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With gauge pressure $p_g$ referenced to atmospheric pressure $p_{\mathrm{atm}}$, absolute pressure is $p=p_g+p_{\mathrm{atm}}$. Thus Boyle's law becomesEven a constant atmospheric pressure cannot be dropped from these sums. For the calculated $100\mathrm{kPa}$ to $200\mathrm{kPa}$ example, take $p_{\mathrm{atm}}=100\mathrm{kPa}$. The initial and final gauge pressures are $0$ and $100\mathrm{kPa}$; their products with the respective volume are $0$ and $5000\mathrm{kPa}\mathrm{c}{\mathrm{m}}^3$, which are unequal. Zero gauge pressure means balance with the surroundings, not absence of gas or absence of wall momentum transfer.

For a frictionless piston of surface area $A$ with its outer face exposed to atmospheric pressure, the additional inward force required for mechanical equilibrium is $F=(p-p_{\mathrm{atm}})A=p_{g}A$. This is why a mechanical load can reveal a pressure difference even though the equation of state requires absolute pressure. With piston friction, the load alone no longer determines $p_{g}A$; a direct pressure sensor avoids that particular inference.

Adsorption is accumulation of a substance at an interface. Binding a solute to a packing surface can delay its passage through a chromatography column. Ideal size-exclusion chromatography separates through SEC pore access without such binding; extra retention from adsorption requires a different interpretation.

For a gas contacting a solid, adsorption concentrates molecules at the boundary; absorption takes material into the bulk. A porous solid can have a large internal surface area: binding on pore walls is still adsorption, even though those walls lie inside the particle. Both processes can occur together. Desorption returns adsorbed material to the surrounding phase. The useful question is therefore not just how much material disappeared from the gas, but where it went and whether it returns when the pressure falls.

An adsorption site model turns this spatial distinction into counting. With twelve identical adsorption sites and three bound molecules, the surface coverage is $\theta =3/12=1/4$. Filling six gives $\theta =1/2$; filling twelve gives a complete monolayer. These are configurations illustrating the count, rather than measured arrangements or claims that all real surfaces have twelve sites.

An axisymmetric cylindrical membrane deformation is invariant under rotations about the cylinder axis, so its radius depends only on the axial coordinate. Its surface area is $2\pi \int R\sqrt{1+R_z^2}dz$, and its enclosed volume is $\pi \int R^{2}dz$. These scalar shape constraints are the basis of the fixed-volume capillary spectrum of a cylindrical membrane.

Boyle's law states that a fixed amount of substance of an ideal gas at constant absolute thermodynamic temperature has constant product of absolute pressure and total volume:Here $p$ is measured relative to a vacuum, not relative to the surrounding atmospheric pressure. Halving the volume doubles the absolute pressure; doubling the volume halves it. A real gas approximately follows Boyle's law when it is sufficiently dilute and far from a phase transition.^[ref]

For an original calculated example, suppose a sealed gas sample has $p_1=100\mathrm{kPa}$ and $V_1=100\mathrm{c}{\mathrm{m}}^3$. After isothermal compression to $V_2=50\mathrm{c}{\mathrm{m}}^3$,Both states have $C=10000\mathrm{kPa}\mathrm{c}{\mathrm{m}}^3=10\mathrm{J}$, since $1\mathrm{kPa}\mathrm{c}{\mathrm{m}}^3=10^{-3}\mathrm{J}$. The volume ratio is dimensionless, so using the same volume units in numerator and denominator leaves the pressure in kilopascals. Although $pV$ has units of energy, its value is not the compression work: isothermal ideal-gas work involves an integral along the path.

The blue curves show calculated ideal gas states, not measurements. The left panel is a branch of a hyperbola; the right panel plots exactly the same absolute pressures against $1/V$. The dashed line uses gauge pressure with a stipulated constant atmospheric pressure of $100\mathrm{kPa}$. Its displaced intercept explains why absolute pressure in Boyle's law matters.

A short derivation starts from the ideal gas law, $pV=nRT$. Fixing the amount of substance $n$ and absolute thermodynamic temperature $T$ fixes $C=nRT$, where $R$ is the molar gas constant. The microscopic picture in ideal gas pressure from wall collisions is equally appealing: at fixed temperature, the typical particle kinetic energy stays fixed, while compression increases the number density and wall momentum transfer per unit surface area and time. A larger absolute pressure follows without requiring hotter particles.

The constraints are essential. A leak changes the amount of substance. Rapid compression can raise the temperature before heat transfer restores thermal equilibrium, as in the isothermal gas syringe experiment. Molecular size and interactions in a dense real gas invalidate the ideal gas approximation; the Van der Waals equation describes one such correction. Equal endpoint temperatures allow an endpoint test of Boyle's law, but do not establish that every intervening state lies on an isothermal process.

For fixed mass $m$, mass density satisfies $\rho =m/V=mp/C$, so it is proportional to absolute pressure. In a dilute bubbly flow, Boyle's law allows bubble volume and its contribution to buoyancy to increase as hydrostatic pressure decreases, provided gas mass is retained, dissolution is negligible, and the temperature stays approximately fixed. Its usefulness is that a simple ratio predicts a substantial size change without tracking individual molecules.

Consider an ideal pressure-driven closure plug with a large chamber-facing projected area $A_p$. Its opposite side seats against a smaller annular packing surface area $A_{\mathrm{seal}}$. The remaining projected part $A_u=A_p-A_{\mathrm{seal}}>0$ is the unsupported area: it has no opposing packing contact in that plane. Here $p$ is the chamber pressure relative to the exterior. Let $q$ denote the mean axial compressive contact stress on the packing, not the chamber's hydrostatic stress.

Neglect friction, the plug's weight, and other axial reactions. The plug's Newton's second law in static balance isThe load is the same on both sides; its smaller supporting surface area raises the packing's mean contact stress. For a stipulated ratio $A_p/A_{\mathrm{seal}}=1.25$ and $p=1\mathrm{GPa}$, the ideal seating stress is $q=1.25\mathrm{GPa}$. Its excess over chamber pressure is $0.25\mathrm{GPa}$. Reducing $A_u$ to zero removes this amplification. This is the elementary force balance behind self-energizing seating, consistent with the lecture's Figure 1 caption.^[ref]

The relation is a model, not a universal sealing law. The mean axial contact stress need not equal the local contact stress along every possible leakage path. Confinement and plasticity in continuum mechanics determine whether the packing closes a gap or instead extrudes into it. Friction diverts load into other surfaces; material plasticity in continuum mechanics changes contact geometry; the vessel must carry its own stress. A finite initial seating load may also be needed before chamber pressure can develop. An arbitrarily small $A_{\mathrm{seal}}$ does not yield arbitrarily useful sealing: the assumptions fail when local deformation, extrusion or structural failure takes over.

Consider an original geometric model of $N$ separate nonporous solid spheres, each with radius $r$. Hold their total volume $V_s$ fixed. The sphere formulas giveAt fixed mass density, fixed $V_s$ also means fixed total mass. Halving $r$ requires eight times as many particles and doubles their total external surface area $S$; this follows by eliminating $N$, without any assumption about chemical kinetics.

For example, take one sphere of radius $1.00\mathrm{mm}$, so $V_s=4\pi /3\mathrm{m}{\mathrm{m}}^3$. Eight spheres of radius $0.50\mathrm{mm}$ have the same $V_s$. Their total surface areas are respectively $4\pi \mathrm{m}{\mathrm{m}}^2$ and $8\pi \mathrm{m}{\mathrm{m}}^2$, approximately $12.57$ and $25.13\mathrm{m}{\mathrm{m}}^2$. Sixty-four particles of radius $0.25\mathrm{mm}$ would instead give $16\pi \mathrm{m}{\mathrm{m}}^2$, four times the original area.

To obtain a kinetic prediction, add assumptions explicitly. Let $f$ be the accessible fraction of external surface area, $\sigma$ the amount of substance of catalytic surface sites per unit accessible area, and $\nu$ the mean number of completed conversions per site per unit time under the chosen conditions. Then the model's reaction product formation rate isHere $\nu$ has units ${\mathrm{s}}^{-1}$ and $\sigma$ has units $\mathrm{mol}{\mathrm{m}}^{-2}$, giving $\mathrm{mol}{\mathrm{s}}^{-1}$ overall. Halving $r$ doubles this predicted rate only if $f$, $\sigma$ and $\nu$ remain unchanged. Particle contact can hide area; pores can add area; diffusion, altered surface structure and catalyst deactivation can change the other factors. Consequently, twice the calculated external area is not a universal promise of twice the measured reaction rate. The geometric argument changes contact, not the net Gibbs free energy or the equilibrium constant of the same overall chemical reaction.

A catalytic surface site is a location on a solid surface capable of participating in a catalytic route. The schematic notation $\ast$ denotes a vacant site; $\mathrm{A}\ast$ denotes A bound there. A minimal illustrative catalytic cycle isThe first step represents adsorption; the second combines conversion and release. Cancelling $\ast$ and $\mathrm{A}\ast$ gives $\mathrm{A}\rightleftharpoons \mathrm{B}$. The site is regenerated even though its occupancy changes during the cycle. This is an original bookkeeping model, not a measured reaction mechanism for manganese dioxide.

If the only site states are vacant and occupied by A, the site inventory obeys $n_{\ast }+n_{\mathrm{A}\ast }=n_{\mathrm{sites}}$. A blocked or inaccessible part of a particle does not contribute available catalytic surface sites merely because it contributes mass. Different surface structures can have different abilities to bind and convert reactants; equal geometric surface area need not mean equal catalytic activity.

For two independent uniform points on the surface of a sphere, rotational invariance permits fixing the first at the north pole. The surface area of a latitude band of angular width $d\theta$ is $2\pi R^2\sin \theta d\theta$, yielding the displayed density for their central angle. Equivalently their unit position vectors have a dot product uniformly distributed on $[-1,1]$. The law is independent of sphere radius.

For a real three-dimensional linear map $A$, preservation of every surface area is equivalent to being an orthogonal matrix. For necessity, diagonalize the symmetric matrix $A^{T}A$ in an orthonormal basis, with nonnegative eigenvalues ${\lambda }_i$. The three unit coordinate parallelograms have image areas $\sqrt{{\lambda }_1{\lambda }_2}$, $\sqrt{{\lambda }_1{\lambda }_3}$, and $\sqrt{{\lambda }_2{\lambda }_3}$. Setting all three equal to $1$ forces every ${\lambda }_i=1$. Sufficiency follows from preservation of dot products and $\Vert \mathbf{u}\times \mathbf{v}{\Vert }^2=\Vert \mathbf{u}{\Vert }^2\Vert \mathbf{v}{\Vert }^2-(\mathbf{u}\cdot \mathbf{v}{)}^2$. This specifically uses dimension three; in dimension two, preservation of planar area alone only requires $|\det A|=1$.

The lateral surface obtained by rotating a straight segment in a plane containing the rotation axis has endpoint radii $a,c$ and slant length $\ell$. Its surface area is $\pi (a+c)\ell$: the radius is linear in the segment's arc length, so the surface area of a surface of revolution is $2\pi {\int }_0^{\ell }\rho (s)ds=2\pi \ell (a+c)/2$. This formula excludes the end disks and assumes that the segment's radial direction is fixed. A straight segment whose endpoint horizontal directions differ generally sweeps a different surface of revolution.

For concentration $C$ per unit surface area, conservation on a moving material interface givesThe material derivative follows the surface particles. The two dilution terms are tangential expansion and curvature-induced area change from normal velocity $u_n$; the last term is surface diffusion.

Coulomb's law gives the mutual electric force between stationary point charges in vacuum. Put electric charges $q_1$ and $q_2$ at ${\mathbf{r}}_1$ and ${\mathbf{r}}_2$, and define the displacement from charge 1 to charge 2 by $\mathbf{r}={\mathbf{r}}_2-{\mathbf{r}}_1$, with separation $r=|\mathbf{r}|>0$ and unit vector $\hat{\mathbf{r}}=\mathbf{r}/r$. In SI units, the force on charge 2 due to charge 1 isHere ${\epsilon }_0$ is the vacuum permittivity. The magnitude is $k|q_1{q}_2|/r^2$. If $q_1{q}_2>0$, the force on charge 2 points away from charge 1; if $q_1{q}_2<0$, it points toward charge 1. Interchanging the charges gives ${\mathbf{F}}_{1\leftarrow 2}=-{\mathbf{F}}_{2\leftarrow 1}$, consistent with Newton's third law in this electrostatic setting. ^[ref].

For a fully computed example, stipulate $q_1=+2\mathrm{nC}$, $q_2=-3\mathrm{nC}$, ${\mathbf{r}}_1=0$, ${\mathbf{r}}_2=0.20\mathrm{m}\hat{\mathbf{x}}$ and the approximate value $k=8.99\times 10^9\mathrm{N}{\mathrm{m}}^2/{\mathrm{C}}^2$. ThenThe force is attractive, of magnitude $1.3485\times 10^{-6}\mathrm{N}$. The source electric field at charge 2 is ${\mathbf{E}}_1({\mathbf{r}}_2)=449.5N/C\hat{\mathbf{x}}$, pointing away from the positive source. Multiplication by the negative $q_2$ reverses the direction to give the attractive force. This field excludes charge 2's own self-field.

Holding the electric charges fixed while changing the separation from $R$ to $2R$ or $3R$ gives the normalized inverse-square law valuesFor the stipulated example, separations $0.20$, $0.40$ and $0.60\mathrm{m}$ give magnitudes $1.3485\times 10^{-6}$, $3.37125\times 10^{-7}$ and $1.498333\dots \times 10^{-7}\mathrm{N}$. The diagram computes these ratios and uses one common arrow-length scale for both sign choices.

The inverse-square law has a geometric explanation through Gauss's law: a source point charge gives fixed total electric flux through a concentric sphere, whose surface area is $4\pi r^2$. Dividing that flux by the area gives a radial electric field proportional to $1/r^2$. This connects a pairwise force calculation to a global conservation-like accounting of field sources. For several stationary point charges, the superposition principle adds the individual electric fields as vectors, and ${\mathbf{F}}_i=q_i{\mathbf{E}}_{\mathrm{other}}({\mathbf{r}}_i)$ uses all sources except charge $i$ itself. The Coulomb potential energy gives an alternative route when the desired quantity is work rather than instantaneous force.

A crista is a fold of the inner mitochondrial membrane into the mitochondrial matrix. Its membrane remains continuous with the surrounding inner mitochondrial membrane; it is not a separately inserted sac like a thylakoid.^[ref]

The diagram idealizes access through the fold's opening. Connectivity alone does not imply instant mixing or identical concentrations throughout the intermembrane space. Narrow geometry and molecular interactions can matter to transport. Increasing surface area also does not, by itself, prove a proportional increase in ATP production: the supply of substrates, available oxygen and the state of the coupling machinery must also support the flux.

For a sphere of radius $R>0$, the spherical cap above $z=R-h$, with $0\le h\le 2R$, has curved surface area $2\pi Rh$. On the surface use $r(z)=\sqrt{R^2-z^2}$. The surface of revolution area element is $2\pi r\sqrt{1+(dr/dz{)}^2}dz=2\pi Rdz$. Integrating from $R-h$ to $R$ proves the formula. It includes only the spherical surface; if the planar closing disk is required, add $\pi (2Rh-h^2)$. The coordinate expression is interpreted by a limit at a pole.

A diamond anvil culet is the small flat contact face at the tip of a diamond anvil. Its diameter $d_c$ determines its nominal surface area $A_c$. The gasket hole has a different surface area $A_h$; neither should be confused with the much larger drive mechanism. For an ideal static axial balance, $F$ is the compressive force on either anvil. The two equal opposing loads do not produce $2F$ through a single contact.

For an original chosen geometry, $F=1000\mathrm{N}$ and $d_c=0.300\mathrm{mm}$ giveHere $\bar{q}$ is the mean axial contact stress over the culet, not a measured chamber pressure. Using a $0.100\mathrm{mm}$ hole instead would give $F/A_h=127.324\mathrm{GPa}$, nine times larger. That substitution falsely assigns the entire applied force to the hole.

To see the missing load path, idealize the culet as a coplanar hole region plus a surrounding contact annulus. With a hydrostatic chamber pressure $p$ and axial resultant $F_g$ carried by the gasket, its force balance isIf the stipulated chamber pressure is $5.40\mathrm{GPa}$, the hole carries only $42.4\mathrm{N}$ and the gasket carries $957.6\mathrm{N}$. These are model loads, not measurements. Real gasket deformation and anvil elasticity alter the local contact stress; the scalar balance cannot predict $p$ without additional information. This is why internal pressure calibration matters even when applied force is accurately known.

Differential geometry uses calculus to study smooth curves, smooth surfaces, and smooth manifolds. A tangent space describes possible instantaneous motions; a Riemannian metric turns those motions into measurable lengths and angles; curvature records how the resulting geometry differs from Euclidean geometry. Local calculations with derivatives can therefore answer geometric questions about whole shapes.

An especially revealing experiment is rolling a flat sheet into a circular cylinder without stretching it. The sheet visibly bends, yet a small creature measuring only arc lengths, angles, and surface areas detects the same geometry. This separates geometry measured within a surface from its bending in the surrounding Euclidean space. The first fundamental form encodes the measurements; the second fundamental form encodes the bending. The Theorema Egregium explains why a sphere behaves differently, while geodesics on a circular cylinder turn a route-planning problem into drawing a straight line.

Electric flux measures how much electric field crosses a chosen oriented surface, with a sign distinguishing the two crossing directions. A strong electric field running along a surface contributes nothing to its electric flux: only the normal component counts. This lets Gauss's law turn a potentially complicated electric field into one number recording the net enclosed electric charge.

For a surface $S$ with chosen unit normal $\mathbf{n}$, the electric flux is the surface integralHere $dA$ is positive surface area, while $d\mathbf{S}$ is the vector area element. The dot product makes electric flux a signed scalar, rather than a vector like the electric field. Its SI unit is $\mathrm{N}{\mathrm{m}}^2/C$: electric field measured using the newton per coulomb, multiplied by surface area in square metres. Reversing the orientation of a surface replaces $\mathbf{n}$ by $-\mathbf{n}$ and reverses the electric flux. Calling it electric flux does not imply that electric charge is moving through the surface. ^[ref].

For an open patch in a plane, with surface area $A$ and a constant electric field, the unit normal is constant and the surface integral simplifies toThe angle $\theta$ is between the electric field and the chosen unit normal, not between the electric field and the plane. The cosine describes the signed projected surface area seen by the electric field. Thus electric flux is positive for $0\le \theta <90^{\circ }$, zero at $90^{\circ }$, and negative for $90^{\circ }<\theta \le 180^{\circ }$. ^[ref].

For example, take $E=100N/C$ and $A=0.02{\mathrm{m}}^2$, so $EA=2\mathrm{N}{\mathrm{m}}^2/C$. Three choices of angle giveKeeping the $60^{\circ }$ patch fixed and reversing its unit normal gives $120^{\circ }$ and changes $+1$ to $-1\mathrm{N}{\mathrm{m}}^2/C$. The electric field and physical patch have not changed. If the electric field is instead tangent to the patch, its dot product with the unit normal vanishes: electric flux is zero even though $E=100N/C$.

For a closed surface enclosing a volume $V$, choose the outward unit normal. Gauss's law givesHere ${\epsilon }_0$ is the vacuum permittivity, and $Q_{\mathrm{enclosed}}$ is the signed total electric charge inside. The electric field in the surface integral includes all sources, including external electric charges. Assume no ideal point charge lies on the integration surface. Zero enclosed electric charge implies zero total electric flux; it does not imply zero electric field everywhere on the closed surface or inside it. Positive and negative contributions to the surface integral can cancel. ^[ref].

To see the cancellation with sources inside, place point charges $+q$ at $(-a,0,0)$ and $-q$ at $(a,0,0)$, where $q>0$, $a=0.30\mathrm{m}$, and both lie inside the sphere of radius $R=1\mathrm{m}$ centered at the origin. This separated pair has the electric dipole geometry. Its enclosed electric charge is $q-q=0$, so its total outward electric flux is zero. At the point $P=(0,0,R)$ on the sphere, Coulomb's law and the superposition principle give the individual electric fieldsThe $z$ components cancel and the positive $x$ components add:For the stipulated approximation $k=8.99\times 10^9\mathrm{N}{\mathrm{m}}^2/{\mathrm{C}}^2$ and $q=1\mathrm{nC}$, this is about $4.740N/C$ along positive $x$. At $P$, the outward unit normal is $\hat{\mathbf{z}}$, so this nonzero electric field is tangent to the sphere. At other points of the sphere, its normal component can be positive or negative. The source-field formula applies away from the two point charges, where the electric field has zero divergence; it is singular at the sources. ^[ref].

The diagram shows streamlines of the three-dimensional electric field in the section $y=0$. The circle is a cross-section of the sphere, not the integration surface: Gauss's law uses the entire three-dimensional surface integral. A two-dimensional line integral of outward field around that circle would be a different quantity, with units $\mathrm{N}m/C$ rather than the electric flux units $\mathrm{N}{\mathrm{m}}^2/C$.

The electrostatic boundary conditions at a conductor relate the one-sided total electric fields at a smooth metal–vacuum interface. Choose a unit normal $\mathbf{n}$ from metal into the adjacent vacuum, let ${\mathbf{E}}_{\mathrm{metal}}$ and ${\mathbf{E}}_{\mathrm{vac}}$ be the two limiting electric fields, and let $\sigma$ be the signed surface charge density. A thin Gaussian pillbox with cap area $A$ has outward cap normals $+\mathbf{n}$ on the vacuum side and $-\mathbf{n}$ on the metal side. As its height tends to zero, its side electric flux tends to zero for bounded local electric fields, leavingThe minus sign comes from the second outward cap normal, not from the sign of the electric charge. A shrinking rectangular loop across the interface and the zero closed line integral in electrostatics give continuity of the tangential component:For a conductor in electrostatic equilibrium, ${\mathbf{E}}_{\mathrm{metal}}=0$, henceThus positive surface charge density gives an electric field away from metal and negative surface charge density gives an electric field toward metal. The derivative uses the same chosen unit normal, from the vacuum side. The ideal surface layer has no unique electric field value on the layer itself; these are limiting values. Ordinary surface electric charge preserves continuity of electric potential; an ideal electric dipole sheet would require a different model. ^[ref].

Here “exterior” means just outside the metal, including the vacuum within a cavity in a conductor. On a spherical inner wall of radius $a$, its unit normal from metal into the cavity in a conductor is $-\hat{\mathbf{r}}$. A central positive point charge $q$ gives ${\mathbf{E}}_{\mathrm{vac}}=q\hat{\mathbf{r}}/(4\pi {ϵ}_0{a}^2)$ there, soThe outward normal of the cavity in a conductor as a volume is instead $+\hat{\mathbf{r}}$; confusing these two unit normals reverses the inferred sign. For a stipulated spherical enclosure with inner radius $a$, outer radius $b$, no external sources and total isolated metal electric charge $Q_c$, spherical symmetry and Gauss's law give the analytical modelIts outer surface charge density is $(Q_c+q)/(4\pi b^2)$. Grounding to $V_c=0$ with zero electric potential at infinity instead gives $Q_c=-q$ and zero exterior electric field, while preserving the same nonzero electric field inside the charged cavity in a conductor.

As a numerical sign check, stipulate $q=1\mathrm{nC}$, $a=0.10\mathrm{m}$, $b=0.20\mathrm{m}$ and $1/(4\pi {ϵ}_0)\simeq 8.99\times 10^9\mathrm{N}{\mathrm{m}}^2/{\mathrm{C}}^2$. The inner electric field is approximately $899N/C$ along $+\hat{\mathbf{r}}$, but ${\mathbf{E}}_{\mathrm{vac}}\cdot \mathbf{n}=-899N/C$ and ${\sigma }_{\mathrm{inner}}\simeq -7.96nC/{\mathrm{m}}^2$. An isolated neutral enclosure has $+1\mathrm{nC}$ on the outer wall, approximately $+1.99nC/{\mathrm{m}}^2$ and $225N/C$ just outside. Multiplying each surface charge density by its spherical surface area recovers the signed wall electric charges.

The local result uses the total electric field, including all induced and imposed sources. The factor $\sigma /{ϵ}_0$ follows from the zero metal-side electric field; a symmetric isolated sheet in vacuum instead has magnitude $|\sigma |/(2{ϵ}_0)$ on each side. ^[ref]. At sharp corners, limiting electric fields can be singular and the smooth local pillbox argument needs care. In material outside the metal, use the normal free-charge jump of the electric displacement field and the material's permittivity, as in electrostatic interface boundary conditions. ^[ref].

Two related properties of an extensive quantity must be distinguished. Additivity says that a quantity $X$ for independent subsystems A and B obeysThe subsystems may be in different states. Extensivity expressed through a homogeneous function says that the equilibrium fundamental relation obeyswith composition held fixed. This is degree-one homogeneity along a replication ray; it does not say that internal energy is a linear function of arbitrary independent changes of entropy, volume and amount of substance.

Additivity gives integer replication by repeated addition. Subdividing a uniform bulk sample into $b$ equivalent parts gives rational scaling $a/b$; continuity then extends it to positive real factors. This argument requires negligible boundary contributions and a consistently prepared state. Conversely, a degree-one homogeneous function need not be additive between different rays: $f(x,y)=\sqrt{x^2+y^2}$ has $f(\lambda x,\lambda y)=\lambda f(x,y)$, but $f(1,0)+f(0,1)=2\ne \sqrt{2}=f(1,1)$. Thus a scaling identity alone cannot justify summing arbitrary states.

For finite samples, the surface correction to thermodynamic free energy can matter. Under geometric enlargement with volume multiplied by $\lambda$, surface area grows as ${\lambda }^{2/3}$, so a surface energy term is not a bulk extensive quantity. Replicating separated droplets retains every original surface, whereas merging them changes the total surface area. Likewise, appreciable mutual gravitational potential energy makes interacting subsystems fail the independent-copy premise. The thermodynamic limit explains when bulk scaling becomes accurate; it does not remove these qualifications for every small or strongly interacting system.

The first fundamental form of a smooth surface $S\subseteq {\mathbb{R}}^3$ is the restriction of the ambient inner product to each tangent space $T_{p}S$. For a regular parametric surface $X(u,v)$, it is

Here $X_u^2$ means $X_u\cdot X_u$, not a componentwise square. The coefficients form the Gram matrix of the two coordinate tangent vectors. Regularity makes this a positive-definite matrix, so $E>0$ and $EG-F^2>0$.

To calculate the length of a curve $\gamma (t)=X(u(t),v(t))$, apply the chain rule and integrate its speed:The same inner product gives the angle $\alpha$ between nonzero tangent vectors $V,W$ through $\cos \alpha =I(V,W)/\sqrt{I(V,V)I(W,W)}$, and gives the area element $dA=\sqrt{EG-F^2}dudv$ as a positive density. One compact expression therefore supplies rulers, protractors, and surface area measurements.

For the lateral surface of a circular cylinder, take the circumferential arc length $u$ rather than the angle:Direct inner products give $E=G=1,F=0$, hence $I=d{u}^2+d{v}^2$. This is exactly the Euclidean metric on the unrolled plane. For instance, a path with $u=3t,v=4t$, $0\le t\le 1$, has length $5$ on either surface. Every corresponding path retains its length, not merely this example, so sufficiently small neighborhoods retain their intrinsic distances. The local isometry preserves angles and surface areas too. Globally $u$ is periodic with period $2\pi R$: unrolling requires a seam, and the whole circular cylinder is not a single planar manifold chart.