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For a fixed geometry, the steady Stokes equation maps $(\mathbf{u},p^{\prime },\mathbf{f})$ to $(-\mathbf{u},-p^{\prime },-\mathbf{f})$ when all imposed boundary velocities and disturbance forces are reversed; $p^{\prime }$ is pressure relative to any unchanged hydrostatic pressure. Linearity of Stokes flow and Uniqueness of Stokes flow therefore reverse the entire velocity field. If the driving schedule and evolving geometry are retraced, fluid elements retrace their paths too. Reversing just one drive while leaving another applied force unchanged does not meet these conditions.

This reversal concerns the paths, while viscous dissipation remains positive: for a sphere, the required power $6\pi \mu a{U}^2$ is unchanged by $U\mapsto -U$. A reversed experiment must supply fresh energy. Inertia, Brownian motion and irreversible contact can prevent perfect retracing. Edward M. Purcell illustrated the mechanism with a viscous fluid flow demonstration in Life at low Reynolds number, American Journal of Physics 45, 3–11 (1977).

For two well-separated spheres with isolated downward speeds $V_i$ and radii $a_i$, let $\mathbf{r}={\mathbf{x}}_1-{\mathbf{x}}_2$, $r=|\mathbf{r}|$, $\mathbf{n}=\mathbf{r}/r$, and let $\hat{\mathbf{z}}$ point upwards. Their leading Stokeslet interactions giveEach sphere's Stokes drag law supplies force $-6\pi \mu a_i{V}_i\hat{\mathbf{z}}$ on the fluid. Evaluate the other's Stokeslet at its centre and subtract the two velocities to obtain the equation. Corrections from finite size and potential dipoles are higher order in radius divided by separation. The approximation must remain well separated throughout the segment of trajectory to which it is applied.

The lensmaker's equation connects focal length to the refractive index and shape of a thin lens. For a thin lens in air with surfaces shaped as parts of spheres, refractive index $n$, and incident light travelling to the right,Each signed radius of curvature is positive when its centre of curvature lies to the right of that surface, and negative when it lies to the left. A symmetric biconvex thin lens with $n=1.5$, $R_a=1\mathrm{m}$, and $R_b=-1\mathrm{m}$ therefore has $1/f=0.5(1+1){\mathrm{m}}^{-1}=1{\mathrm{m}}^{-1}$, giving $f=1\mathrm{m}$ and optical power $+1\mathrm{D}$. This is an ideal shape example consistent with the thin lens diagrams, not a specification of their finite thickness.

In a uniform surrounding medium of refractive index $n_s$ on both sides, replace $n-1$ by $n/n_s-1$. Thus the same shape reverses its focusing sign if its refractive index is lower than its surroundings: for $n=1.5$ and $n_s=2$, the same radii give $1/f=-0.5{\mathrm{m}}^{-1}$, or $f=-2\mathrm{m}$. Positive focal length therefore cannot be inferred from convex shape in every environment. A proof outline applies Snell's law with the small-angle approximation at each surface, adds their direction changes, and neglects the distance between them. ^[ref].

Two identical well-separated spheres falling at the same speed share one laboratory volume flux, rather than producing twice the single-sphere piston flux. In the middle of the three drag regimes for a nearly occluding sphere, the same overall Hagen-Poiseuille flow pressure drop is divided equally between them, halving each sphere's drag. In the local-shear and local-gap-pressure regimes each sphere retains its own leading resistance, so the drag per sphere is unchanged. This conclusion assumes nonoverlapping lubrication regions and sufficient separation from the tube ends.

For smooth no-slip spheres of radius scale $a$ with minimum gap $\delta \ll a$, the lateral gap scale is $(a\delta {)}^{1/2}$. Squeeze-film lubrication theory gives pressure $p\sim \mu a(-\dot{\delta })/{\delta }^2$ and the displayed normal resistance. Bounded forcing therefore gives $-\dot{\delta }=O(\delta )$, so the gap cannot vanish in finite time. The conclusion uses smooth surfaces, no slip and continuum creeping flow; it is not a contact model for rough particles.

A sphere almost filling a tube has local parabolic lubrication gap $h\simeq a[ϵ+(z/a{)}^2]/2$. The resistance integrals are $I_1=2\pi {ϵ}^{-1/2}$, $I_2=(2\pi /a){ϵ}^{-3/2}$ and $I_3=(3\pi /a^2){ϵ}^{-5/2}$, obtained with $z=a\sqrt{ϵ}\xi$ and $\int (1+{\xi }^2{)}^{-n}d\xi$. Moving-boundary lubrication flux weights shear-driven and pressure-driven transport by different powers of the gap, explaining the three distinct divergence rates.

The lumen is the International System of Units unit of luminous flux: $1\mathrm{lm}=1\mathrm{cd}\mathrm{sr}$. A source with constant luminous intensity $1\mathrm{cd}$ within a cone of solid angle $1\mathrm{sr}$ supplies $1\mathrm{lm}$ into that cone. A source with the same luminous intensity in every direction supplies $4\pi \mathrm{lm}$ over a full sphere. ^[ref]

The lunar illuminated fraction is the directly sunlit projected area of the apparent lunar disk divided by its total projected area. For a sphere with parallel sunlight and a distant observer, outside a lunar eclipse, it is $k=(1+\cos \alpha )/2$, with phase angle $\alpha$ at the Moon. It measures geometry rather than the fraction of full-Moon radiant flux. ^[ref].

A lunar phase is the appearance of the sunlit portion of the Moon as seen from Earth. Outside a lunar eclipse, the distant Sun illuminates approximately one hemisphere of the Moon, but an observer sees a changing overlap of that hemisphere with the hemisphere facing Earth. The lunar phase therefore changes even though synchronous rotation keeps approximately the same near side of the Moon facing Earth. ^[ref].

The cycle runs through new Moon, waxing crescent, first quarter, waxing gibbous, full Moon, waning gibbous, last quarter and waning crescent. “Waxing” means that the lunar illuminated fraction is increasing; “waning” means it is decreasing. A crescent has less than half its apparent disk sunlit, and a gibbous lunar phase has more than half. First and last quarter each show half a disk: “quarter” refers to progress around the cycle, rather than a quarter of the disk being lit. ^[ref].

An instructive circular orbit model puts Earth at the origin, the Sun far away along $+x$, and the Moon at $R(\cos \theta ,\sin \theta )$. The sunward position $x>0$, $\theta =0$, gives new Moon; the opposite position $x<0$, $\theta =\pi$, gives full Moon. Counterclockwise motion gives first quarter at $\theta =\pi /2$ and last quarter at $\theta =3\pi /2$. The Sun's direction is fixed during this snapshot of the cycle. Every small globe in the orbital view is illuminated on its $+x$ half; the adjacent apparent disks are separate views from Earth, rather than globes viewed from above.

The disk views choose north up, with the upward direction perpendicular to the orbital plane. In this convention waxing lunar phases are bright on the right and waning lunar phases on the left. Local horizon orientation and viewing from different locations on Earth can rotate the apparent disk; its lit edge always points toward the projected direction of the Sun. The lunar illuminated fraction is unaffected by this rotation.

The useful angle is the phase angle $\alpha =\angle SME$, with $S$ the Sun, $M$ the Moon and $E$ the observer at Earth's center: its vertex is at the Moon. Full Moon has $\alpha =0$, quarter has $\alpha =\pi /2$, and new Moon has $\alpha =\pi$ in the ideal model. By contrast, the elongation $ϵ=\angle SEM$ has its vertex at Earth. For a distant Sun, its rays are parallel, soHere $k$ is the lunar illuminated fraction: a fraction of projected disk area, not a fraction of the Moon's total surface area or its received radiant flux relative to full Moon. The phase angle and unsigned elongation each lie between $0$ and $\pi$, so either alone gives the illuminated amount without distinguishing waxing from waning. The orbital direction resolves that ambiguity.

The lunar illuminated fraction has an elegant geometric derivation. Model the Moon as a sphere of unit radius and view it along the $z$ direction. Take the Sun's unit direction to be $(\sin \alpha ,0,\cos \alpha )$. A visible surface point above apparent disk coordinates $(x,y)$ has $z=\sqrt{1-x^2-y^2}$. Direct illumination requires the outward normal's dot product with the sunward direction to be positive. The day-night terminator therefore obeys $x\sin \alpha +z\cos \alpha =0$. At a fixed $y$, set $q=\sqrt{1-y^2}$; the visible day-night terminator is $x_t=-q\cos \alpha$, and the illuminated strip extends from $x_t$ to the right edge $q$. Integrating its width givesThe integral is the area of half a unit circle. Thus a three-dimensional visibility problem reduces to a circle's area and a cosine. The derivation assumes parallel sunlight, a distant observer and a sphere, and ignores a lunar eclipse. Real observed brightness additionally depends on surface scattering of light and observing geometry, so $k=1/2$ does not promise half the full-Moon radiant flux.

For a worked example, elongation $ϵ=60^{\circ }$ gives phase angle $\alpha =120^{\circ }$ and $k=(1+\cos 120^{\circ })/2=1/4$: a crescent with $25\%$ of its apparent disk sunlit. At elongation $120^{\circ }$, the phase angle is $60^{\circ }$ and $k=3/4$: a gibbous lunar phase. At $90^{\circ }$, $k=1/2$. In each case the same $k$ occurs once while waxing and once while waning in the ideal cycle.

Ordinary lunar phases are not caused by Earth's shadow. A lunar eclipse occurs only when the Moon actually enters that shadow near full Moon. The orbital inclination of the Moon is about $5^{\circ }$ relative to Earth's orbital plane around the Sun; most full Moons pass above or below the shadow. This is why full Moon does not imply a lunar eclipse every month. ^[ref].

The phase cycle is a synodic month, about $29.5$ days, whereas the Moon's orbital period relative to distant stars is a sidereal month, about $27.3$ days. During one sidereal month, Earth advances around the Sun, so the Moon must move farther to repeat the sunward alignment. For coplanar circular orbits traversed in the same direction, subtract their mean motions: $n_{\mathrm{syn}}=n_{\mathrm{sid}}-n_{\mathrm{year}}$, with $n=2\pi /P$. HenceUsing $P_{\mathrm{sid}}=27.321661\mathrm{days}$ and $P_{\mathrm{year}}=365.256\mathrm{days}$ gives $P_{\mathrm{syn}}=29.530591\mathrm{days}$, consistent with the mean synodic month of $29.53059$ days. After one sidereal month, the sunward direction has advanced about $26.93^{\circ }$; catching up takes about $2.21$ more days. These are mean-period estimates: orbital eccentricity and changing geometry make individual synodic months vary. ^[ref]. ^[ref].

The lunar phase is also a practical clock for observing. Full Moon is opposite the Sun and rises roughly at sunset; first quarter is about a right angle from the Sun and is convenient in the evening, while last quarter is convenient before dawn. These are rough guides, not exact rise-time predictions for every place and season. ^[ref]. Historically, repeated lunar phases provided the month in lunar calendars, while half-lit disks gave Aristarchus of Samos a way to infer astronomical distances. ^[ref].

For a sphere of radius $R>0$, its outward unit normal is $N(p)=p/R$. Applying the shape operator gives $S(v)=-v/R$, so the two principal curvatures coincide:For the lateral surface of a circular cylinder with outward unit normal, moving around a circle turns $N$ at rate $1/R$, while moving along the axis leaves $N$ constant. Thus the circumferential and axial principal curvatures giveHere $K$ is Gaussian curvature. The circular cylinder bends in only one principal direction, so its mean curvature is half that of the equally sized sphere. These formulas concern the smooth lateral surface, excluding any edges where end caps meet it.

The common length scale is $\ell$, equal to the radius $R$ for the sphere and circular cylinder, and to the waist radius $a$ for the catenoid. The mean curvature marker is halfway between the two principal curvature markers. Gaussian curvature has a different unit and is listed separately beneath each panel.

For an $n$-dimensional embedded submanifold, let $A$ denote its vector-valued second fundamental form and $e_1,\dots ,e_n$ an orthonormal basis of its tangent space. This branch uses the trace mean curvature vectorAn alternative averaged mean curvature vector is $\mathbf{H}/n$; on a smooth surface in three-dimensional Euclidean space, these are $2HN$ and $HN$, respectively. Reversing $N$ reverses $H$, so their product is independent of the chosen orientation of a surface. With the trace convention, the first variation of area formula has no additional factor $n$. For an outward-oriented sphere of radius $R$, $\mathbf{H}=-2N/R$: it points toward the centre even though the scalar's sign depends on conventions.

For equal spheres of radius $a$ separated by $R\gg a$, one held fixed and the other forced with $\mathbf{F}=6\pi \mu a\mathbf{V}$, put $\zeta =6\pi \mu a$, $\mathbf{n}=\mathbf{X}/R$ and $\mathsf{A}=3a(I+\mathbf{nn})/(4R)$. The method of reflections for Stokes flow gives $\mathbf{U}=\mathbf{V}-{\mathsf{A}}^2\mathbf{V}+O(V{a}^4/R^4)$, since the fixed sphere creates the reflected Stokeslet with force $-6\pi \mu a\mathsf{A}\mathbf{V}$. The leading hydrodynamic mobility matrix correction therefore reduces longitudinal motion more strongly than transverse motion.