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Ordinary solid sodium chloride has the rock salt structure. Choose a conventional cube of side $a$ with chlorine anions at the corners and face centers, and sodium cations at the edge centers and cube center. Corner sites are shared by eight cubes, face sites by two and edge sites by four. The unit cell counts are thereforeThere are four formula units per conventional unit cell, rather than four separate NaCl molecules.^[ref]

The Bravais lattice is a face-centered cubic lattice. One choice of primitive lattice translation vectors and atomic basis isThe determinant gives a primitive unit cell volume of $a^3/4$; that cell represents one cation and one anion. Four such volumes make the conventional cube. The union of all ion positions resembles a primitive cubic lattice of step $a/2$ only if the chemical labels are discarded: translation by $a(1,0,0)/2$ exchanges unlike ions and fails to preserve the chemical structure.

Each ion has six nearest unlike neighbors at distance $a/2$, along the positive and negative directions of the three cube axes. Thus the coordination number is six. Within either same-species face-centered cubic lattice, the nearest same-ion neighbors instead number twelve: their displacements are permutations of $(\pm a/2,\pm a/2,0)$, all of length $a/\sqrt{2}$. Since $a/\sqrt{2}>a/2$, those twelve are a more distant shell in the complete sodium chloride structure. The number twelve describes the same-ion FCC lattice neighborhood; six describes the nearest unlike-ion neighborhood.

Let the net cooling per unit mass of an ideal gas be $\mathcal{L}=A\rho T^{\alpha }-H$, with positive constants $A,H$. At fixed pressure and composition, $\rho T$ is constant, so $\mathcal{L}=C{T}^{\alpha -1}-H$. At a thermal equilibrium $T_0$, $C{T}_0^{\alpha -1}=H$. The specific heat capacity at constant pressure gives $c_{p}d(\delta T)/dt=-(\partial \mathcal{L}/\partial T{)}_p\delta T$. Thus $\delta T\propto e^{st}$ with the displayed growth rate. The equilibrium is stable for $\alpha >1$, unstable for $\alpha <1$, and marginal at linear order for $\alpha =1$. In particular, optically thin Bremsstrahlung with $\alpha =1/2$ is unstable under this heating law and the isobaric constraint. Fixed-density perturbations have a different criterion.

A binary black hole is a binary system whose two components are black holes. Gravitational waves carry away orbital energy and angular momentum, gradually reducing the separation; sufficiently close systems progress through inspiral, merger and the settling of the final black hole. The increasing orbital angular velocity produces a signal whose gravitational-wave frequency rises, called a chirp. In GW150914, two LIGO interferometers recorded a signal sweeping from about 35 to 250 hertz in September 2015. Comparing that measured gravitational wave strain with general relativity waveforms inferred the component masses and the merger's properties. This permits the study of systems that need no luminous surrounding gas to be detected.^[ref]

An arithmetic series adds the sequence terms of an arithmetic progression. Its $n$-th partial sum is $S_n={\sum }_{k=1}^n{a}_k$, with the empty sum convention $S_0=0$. For $n\ge 1$,Thus the total is the number of sequence terms multiplied by the arithmetic mean of the two endpoints. The formula also covers decreasing and constant arithmetic progressions.^[ref]

For the eight-entry example, the sum is $8(7+28)/2=140$, and the arithmetic mean of all eight entries is $17.5$. A decreasing example $11,9,7,5,3,1$ gives $6(11+1)/2=36$. Negative values cause no algebraic difficulty: $3,1,-1,-3,-5$ has sum $5(3-5)/2=-5$.

The triangular number $T_n=1+2+\cdots +n$ is the special case $a_1=d=1$, giving $n(n+1)/2$. More generally, an equal increase in each successive allocation makes the total grow as a quadratic function of the number of allocations. The eight-entry example hasIts first four entries add to $46$, its first eight to $140$, and its first twelve to $282$. The sequence terms lie on a straight line when plotted against their sequence indices, while the partial sums follow a quadratic function.

Starting at $a_1$ and repeatedly adding a nonzero common difference $d$, a proposed final value $z$ is reached at positionif and only if this expression is a positive integer. For $a_1=7$, $d=3$, and $z=28$, it gives $n=8$. The quotient $21/3=7$ counts increments, while the initial entry adds one to the number of sequence terms. The proposed value $27$ gives $n=1+20/3$, which is not an integer, so it is never reached by these updates. The test also handles decreasing arithmetic progressions; when $d=0$, the repeated value alone cannot determine a length.

For the electronic ${}^2{S}_{1/2}\to {}^2{P}_{1/2}$ transition, neglect hyperfine structure and use weak-field LS coupling. The Landé g-factors are $g_g=2$ and $g_e=2/3$, so a transition from $m_g$ to $m_e$ shifts by $({\mu }_{B}B/h)(g_e{m}_e-g_g{m}_g)$.

With propagation perpendicular to the magnetic field, linear polarization parallel to the field drives the $\pi$ electric-dipole selection rule $\Delta m=0$. The two shifts are $\pm 2{\mu }_{B}B/(3h)$. Perpendicular linear polarization is an equal-amplitude combination of the two field-axis circular components, driving ${\sigma }^{\pm }$ transitions with $\Delta m=\pm 1$ and shifts $\pm 4{\mu }_{B}B/(3h)$. Each geometry therefore selects a doublet, with twice the separation in the perpendicular-polarization case. Equal line heights within a pair require equal lower-state populations and an unsaturating probe; line positions do not require that extra population assumption.

Neglect hyperfine structure and use a magnetic field weak compared with the spin-orbit interaction. The sodium D doublet has transitions $3p{}^2{P}_{1/2}\to 3s{}^2{S}_{1/2}$ and $3p{}^2{P}_{3/2}\to 3s{}^2{S}_{1/2}$. Their Landé g-factors are $2/3$, $4/3$ and $2$ respectively. With $m_u,m_l$ denoting upper and lower magnetic labels, the photon shift isThe first transition has four allowed shifts ${\mu }_{B}B\{\pm 2/3,\pm 4/3\}$; the second has six, ${\mu }_{B}B\{\pm 1/3,\pm 1,\pm 5/3\}$. The most blue-shifted component of the second is $m_u=1/2\to m_l=-1/2$, with shift $5{\mu }_{B}B/3$. The apparently larger shift $m_u=3/2\to m_l=-1/2$ is excluded by the electric-dipole selection rule. The $3p{}^2{P}_{3/2}$ level lies above $3p{}^2{P}_{1/2}$, so in this weak-field model the $5/3$ component also has the shortest wavelength of the whole doublet. ^[ref]