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Gauss's law relates the electric field on a closed surface to the net electric charge inside it. In SI units, with outward unit normal $\mathbf{n}$,Here $\rho$ describes a volume charge density, and ${\epsilon }_0$ is the vacuum permittivity; for ideal point or surface charges, use the enclosed electric charge directly rather than an ordinary volume-density integral. The left side is electric flux: the surface integral of the outward normal component of the electric field. Field directed outward contributes positively; field directed inward contributes negatively. The equality uses net signed electric charge, so positive and negative enclosed charges can cancel. The Gaussian surface is an imagined accounting boundary, not necessarily a material object. See OpenStax, explaining Gauss's law.

Where the electric field and charge density are smooth, the local statement is $\nabla \cdot \mathbf{E}=\rho /{\epsilon }_0$. The divergence theorem connects this divergence to the surface integral. This is the satisfying connection between local sources and global accounting: a surface can register how much net electric charge it encloses without revealing where that charge is located within it. A point charge requires care with the singular field at its position; the integral formulation remains useful with a boundary that avoids the charge.

The law fixes total electric flux, rather than every value of the electric field. In an uncharged region, a uniform electric field $\mathbf{E}=(E_0,0,0)$ crosses a cube of side $\ell$. Its left and right faces contribute $-E_0{\ell }^2$ and $+E_0{\ell }^2$; the other four faces contribute zero. Thus the net electric flux is zero while the electric field is nonzero everywhere. Charges outside a Gaussian surface can change the field on it without changing its net electric flux. Conversely, fields cannot generally be reconstructed from enclosed electric charge alone. Electrostatics also requires the electric field to be the negative gradient of an electric potential, with appropriate boundary information. This distinction is emphasized in Feynman lectures II.5, application of Gauss' law.

Spherical symmetry supplies the extra information needed for a particularly simple calculation. An isolated spherically symmetric charge distribution in vacuum produces a radial electric field, constant in magnitude on each concentric sphere. Its electric flux is therefore $4\pi r^2{E}_r(r)$, givingOutside the distribution, doubling $r$ quadruples the surface area and divides the radial electric field by four, while the electric flux stays $Q/{\epsilon }_0$. The inverse-square dependence is an area-dilution result. The electric field of a uniformly charged ball and electric field of a uniformly charged spherical shell show that the same total charge can give different interior fields. These symmetry arguments and their limits are developed in OpenStax, applying Gauss's law.

In heterogeneous catalysis, the catalyst and reactants occupy different thermodynamic phases; a solid surface can furnish the catalytic route.

For example, specify dissolved hydrogen peroxide as the reactant, water as the solvent, and solid manganese dioxide as the catalyst. The solid and liquid chemical solution are separate thermodynamic phases. A dispersion that looks thoroughly mixed still contains those separate phases; mixing distributes contact, rather than making the solid molecularly dissolved. The manganese dioxide peroxide contact demonstration makes the arrival of the liquid at the solid tangible.

A useful surface-route picture is approach from the liquid, adsorption at a catalytic surface site, conversion through surface-bound reaction intermediates, and release of reaction products. ^[ref] This sequence organizes possible roles; it is not a verified sequence of elementary reactions for every catalyst, or for the particular manganese dioxide demonstration. The net chemical equation supplies neither the binding geometry nor the identity of the working catalytic surface site.

This contact requirement makes surface area and diffusion practically important. More exposed solid can provide more catalytic surface sites, whereas diffusion through the surrounding liquid can limit their supply of reactants. Changing stirring or particle size can therefore change the observed reaction rate without identifying the microscopic rate constant. A recoverable solid is attractive for reuse, but separation alone does not establish retained activity: compare with catalyst deactivation.

With tension taken positive, hydrostatic stress is the stress tensor $\mathbf{\sigma }=-p\mathbf{I}$, where $\mathbf{I}$ is the identity matrix. Every plane has the same compressive normal stress, and no tangential traction. More generally,where $\mathbf{s}$ is deviatoric stress. Two samples can share the same mean pressure while experiencing different deviatoric stresses. A stationary liquid cannot sustain equilibrium shear stress; a solid transmitting medium can support it. Consequently, a load divided by surface area does not by itself establish hydrostatic sample conditions. A transmitting medium's state at the actual pressure and temperature, and contact between sample and walls, matter.

Bridgman's Figure 4 compares mild steel fractured in tension at atmospheric pressure with steel stretched further in a liquid at $25000\mathrm{kgf}\mathrm{c}{\mathrm{m}}^{-2}$.^[ref] The imposed tension remains a directional load: the experiment combines confinement with deformation, rather than measuring a hydrostatic stress alone. Its comparison motivates recording the loading path and the surrounding medium as well as pressure; it does not establish a universal fracture threshold for all steels.

Consider a particle of mass $m$ in a rigid rectangular box of length $L$ along $x$ and wall surface area $A$, so $V=AL$ is the volume. Choose $+x$ toward the right wall. For an incoming normal velocity $v_x=v>0$, an elastic collision reverses the normal momentum:The impulse delivered to the wall points outward, toward $+x$, by Newton's third law. The particle's opposite momentum change is not the wall's impulse. A tangential velocity component is unchanged by the ideal smooth-wall collision.

Ignoring intervening particle-particle collisions in this counting model, the particle travels a normal distance $2L$ before hitting the same wall again. Its hit rate is $|v_x|/(2L)$; multiplying by the impulse per hit gives the time-averaged normal force and its contribution to pressure:For example, stipulate $m=4\times 10^{-26}\mathrm{kg}$, $|v_x|=500\mathrm{m}{\mathrm{s}}^{-1}$, $L=0.10\mathrm{m}$, and $A=0.010{\mathrm{m}}^2$. Then $V=0.0010{\mathrm{m}}^3$, the impulse per hit is $4\times 10^{-23}\mathrm{N}\mathrm{s}$, the hit rate is $2500{\mathrm{s}}^{-1}$, and $P_1=10^{-17}\mathrm{Pa}$. This is one particle's tiny time-averaged contribution to one wall's pressure, not an isotropic equilibrium gas by itself.

For $N$ particles of equal mass, add their force contributions and define $⟨v_x^2⟩=N^{-1}{\sum }_{i=1}^N{v}_{x,i}^2$. In the gas's rest frame, isotropy gives $⟨v_x^2⟩=⟨v_y^2⟩=⟨v_z^2⟩=⟨v^2⟩/3$, where $v^2=v_x^2+v_y^2+v_z^2$. HenceThe final equality uses the classical equipartition theorem. The kinetic pressure of an isotropic gas therefore recovers the ideal gas law from momentum transfers. Doubling the normal speed would double both the impulse per hit and the hit rate, giving four times the pressure contribution; pressure measures a mean squared speed. Interparticle elastic collisions redistribute velocities and help establish isotropy; the individual round-trip times need not remain $2L/|v_x|$, but the equilibrium momentum flux result survives. ^[ref]

The inner mitochondrial membrane separates mitochondrial matrix from intermembrane space. Its selective permeability permits distinct chemical conditions, while embedded mitochondrial respiratory chain proteins and ATP synthase support oxidative phosphorylation. Cristae increase its surface area.^[ref]

A selective boundary and a productive enzyme have different jobs. Even active respiratory enzymes cannot sustain a useful electrochemical gradient if protons return freely across the boundary. Conversely, an intact boundary without an electron donor does not supply the energy for continued pumping. The isolated mitochondria respiration experiment separates some of these dependencies by additions to the same preparation.

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 closed right circular cylinder with radius $r>0$, height $h>0$, and total surface area $A=2\pi r(r+h)$, inscription in a sphere of radius $R$ gives $r^2+h^2/4=R^2$. The Lagrange multiplier equations imply $h^2+2rh-4r^2=0$, so the positive aspect ratio is $h/r=\sqrt{5}-1$. The dimensions are $r^2=R^2(5+\sqrt{5})/10$ and $h^2=2R^2(5-\sqrt{5})/5$.

There is also an eigenvalue proof of global maximality. Set $w=h/2$, so the constraint is $r^2+w^2=R^2$, and write $A/(2\pi )=r^2+2rw$. This quadratic form has symmetric matrix $\left(\begin{array}{cc}1& 1\\ 1& 0\end{array}\right)$ with largest eigenvalue $(1+\sqrt{5})/2$, the golden ratio. Its maximizing eigenvector lies in the positive quadrant, yielding the displayed area and aspect ratio.

For a lidless rectangular parallelepiped with positive dimensions $l,w,h$ and prescribed external surface area $S=lw+2lh+2wh$, the largest volume is attained at $l=w=2h=\sqrt{S/3}$. The arithmetic-geometric mean inequality applied to $lw,2lh,2wh$ giveswith equality exactly when these three face-area terms are equal. This proves both the claimed global bound and its attainable equality case. Lagrange multipliers produce the same dimensions from the interior stationarity equations. The weighted open-box minimization is the corresponding fixed-volume formulation with general positive face weights.

Mean curvature measures the average signed bending of a smooth surface. Unlike Gaussian curvature, it adds the two principal curvatures rather than multiplying them. A circular cylinder therefore has nonzero mean curvature despite zero Gaussian curvature, whereas a catenoid has zero mean curvature despite negative Gaussian curvature. The cancellation on a catenoid explains how a curved minimal surface can be stationary for surface area.

Choose a unit normal $N$ and the shape operator $S=-dN$, as in the second fundamental form convention $II(v,w)=⟨Sv,w⟩$. For a smooth surface in Euclidean space of dimension three, define the signed scalar mean curvature byHere $E,F,G$ are the coefficients of the first fundamental form, and $e,f,g$ those of the second fundamental form. The factor $1/2$ makes $H$ an average; the scalar trace convention instead calls $k_1+k_2=2H$ the mean curvature. Authors who use $S=+dN$ reverse both signs. A formula involving mean curvature is therefore incomplete unless its shape operator, unit normal, and averaging convention are specified.

In orientation reversal of surface curvature, both principal curvatures and $H$ change sign, while Gaussian curvature and the mean curvature vector do not. The units of mean curvature are inverse length, compared with inverse squared length for Gaussian curvature. The distinction matters in soap film pressure balance: pressure couples to the sum of the signed principal curvatures.

A minimal surface has vanishing mean curvature vector. For a smooth surface in three-dimensional Euclidean space, this means $H=0$, equivalently $k_1+k_2=0$. Thus Gaussian curvature is $K=-k_1^2\le 0$, but neither principal curvature needs to vanish. A plane and a catenoid both satisfy the condition, with very different Gaussian curvatures.

By the first variation of area formula, a minimal surface is a stationary point of surface area for every smooth variation that fixes its boundary, or has compact support in its interior. The word “minimal” specifies this differential equation; it does not guarantee a global minimum of surface area, or even a local minimum. A stable minimal surface imposes a further condition on the second variation.

For a regular parametric surface $\mathbf{r}:U\to X$, the displayed surface integral counts each image point as many times as its preimages in $U$, apart from sets of zero surface area. It equals the ordinary integral over the image when the parametrization is injective except on such sets. A periodic parametrization integrated over more than one period can give a different answer from integration over its image set. For example, $(u,v)\mapsto (\cos u,\sin u,v)$ with $0\le u\le 4\pi$ and $0\le v\le 1$ gives parameter area $4\pi$, whereas the image circular cylinder has surface area $2\pi$.

For a steady point-like astronomical source with isotropy of emission, in transparent space with negligible relativistic or cosmological effects, conservation of energy spreads its luminosity over a sphere of surface area $4\pi d^2$. Its bolometric observed radiative flux isThis inverse-square law reflects geometric dilution: the same emitted power crosses increasingly large spheres. It does not require light to lose energy on the journey. For a fixed luminosity, doubling Euclidean distance quarters the observed radiative flux. ^[ref].

With $L_0=10^{26}\mathrm{W}$ and $d_0=20$ parsecs, the result is $F_0\simeq 2.09\times 10^{-11}\mathrm{W}{\mathrm{m}}^{-2}$. At $40$ parsecs, the same source gives $F_0/4\simeq 5.22\times 10^{-12}\mathrm{W}{\mathrm{m}}^{-2}$. An ideal collecting area of $1{\mathrm{m}}^2$ at $20$ parsecs intercepts approximately $2.09\times 10^{-11}\mathrm{W}$, before detector losses.

A closed circular cylinder of radius $a>0$ and length $\ell >0$ has volume $\pi a^2\ell$ and total surface area $2\pi a(a+\ell )$, including both end disks. At fixed area $A$, Lagrange multipliers for maximizing the volume give $\ell =2a$, henceEquivalently, eliminate $\ell =A/(2\pi a)-a$ and maximize $(A/2)a-\pi a^3$ on $0<a<\sqrt{A/(2\pi )}$. The derivative changes from positive to negative at the displayed radius, and the volume tends to zero at both ends.

For a cylinder inscribed in a sphere of radius $R$, the constraint is $a^2+{\ell }^2/4=R^2$. Maximizing $a^2+a\ell$ by Lagrange multipliers gives ${\ell }^2+2a\ell =4a^2$, so the positive aspect ratio is $\ell /a=\sqrt{5}-1$. It is independent of $R$. With $q=\ell /a$, the area is $2\pi R^2(1+q)/(1+q^2/4)$; differentiating this expression confirms the unique maximum on $q>0$.

For a coaxial cylinder standing on the base of a right circular cone of height $h>0$ and semi-vertical angle $0<\alpha <\pi /2$, similar triangles give radius $r=(h-x)\tan \alpha$ at cylinder height $x$. Its volume is $V=\pi x(h-x{)}^2{\tan }^2\alpha$. Differentiation gives $V^{\prime }=\pi {\tan }^2\alpha (h-x)(h-3x)$, so the unique positive-volume maximum has $x=h/3$ and occupies $4/9$ of the cone's volume. Its curved surface area is instead $A=2\pi x(h-x)\tan \alpha$, maximized at $x=h/2$. Changing the optimized quantity changes the optimal shape, even though the geometric constraint is identical.

Orientability asks whether local choices of handedness can agree everywhere on a manifold. On a smooth manifold, an orientation of a smooth manifold chooses an orientation of a vector space for each tangent space, continuously with position. An ordered basis describes the same handedness as another one when the change-of-basis determinant is positive. An oriented atlas implements this condition through positive Jacobian determinants of the smooth transition maps.

A paper strip makes the obstruction visible. Join its short ends without twisting: the resulting untwisted band admits a consistent choice of unit normal. Add a half-turn before joining: on the Möbius band, following the core circle of a Möbius band reverses that choice when you first return to the same point. A local decision about handedness has encountered a global contradiction. The normal reversal around the Möbius core makes this precise for an explicit model.

Both rectangles have coordinates $(u,v)\in [0,1]\times [-1,1]$. The untwisted band identifies $(0,v)$ with $(1,v)$; the Möbius band identifies $(0,v)$ with $(1,-v)$. Across a seam the corresponding local coordinate changes preserve or reverse $v$, giving determinants $+1$ and $-1$. This sign explains the handedness change; the drawing also tracks the two boundary circles of the untwisted band and the single boundary circle of the Möbius band.

For a smooth topological surface embedded in ${\mathbb{R}}^3$, choosing an orientation of a surface is equivalent to choosing a continuous unit normal. This picture is specific to surfaces in three-dimensional space: the general definition uses compatible local orientations, without requiring an embedding. Orientability is a topological invariant, so a homeomorphism cannot turn a Möbius band into a untwisted band. The sphere and torus are orientable surfaces; the Klein bottle is not. ^[ref]

The distinction matters when a sign has geometric meaning. A signed flux surface integral needs an orientation of a surface, and Stokes theorem relates that choice to the outward-normal-first boundary orientation. Ordinary unsigned surface area does not require orientability. Thus a Möbius band still has a perfectly meaningful area even though it has no continuous global unit normal.

A connected orientable manifold of positive dimension has two possible orientations: reversing every local choice gives the other one. If a manifold has several connected components, each component can be oriented independently when it is orientable. For topological surfaces, a useful obstruction criterion is the presence of a Möbius band: a surface is nonorientable exactly when it contains such a band. ^[ref]

Parametrize the ellipsoid by $\mathbf{x}=(b\sin \theta \cos \varphi ,b\sin \theta \sin \varphi ,a\cos \theta )$, with positive semiaxes $a,b$, $0\le \theta \le \pi$ and $0\le \varphi <2\pi$. The cross product ${\mathbf{x}}_{\theta }\times {\mathbf{x}}_{\varphi }$ gives the displayed outward vector area element. For $0<\theta <\pi$, it is the positive multiple $a{b}^2\sin \theta$ of the outward normal direction $(x/b^2,y/b^2,z/a^2)$, fixing its orientation. A flux integral can therefore be evaluated directly as $\int \int \mathbf{F}(\mathbf{x})\cdot ({\mathbf{x}}_{\theta }\times {\mathbf{x}}_{\varphi })d\theta d\varphi$, without computing the length of the normal separately. The scalar surface area element is the Euclidean norm of this vector element.

For $h,R>0$ and $0<\alpha \le 2\pi$, the curved sector $z=h(1-r^2/R^2)$, $0\le r\le R$, $0\le \theta \le \alpha$ has volume $\alpha h{R}^2/4$ beneath it. Its surface area isTo derive this, parametrize the surface by $(r\cos \theta ,r\sin \theta ,h(1-r^2/R^2))$. The upward vector area element is $(2h{r}^2\cos \theta /R^2,2h{r}^2\sin \theta /R^2,r)drd\theta$; its magnitude integrates to the stated surface area. Its integral gives the vector areaIn the shallow limit $h/R\to 0$, the surface area is $\alpha R^2/2+\alpha h^2/2+O(h^4/R^2)$, approaching the area of its planar sector projection.

Including both flat ends of the cylindrical cell, its membrane surface area isThe capacitance is $C=(0.01\mathrm{F}{\mathrm{m}}^{-2})A=6.34602\mathrm{pF}$. The electric charge needed for the specified depolarization is thereforeIgnoring the ends gives $3.33\times 10^6$, only about one percent smaller. Inward movement of positive ions, or outward movement of negative ions, produces the stated positive change in the inside-minus-outside membrane potential.

The cell volume is $\pi r^2\ell =3.14159\times 10^{-16}{\mathrm{m}}^3$. Distributed over this volume, the transferred number would correspond to only $0.0177955\mathrm{mM}$. Relative to an illustrative, not source-specified, $100\mathrm{mM}$ internal ionic pool, that is a fraction $1.78\times 10^{-4}$. Thus a substantial change in membrane potential requires movement of only a tiny fraction of bulk ions; it does not require wholesale depletion of the intracellular solution. The separated electric charge is concentrated near the two membrane surfaces.

Spatial geometry. An embedding of the three-sphere isTaking differentials and collecting terms gives the induced metric tensorPut $\bar{r}=A\sin \chi$. Then $d\bar{r}=A\cos \chi d\chi$, soThe areal radius $\bar{r}$ is not a global radial coordinate on the entire three-sphere: its maximum occurs at the equator and two hemispheres share each smaller value. Use separate charts there; the coordinate singularity at $\bar{r}=A$ is not a physical singularity.

For Euclidean space and hyperbolic space of curvature $-1/A^2$, respectively,For the latter, $\bar{r}=A\sinh \chi$ gives $d{\ell }_{-}^2=A^2[d{\chi }^2+{\sinh }^2\chi d{\Omega }^2]$. In all three cases the induced two-metric at fixed $\bar{r}=\bar{R}$ is ${\bar{R}}^{2}d{\Omega }^2$. Its surface area is thereforeThis equality compares the same areal radius, not the same proper radial distance. For a proper radial distance $\ell$, the corresponding areas are $4\pi A^2{\sin }^2(\ell /A)$, $4\pi {\ell }^2$, and $4\pi A^2{\sinh }^2(\ell /A)$.

Homogeneity, isotropy and redshift. The FRW metric has constant-time spatial slices of constant curvature $k/a^2(t)$, with $k=+1,0,-1$ selecting the three geometries above. Their isometries act transitively on points and isotropically about every point, implementing the cosmological principle. Multiplication by one common time-dependent scale factor preserves these properties. With the source's $k=\pm 1$ normalization, $r$ is dimensionless and $a$ has units of length; using a dimensionless scale factor instead rescales the spatial curvature parameter. The origin of the spherical coordinates is arbitrary; it does not identify a physical centre. Cosmic time is the proper time of the comoving observers, and the absence of time-space cross terms makes their worldlines orthogonal to the homogeneous slices.

For a radial null geodesic between comoving endpoints,The next wave crest travels between the same endpoints. Subtracting the two path integrals, for wave periods short compared with the expansion time, gives $\delta t_0/a_0=\delta t_E/a_E$. Thus the cosmological redshift isBoth observers' clocks are cosmic proper-time clocks. Peculiar velocities and local gravitational-potential shifts have been left out of this homogeneous calculation.

Angular size and brightness. For a galaxy small compared with its distance, the angular diameter distance is $D_A=a_{E}R$ and its angular diameter isHere the emitted proper diameter is measured transverse to the line of sight; radial proper distance is a different integral when $k\ne 0$. Ignoring absorption, lensing caustics and intrinsic evolution, photon-number conservation spreads an isotropic-equivalent luminosity across the sphere $4\pi a_0^2{R}^2$. Each photon loses energy by $1+z$, and its arrival interval is stretched by the same factor. The received bolometric flux isThe disc's proper projected area is ${\mathcal{A}}_d=\pi d^2/4$, and its observed solid angle is ${\Omega }_d\simeq {\mathcal{A}}_d/D_A^2$. Therefore its surface brightness isThis is cosmological surface-brightness dimming. The normalization uses $L$ as isotropic-equivalent luminosity along the viewing direction; another disc emission law changes that normalization but not the redshift power. This is a bolometric result. In a specified observed frequency band the emitted spectrum is sampled at a shifted frequency and requires a spectral correction.

The distinction in telescope aperture and resolved surface brightness matters here. A larger telescope collects more photons and improves angular resolution, but cannot restore the incoming radiance lost by cosmological dimming. For aperture diameter $D$ and a diffraction resolution element of solid angle ${\Omega }_{\mathrm{res}}\sim (\lambda /D{)}^2$, the collected power from uniform resolved brightness scales asindependent of $D$. This is consistent with conservation of etendue. It explains why increasing aperture while keeping every gain in resolution does not increase counts per native resolution element from a smooth extended source. It does not make larger telescopes useless: at a fixed angular bin the counts grow as $D^2$, so a photon-noise-limited or uniform-sky-noise-limited signal-to-noise ratio grows roughly as $D\sqrt{t_{\exp }}$. They also resolve smaller structures and improve point-source sensitivity. Aperture, exposure and angular binning help measure a dim galaxy, without changing its intrinsic $(1+z{)}^{-4}$ brightness penalty.

Set $C=\cos u$, $S=\sin u$, and $D=1+S^2=2-C^2$. At $v=0$, the first partial derivatives areTheir dot products give $E=1$, $F=1+C$, $G=3+2C$. HenceSince $EG-F^2=3+2C-(1+C{)}^2=D>0$, the parametric surface is regular everywhere. Its cross product and unit normal at $v=0$ areThe second partial derivatives at $v=0$ areTheir dot products with $N$ give $e=f=-1/\sqrt{D}$ and $g_2=-(1+C)/\sqrt{D}$. ThusPart (b) transports these first fundamental form and second fundamental form coefficients to every $v$. Finally,The last expression is the surface area element from $\sqrt{EG-F^2}$. In particular the Gaussian curvature changes sign where $\cos u$ does; it vanishes at $u=\pi /2$ modulo $\pi$.