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The chemical potential is the change in internal energy when one particle is added at fixed entropy and volume:Equivalently, $\mu =(\partial F/\partial N{)}_{T,V}$ for the Helmholtz free energy.

Let a microstate $s$ have energy $E_s$ and particle number $N_s$. Maximizing the Gibbs entropy subject to normalization and fixed mean values of $E$ and $N$ is the maximum-entropy derivation of equilibrium ensembles. With Lagrange multipliers $\alpha$, $\beta$, and $-\beta \mu$, varyThe stationarity equation is $-\log p_s-1-\alpha -\beta (E_s-\mu N_s)=0$. Normalization therefore gives the grand canonical ensembleHere $\mathcal{Z}$ is the grand canonical partition function and $k_B$ is the Boltzmann constant.

For one fermionic quantum state of energy $\epsilon$, the Pauli exclusion principle permits occupation numbers only $n=0,1$. Its two grand-canonical weights are $1$ and $e^{-\beta (\epsilon -\mu )}$, so its mean occupation is the Fermi-Dirac distribution

For a free nonrelativistic particle, $\epsilon ={\hslash }^2{k}^2/(2m)$. In a region of area $A$, the number of wave-vector states in the annulus from $k$ to $k+dk$, including the two spin angular momentum states, isSince $kdk=(m/{\hslash }^2)d\epsilon$, the two-dimensional free-electron density of states is constant:

We now use units in which $k_B=1$, as in the question. At zero temperature, the Fermi-Dirac distribution is a step function, and henceAt positive temperature the same fixed particle number satisfies the exact relationThusThis is the low-temperature particle-number cancellation for constant density of states. Therefore, for $T\ll {\epsilon }_F$, the mean number of particles in the energy interval $[\epsilon ,\epsilon +d\epsilon ]$ is

Finally, compare the internal energy with its zero-temperature value. The thermally excited particles above ${\epsilon }_F$ and holes below ${\epsilon }_F$ have equal leading particle numbers, so their terms proportional to ${\epsilon }_F$ cancel. With $x=|\epsilon -{\epsilon }_F|/T$,where the integral follows from the fermion-to-boson thermal integral ratio. Differentiation gives the linear low-temperature heat capacity of a two-dimensional Fermi gas:The requested power law is therefore linear in $T$, with exponent one.

The University of Oklahoma K20 Center's handout describes three sealed vial-and-pipette respirometers in each temperature bath: soaked, germinating peas; nongerminating peas plus inert plastic beads; and beads alone. Displaced water establishes equal occupied volumes. Absorbent cotton holds potassium hydroxide, which removes carbon dioxide; nonabsorbent cotton separates it from the peas. The source equilibrates the vials before submersion and again before the initial reading. Colored water makes the boundary visible in each calibrated pipette. Readers then record positions every five minutes for twenty minutes.^[ref]

Bath temperature, vial position and seal integrity matter. The beads-only respirometer detects common temperature or atmospheric pressure drift; its change is subtracted from each pea reading. Nongerminating peas provide a biological comparison, not an assumption of zero cellular respiration. A larger corrected uptake slope for germinating peas would support greater molecular oxygen consumption by that sample under the tested conditions.^[ref]

Equal displaced volumes help match the gas spaces; they do not establish equal living mass or equal developmental state. Comparing uptake per seed and uptake per dry mass asks different questions. Leaks, incomplete carbon dioxide absorption, gas dissolution and changing water-vapour contributions can distort the respirometer inference. A blank correction cannot repair a sample-specific leak. These are consequences of the respirometer gas balance, rather than measurements reported in the handout.

Permittivity describes how an electric field and a material's electric polarization combine into the electric displacement field. In a static, local response with linearity and isotropy, and no electric polarization at zero electric field, the constitutive equation isHere ${ϵ}_0$ is vacuum permittivity, while ${ϵ}_r$ is the relative permittivity, a dimensionless quantity. The SI unit of permittivity is farad per metre, $F/m$: multiplying it by an electric field in $V/m$ gives an electric displacement field in $C/{\mathrm{m}}^2$. A larger positive permittivity lets the same parallel-plate capacitor hold more plate electric charge at the same electric potential difference.

The material contribution is electric polarization $\mathbf{P}$, the electric dipole moment per unit volume. The electric displacement field and electric susceptibility ${\chi }_e$ satisfySubstitution gives the first constitutive equation. The electric field here is the averaged, total field inside the material, not just the field of the external plate electric charges. The continuum approximation averages over many chemical atoms or molecules; locality means that the electric polarization at a position depends on the electric field there. Static equilibrium, fixed material conditions and sufficiently small electric fields are needed for this scalar proportionality. ^[ref].

For an ideal parallel-plate capacitor completely filled by a uniform dielectric, hold the free plate electric charge magnitude $Q$ fixed and neglect the fringing field of a capacitor. Writing $A$ for the plate area, Gauss's law gives $D=Q/A$. ThusThe electric dipoles point along the electric field. Their surface electric charges, described by polarization density, oppose the field from the free plate electric charges. For a hypothetical dielectric with ${ϵ}_r=4$, the internal electric field is one quarter of its vacuum value, while $P=3D/4$. At fixed electric potential difference instead, the internal electric field stays approximately $V/d$ and the plate electric charge grows fourfold. These are different experimental constraints. ^[ref].

Permittivity is a useful material summary, not a universal constant of the material under every condition. Temperature, frequency and electric field strength can change the response. In a linear anisotropic dielectric, permittivity is a tensor, $D_i={\sum }_j{ϵ}_{ij}{E}_j$, so the electric displacement field need not be parallel to the electric field. At nonzero frequency, delayed electric polarization requires complex permittivity and may produce dielectric loss. ^[ref].

A power law is a relation $f(x)=C{x}^p$, with fixed constants $C$ and $p$. For positive $x$, any real number $p$ is allowed. Its useful feature is a simple scaling rule:The same proportional change of input produces the same proportional change of output, independently of the starting value. The exponent controls that response; the coefficient fixes the magnitude.

For example, the area of a circle is $A(r)=\pi r^2$. Doubling the radius multiplies its area by four, and tripling it multiplies the area by nine. The volume of a sphere is $V(r)=4\pi r^3/3$: the same changes multiply its volume by eight and twenty-seven. The distinction explains why similarly shaped objects do not preserve their ratio of surface area to volume when their size changes.

When the output is positive, a logarithm turns the power law into a straight-line relation. Using positive reference values $x_0$ and $f_0=f(x_0)$,These ratios are dimensionless quantities, so the formula also applies to physical measurements with units. Plotting both ratios on a logarithmic scale gives a line of slope $p$. Changing a unit changes the numerical coefficient, while the exponent stays the same. Positive $p$ describes increasing values; negative $p$ describes decreasing values; $p=0$ gives a constant. The zero coefficient is a degenerate case with no useful logarithmic slope.

Exact scaling is stronger than regular variation. The function $x^2\log x$ has a doubling ratio tending to four, but its logarithmic correction keeps growing. The diagram contrasts it with the exact power law $x^2$.

A power law for a probability density and a power law for a tail probability have different exponents. For a Pareto distribution with $\alpha >0$ and lower endpoint one,The tail probability has exponent $-\alpha$; differentiating it and changing sign gives the probability density exponent $-\alpha -1$. Confusing them changes the interpretation of a fitted exponent. The normalizing coefficient and lower endpoint also matter: a decreasing power law is not automatically a valid probability density over the entire positive real line.

The appearance of a line over a finite logarithmic range is evidence to investigate, rather than a proof of an exact power law or its mechanism. In their study of power-law distributions, Clauset, Shalizi and Newman combine maximum likelihood estimation, goodness-of-fit tests and comparisons with alternative models; their abstract explains why ordinary least-squares fitting can give misleading results for tail data. See Aaron Clauset, Cosma Rohilla Shalizi and M. E. J. Newman, Power-law distributions in empirical data, SIAM Review 51 (2009), 661–703. The scaling rule is mathematically exact when it defines a model; whether a dataset supports that model is a question of statistical inference.

Consider an original toy AB salt model with $K_{\mathrm{sp},c}=10^{-8}{\mathrm{M}}^2$ stipulated at fixed temperature and fixed activity coefficient product. This is a conditional concentration constant, not measured data for calcium carbonate or any identified material. A stock contains A cations and X anions; another contains Y cations and B anions. All four ions have unit electric charge magnitude. Only AB can form a solid; X and Y remain spectator ions. Assume full dissociation, no other chemical reactions, and a mixed volume equal to the sum of the initial volumes.

Mix two stocks, each of volume $V$,, each at molar concentration $C$ of its indicated salt. Before any precipitation, the mixed volume is $2V$, soMultiplying the two stock concentrations would overestimate the mixed ion product by a factor of four. The saturation threshold is $C=2\sqrt{K_{\mathrm{sp},c}}=2\times 10^{-4}\mathrm{M}$.

If $C=10^{-4}\mathrm{M}$, mixing gives $Q_{c,0}=2.5\times 10^{-9}{\mathrm{M}}^2<K_{\mathrm{sp},c}$: no AB precipitate is favored. If $C=4\times 10^{-4}\mathrm{M}$, mixing gives $Q_{c,0}=4\times 10^{-8}{\mathrm{M}}^2>K_{\mathrm{sp},c}$. Suppose nucleation occurs and bulk AB solid remains at chemical equilibrium. Equal starting inventories and one-to-one stoichiometry require equal remaining concentrations, henceFor two $10\mathrm{mL}$ portions in the second case, each reacting ion starts with $4\mu \mathrm{mol}$; the final $20\mathrm{mL}$ chemical solution retains $2\mu \mathrm{mol}$ of each, and $2\mu \mathrm{mol}$ of AB enters the solid. The respective X and Y inventories stay at $4\mu \mathrm{mol}$. Exactly half of each reacting inventory precipitates in this toy model.

The chemical equilibrium calculation adds a phase constraint to the inventory balance. Write $x=n_{\mathrm{AB},s}/(2V)$ and $c=C/2-x$. In the equilibrated bulk model,When $x>0$, the last condition requires saturation; when $x=0$, the inventory can leave an undersaturated chemical solution. Together these give $x=\max \{0,C/2-\sqrt{K_{\mathrm{sp},c}}\}$. This excludes a persistent supersaturated solution by its explicit chemical equilibrium assumption; it does not model a nucleation delay.

A uniform-depth particle-laden gravity current in fixed triangular cross section has conserved volume proportional to $h^{2}X$. A gravity-current front condition and absorbing-boundary particle deposition flux give $\dot{X}=\mathrm{Fr}\sqrt{G\varphi h}$ and $\dot{\varphi }=-2w(\varphi )\varphi /h$. The box describes late-time bulk evolution; it does not resolve the nose or immediate dam-break flow.

Rate-zonal centrifugation resolves an initially narrow sample zone by differences in sedimentation coefficient during a finite interval. Its bands remain transient: extending the interval can carry material into a centrifugation pellet. Thus a density gradient need not imply an equilibrium buoyant density measurement.^[ref]

A typical preparation layers a narrow sample above a preformed density gradient, often containing sucrose. The initial zone must be narrow enough that differences in starting position do not overwhelm the later separation. The run ends while useful bands remain inside the liquid. In the simple case where all particles are denser than every part of the column, none has an interior isopycnic centrifugation position.^[ref][ref]

The sedimentation coefficient $s=(m-{\rho }_{f}V)/f$ explains why separation is not an ordering by mass alone. At fixed local centrifugal acceleration, increasing displaced volume can increase buoyancy, while a less compact shape can increase the linear friction coefficient and reduce velocity. Even for a sphere, Stokes's law gives $s=2b^2({\rho }_p-{\rho }_f)/(9\eta )$, so both the density difference and the dynamic viscosity matter. The resulting sedimentation coefficient can change along a density gradient.

Rayleigh scattering is elastic scattering of electromagnetic radiation by particles much smaller than the incident wavelength. In the stationary particle approximation, the scattered electromagnetic radiation keeps the incident frequency: its direction changes, rather than its color. The incident electric field induces an oscillating electric dipole moment, which produces electric dipole radiation. This is a useful example of how a microscopic response can explain a familiar large-scale sight, why the sky is blue. ^[ref].

The smallness condition for Rayleigh scattering by a small sphere of radius $a$ is $ka\ll 1$, where the wavenumber is $k=2\pi /\lambda$ and $\lambda$ is the surrounding medium's wavelength. For ordinary materials with modest refractive index, the incident electric field then varies little across the particle, making the electric-dipole approximation for radiation useful. For very large refractive index, also check the internal size $|m|ka\ll 1$ before applying the simple electric dipole model. Once these conditions fail, use Mie scattering for a sphere instead of extrapolating the Rayleigh scattering wavelength dependence.

Historically, the 1871 paper ^[ref] connected the color and polarization of an electromagnetic wave of scattered skylight with particles smaller than the wavelength. Its dimensional analysis anticipates scattered light intensity proportional to particle volume squared and to ${\lambda }^{-4}$. The original mechanical picture is historical; the modern explanation uses induced electric dipole radiation.

For one chemical reaction in a closed system of fixed volume $V$, with uniform concentrations and no competing reactions or appreciable accumulation of unlisted species, choose the chemical equationBrackets denote amount of substance per volume, for example $[\mathrm{A}]=n_{\mathrm{A}}/V$. The stoichiometry gives one instantaneous reaction ratewhere $\xi$ is the existing extent of reaction. The minus signs make forward consumption positive; dividing by the coefficients makes measurements of different chemical species agree. Units can be $\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$. See IUPAC Gold Book: rate of reaction and Flowers et al., Chemistry 2e, §12.1: chemical reaction rates.

For the illustrative chemical equation $2\mathrm{A}+\mathrm{B}⟶3\mathrm{C}$, a measured $d[\mathrm{C}]/dt=0.060\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$ implies $v=0.020\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$, $d[\mathrm{A}]/dt=-0.040\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$, and $d[\mathrm{B}]/dt=-0.020\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$. These numbers express stoichiometry, without supplying a rate law.

Each derivative is the slope of a concentration curve at one instant. Over a finite interval, $-\Delta [\mathrm{A}]/(a\Delta t)$ instead gives an average reaction rate. For A converting reversibly to B, the net reaction rate is $v=d[\mathrm{B}]/dt=J_f-J_r$; both directional reaction rates can remain positive when $v=0$ at chemical equilibrium.

For an original idealized mathematical model, let $n_{{\mathrm{O}}_2}(t)$ be the amount of substance of molecular oxygen consumed since the initial reading and $n_{{\mathrm{CO}}_2}(t)$ the carbon dioxide produced. Suppose a fraction $f$ of that carbon dioxide is removed from the gas, all other gas inventories remain fixed, and temperature $T$ and gas pressure $P$ stay constant. The ideal gas law gives the positive gas contractionwhere $R$ is the molar gas constant and $V$ the gas volume. With complete absorption, $f=1$, the contraction reflects molecular oxygen uptake regardless of the amount of carbon dioxide made. For a uniform pipette bore of area $A$, an inward boundary displacement $x$ gives $D=Ax$. A graduated pipette can instead supply the displaced volume directly.

Let $D_s$ denote sample contraction and $D_b$ the matched blank contraction. Define $D_{\mathrm{corr}}=D_s-D_b$ with both taken relative to their own initial readings. Under the preceding assumptions,The slope of corrected volume against time estimates uptake over the chosen interval. The approximate sign acknowledges that a liquid boundary also responds to hydrostatic and capillary pressure, while the gas can exchange water vapour. A fixed-volume apparatus would instead show a pressure decrease; its ideal conversion is $n_{{\mathrm{O}}_2}\simeq V(P(0)-P(t))/(RT)$ after complete absorption and appropriate blank correction.

The ideal gas law also exposes drift without any biology: for fixed gas inventory, $V\propto T/P$. A one-percent rise in absolute temperature at fixed pressure produces a one-percent expansion. Matching the initial gas spaces makes such common changes easier to subtract. Subtracting unmatched blanks can leave a physical signal that looks biological.

Here is an explicitly stipulated elementary kinetic model, not experimental data. A well-mixed ideal closed system has constant temperature and volume. Alongside an uncatalysed elementary reaction, allow an illustrative catalytic encounter that returns C immediately:Both channels obey the law of mass action. In the second, the forward reaction rate is ${\alpha }_f[\mathrm{A}][\mathrm{C}]$ and the reverse reaction rate is ${\alpha }_r[\mathrm{B}][\mathrm{C}]$. C cancels in each encounter, so its free concentration $c_C$ stays constant without an external supply. This toy elementary reaction encodes regeneration without binding; it is not a claim that real catalytic cycles universally consist of one encounter. A binding cycle can accumulate a reaction intermediate, in which case constant total C does not imply constant free $c_C$.

Adding the parallel reaction rates gives effective first-order reaction rate constantsBecause $k_f^u/k_r^u={\alpha }_f/{\alpha }_r=3$, each channel balances at the same $[\mathrm{B}]/[\mathrm{A}]=3$, and so does their sum. At $c_C=0$ the pair is $(0.030,0.010){\mathrm{s}}^{-1}$; at $c_C=0.001\mathrm{mol}{\mathrm{L}}^{-1}$ it is $(0.300,0.100){\mathrm{s}}^{-1}$. Both give $K=3$ with ideal thermodynamic activities and a common concentration standard state. At $T=300\mathrm{K}$ their compatible standard reaction change in Gibbs free energy is ${\Delta }_r{G}^{\circ }=-RT\ln 3\approx -2.7403\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. Choosing an unrelated reverse rate constant would break this consistency.

Let $S=[\mathrm{A}]+[\mathrm{B}]$ be the conserved concentration, and define the dimensionless quantity $x=[\mathrm{B}]/S$. ThenSolving this first-order linear differential equation givesThe chemical relaxation time is $25\mathrm{s}$ without C and $2.5\mathrm{s}$ with the chosen C. From either $x_0=0$ or $x_0=1$, the exponential decay of the deviation leads to the same chemical equilibrium $x=3/4$. Reducing that deviation to one percent takes $\ln (100)\tau$: about $115.13\mathrm{s}$ versus $11.51\mathrm{s}$.

At this chemical equilibrium, conversion continues in both directions. For $S=1\mathrm{mol}{\mathrm{L}}^{-1}$, the uncatalysed channel has equal forward and reverse reaction rates $0.0075\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$. Adding C gives total directional reaction rates $0.075\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$ each, yet zero net reaction rate. Faster molecular exchange and unchanged composition coexist. This exact calculation explains how catalysis can matter enormously within an observation window while leaving the final equilibrium constant intact.

In his 1883 paper, Osborne Reynolds investigated how the character of pipe flow depends jointly on velocity, pipe diameter and viscosity. His final dye apparatus drew water from a large glass tank through a horizontal glass tube with a smooth trumpet-shaped entrance. A separate reservoir supplied strongly colored water just upstream of the entrance; a valve below the tank controlled the outflow. The glass tube was immersed in the tank, so its internal dye band could be watched. Reynolds repeated the trials with different tube diameters and water temperatures. ^[ref]

The volumetric flow rate was measured independently of the appearance of the dye. Reynolds timed a one-inch fall of the tank's water level, corresponding to a calibrated discharged volume. In modern notation, if this volume is $V$ and the elapsed time is $\Delta t$, then $Q=V/\Delta t$ and $U=4V/(\pi D^2\Delta t)$. He determined the glass tubes' internal diameters by filling them with water. For a uniform bore, a known filled length $\ell$ and measured water volume $V_{\mathrm{tube}}$ determine $D=\sqrt{4V_{\mathrm{tube}}/(\pi \ell )}$; checking bore uniformity matters because the same volume would only give an average cross-section in a tapered tube. Thermometers at different tank levels checked both the temperature and its uniformity. Reynolds compared the viscosity at different temperatures using Poiseuille's results and checked his own water by separate experiments of that type. Thus the Reynolds number estimate did not come from judging the dye band or from assuming that transition occurred at a preset value. ^[ref]

A practical repetition can instead collect a known outflow volume in a measuring vessel and time its collection, or weigh the collected water and divide its mass by mass density. Measure the internal diameter, rather than the external diameter, and obtain kinematic viscosity at the measured water temperature from independently established properties or a separate viscosity measurement. Record the dye appearance and observation position alongside $Q$, $D$, temperature and the resulting Reynolds number. These are alternative measurement procedures following $Q=V/\Delta t$, not additional historical claims about Reynolds' dye apparatus.

Begin with settled water and a small dye supply, then open the outlet valve in small steps. Reynolds saw a thin straight band at sufficiently low speeds. With a tank that had not fully settled, the band could wander while retaining its coherence. At higher discharge it could abruptly spread into the surrounding water downstream of the entrance. Short illumination by an electric spark resolved curls within what otherwise looked like a colored cloud. This connected the visible breakup with internal irregular motion rather than merely the appearance of an opaque pipe. ^[ref]

The controls are part of the result. Reynolds used a smooth entrance and let the tank rest to reduce inlet disturbances. He found that a heated tank could retain slow circulation as it cooled, so measuring one temperature was insufficient: thermal uniformity and residual motion also mattered. His wire-coil trial broke up the band below the previous onset speed, but he explicitly cautioned that once the dye was mixed he could not tell from that dye alone whether the motion became orderly again farther along the tube. In a repetition, keep the dye supply small enough to act approximately as a passive scalar, avoid a strong injection jet, and record the inlet condition and downstream observation distance. If also testing the Hagen-Poiseuille equation, allow the profile to develop and measure pressure drop between positions within the developed region; do not treat the entrance and outlet losses as part of its uniform pressure gradient. The dye experiment by itself does not establish fully developed flow. ^[ref]

The experiment's usefulness is that it compares a visible change in the motion with independently measured scales. Repeating it with several diameters and temperatures, while controlling disturbances, tests the organizing role of the Reynolds number. Its limitations are equally instructive: dye is a tracer of transport, diffusion still acts in laminar flow, and a local dye observation cannot by itself prove that turbulence persists throughout an arbitrarily long pipe.