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For $N=300$, the critical polymer volume fraction isIf $v_0$ is the volume assigned to one amino acid in the lattice model, the critical number densities are exactlyThe extra physical approximation needed to use the given radius of gyration is that a chain in a poor solvent is a compact polymer globule. Its size then scales as $R_g\sim b{N}^{1/3}$ with $v_0\sim b^3$, so $N{v}_0\sim R_g^3$. This gives the standard scaling estimatesWith $R_g=3\mathrm{nm}$, numerical substitution givesUsing only the leading large-$N$ value ${\varphi }_c\sim N^{-1/2}$ gives $6.4\times 10^{26}$ and $2.1\times 10^{24}{\mathrm{m}}^{-3}$, respectively. These are scaling estimates, not geometrically exact densities.

For comparison, a specifically uniform spherical compact protein has the radius of gyration of a uniform sphere, $R_g^2=3R^2/5$, so its physical volume is $V_p=(4\pi /3)(5/3{)}^{3/2}{R}_g^3$. Taking $N{v}_0=V_p$ instead gives $n_{m,c}\simeq 6.7\times 10^{25}{\mathrm{m}}^{-3}$ and $n_{p,c}\simeq 2.2\times 10^{23}{\mathrm{m}}^{-3}$. The factor between these conventions is geometric. The PDF does not give a monomer volume, globule density or shape model, and a radius of gyration alone does not determine an exact molecular volume. The model-independent conclusions are the formulas in terms of $v_0$ and the 300:1 ratio of monomer to protein number density.

The particle carries positive surface charge density $\sigma =Q/(4\pi a^2)$. Its surrounding electric double layer has an excess of negative chloride counterions and a deficit of positive magnesium coions. Thus $\varphi (r)>0$ decreases to zero, $n_{\mathrm{Cl}}(r)$ decreases towards its bulk value, and $n_{\mathrm{Mg}}(r)$ increases towards its bulk value. The diffuse charge density is negative and tends to zero.

If $c$ denotes the bulk number of dissolved salt formula units per unit volume, the far-field number densities are $n_{\mathrm{Mg},\infty }=c$ and $n_{\mathrm{Cl},\infty }=2c$. The bulk is neutral even though these number densities differ. In the dilute mean-field picture the total diffuse electric charge is $-Q$, ensuring overall electroneutrality. A molecularly bound compact layer would require additional surface-chemistry information and is not assumed in the supplied potential.

Let $\epsilon ={\epsilon }_0{\epsilon }_r$ be the permittivity, $\beta =(k_{B}T{)}^{-1}$, $e>0$ the elementary charge, and $c$ the bulk salt formula-unit number density. The Boltzmann distribution givesConsequently the exterior Poisson-Boltzmann equation isFor small $Q$, meaning $2\beta e|\varphi (a)|\ll 1$, linearize the exponentials. The Debye–Hückel approximation givesThis is also ${\kappa }^2=\beta e^2{\sum }_i{z}_i^2{n}_{i,\infty }/\epsilon$: the divalent cation contributes $4c$ and the two monovalent anions contribute $2c$. In terms of molar salt concentration $C$ in $\mathrm{mol}{\mathrm{m}}^{-3}$, replace $c$ by $N_{A}C$; if $C$ is in $\mathrm{mol}{\mathrm{L}}^{-1}$, use $c=10^3{N}_{A}C$. The molar ionic strength is $3C$ in the same concentration units.

The boundary conditions are $\varphi (\infty )=0$ and, by Gauss's law,Here the surface charge is uniform and the spherically symmetric interior field is zero. Set $u=r\varphi$; the radial equation reduces to $u^{\prime \prime }={\kappa }^{2}u$. Decay excludes the growing exponential, so write $\varphi =B{e}^{-\kappa (r-a)}/r$. Then ${\varphi }^{\prime }(a)=-B(1+\kappa a)/a^2$, givingIt solves the linearized Poisson-Boltzmann equation, not the full nonlinear equation at finite $Q$. The radial ion profiles, consistently to first order, areThe exponential Boltzmann distribution formulas above express the full mean-field relation, but substituting a linearized potential into them is accurate only to the retained order. A useful electroneutrality check isThe result is the Debye-Hückel potential of a charged sphere.

Use thermal de Broglie wavelength ${\lambda }_T=h/\sqrt{2\pi m{k}_{B}T}$ and spacing $a=n^{-1/3}$. Degeneracy onset from thermal wavelength occurs atThe numerical factor depends on the convention for a typical momentum, but the scale ${\hslash }^2{n}^{2/3}/(m{k}_B)$ does not. For fermions the Fermi temperature $T_F={\hslash }^2(3{\pi }^{2}n{)}^{2/3}/(2m{k}_B)$ gives a comparable scale for spin degeneracy two.

Representative estimates, with the assumed densities stated explicitly, are:

Here $u$ is the atomic mass constant. Thus the characteristic orders are $10^5\mathrm{K}$ for electrons, sub-microkelvin for a dilute heavy atomic gas, and $10^{12}\mathrm{K}$ for neutrons. There is no density-independent quantum-degeneracy temperature for atomic gases: for example, taking $n=2.5\times 10^{25}{\mathrm{m}}^{-3}$ and mass $4u$ instead gives about $0.065\mathrm{K}$. These are illustrative input estimates, not values supplied by the paper. Denser neutron-star cores require relativistic and interaction corrections; the estimate identifies the degeneracy scale rather than a precise stellar temperature.

Use a three-dimensional ideal gas at the same number density n for both isotopes. An unpolarized helium-3 gas has spin degeneracy two, henceHelium-4 has a single spin state, and the ideal Bose-gas saturation condition givesEliminating n, using $m_3/m_4\simeq 3/4$ and $\zeta (3/2)=2.612$, yieldsThe fermionic spin degeneracy must be included. This is an ideal-gas estimate; strong interactions in liquid helium prevent identifying it with an exact experimental transition temperature.

A non-ideal gas departs from $p=n{k}_{B}T$ because interactions and finite particle size affect its pressure. At sufficiently low number density $n=N/V$, the virial expansion separates these corrections:The second virial coefficient has dimensions of volume per particle, while $B_3$ has dimensions of volume squared. These are density-series coefficients, not the coefficients of a pressure-series expansion with the same names.

For classical particles interacting through a pair potential $u(r)$,Repulsion contributes positively and attraction negatively. A hard core of diameter $d$ gives $B_2=2\pi d^3/3$, four times the geometric volume of one sphere. The factor reflects the volume excluded to the centre of a second particle together with pair counting; it is not simply the occupied geometric volume.

Expanding the Van der Waals equation at $nb\ll 1$ givesThe virial coefficients of the van der Waals gas are therefore $B_2=b-a/(k_{B}T)$ and $B_3=b^2$. The Boyle temperature, $T_B=a/(k_{B}b)$, makes the leading correction vanish, but does not remove the higher coefficients or turn the gas exactly ideal. More realistic potentials give temperature-dependent higher coefficients that need not equal the van der Waals values.

A useful measurement is a sequence of dilute isothermal pressure–volume observations. Plotting $[p/(n{k}_{B}T)-1]/n$ against $n$ exposes $B_2$ as the zero-density intercept and $B_3$ as its initial slope. Control temperature and density calibration, check equilibration and adsorption on the container walls, and vary the fitted density interval to test truncation error. One measured virial coefficient cannot uniquely reconstruct the pair potential. Near coexistence or a critical point a short density series may fail, and at quantum-degenerate densities exchange statistics add corrections even without interactions.

Ensemble and applications. A grand canonical ensemble describes a system exchanging both energy and particles with a large reservoir. Its fixed control parameters are $T,V,\mu$, while microstates with different $N$ have probabilitiesWeak coupling allows the system's energy levels to remain meaningful; equilibrium fixes the reservoir temperature and chemical potential. The grand potential is $\Phi =-k_{B}T\ln \Xi =U-TS-\mu N$, with $N$ now the mean population. It generates entropy, pressure and mean particle number by differentiation. This ensemble is convenient for gases in contact with particle reservoirs, independently occupied quantum levels and surface adsorption; its particle-number fluctuations are part of the description, not an error in counting.

Classical single-level limit. For the single-level grand partition function in the dilute limit, put $x_k=e^{-\beta ({ϵ}_k-\mu )}$. Empty and singly occupied levels contribute $1$ and $x_k$. In the dilute classical limit $x_k\ll 1$, higher occupancies contribute only at order $x_k^2$ or beyond:This truncation does not impose the Pauli principle on a classical gas. A fermion state has exactly $1+x_k$, a boson state has $(1-x_k{)}^{-1}$, and Maxwell-Boltzmann occupancy weights give $e^{x_k}$; they all agree to this order. In particular, $\ln {\Xi }_k=x_k+O(x_k^2)$, with the classical Maxwell-Boltzmann result exactly $x_k$.

grand potential and chemical potential. With $z=e^{\beta \mu }$, summing the single-level logarithms and using the stated density of states givesSubstitute $ϵ=k_{B}T{x}^2$, for whichCombining this with the density-of-states prefactor gives $\ln \Xi =zV/{\lambda }^3$. ThusDifferentiating at fixed $T,V$,Similarly $p=-({\partial }_V\Phi {)}_{T,\mu }=k_{B}Tz/{\lambda }^3$, so $pV=N{k}_{B}T$. This is the grand-canonical chemical potential of a classical ideal gas. The condition $N{\lambda }^3/V\ll 1$ states the classical regime. The supplied density of states counts one internal state; extra internal degeneracies would have to be included consistently rather than inserted at only one stage.

One adsorption site. Take the free-vapour energy reference as zero and the binding energy as $-ϵ$. A site is either empty or singly occupied; the latter has vibrational partition function $z_s(T)$. Its adsorption-site grand partition function and mean occupation areThe chemical potential of an adsorbed species on equivalent sites follows by inverting the occupation ratioThe factor $1-\theta$ is the vacant-site probability. The binding reference and the energies included in $z_s$ must be chosen consistently, so that vibrational zero-point energy is not counted twice.

Isotherm. The vibrational factor in a Langmuir equilibrium constant is fixed by adsorption equilibrium, ${\mu }_s={\mu }_{\mathrm{vap}}$. Neglecting vapour internal degrees of freedom and writing its number density as $p/(k_{B}T)$ gives $e^{\beta {\mu }_{\mathrm{vap}}}=p{\lambda }^3/(k_{B}T)$. ThereforeThis is the Langmuir isotherm: initially linear, increasing monotonically and concave down, approaching unit coverage. Half coverage occurs at $p_{1/2}=1/b_s(T)$. The coefficient has inverse-pressure units. It assumes equivalent independent sites, at most one molecule per site and no lateral interaction or multilayer adsorption. The molecular mass appropriate to the vapour determines $\lambda$.

A real uptake measurement must separate adsorbed molecules from free gas in the apparatus. Langmuir's original mica experiment compared a mica-containing bulb with a similar blank and inferred uptake from calibrated gas volumes and pressures. His stated blank contribution and bath-temperature drift limited the inferred adsorption; a saturating curve alone does not establish ideal equivalent sites.^[ref]

Structure through pair correlations. For a homogeneous isotropic monatomic liquid of number density $n=N/V$, the radial distribution function $g(r)$ is defined byIt compares the conditional density around a tagged particle with the bulk density. A repulsive core makes close neighbours unlikely; a first peak identifies a preferred coordination shell, and damped peaks at larger radii describe short-range order. Far away, $g(r)\to 1$ in the bulk limit. Unlike a crystal, an ordinary liquid has no indefinitely repeated positional order. An ideal uncorrelated gas has $g(r)=1$ away from the tagged particle. The shell factor $4\pi r^2$ must not be confused with $g(r)$ itself.

The same pair structure enters the static structure factor measured through scattering. In the bulk isotropic convention,Thus real-space coordination and reciprocal-space scattering peaks describe the same correlations.

Energy from the structure. If interactions are sums of a central pair potential $v(r)$, the equipartition theorem gives the classical kinetic energy $3N{k}_{B}T/2$. Count the interaction of a tagged particle with its neighbours and divide by two to avoid counting each pair twice. The pair-correlation expression for internal energy isThis makes structure thermodynamically useful: short-distance repulsion and the attractive region contribute with different weights. Additional internal molecular modes or many-body interactions require extra terms. The formula is a classical bulk pair-potential result, not a universal energy formula for every liquid.

Pressure and the dilute expansion. The mechanical virial theorem relates the pair forces to pressure. The virial pressure from pair correlations isAt low density, $g(r)\simeq e^{-\beta v(r)}$, giving the virial expansionThe second virial coefficient has units of volume per particle; higher coefficients encode additional cluster correlations. A hard-sphere exclusion diameter $d$ gives $B_2=2\pi d^3/3>0$. Attractions contribute negative terms and can make $B_2$ negative. Although the pair-correlation formulas can describe a dense liquid when its actual $g(r)$ is known, truncating the virial series after a few terms is normally a dilute-fluid approximation. It should not be silently extrapolated through condensation or into a dense liquid.

Component chemical potentials. Define the per-spin quantum concentration $n_Q={\lambda }_T^{-3}=(m{k}_{B}T/(2\pi {\hslash }^2){)}^{3/2}$. For a classical component with one-particle potential energy $U_{\sigma }(\mathbf{r})$, the local Helmholtz free energy density is $k_{B}T{n}_{\sigma }[\ln (n_{\sigma }/n_Q)-1]+n_{\sigma }{U}_{\sigma }$. Its derivative with respect to number density gives the full local chemical potential, including the magnetic energy. With the up state defined as the lower level,The Bohr magneton ${\mu }_B$ is a magnetic moment; ${\mu }_{B}B$ is an energy. The factor two for the two spin states is not already included in $n_Q$.

Total equilibrium chemical potential. If spin conversion is allowed, a particle can be added in either spin state at the same equilibrium free-energy cost. Thusnot their sum or zero. The total particle number, or a reservoir, fixes this common value; the supplied information does not specify a numerical chemical potential. A useful local expression is $\mu =k_{B}T\ln [n/(2n_Q\cosh x)]$, with $x={\mu }_{B}B/(k_{B}T)$, whose density dependence is found below.

Why it is uniform and common. Moving a small number $dN$ of one species from ${\mathbf{r}}_1$ to ${\mathbf{r}}_2$ changes the free energy by $[{\mu }_{\sigma }({\mathbf{r}}_2)-{\mu }_{\sigma }({\mathbf{r}}_1)]dN$. If this difference were nonzero, a transfer could lower the free energy. Equilibrium under diffusion therefore requires each full chemical potential to be spatially constant. A spin flip transferring $dN$ from down to up similarly changes free energy by $({\mu }_{\uparrow }-{\mu }_{\downarrow })dN$, proving equality when spin relaxation is permitted. This is chemical equilibrium of a classical spin gas. Mere thermal contact at fixed separate spin populations would not enforce the latter equality; full spin equilibrium is the necessary assumption. A spatially varying magnetic field is balanced by a spatially varying density, not by a varying full chemical potential.

Magnetization. Invert the component equations at the common $\mu$:The two levels have moments $+{\mu }_B$ and $-{\mu }_B$ along the field. The magnetization, magnetic moment per unit volume, is thereforeIts SI units are $\mathrm{A}{\mathrm{m}}^{-1}$. This expression neglects interactions and magnetic feedback of the gas on the specified external field.

High-temperature response. For $|x|\ll 1$, $\sinh x\simeq x$ and $\cosh x\simeq 1$, yieldingAt fixed number density, this is the Curie law: the susceptibility with respect to $B$ is $n{\mu }_B^2/(k_{B}T)$. The conventional dimensionless susceptibility with respect to $H$ is approximately ${\chi }_v={\mu }_{0}n{\mu }_B^2/(k_{B}T)$ when $B\simeq {\mu }_{0}H$. At fixed $\mu$, however, $n_Q{e}^{\beta \mu }$ changes with temperature, so $M$ itself need not scale simply as $1/T$. The moment per particle has the Curie behaviour; the density constraint must be stated.

Total density. Adding the two populations givesFor a finite container and fixed $N$, normalization gives $e^{\beta \mu }=N/[2n_Q\int \cosh x(\mathbf{r})d^{3}r]$. In a weak field the leading density is uniform, with changes only of order $x^2$. Particles preferentially occupy regions where the magnitude of the field lowers one of the spin energies, even though the common chemical potential remains uniform.

Classical translational statistics. Write the number density as $n=N/V$ and the thermal de Broglie wavelength as $\lambda =\sqrt{2\pi {\hslash }^2/(m{k}_{B}T)}$. Quantum exchange is negligible when wave packets rarely overlap:The equivalent state-counting statement is that the mean occupation of each thermally relevant one-particle state is much smaller than one. The equations in this question count one internal species; with $g$ equally populated spin states the relevant density per species is $n/g$.

Let $z=e^{\beta \mu }$ be the dimensionless quantum gas fugacity, with the lowest one-particle energy chosen as zero. Both quantum distributions can be writtenIf $z\ll 1$, then $n_{\eta }(E)\simeq z{e}^{-\beta E}\ll 1$ for all $E\ge 0$, recovering Maxwell–Boltzmann statistics. Conversely, low occupation even at the spectrum's bottom requires $z/(1-z)\ll 1$ for bosons or $z/(1+z)\ll 1$ for fermions; either condition impliesA single high-energy state can be sparsely occupied even in a degenerate gas, so the condition must hold across the relevant spectrum, not just for one arbitrarily chosen energy. Also, classical translational statistics does not imply that all internal excitation energies or other quantum effects can be ignored.

Grand partition function, density and pressure. Noninteracting particles have independent occupations of each one-particle state. Put $x_{\mathbf{k}}=z{e}^{-\beta {\epsilon }_{\mathbf{k}}}$. Summing the allowed occupations givesThereforeThe grand potential is ${\Omega }_G=-k_{B}T\ln \Xi =-pV$ for this homogeneous gas. Replacing the sum by $V\int d^{3}k/(2\pi {)}^3$ and using ${\epsilon }_{\mathbf{k}}={\hslash }^2{k}^2/(2m)$ givesThe mean number is $N=z{\partial }_z\ln \Xi$, givingFor bosons, the grand sum requires $z<1$ when the ground energy is zero. A macroscopic condensate at the limiting $z\to 1$ must be separated from the continuum integral in the condensed phase. The following small-$z$ calculation is in the normal, nondegenerate regime, so that complication does not enter.

Density expansion and inversion. Expand the occupation as a geometric series:The Gaussian wavevector integral for its $\ell$th term isIt follows thatFor the inverse series, put $y=n{\lambda }^3$ and try $z=y+c{y}^2+O(y^3)$. Substitution gives $0=(c+\eta /2^{3/2})y^2+O(y^3)$, henceThe leading equality $n{\lambda }^3\simeq z$ directly connects the density and occupancy criteria for classical behavior.

Pressure and its first quantum correction. Expanding the logarithm in the pressure expression gives one extra factor $1/\ell$ compared with the density series:Insert the inverse series. To second order in $y=n{\lambda }^3$,ThusThe first quantum correction at fixed density and temperature isIt is negative for bosons and positive for fermions, and vanishes in the classical density limit.

Virial coefficient. Define the number-density virial expansion byThe coefficient $B_j$ has units ${\mathrm{volume}}^{j-1}$ in this convention. Comparing terms givesThis is the exchange contribution to the second virial coefficient. It arises without a pair-interaction potential; the underlying gas is noninteracting. A molar-density convention would multiply this $B_2$ by the Avogadro constant. For $g$ equally populated internal species, exchange occurs within each species and the coefficient in terms of total density is smaller by a factor $g$.

Why the two signs differ. Bosonic exchange symmetry enhances the probability of nearby identical bosons, whereas fermionic exchange symmetry suppresses nearby identical fermions in the same internal state. In the dilute regime, the normalized equal-time pair detection correlation isThe exchange part follows by squaring the Gaussian one-particle coherence factor $e^{-\pi r^2/{\lambda }^2}$. At zero separation this gives $g^{(2)}(0)=2$ for bosons and $0$ for one-component fermions. Integrating the exchange pair weight provides a direct coefficient check:Bunching therefore acts like an effective statistical attraction in the dilute equation of state, lowering the pressure needed at fixed density. Exclusion acts like an effective statistical repulsion, raising it. Neither description asserts an actual interparticle force. Different internal fermion species do not exclude each other from a common spatial location in the same way, explaining the importance of the species convention.

Use the Gibbs-Duhem equation for a homogeneous, extensive, single-component thermodynamic phase:At fixed temperature, this becomes $dp=nd\mu$, with number density $n=N/V$. Separately, at fixed particle number,The chain rule along the same isothermal process in the equation of state now givesConsequentlyThis is isothermal compressibility from chemical-potential response. Since chemical potential has units of energy per particle, the right-hand side has units of volume per energy, or $\mathrm{P}{\mathrm{a}}^{-1}$, as required. The derivation assumes a smooth bulk equation of state; changing the fractions of coexisting thermodynamic phases at fixed pressure needs separate treatment.

Measure single-particle energy from the lowest state and write ${\epsilon }_k=c{k}^{\alpha }$, with $c>0$. In a homogeneous system, the number of wavevector states in a thin spherical shell is proportional to $k^{d-1}dk$. Changing variables from $k$ to $\epsilon$ gives the density of states per unit spatial volume,where $S_{d-1}$ is the area of the unit sphere in $d$ dimensions; an additional internal state degeneracy would multiply this expression by a constant.

The Bose-Einstein distribution has $\mu \le 0$. At saturation, $\mu \to 0^{-}$, the maximum excited-state number density isOnly the lower endpoint can cause divergence. There the integrand is proportional toThe improper integral converges precisely when its power exceeds $-1$. ThusWhen $d=\alpha$, the divergence is logarithmic; when $d<\alpha$, it is a power divergence. In either case the excited states can accommodate any fixed finite density at every positive temperature, so there is no bulk Bose-Einstein condensation at nonzero temperature.

For completeness, putting $s=d/\alpha >1$ and using the Gamma function and Riemann zeta function givesAt fixed density $n$, the right-hand side reaches $n$ at $T_c$; below it the excess particles occupy the lowest state, with condensate fraction $1-(T/T_c{)}^s$. This is Bose condensation with power-law dispersion in the homogeneous thermodynamic limit. A finite vessel or a trapping potential changes the low-energy state count and must not be silently substituted for that setting.

For attractive nearest-neighbour interactions, $J>0$, the transformed model is a ferromagnetic Ising model. In three dimensions and at $T<T_c$, its infinite-volume spontaneous magnetization has two branches. As the effective field crosses zero, the equilibrium mean spin changes discontinuously from a negative value to a positive value. Since $(\partial h/\partial \mu {)}_T=1/2$, varying the physical chemical potential crosses this same first-order phase transition.

The particle-hole-symmetric coexistence line isThe mean Ising spin is a rescaled occupation density. Writing $n=⟨N⟩/V$ for the physical number density,The negative-spin phase is a dilute gas and the positive-spin phase is a dense, liquid-like phase. If the spontaneous spin magnitude at coexistence is $m_0(T)>0$, their number densities areThus the physical order parameter is a density or occupation imbalance, not an actual magnetic moment of these particles. In a closed system with an intermediate mean density, the same transition appears as phase coexistence between the two densities. At $T=T_c$ the density contrast vanishes continuously; above $T_c$, sweeping the chemical potential gives no coexistence jump. A true discontinuity requires the thermodynamic limit; a finite sum over finitely many cells remains analytic.

A colloidal grain in a liquid is large enough to see with a microscope, yet small enough for thermal motion to compete with its tendency to settle. In the Perrin sedimentation experiment, the equilibrium number density falls with height: a visible miniature atmosphere. Counting grains at different heights measures an energy scale; tracking their Brownian motion measures a transport rate. Connecting both to the same Avogadro constant makes the experiment a quantitative test of kinetic theory. The Wiener measure describes an idealized path law; these measurements instead test a physical model and its material parameters.

Jean Baptiste Perrin's 1909 memoir traces his sedimentation experiments to May 1908. It reports relative counts $100$, $47$, $22.6$, $12$ at heights $5$, $35$, $65$, $95\mu \mathrm{m}$, and compares radius estimates from Stokes settling velocity, counted grain mass, and rows of touching grains.^[ref]

The 1926 lecture describes resin precipitated from alcohol into water, washed and fractionated by centrifugation to obtain nearly monodisperse gamboge or mastic grains. Diameter and density were measured separately. A sealed slide-and-cover-glass chamber prevented evaporation; a vertical microscope resolved roughly $2\mu \mathrm{m}$ layers through a roughly $0.1\mathrm{mm}$ depth. Refocusing and photographing compared layer populations after equilibrium; changing temperature reversibly changed the distribution.^[ref]

In Les Atomes (1913), Jean Baptiste Perrin restricts the eyepiece view to fewer than five or six grains and alternates observations between heights. Focus displacement requires a refractive index correction to recover actual layer separation. Horizontal positions were marked with a camera lucida, usually every $30\mathrm{s}$, and distances calibrated with a micrometer; the polygon joins sampled positions rather than recording the intervening trajectory.^[ref][ref]

The dimensionless occupancy scale $n{\lambda }_T^3$, with number density $n$ and thermal de Broglie wavelength ${\lambda }_T=h/\sqrt{2\pi m{k}_{B}T}$. Increasing it approaches the regime where quantum statistics matters. This use of degeneracy must not be confused with degeneracy of an individual energy eigenspace.

A photochemical steady state has production and loss rates balanced for each short-lived species even though individual photochemical reactions continue.

For photons in thermal equilibrium at temperature $T$, Planck's law gives the photon number density per frequency interval. In terms of the Planck function, it is $dn/d\nu =4\pi B_{\nu }(T)/(ch\nu )$, hence

For carrier number density $n$, charge $q$ and effective mass $m$, the plasma frequency is ${\omega }_p^2=n{q}^2/({ϵ}_{0}m)$. Combining the second London equation with the magnetostatic Ampère law and Gauss's law for magnetism gives ${\nabla }^2\mathbf{B}=({\omega }_p^2/c^2)\mathbf{B}$. Therefore the decaying planar solution has length ${\lambda }_L=c/{\omega }_p=\sqrt{m/({\mu }_{0}n{q}^2)}$.

The overlap concentration is the concentration at which distinct dilute polymer coils first substantially interpenetrate. It corresponds to roughly one chain per coil volume. If a chain contains $N$ repeat units, has molar mass $M$ and occupies a coil volume $V_{\mathrm{coil}}$, then its chain number density is approximately $1/V_{\mathrm{coil}}$, its repeat-unit number density is $N/V_{\mathrm{coil}}$, and its mass concentration is $M/(N_A{V}_{\mathrm{coil}})$. A scaling convention uses $V_{\mathrm{coil}}\sim R^3$; a commonly chosen geometric convention uses $V_{\mathrm{coil}}=4\pi R_g^3/3$. These give different order-one prefactors because a coil has no sharp surface. For an ideal chain, $R\sim b{N}^{1/2}$ and a site-volume estimate $v_0\sim b^3$ gives overlap volume fraction ${\varphi }^{\ast }\sim N^{-1/2}$. A good solvent chain with the Flory approximation $R\sim b{N}^{3/5}$ instead gives ${\varphi }^{\ast }\sim N^{-4/5}$.

For $A,b>0$ and $\alpha >1/2$, the projected profile $F(R)=A(R^2+b^2{)}^{-\alpha }$ has the displayed spherical Abel transform inverse. Indeed its composed transform is $A(r^2+b^2{)}^{1/2-\alpha }{I}_{\alpha }$, with $I_{\alpha }={\int }_1^{\infty }{s}^{-\alpha }(s-1{)}^{-1/2}ds=B(1/2,\alpha -1/2)$. Differentiate in $r$, divide by $-2\pi r$, and use the Gamma function recurrence to obtain the coefficient. The same kernel reconstructs tracer number density or isotropic random pressure from their respective projected profiles.