 1657 De ratiociniis in ludo aleae, Christiaan Huygens assigns the value $(a+b)/2$ to equally likely rewards $a,b$, and $(pa+qb)/(p+q)$ when they have $p,q$ equally likely chances. These are the finite weighted-average formulas for expected value. His opening discussion asks how to divide the stakes when a game stops before completion, connecting probability to a concrete question about an unfinished contest. ^[ref].

A Frenkel exciton is an exciton whose internal excitation is concentrated on the scale of one molecule or unit cell. The local excitation can transfer between sites; localized internal structure does not require an immobile center of mass. The large-radius Wannier-Mott exciton approximation cannot simply be extrapolated to this limit.^[ref]

For an ordinary stationary laboratory sample, a useful convention isThe bulk kinetic energy $K_{\mathrm{bulk}}$ includes organized whole-body motion; for pure translation it is $M{v}_{\mathrm{cm}}^2/2$, with total mass $M$ and center of mass speed $v_{\mathrm{cm}}$. The external potential energy ${\Phi }_{\mathrm{ext}}$ includes elevation in an imposed gravitational field, approximately $Mgz$ in a uniform field. Accelerating a sealed bottle or lifting it without otherwise changing its state changes these mechanical terms, rather than its internal energy. Molecular kinetic energy is measured relative to the chosen bulk motion.^[ref]

State the convention rather than applying this separation blindly. If an experiment separately tracks a rotating apparatus or an organized flow, its mechanical kinetic energy must not also be counted in $U$. In Joule's paddle-wheel experiment, waiting for the apparatus and fluid to come to rest removes such an endpoint ambiguity. Including the vessel and paddle changes whose internal energy is measured; including the falling weights additionally puts their changing gravitational potential energy inside the enlarged system's total energy balance.

For particles with masses $m_j$ and positions $r_j$, Jacobi coordinates consist of the total center of mass and successive relative coordinates ${\rho }_j=r_{j+1}-({\sum }_{l=1}^j{m}_l{r}_l)/({\sum }_{l=1}^j{m}_l)$. They separate translation from internal motion and diagonalize the kinetic energy with the corresponding reduced masses.

Writing particle positions relative to the center of mass as ${\rho }_i=r_i-R$ makes ${\sum }_i{m}_i{\rho }_i=0$, so the cross term in the squared velocities vanishes. The total kinetic energy splits into center-of-mass translation and relative motion as displayed. For a rigid body, the latter is $\frac{1}{2}I{\omega }^2$, where $I$ is the moment of inertia about the instantaneous rotation axis through the center of mass.

For equal masses and relative position $\mathbf{r}$, the individual positions are ${\mathbf{r}}_A=\mathbf{R}-\mathbf{r}/2$ and ${\mathbf{r}}_B=\mathbf{R}+\mathbf{r}/2$, where the center of mass has uniform velocity in an isolated system. Bound inverse-square force motion makes $\mathbf{r}$ an ellipse; each mass follows a half-size ellipse in the center-of-mass frame. Adding a nonzero center-of-mass drift generally makes their laboratory trajectories nonclosed. The relative orbital period is not necessarily a period of either laboratory position.

Equal product masses imply equal kinetic energies and opposite momenta in the center of mass frame. The laboratory frame has nonzero initial momentum because the proton is moving. Conservation of momentum therefore forbids the two laboratory momenta from always being equal and opposite. The products are equal in laboratory kinetic energy only for particular emission directions, not for every event.

An exact two-body calculation makes this distinction quantitative. Write $a=m_p{c}^2$, $b=m_{\mathrm{Li}}{c}^2$, $h=m_{\alpha }{c}^2$, and $T$ for incident kinetic energy. Express every quantity below in the same energy units; $\mathcal{P}=pc$ is momentum times the speed of light. For a stationary target,Here $W$ is the center-of-mass energy, and $\beta$ is the center-of-mass speed divided by the speed of light. Each outgoing alpha particle has center-of-mass total energy $E_{\ast }=W/2$ and momentum-times-$c$ magnitude ${\mathcal{P}}_{\ast }=\sqrt{E_{\ast }^2-h^2}$. A Lorentz transformation gives the paired laboratory kinetic energies, with ${\theta }_{\ast }$ measured from the incident beam in the center of mass frame:Their sum is independent of ${\theta }_{\ast }$ and equals $Q+T$. Their difference is $2\gamma \beta {\mathcal{P}}_{\ast }\cos {\theta }_{\ast }$; the symmetry belongs to the product interchange and the center of mass frame.

For the chosen $T=0.500\mathrm{MeV}$ model, subtract $Z$ electron rest masses from each neutral-atom mass, neglect electronic binding energies, and use $m_e{c}^2=0.51099895069\mathrm{MeV}$.^[ref] At ${\theta }_{\ast }=0$, the paired laboratory kinetic energies are approximately $9.979$ and $7.867\mathrm{MeV}$; at ${\theta }_{\ast }=90^{\circ }$ both are approximately $8.923\mathrm{MeV}$. In the latter case both products have a forward laboratory momentum component, so even equal laboratory kinetic energies do not imply back-to-back laboratory trajectories. A real thick target and finite detector acceptance would broaden or select these ideal kinematic possibilities.

Consider a nonrelativistic particle of mass $m$ undergoing an elastic collision with a stationary free target of mass $M=Am$, where $A\ge 1$. Assume the outgoing direction is uniformly distributed over solid angle in the center of mass frame; this assumption concerns the scattering angle distribution and need not hold for every target or energy. Let $\mu$ be the cosine of that frame's scattering angle. Transforming the outgoing velocity back to the laboratory givesThe uniform distribution of $\mu$ on $[-1,1]$ makes $x$ uniform on $[\alpha ,1]$. For $A>1$, integration yieldsFor equal masses, take the limit $\alpha \to 0$: $\xi =1$. The expected value of the fractional energy after a collision is instead $\mathbb{E}[x]=(1+\alpha )/2$. These are different averages: averaging a logarithm does not give the logarithm of an average.

For independent collisions with the same distribution, the logarithms of successive energy ratios add. ThusHere $E_{\mathrm{geom}}$ is the geometric-mean energy, and $n_{\log }$ is a characteristic collision count for reaching an energy scale $E_{\ast }$. It is not an exact first-passage expected value. Individual histories fluctuate, and neither the time between collisions nor absorption is specified by this model.

In an original chosen example, set $E_0=10^6\mathrm{eV}$ and $E_{\ast }=0.025\mathrm{eV}$. Approximate hydrogen by $A=1$ and carbon by $A=12$. ThenThese are illustrative model values, not measured collision counts from 1934. The energy scale marks where neglected thermal motion matters; the calculation does not assert continued cooling to zero energy.