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With the time convention $e^{-i\omega t}$, pressure amplitude $p$ and background density ${\rho }_0$, the acoustic energy flux is $I_x=(2{\rho }_0\omega {)}^{-1}\mathrm{Im}(\overline{p}{\partial }_{x}p)$. A transverse plane wave mode with positive axial wavenumber $\kappa$ carries flux $\kappa |p{|}^2/(2{\rho }_0\omega )$. For the reduced pressure $p=e^{ikx}E$, the parabolic wave equation gives $\kappa =k-{\nu }^2/(2k)$, approximated by $k$ at leading paraxial order. Stationary illumination uses the infinite-aperture spectral normalization: replace $|p{|}^2$ by the corresponding spectral power per unit transverse length. Other acoustic amplitude conventions change the dimensional prefactor.

The spectral arc length $\theta (U)$ is the length of the shortest closed unit-circle arc containing every eigenvalue of $U$.

For a nutrient-consuming chemotactic travelling band in a tube of cross-sectional area $a$, let $N=a{\int }_{\mathbb{R}}B(z)dz$. Integrating $v{C}^{\prime }=kB$ between depleted nutrient behind and $C_{\infty }$ ahead gives $v{C}_{\infty }=kN/a$, hence $v=Nk/(a{C}_{\infty })$. This is a nutrient budget: advancing by a unit length supplies $a{C}_{\infty }$ nutrient, while the bacteria consume at total rate $kN$. Diffusion and chemotactic sensitivity affect the band shape and admissibility, but not this speed at fixed $N,k,a,C_{\infty }$.

The sperm number compares filament length with an elastohydrodynamic penetration length. For periodic forcing at angular frequency $\omega$, the usual convention isIts fourth power is a viscous-to-bending ratio on the full filament length. In steady cross-flow problems a different convention also occurs: $\mathrm{Sp}={\zeta }_{\perp }{L}^{3}U/A$, where $U$ is flow speed and $A$ the filament bending modulus. The defining formula distinguishes these conventions.

Spherical geometry studies lengths and angles measured along the surface of a sphere, rather than through its interior. It is a useful model whenever travel is constrained to a round surface. Its straightest paths are great circle arcs: they are geodesics, even though they appear curved in the surrounding Euclidean space. Two distinct great circles meet at an antipodal pair, so the parallel-line picture of Euclidean geometry does not apply.

For a sphere of radius $R>0$, write two positions as $Ru,Rv$, with $u,v$ unit vectors. Their shorter surface route has great-circle distancewhere the dot product gives the central angle in radians. The straight chord through the surrounding Euclidean space instead has length $2R\sin (d/(2R))$. For a quarter-circle separation these are $\pi R/2\approx 1.571R$ and $\sqrt{2}R\approx 1.414R$: specifying which distance we mean has a measurable effect. On the unit sphere, $R=1$ and great-circle distance equals the central angle.

Consider the spherical triangle with vertices $(R,0,0)$, $(0,R,0)$ and $(0,0,R)$, joined by the shorter great circle arcs. Each side has length $\pi R/2$. At each vertex the two outgoing tangent vectors point along the other coordinate axes and are orthogonal, so all three interior angles are $\pi /2$. This is the positive coordinate octant of the sphere, not a flat triangle drawn inside it.

For a convex minor spherical triangle with interior angles $\alpha ,\beta ,\gamma$ in radians, the spherical excess formula statesOur octant therefore has area $R^2(3\pi /2-\pi )=\pi R^2/2$. Independently, eight congruent octants partition the sphere, giving $4\pi R^2/8$ again. Thus angles alone determine area when $R$ is fixed, unlike Euclidean geometry, where similar triangles can have arbitrarily different areas. The Gauss-Bonnet theorem explains the connection: the sphere's constant Gaussian curvature $1/R^2$ times the area equals this excess. For spherical triangles whose area is small compared with $R^2$, the excess is small, recovering the familiar flat picture.

A simple closed curve of length $L$ on the unit sphere enclosing the smaller area $A$ satisfiesEquality holds for a circle. A variational proof uses Gauss--Bonnet to turn area maximization at fixed length into geodesic-curvature minimization, proves that an extremizer has constant geodesic curvature and is planar, and evaluates its spherical cap.

For a nondegenerate convex minor spherical triangle, the spherical law of cosines relates two angular sides $b,c$, their included interior angle $A$, and the opposite angular side $a$:Relabelling the vertices gives the corresponding formulas for $b,B$ and $c,C$. As with the spherical law of sines, physical side lengths must first be divided by the sphere's radius. The spherical law of cosines determines $a\in (0,\pi )$ uniquely from $b,c,A$ when the data describe such a spherical triangle. It also recovers each interior angle from three sides, for exampleUsing the cosine and the range $0<A<\pi$ selects a unique angle, avoiding the supplementary-angle ambiguity of an inverse sine calculation.

To derive the spherical law of cosines, let $u,v,w$ be the vertex unit vectors at $A,B,C$, so $u\cdot v=\cos c$, $u\cdot w=\cos b$ and $v\cdot w=\cos a$. The outgoing unit tangent vectors at $u$ are $t_B=(v-\cos cu)/\sin c$ and $t_C=(w-\cos bu)/\sin b$. Their dot product is both $\cos A$ and $(\cos a-\cos b\cos c)/(\sin b\sin c)$, proving the identity. This gives a geometric explanation: spherical trigonometry compares angles in the tangent plane using vectors in the surrounding Euclidean space.

For a worked example on the unit sphere, take the verticesAt $A$, the outgoing unit tangent vectors are $(1,0,0)$ and $(0,1,0)$, so the interior angle is a right angle. The vertex dot products give $b=c=\pi /3=60^{\circ }$. The spherical law of cosines then reduces to the right-angle identity $\cos a=\cos b\cos c$, givingTo find $B$, use the relabelled spherical law of cosines:The physical side lengths on a sphere of radius $R$ are ${\ell }_a\approx 1.318116R$ and ${\ell }_b={\ell }_c=\pi R/3\approx 1.047198R$. For example, on a toy sphere with $R=10$ metres they are approximately $13.18116$, $10.47198$ and $10.47198$ metres; the interior angles stay the same. Applying the planar Pythagorean theorem directly to the angular sides would instead give $\sqrt{60^2+60^2}\approx 84.852814^{\circ }$. That flat right triangle calculation does not give the great-circle distance between $B$ and $C$.

The familiar law of cosines reappears as a local approximation. With $a,b,c$ all small in radians, substitute the small-angle approximations $\sin x\approx x$ and $\cos x\approx 1-x^2/2$ into the spherical law of cosines and keep terms through second order:Multiplication by $R^2$ gives the planar law of cosines for the physical side lengths. For a fixed small surface triangle on a progressively larger sphere, the angular sides shrink and this approximation improves. Thus the curved and flat formulas fit together: the Pythagorean theorem is recovered locally when $A$ is a right angle.

For a nondegenerate convex minor spherical triangle, with angular sides $a,b,c$ opposite interior angles $A,B,C$, the spherical law of sines statesThe sines of sides refer to central angles, so a physical side length ${\ell }_a$ on a sphere of radius $R$ must enter as $\sin ({\ell }_a/R)$. All six angles lie in $(0,\pi )$, and the denominators are positive. The reciprocal form, $\sin a/\sin A=\sin b/\sin B=\sin c/\sin C$, is equivalent. This is Tatum's sine formula, equation 3.5.1, in Celestial Mechanics, section 3.5: spherical sine and cosine formulas.

A short proof explains the symmetry of the spherical law of sines. Let $u,v,w$ be the vertex unit vectors at $A,B,C$. At $u$, the outgoing unit tangent vectors towards $v,w$ are $t_B=(v-\cos cu)/\sin c$ and $t_C=(w-\cos bu)/\sin b$. Their angle is $A$, so the magnitude of their cross product is $\sin A$. Since both tangent vectors are orthogonal to $u$, the scalar triple product obeysDividing by $\sin a\sin b\sin c$ proves the spherical law of sines. Its beauty is that the same three-dimensional scalar triple product can be calculated at any vertex, producing all three ratios at once.

For the spherical triangle worked out under the spherical law of cosines, $A=90^{\circ }$, $b=c=60^{\circ }$ and $\cos a=1/4$. Hence $\sin a=\sqrt{15}/4$, and the spherical law of sines givesThe inverse sine gives $B\approx 63.434949^{\circ }$, but the supplementary angle $116.565051^{\circ }$ has the same sine. The spherical law of sines alone therefore leaves two candidates for this angle. The spherical law of cosines, using the known three sides, gives $\cos B=1/\sqrt{5}>0$ and selects $63.434949^{\circ }$; likewise $C=B$. When only two sides and an opposite angle are given, two distinct valid spherical triangles can remain. A cosine relation resolves an inverse sine branch only when enough additional data are known; it does not make every ambiguous data set unique.

For an attractive spherical square well, let $\kappa =\sqrt{2m{V}_0}/\hslash$. At zero energy, the outer reduced radial wave is proportional to $r-a_s$ and the regular inner wave to $\sin (\kappa r)$. Matching their logarithmic derivatives gives $a_s=a-\tan (\kappa a)/\kappa$. Away from a threshold and at sufficiently low energy, the total S-wave scattering scattering cross-section tends to $4\pi a_s^2$. The length can be negative, zero or divergent.

Spherical trigonometry calculates the sides and interior angles of spherical triangles using trigonometric functions. Its sides must be shorter great circle arcs, which are geodesics on a round sphere. A curve at constant nonzero latitude does not qualify. The spherical law of sines and spherical law of cosines play the roles of the familiar flat-triangle rules, but account for the sphere's curvature.

Use uppercase $A,B,C$ both for the vertices and for their interior angles, and lowercase $a,b,c$ for the opposite angular sides. For example, $a$ is the central angle subtended by the arc from $B$ to $C$. On a sphere of radius $R$, its physical arc length is ${\ell }_a=Ra$ when $a$ is in radians; similarly ${\ell }_b=Rb$ and ${\ell }_c=Rc$. Only on the unit sphere do the angular and physical side lengths have the same numerical values. See J. B. Tatum, Celestial Mechanics, chapter 3, section 3.5: great-circle sides and angular units.

One useful application of spherical trigonometry is finding great-circle distance from spherical coordinates. Write a unit vector as $u(\varphi ,\lambda )=(\cos \varphi \cos \lambda ,\cos \varphi \sin \lambda ,\sin \varphi )$, where $\varphi$ is latitude and $\lambda$ is the equatorial azimuth, or longitude, measured from a chosen reference direction. The dot product of two such unit vectors givesHere $\delta \in [0,\pi ]$ is the central angle and $d$ is the great-circle distance. The same identity follows from the spherical law of cosines by adding the north pole as the third vertex, with angular sides $\pi /2-{\varphi }_1$ and $\pi /2-{\varphi }_2$. This connects spherical geometry to position calculations in astronomy and to navigation on a spherical model of Earth. Choosing the surface model matters: Paula Neeley and Anthony Narkawicz, Map Projection Induced Variations in Locations of Polygon Geofence Edges, NASA/TM-2017-219675 (2017), sections 2-3 distinguishes spherical and ellipsoidal geodesics and gives this coordinate identity.

For a toy example, put two points at latitude $30^{\circ }$, with equatorial azimuths $-60^{\circ }$ and $60^{\circ }$. The great-circle distance and the arc length along constant latitude areThe great circle takes a shorter route by bending towards the pole in these spherical coordinates. To locate its highest latitude, the endpoint unit vectors are $p=(\sqrt{3}/4,-3/4,1/2)$ and $q=(\sqrt{3}/4,3/4,1/2)$. Their great circle lies in the plane through the origin with $z=2x/\sqrt{3}$. On its shorter arc, the maximum latitude occurs at $y=0$ and is $\arctan (2/\sqrt{3})\approx 49.106605^{\circ }$. Thus a geodesic can change latitude substantially even when its endpoints have equal latitude. These are lengths on an ideal sphere; the example illustrates the geometry rather than specifying an actual flight route.

Historically, spherical trigonometry was prominent in teaching as well as in astronomy. John Conway and Alex Ryba note its presence in early twentieth-century school textbooks in Remembering spherical trigonometry, The Mathematical Gazette 100(547), March 2016, pages 1-8, DOI 10.1017/mag.2016.3. Their introductory observation documents that teaching history; it does not identify a single inventor of the spherical law of sines or spherical law of cosines.

Spin decimation sums over spins at selected lattice sites while retaining the others. For alternate-site elimination in a nearest-neighbour periodic chain of even length, the coarse spin-chain transfer matrix is $W^2$, because its $(s,t)$ entry sums $W_{sa}{W}_{at}$ over the eliminated middle spin. Exact preservation $\mathrm{tr}(W^2{)}^{N/2}=\mathrm{tr}{W}^N$ requires keeping the scalar normalization as well as the entry ratios.

In spontaneous parametric down-conversion, a pump field drives a nonlinear optics process producing correlated pairs of photons.

A linear spring stores elastic energy $k{x}^2/2$ when displaced by $x$ from its relaxed position. The external force maintaining that displacement is $kx$; its restoring force on the moving attachment has the opposite sign. Its spring constant $k$ has units force per length.

A two-dimensional Bravais lattice with perpendicular primitive vectors of the same length $\ell$. An orthogonal matrix $R$ gives its orientation, and its primitive area is ${\ell }^2$. Its nearest-neighbor graph is a square lattice; the set of crystal translations and the graph with chosen adjacency are related but distinct objects.

A specimen-stage reference bearing a scale of known length, imaged through the microscope objective during microscope scale calibration.

An effective step length describing a specified chain-count convention: an ideal polymer with $N$ counted units has mean-squared end-to-end distance of a polymer chain $⟨R^2⟩=N{b}^2$. If $N$ counts freely jointed coarse-grained units, $b$ is their Kuhn length. If $N$ counts chemical repeat units, the effective $b$ incorporates local conformational correlations and is not automatically a physical monomer diameter. A repeat-unit volume $v_0$ is independent information; setting $v_0\sim b^3$ is a scaling approximation.

In the steady nondiffusive gravity–Marangoni model with capillarity neglected, zero surfactant flux and positive concentration make the surface velocity zero. With $Q=-H^2{\Gamma }_X/2-H^3{H}_X/3$ and $J=-H\Gamma {\Gamma }_X-H^2\Gamma H_X/2$, integration gives $H^4=H_0^4-48QX$ and $\Gamma =1+(H_0^2-H^2)/4$. A concentration drop $\delta$ over unit length therefore selects $Q=-\delta (H_0^2+2\delta )/6$. The liquid flows opposite to the imposed Marangoni traction, with velocity proportional to $H_{X}Y(Y-H)/2$ and an immobilized surface. This reduction is singular at a dry endpoint; its small-slope assumption does not resolve that edge.

For a convective star, the activity convention compares its rotation period with a convective turnover time. It differs by fixed factors from the fluid Rossby number estimated from speed, length and angular velocity. Small ${\mathrm{Ro}}_{\ast }$ denotes strong rotational influence on convection. Magnetic activity usually rises as this ratio decreases in the unsaturated regime, but levels off in a saturated rapid-rotation regime.

Stokes' second problem describes the oscillatory shear flow generated by a tangentially oscillating flat plate in a semi-infinite viscous fluid flow. The velocity satisfies $u_t=\nu u_{xx}$, with $u(0,t)=U_0\cos \omega t$ and decay away from the plate. For $\omega >0$, the displayed time-periodic solution follows by taking the real part of $U_0{e}^{i\omega t-(1+i)x/\delta }$. The length $\delta$ is the penetration depth: both amplitude attenuation and phase lag grow with $x/\delta$. Boundary data alone also allow homogeneous transients; the periodic solution describes the established oscillation after those transients decay.