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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.

For an original mathematical model, imagine a sustained, imposed release pathway, with no store refill during the plotted interval. Let $S$ be releasable store amount of substance, $X$ excess freely dissolved cytosolic calcium amount of substance, and $Q$ the cumulative amount removed from $X$ into a separate sink. Choose $S(0)=1\mathrm{amol}$, $X(0)=Q(0)=0$, cytosolic volume $V_c=1\mathrm{pL}$, release rate constant $k=0.4{\mathrm{s}}^{-1}$ and clearance rate constant $h=0.8{\mathrm{s}}^{-1}$. These are stipulated values, not physiological measurements or a model of IP3 receptor gating.

Assume fixed volume, uniform mixing, no binding, no reverse transfer, no outside entry and first-order loss from each donating compartment. ThenThe last equality is conservation of mass for the tracked calcium across the store, mobile excess and sink together; it does not assert conservation inside a biological cell if the sink represents export. A constant background $c_b=0.1\mu \mathrm{M}$ is held separate from this excess bookkeeping. Since $1\mathrm{amol}/1\mathrm{pL}=1\mu \mathrm{M}$, the response is $c=c_b+X/V_c$.

Solving these ordinary differential equations givesRelease $kS$ declines from the start, yet $X$ initially rises because delivery exceeds clearance. Its peak occurs when $kS=hX$, at $t_{\ast }=\log (h/k)/(h-k)=\log (2)/0.4\approx 1.73\mathrm{s}$. There $X=0.25\mathrm{amol}$ and $c=0.35\mu \mathrm{M}$. Ultimately $S$ and $X$ tend to zero, $Q$ tends to $1\mathrm{amol}$ and $c$ returns to $c_b$. The delayed peak is a consequence of material balance, even though this model has no delayed channel activation.

This distinction guides interpretation: a large released amount need not mean a large simultaneous free cytosolic concentration. Clearance and volume intervene. Real biological cells add binding, store refilling, IP3 turnover and IP3 receptor regulation; the two plotted curves isolate one finite-store effect rather than explaining every calcium response.

For a disk with thickness $z_0$, random relative speeds $u=\lambda v$ and small-impact cutoff $b_0\ll z_0\ll R$, the leading three-dimensional logarithmic encounter term has upper cutoff of order $z_0$. At fixed mass and radius, its number volume density exceeds the spherical density by order $R/z_0$. Hence $t_{\mathrm{disk}}/t_{\mathrm{sphere}}\sim {\lambda }^3(z_0/R)\log (R/b_0)/\log (z_0/b_0)$ when the same cutoff is used for comparison. Different speed-dependent cutoffs must be distinguished in numerical applications. Distant planar encounters contribute a convergent, non-logarithmic correction. This scaling ceases to apply if thickness is comparable to the cutoff.

The first law applies conservation of energy to a thermodynamic system. For a closed system with unchanged bulk kinetic energy and potential energy, writeHere $U$ is internal energy, $Q>0$ is heat entering the system and $W>0$ is work done by the system on its surroundings. Compression therefore gives negative $W$. Some references define work positive when done on the system; their plus sign represents the same balance with a different convention. If bulk kinetic energy or potential energy changes, include it in the total energy balance. ^[ref]

Supplying $1000\mathrm{J}$ of heat while a gas does $300\mathrm{J}$ of work increases its internal energy by $700\mathrm{J}$. Compressing it with $200\mathrm{J}$ of input work in an adiabatic process instead gives $Q=0$, $W=-200\mathrm{J}$ and $\Delta U=200\mathrm{J}$.

The distinction between $d$ and $\delta$ records state versus path: internal energy is a state function, whereas heat and work depend on the process. For an ideal gas expanding reversibly at constant temperature, $W=N{k}_{B}T\log (V_2/V_1)$ and $Q=W$, since internal energy of an ideal gas depends only on temperature. Choose $N{k}_{B}T=1000\mathrm{J}$ and $V_2=2V_1$: then $Q=W=693.15\mathrm{J}$. An insulated free expansion into vacuum between the same equilibrium endpoints has $Q=W=0$. Both have $\Delta U=0$, yet their transfers differ. Their common entropy increase is $N{k}_B\log 2$. ^[ref]

For a simple compressible equilibrium system of fixed composition, combining the laws givesHere $p$ is pressure and $V$ is volume. This identity relates neighboring equilibrium states. Identifying $TdS$ with actual heat and $pdV$ with actual work requires a suitable reversible thermodynamic process; an irreversible piston calculation uses the external resisting pressure for boundary work. This distinction prevents applying a reversible ideal-gas adiabat to an irreversible adiabatic piston compression.

A freely floating body at rest has no supporting contact or applied holding force. Neglecting surface tension and the buoyancy of air, its upward buoyancy balances its downward force from gravity:For a body of uniform mass density ${\rho }_{\mathrm{body}}$ and total volume $V_{\mathrm{body}}$, its mass is $m={\rho }_{\mathrm{body}}{V}_{\mathrm{body}}$. Substitution into Archimedes' principle givesPartial immersion is possible when ${\rho }_{\mathrm{body}}<{\rho }_f$. A nonuniform body uses its mean mass density $m/V_{\mathrm{body}}$ instead; for a hollow body this volume must count the enclosed space that excludes water. This explains how the amount of displaced water can support a vessel whose walls alone are denser than water. ^[ref]

For the $600\mathrm{kg}{\mathrm{m}}^{-3}$ cube in water of mass density $1000\mathrm{kg}{\mathrm{m}}^{-3}$, the displaced fraction is $600/1000=0.60$: sixty percent of its volume lies below the surface. With a horizontal top and bottom, the immersed depth $d$ satisfies $V_{\mathrm{displaced}}=Ad$, soIts remaining height is $0.040\mathrm{m}$, and the buoyancy is $1000\times 9.81\times 0.000600=5.886\mathrm{N}$, exactly balancing its downward force from gravity.

Fluid mechanics studies the motion and forces of liquids and gases. Its central simplification is the continuum approximation: replace individual molecules by bulk properties defined at every position and time. The averaging region must contain many molecules while remaining small compared with the distances over which those properties vary. A fluid element is such a small material parcel, rather than a single molecule. This connects fluid mechanics to continuum mechanics and makes calculus useful for describing flow. ^[ref]

Three basic fields are the velocity field $\mathbf{u}(\mathbf{x},t)$, the mass density $\rho (\mathbf{x},t)$ and the pressure $p(\mathbf{x},t)$. The first gives a parcel's instantaneous velocity; the second gives mass per unit volume. The pressure contributes an inward normal traction $-p\mathbf{n}$ across a surface with outward unit normal $\mathbf{n}$. Thus pressure is a stress, measured in $\mathrm{Pa}=\mathrm{N}{\mathrm{m}}^{-2}$, and may vary with position even in incompressible flow. Additional viscous stresses can act tangentially. ^[ref]

The guiding questions are what must be conserved and which forces matter. Mass conservation constrains the velocity field; momentum conservation determines its response to pressure gradients, viscosity and body forces. An incompressible flow model, a Reynolds number estimate and Bernoulli's principle answer different parts of a problem. Their appeal is that a few carefully stated assumptions can determine useful quantities before the entire velocity field is known.

Placement of geometric objects without interior overlap. A packing fraction measures how much area or volume they occupy. Circle packing and sphere packing are examples.

A hard-sphere gas is a classical model of identical impenetrable spheres, each of diameter $\sigma$ and radius $\sigma /2$. Its pair potential is $u(r)=+\infty$ for centre separation $r<\sigma$ and $u(r)=0$ for $r\ge \sigma$; the convention at exact contact has no effect on the integral. It isolates the effect of particle size without an attractive potential energy.

The upper-left panel is a cross-section of a three-dimensional forbidden region: the second centre cannot enter a sphere of radius $\sigma$ around the first centre. The particles themselves have radius $\sigma /2$. Consequently, with particle volume $v_{\mathrm{s}}$ and pair excluded volume $v_{\mathrm{ex}}$,The last equality follows directly from the Mayer f-function: $f=-1$ throughout the forbidden volume and zero outside it. Confusing particle radius with contact separation changes the answer by a factor of eight; omitting the unordered-pair factor changes it by a further factor of two.

Define the packing fraction $\eta =n{v}_{\mathrm{s}}$, where $n$ is number density. Then the virial expansion beginsFor example, at the toy value $\eta =0.01$, the leading correction predicts $Z\approx 1.04$; the omitted terms still matter if accuracy beyond this approximation is needed. The classical second virial coefficient is independent of temperature because the geometry of forbidden configurations does not change when the gas is warmed. The pressure itself still scales with temperature at fixed number density.

This model makes the positive correction particularly clear: eliminating overlapping configurations changes the dependence of the available configuration space on volume, even though the allowed configurations all have zero interaction energy. The figure's other panels add a toy attractive shell and show how its Boltzmann factor competes with that geometry. The hard-sphere gas is a useful reference model, while a low-order virial expansion remains a dilute-gas approximation rather than a dense sphere packing formula.

Two convex bodies are homothetic convex bodies if one is a translation of a positive uniform dilation of the other. Their lengths scale by $s$ and their volumes by $s^n$. Closed disks of positive radii are homothetic convex bodies; axis-aligned positive-sided rectangles are homothetic convex bodies exactly when their aspect ratios agree. The Brunn-Minkowski theorem characterizes this relation by equality of root-volumes in Minkowski addition.

An incompressible flow preserves the volume of each moving fluid element. Its local condition is zero velocity divergence, $\nabla \cdot \mathbf{u}=0$. Combined with mass conservation, it says that each parcel's mass density stays unchanged as it moves. This does not require every parcel to have the same mass density: an initially nonuniform distribution can be transported without compression. Nor does incompressibility require a constant velocity or constant pressure.

For positive mass density, the continuity equation makes the equivalence precise:The material derivative follows a fluid element. Thus $D\rho /Dt=0$ if and only if the velocity field is a divergence-free vector field. The distinction between following a parcel and observing a fixed position prevents confusing incompressibility with an unchanging density field. ^[ref]

For a rigid pipe with impermeable walls, no leaks and steady flow, mass conservation at constant mass density implies a constant volumetric flow rate:Here $\mathbf{n}$ points downstream, $A$ is the section's area and $\overline{U}$ is the mean normal velocity, not necessarily the centreline speed. No assumption of inviscid flow or a uniform velocity profile is needed. If unequal amounts entered and left a fixed segment, its mass would accumulate, contradicting steady flow.

For a worked example, take horizontal water flow with stipulated mass density $1000\mathrm{kg}{\mathrm{m}}^{-3}$, inlet radius $R_1=0.010\mathrm{m}$, outlet radius $R_2=0.005\mathrm{m}$ and inlet mean velocity ${\overline{U}}_1=0.20\mathrm{m}{\mathrm{s}}^{-1}$. The circular areas giveThe radius halves, the area falls to one quarter, and the mean velocity quadruples. The mass flow rate is $\rho Q=0.02\pi \mathrm{kg}{\mathrm{s}}^{-1}$. The same amount of water passes both sections each second; narrowing changes its speed rather than compressing it. The Reynolds number and Bernoulli's principle discussions use this same example to examine viscous effects and the additional assumptions needed to calculate pressure.

An extensive quantity measures how much of a thermodynamic system there is; an intensive quantity describes its state independently of that amount. The clean comparison is replication: take independent identical systems at the same thermodynamic equilibrium, with the same substance, composition and thermodynamic phase. Combining their accounting boundaries multiplies internal energy $U$, entropy $S$, volume $V$, mass $m$ and amount of substance $n$, while leaving temperature $T$, pressure $p$ and chemical potential $\mu$ unchanged.^[ref]

For $\lambda$ copies the ledger isAn integer $\lambda$ counts actual copies; a positive real $\lambda$ describes the corresponding bulk scaling law. Copies can remain separated: replication need not involve removing a partition, changing a container or performing any work. Distinguish $n$, the amount of substance measured in moles, from $N=n{N}_A$, the dimensionless count of specified entities, where $N_A$ is the Avogadro constant in ${\mathrm{mol}}^{-1}$. For example, $2\mathrm{mol}$ contains $1.204428152\times 10^{24}$ specified entities; tripling it gives $6\mathrm{mol}$ and $3.613284456\times 10^{24}$ entities. This uses the exact Avogadro constant $N_A=6.02214076\times 10^{23}{\mathrm{mol}}^{-1}$, which does not triple. In the differential $dU=TdS-pdV+\mu dn$, $\mu$ is a molar chemical potential in $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$; the coefficient of $dN$ instead has units of energy per entity.

As a direct check, the ideal gas law $pV=nRT$ givesFor a monatomic ideal gas with only translational degrees of freedom active, internal energy of an ideal gas gives $U=3nRT/2$, so $U^{\prime }=\lambda U$ at unchanged temperature. Here $R$ is the molar gas constant. The scaling distinction lets a small sample supply an equation of state for a much larger amount: material state and inventory become separate questions.

Replication does not mean multiplying every linear dimension by $\lambda$: a geometrically similar container then has ${\lambda }^3$ times the volume. Nor does it mean combining samples at different states. Equal temperature establishes thermal equilibrium but does not imply equal pressure, composition or mass density. Density of a composite system makes this distinction concrete.

For a constant real square matrix $A$, the autonomous differential equation $\dot{x}=Ax$ has the linear flow ${\phi }_t(x)=e^{tA}x$, where $e^{tA}$ is the matrix exponential. The identity $e^{(t+s)A}=e^{tA}{e}^{sA}$ gives the flow map composition law, and $e^{-tA}$ reverses the evolution. Its determinant is $e^{t\mathrm{tr}A}$, so the matrix trace gives the instantaneous fractional rate of volume change, as illustrated in trace and local volume change. A zero trace preserves volume even when the flow stretches one direction and contracts another.

Why does an isopycnic centrifugation band return after displacement? Choose a dilute mathematical model with outward coordinate $z$, constant local centrifugal acceleration $a>0$, fixed displaced volume $V>0$, linear friction coefficient $f>0$, and an imposed liquid density ${\rho }_f(z)={\rho }_0+Gz$ with $G>0$. Treat the species' buoyant density ${\rho }_b$ as constant in this medium. The sedimentation coefficient force balance gives the ordinary differential equationIts exponential function solution is $z(t)=z_{\ast }+(z(0)-z_{\ast })e^{-kt}$. The sign of the velocity reverses across $z_{\ast }$, proving stable equilibrium when $z_{\ast }$ lies inside the liquid. Changing $V/f$ changes the approach time $1/k$, while $z_{\ast }$ depends on buoyant density and the imposed density gradient. A real centrifuge rotor has radially varying centrifugal acceleration; the constant-$a$ approximation describes a short radial interval. Here the density gradient is held fixed, rather than modeling its formation.

For an original dimensionless example, let $u=z/L$ and $\tau =t/T$, where $L$ is column length and $T$ a chosen time scale. Write $q=z_{\ast }/L$ and $\lambda =kT$, so $u(\tau )=q+(u(0)-q)e^{-\lambda \tau }$. Species A and B have the same $q=0.45$ but different friction-dependent rates: $({\lambda }_A,u_A(0))=(0.4,0.1)$ and $({\lambda }_B,u_B(0))=(1.6,0.95)$. A moves outward and B inward, both toward $u=0.45$. Species C has $(q_C,{\lambda }_C,u_C(0))=(0.8,0.8,0.1)$ and approaches a separate band at $u=0.8$. At $\tau =4$, their positions are approximately $0.3793$, $0.4508$ and $0.7715$: a finite-time photograph would still show A short of its limiting position.

An equilibrium band has finite width because diffusion opposes focusing. In the same local model, let $c(z)$ be concentration and let the constant diffusion coefficient be $D>0$. Combining the restoring drift with Fick's laws, zero net flux requiresDividing by $c$ and integrating proves the Gaussian function profile, whose standard deviation is ${\sigma }_z=\sqrt{D/k}$. This local result assumes that the band is narrow relative to its distance from the boundaries and that $D$, $k$ and the density gradient vary negligibly across it. Equal buoyant densities therefore imply coincident centers in this model, not identical widths. A band outside the available density range has no such interior restoring point.

Linear mass density is mass per unit length, measured in $\mathrm{kg}{\mathrm{m}}^{-1}$. A uniform string of mass $m$ and length $L$ has $\mu =m/L$; with constant tension $T$, its transverse wave speed is $\sqrt{T/\mu }$. Mass density instead measures mass per unit volume.

The luminosity per unit spatial volume. For a spherical source, its enclosed luminosity is $L(r)=4\pi {\int }_0^r\lambda (s)s^{2}ds$, and its projected surface brightness is $\mu (R)=2{\int }_R^{\infty }\lambda (r)rdr/\sqrt{r^2-R^2}$.