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If $N\mid L$ has Poisson mean $L$, conditional expectation and the law of total variance give the display. A positive variance of the random mean prevents the unconditional count from being Poisson. For Poisson parents of density $\lambda$ and offspring mean $\mu$, the count in $A$ has excess variance $\lambda {\mu }^2\int \kappa (A-z{)}^{2}dz$, which is positive for a bounded positive-area set and a nontrivial model.

Occupied area or volume divided by the corresponding total area or volume of a geometric packing. For a periodic packing, it is the total object area or volume within a repeat cell divided by the cell measure. It is distinct from the number fraction of objects of a given size.

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

A parallel first-order reaction mechanism consumes the same reactant through competing routes. Extend the product-split idea in catalytic selectivity by following both reaction products over time. Stipulate irreversible first-order reactions $\mathrm{A}\stackrel{k_P}{\to }\mathrm{P}$ and $\mathrm{A}\stackrel{k_Q}{\to }\mathrm{Q}$ in a uniformly mixed closed system of fixed volume and temperature. Initially only A is present, with concentration $A_0$. With $K=k_P+k_Q$,Here $K$ abbreviates a sum of rate constants, not an equilibrium constant. The rate equations are $-d[\mathrm{A}]/dt=K[\mathrm{A}]$, $d[\mathrm{P}]/dt=k_P[\mathrm{A}]$, and $d[\mathrm{Q}]/dt=k_Q[\mathrm{A}]$. Integrating the last two after solving the first proves the expressions and their conserved total $[\mathrm{A}]+[\mathrm{P}]+[\mathrm{Q}]=A_0$.

Choose two alternative models: model U has $(k_P,k_Q)=(0.30,0.10){\mathrm{s}}^{-1}$; model V has $(0.10,0.30){\mathrm{s}}^{-1}$. Both have $K=0.40{\mathrm{s}}^{-1}$, so every measurement of A alone is identical, even with no measurement error. Yet their P-to-Q concentration ratios at any $t>0$ are $3$ and $1/3$, respectively. This is a precise failure of identifiability from that one observable. A calibrated measurement distinguishing P from Q resolves it within the stipulated two-route model.

At $t=5.00\mathrm{s}$, both models give $[\mathrm{A}]=e^{-2}{A}_0\approx 0.135335\mathrm{mmol}{\mathrm{L}}^{-1}$. Model U gives $([\mathrm{P}],[\mathrm{Q}])\approx (0.648499,0.216166)\mathrm{mmol}{\mathrm{L}}^{-1}$; model V exchanges these. Measuring only the summed reaction product concentration would also fail to distinguish them. Product-specific observations therefore answer a question that total conversion cannot. As with any reaction mechanism, fitting these observables still does not rule out a more elaborate untested route.

The total extinction cross-section per unit volume of particles is a length-inverse coefficient. Its path integral is optical depth. It combines scattering and absorption and differs from the dimensionless imaginary part of a complex refractive index. For liquid droplets with size-independent $Q_{\mathrm{ext}}$, ${\beta }_{\mathrm{ext}}=3Q_{\mathrm{ext}}w/(4{\rho }_w{r}_e)$.

Passive transport is net movement down the transported substance's electrochemical gradient without coupling that movement to another source of free energy. Facilitated diffusion is passive transport through an ion channel or a passive carrier protein. Providing such a route can speed passage, but does not turn an uphill transfer into a downhill one. For an uncharged solute under the ideal conditions above, net passive transport runs from higher to lower concentration; for an ion, the membrane potential also matters. See Cooper, The Cell: transport of small molecules.

In the figure, stipulate eight neutral solute tracers in A and two in B, with equal fixed volumes, identical surroundings, and no imposed electric potential difference. The concentration ratio is $c_{\mathrm{A}}/c_{\mathrm{B}}=8/2=4$, so A-to-B transfer has $\Delta \tilde{\mu }=-RT\ln 4\simeq -1.38629RT$, or $-3.458\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ at the assumed temperature $T=300\mathrm{K}$. If this solute can cross, thermodynamic equilibrium has equal expected concentrations: five tracers per side on average. Individual crossings occur in both directions; an observed ten-particle snapshot need not contain exactly five on each side. These stipulated counts illustrate a driving force, not a measured flow rate or a prediction of the time to equilibrate.

For a separate illustrative ion calculation, keep $c_{\mathrm{A}}/c_{\mathrm{B}}=4$, but assign $z=+1$ and $V_{\mathrm{A}}-V_{\mathrm{B}}=-60\mathrm{mV}$ at $300\mathrm{K}$. Using $R\simeq 8.31446\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ and $F\simeq 96485.33\mathrm{C}\mathrm{mo}{\mathrm{l}}^{-1}$, B-to-A transfer gives $RT\ln 4-0.060F\simeq 3.458-5.789=-2.331\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. Thus an appropriate open ion channel can permit passive transport toward the higher concentration: the electrical contribution outweighs the opposing chemical potential contribution. This stipulated membrane potential is not inferred from the tracer counts.

Use normalized plane-wave spin orbitals $V^{-1/2}{e}^{i\mathbf{k}\cdot \mathbf{r}}{\chi }_{\sigma }$ and fill a separate Fermi sphere for each spin species. Each spin state counts once, soIts kinetic energy per volume isThe spin-independent contact interaction has uncancelled direct energy $E_H/V=\gamma n^2/2$, with $n=n_{\uparrow }+n_{\downarrow }$. By the imposed background assumption this direct contribution cancels. For the Fock term the contact delta function sets the two spatial positions equal; its spin factors then select equal spins. The local one-body density matrix for each spin has diagonal value $n_{\sigma }$, givingCombining these results gives the contact-interaction Hartree-Fock spin energyHere $\gamma$ has units of energy times volume. The result is the model with the stated background subtraction; an isolated contact-interacting gas without that subtraction retains its direct contribution.

Let $V=L^3$ be the box volume. The number operator is the sum of the occupation numbers of all momentum eigenstates, including the zero mode:In the Fourier series for the contact interaction, the two creation operators supply phase $e^{-i({\mathbf{k}}_1+{\mathbf{k}}_2)\cdot \mathbf{r}}$ and the two annihilation operators supply $e^{i({\mathbf{k}}_3+{\mathbf{k}}_4)\cdot \mathbf{r}}$. Fourier orthogonality with periodic boundary conditions givesThe four normalization factors together contribute $V^{-2}$, leaving the momentum-space contact-interaction HamiltonianThe Kronecker delta enforces momentum conservation. An equivalent expression that resolves the momentum transfer isBoth are already in normal ordering; no extra commutator constant is introduced.

The coherent state in the zero mode has mean occupation number $N_0$. The other occupation numbers follow from part (c), giving exactlyImpose the fixed mean particle number $⟨\hat{N}⟩=N$ by choosingHere $D$ is the quantum depletion of a Bose condensate. The approximation requires $0\le D\ll N$, so $N_0$ remains positive and macroscopically large. The specified coherent state and multimode squeezed vacuum are not exact eigenstates of the total number operator; $N$ in this variational construction is their prescribed mean particle number.

Write $P={\sum }_{\mathbf{k}\ne 0}{s}_k{c}_k$. Substituting into the retained part of the interaction energy before truncating givesThe last two terms quadratic in $D,P$ have no explicit large-$N$ factor when the Bogoliubov transformation parameters are held fixed, and are discarded along with the nonzero-mode quartic contribution already omitted in part (c). This is the same truncation that underlies the number-conserving Bogoliubov quadratic Hamiltonian. Thus, defining the physical number density correctly as $\rho =N/V=N/L^3$,Replacing $N_0$ by $N$ at the outset would give the wrong coefficient $2{\sinh }^2{\theta }_k$: expanding the leading condensate interaction energy supplies the necessary subtraction.

The printed definition $\rho =N/L$ is an error. In three dimensions the contact interaction coupling $u$ has units of energy times volume, so $uN/L$ does not even have units of energy. There is a direct counterexample with no depletion: when ${\theta }_k=0$, the specified coherent state has exactly $⟨H_{\mathrm{int}}⟩=u{N}^2/(2L^3)$. In fixed length and energy units, $L=2$, $u=1$ and $N=100$ give $625$, whereas the printed $u\rho N/2$ with $\rho =N/L$ gives $2500$. The requested formula and the minimization in part (e) hold with $\rho =N/L^3$.

Interstellar shocks arise when relative flow is supersonic: stellar winds, supernova ejecta, colliding clouds and gas moving through strong disturbances can drive them. Neighboring compressive characteristics steepen or a fast flow encounters material it cannot decelerate through upstream sound communication. A thin dissipative front then converts directed kinetic energy into thermal energy and increases entropy.

Integrate mass, normal momentum and total energy conservation through a planar steady shock in its rest frame. With no mass source, negligible external impulse across the thin layer and no radiative energy loss, their normal fluxes are respectivelyFor a perfect gas of constant adiabatic index, $h=\gamma p/[(\gamma -1)\rho ]$. Equality of the fluxes on the two sides yields the three Rankine-Hugoniot conditions in the question. Adiabatic means no energy escapes this control volume; it does not make the shock isentropic.

In the strong-shock limit, let $r={\rho }_2/{\rho }_1=u_1/u_2$ and neglect $p_1$. Momentum conservation gives $p_2={\rho }_1{u}_1^2(1-1/r)$. Substitution into energy conservation givesThe nontrivial compressive root isThe $r=1$ root describes no shock. A monatomic gas has the familiar compression four.

For identical colliding clouds in their center-of-momentum frame, reflection symmetry fixes the contact plane and the uniform compressed material on either side at rest. The unshocked clouds still approach at speeds $\pm V$; their supersonic upstream flow has not received a smooth pressure adjustment ahead of the shock. Let the right shock move to the right at speed $S$. In its own frame the upstream and downstream normal speed magnitudes are $V+S$ and $S$. The same shock compression ratio therefore givesThis is the shock speed in a symmetric cloud collision. For $\gamma =5/3$, the shocks move outward at $V/3$, not at the upstream speed or the relative cloud speed $2V$. As a check, the post-shock pressure is $p_2=(\gamma +1){\rho }_1{V}^2/2$; in the monatomic case $p_2/{\rho }_2=V^2/3$, giving specific internal energy $V^2/2$, exactly the initial bulk kinetic energy per unit cloud mass when brought to rest.

When a shock reaches a cloud's rear boundary, its supply of cold upstream gas ends. If the surrounding pressure is low, the hot compressed gas expands outward and a rarefaction travels back into it, converting thermal energy into motion and cooling the gas. A sustained uniform stationary shocked region is no longer the solution. Interaction with an appreciable exterior medium can instead drive another outward disturbance; the boundary pressure determines that part of the evolution. The stated adiabatic model provides no radiative mechanism that would keep a permanently cold dense sheet at the contact.

Follow a fluid element with position $\mathbf{X}(t)$ satisfying $\dot{\mathbf{X}}=\mathbf{u}(\mathbf{X},t)$. The chain rule gives the material derivativeApply Newton's second law to a material element of volume $dV$ and mass $\rho dV$. The net pressure force is $-\nabla pdV$, and gravity gives $-\rho \nabla \varphi dV$. Dividing by the element mass, neglecting viscosity and using the chain rule componentwise gives the Euler equation

For a static spherical isothermal gas, $p=c_s^2\rho$ and $dp/dr=-GM(r)\rho /r^2$. Trial density $\rho =A/r^2$ gives enclosed mass $M(r)=4\pi Ar$, soThis is the singular isothermal sphere. Its central density diverges, although the enclosed mass tends to zero linearly and its total mass inside a finite radius is finite. It is not the finite-central-density regular isothermal solution.

Truncate at $r_0$ with boundary pressure $p_0$. Then $M=4\pi A{r}_0=2c_s^2{r}_0/G$ and $p_0=c_s^2\rho (r_0)$, givingThe external medium is being treated as a prescribed pressure boundary; its own gravitational field is excluded from the cloud self-energy calculation.

For a monatomic perfect gas, the equilibrium is isothermal but the internal-energy relation still uses its adiabatic index $\gamma =5/3$:The gravitational self-energy follows by assembling shells:Both integrals remain finite at the singular center. ConsequentlyThe pressure-confined singular isothermal gas sphere therefore satisfies the virial theorem including the work of the confining surface pressure. One can see its origin directly by integrating $\mathbf{r}\cdot \nabla p$: integration by parts gives $\oint p\mathbf{r}\cdot d\mathbf{S}-3\int pdV$, whose surface term is $4\pi r_0^3{p}_0$. An isolated vacuum-boundary version would omit that term and could not truncate this same positive-temperature isothermal profile at finite radius. The equality expresses equilibrium force and energy balance, rather than proving dynamical stability.

The Mach number is $\mathcal{M}=u/c_s$, comparing the flow speed relative to a disturbance with the local sound speed. For a body moving supersonically, spherical acoustic disturbances emitted earlier have a tangent envelope with Mach cone half-angle $\sin \theta =1/\mathcal{M}$. Signals cannot propagate arbitrarily far upstream. Compressive characteristics can converge and steepen into a shock wave, within which microscopic dissipation prevents a multivalued fluid state; supersonic motion alone does not require every flow to contain a shock.

Work in the shock frame with steady normal velocities $u_1,u_2>0$. Integrating conservation of mass, normal momentum and total energy over a thin pillbox, neglecting gravity across the thin layer, gives the Rankine-Hugoniot conditions for a perfect gas:The energy equation includes pressure work through specific enthalpy. The shock is adiabatic if negligible heat escapes the layer, but its irreversible dissipation means it is not isentropic.

Put $\chi ={\rho }_2/{\rho }_1=u_1/u_2$ and ${\mathcal{M}}_1=u_1/\sqrt{\gamma p_1/{\rho }_1}$. Momentum conservation gives $p_2=p_1+{\rho }_1{u}_1^2(1-1/\chi )$. Substitution in the energy equation produces the unshocked root $\chi =1$ and the compressive rootSince ${\rho }_1{u}_1^2/p_1=\gamma {\mathcal{M}}_1^2$, the strong-shock limit has negligible upstream thermal pressure compared with ram pressure. The Strong-shock Rankine-Hugoniot conditions giveFor unchanged composition, the ideal gas law implies $T_2/T_1=(p_2/p_1)/\chi$, soFor $\gamma =5/3$, the compression tends to four and $T_2\simeq 3\bar{m}{u}_1^2/(16k_B)$, where $\bar{m}$ is mean particle mass.

With constant specific heat $c_v$, entropy production in a perfect-gas shock is measured by the change in specific entropyThe strong-shock expression is thereforeThe exact jump is positive for every compressive shock with ${\mathcal{M}}_1>1$, in accordance with the Second law of thermodynamics; the asymptotic expression is only for ${\mathcal{M}}_1\gg 1$. There is a normalization issue in the printed description: this logarithmic formula with constant $c_v$ describes entropy per unit mass. The true entropy density is $\rho s$, so its difference is ${\rho }_2{s}_2-{\rho }_1{s}_1$, rather than the same logarithmic jump with a constant heat capacity per unit volume. A material parcel's irreversible entropy increase should be assessed using $\Delta s$.

Downstream gas is denser and hotter. For optically thin collisional emission the loss per unit volume is typically $n_e{n}_i\Lambda (T)$, whereas stored thermal energy scales approximately as $nT$. Compression can therefore reduce the radiative gas cooling time; excitation and ionization at the higher temperature can open additional cooling channels. The actual efficiency still depends on the astrophysical cooling function, composition and optical depth. In a radiative shock with separated scales, the temperature first rises across a narrow dissipative layer, then decreases through a much broader cooling region of characteristic length ${\ell }_{\mathrm{cool}}\sim u_2{t}_{\mathrm{cool}}$. Subsonic cooling usually brings further compression with approximately constant pressure. The immediate microscopic shock thickness and the macroscopic cooling length are distinct.

Let state $f$ be the fully cooled gas, with $T_f=T_1$ and the same composition. Set $a^2=p_1/{\rho }_1$, the isothermal sound speed squared. Across the entire shock plus cooling layer, mass and momentum are still conserved, but gas energy alone is not. With ${\chi }_f={\rho }_f/{\rho }_1$,The nontrivial isothermal shock therefore hasIf the upstream Mach number were defined with $a$ instead, this would be ${\rho }_f/{\rho }_1={\mathcal{M}}_{\mathrm{iso}}^2$. Retaining the distinction avoids losing the factor $\gamma$.

The gas's specific entropy rises irreversibly at the hot shock and then decreases as escaping radiation removes heat. Since the final temperature equals the initial temperature,This decrease for the gas is consistent with the Second law of thermodynamics: the emitted radiation and its eventual surroundings carry entropy away. It does not make the microscopic compressive shock reversible. The figure illustrates both the hot entropy jump and the subsequent cooling; its spatial cooling profile is schematic because no cooling law has been specified.

With zero initial messenger RNA and protein concentration, solving the two linear ordinary differential equations in sequence givesThe last expression applies after the short messenger RNA transient when ${\alpha }_R\gg {\alpha }_P$. All three mean curves have this shape. At a fixed steady state, a larger translational burst size requires fewer transcription events: ${\beta }_R={\alpha }_P{X}_{P,\ast }/b$. Thus $b=400$ gives infrequent large upward jumps followed by protein degradation, whereas $b=3$ gives many smaller jumps and a smoother trace.

Noise depends on molecule counts. If the fixed cell volume is $V$, write $\bar{N}=V{X}_{P,\ast }$. In the instantaneous-burst limit, independent Poisson process arrivals of geometrically distributed bursts, with count arrival rate $\lambda =V{\beta }_R$, and unit protein degradation giveIndeed, the moment equations are $d\mathbb{E}N/dt=\lambda b-{\alpha }_P\mathbb{E}N$ and $d\mathrm{var}N/dt=\lambda b(1+2b)+{\alpha }_P\mathbb{E}N-2{\alpha }_P\mathrm{var}N$. With zero initial counts the variance becomes $\bar{N}[(1+b)-e^{-{\alpha }_{P}t}-b{e}^{-2{\alpha }_{P}t}]$. These equations explain both the approach to the steady state and the scatter about it.

For illustration only, $V=1\mu {\mathrm{m}}^3$ gives $\bar{N}=10^4$ at the higher concentration: the two values of the coefficient of variation are $0.020$ and $0.200$. At the lower concentration, $\bar{N}=100$ and $b=3$ gives $0.200$. Decreasing the concentration by a factor of 100 therefore increases relative fluctuations tenfold, despite keeping small bursts. Absolute fluctuation sizes cannot be inferred without a cell volume. The original sketches below stipulate that volume; they are model realizations, not source data.^[ref]

For a fixed perfectly absorbing spherical target, steady diffusion outside radius $r_0$ obeys ${\nabla }^{2}c=0$, $c(r_0)=0$, and $c(\infty )=c_{\infty }$. Spherical symmetry gives $c=A+B/r$, hence $c(r)=c_{\infty }(1-r_0/r)$. The inward diffusion flux integrated over the target is $4\pi r_0^2{D}_A{c}^{\prime }(r_0)=4\pi r_0{D}_A{c}_{\infty }$. Consequently the number-density rate constant isThis Smoluchowski association rate has units ${\mathrm{m}}^3{\mathrm{s}}^{-1}$, as required by the given number-density rate law. If both partners moved independently, their relative diffusion coefficient would be the sum; the chromosomal target is treated as fixed here.

For a spherical RNA polymerase of radius $a=10\mathrm{nm}$, the Stokes–Einstein relation givesFor $N=6000$ free independent, approximately well-mixed searchers in a volume $V$, the small-target encounter hazard is approximately $N{k}_a/V$. ThusThe numerical coefficient explicitly uses a typical illustrative cell volume and an illustrative effective capture radius, not dimensions supplied by the source. A 60-base-pair B-form DNA segment has a contour length of about $20\mathrm{nm}$, motivating an encounter size of this order, but its contour length does not uniquely determine an absorbing sphere's radius. Taking $r_0=10\mathrm{nm}$ instead doubles the estimate. A unique numerical answer requires $V$, $T$, and a capture-geometry convention, none specified in this block.

The estimate also treats all 6000 molecules as freely available and every encounter as productive. Crowding, nonspecific binding, orientation constraints, imperfect reaction, and finite-domain initial transients can alter it. In particular, an arbitrary first-passage distribution's mean does not divide by $N$ exactly; the division follows from the approximate exponential-hazard description.^[ref]

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

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

Use signed ion valence $z$ and define $V={\psi }_1-{\psi }_2$. The molar electrochemical potential in side $i$ of an ideal dilute solution isTransfer of an infinitesimal amount $dn$ from side 1 to side 2 changes the total Gibbs free energy by $({\tilde{\mu }}_2-{\tilde{\mu }}_1)dn$. At equilibrium this derivative is zero, soFor a cell with side 1 inside, this is $E_i=(RT/z_{i}F)\ln (c_{i,\mathrm{out}}/c_{i,\mathrm{in}})$. The concentrations are equilibrium free-ion concentrations, not automatically their exact initial values. For nonideal solutions the ratio must be a ratio of thermodynamic activities. The temperature and solvent must be common to both reservoirs, and an ion must actually be able to exchange for this equilibrium condition to be dynamically established.

Equal compartment volumes are unnecessary for deriving the voltage law, but help specify number conservation if concentration changes are appreciable. For equal volume $\mathcal{V}$ and a monovalent cation transfer $n$ from side 1 to 2, $c_1=c_1^0-n/\mathcal{V}$, $c_2=c_2^0+n/\mathcal{V}$. Combined with the membrane capacitance, this gives an implicit finite-reservoir equilibrium rather than treating both concentrations as externally clamped.

Model the target as a perfectly absorbing sphere of reaction radius $r_0$, with a fixed target and a dilute far-field polymerase density $c_{\infty }$. Steady radial diffusion satisfies ${\nabla }^{2}c=0$, $c(r_0)=0$, $c(\infty )=c_{\infty }$, henceTherefore, for a rate law multiplying number densities,For the earliest diffusive target encounter among independent searchers, with $N_p$ independent well-mixed searchers in volume $V$, the approximate first-arrival rate is $N_p{k}_a/V$, giving ${\tau }_3\simeq V/(N_p{k}_a)$. The target's effective reaction radius includes encounter geometry and is not uniquely determined by its contour length.

A numerical estimate needs temperature, cell volume and a capture-geometry convention, none of which is fixed numerically in the source. State representative estimates $T=300\mathrm{K}$ and $V=1\mu {\mathrm{m}}^3$. The Stokes–Einstein estimate for a sphere of radius $15\mathrm{nm}$ in the stipulated viscosity isUsing the target contour scale $40\times 0.34\mathrm{nm}=13.6\mathrm{nm}$ as an order-of-magnitude $r_0$ gives $k_a\simeq 2.50\times 10^{-19}{\mathrm{m}}^3{\mathrm{s}}^{-1}$ andThis is an idealized $10^{-2}$-second-scale estimate, not a unique geometric prediction. Finite cell shape, crowded motion, accessibility and orientational restrictions can reduce the ideal capture rate.

If nonspecific binding merely sequesters searchers and their free fraction is $0.2$, the three-dimensional capture hazard is reduced fivefold: $\tau \simeq {\tau }_3/0.2\simeq 0.067\mathrm{s}$ for those estimates. If bound molecules instead search productively by sliding, a free-time fraction alone cannot determine the total arrival time or its sign of change. That is the distinct facilitated-search mechanism treated in c.

The stipulated DNA volume is $V_D=20000\mathrm{n}{\mathrm{m}}^3$ and the spherical capsid volume is $V_c=4\pi (20\mathrm{nm}{)}^3/3=33510\mathrm{n}{\mathrm{m}}^3$. ThusThe packed DNA occupies a substantial fraction of the available space: the remaining nominal fraction is $40.3\%$. Compared with the paper's stated $0.71$ reference, it reaches about $84\%$ of that value. This indicates dense confinement, but does not prove crystalline order or specify interstrand spacing. DNA is a connected elongated polymer, rather than a collection of independent hard spheres.

There is also a numerical qualification: the exact densest equal-sphere packing fraction is $\pi /(3\sqrt{2})\simeq 0.74048$, not $0.71$. Using it changes the comparison to about $81\%$ without changing the calculated $0.597$ volume fraction or its physical interpretation.

Adding $40000$ monovalent counterions makes the combined genome-plus-ion electric charge nominally zero. It does not make the local electric field or free energy vanish: opposite charges have spatial distributions, and confining mobile ions costs entropy.

A definite screened continuation of b replaces the bare Coulomb potential by $e^{-\kappa r}/(4\pi ϵr)$, with $\kappa =1/{\lambda }_D$ and ${\lambda }_D=1\mathrm{nm}$. In the Debye–Hückel approximation, a uniformly distributed fixed charge density $\rho =Q/V_c$ satisfies $({\nabla }^2-{\kappa }^2)\phi =-\rho /ϵ$ inside, with zero fixed charge outside and matching potential and radial field at $R$. Its interior solution isThe screened charging free energy $U_{\kappa }=\frac{1}{2}\int \rho \phi dV$ is thereforeFor $R\gg {\lambda }_D$, the bracket approaches one: only charges within roughly a screening length contribute appreciably, yielding the local estimate $U_{\kappa }\simeq Q^2{\lambda }_D^2/(2ϵV_c)$. With the same ${ϵ}_r=80$ as before, $x=20$ gives bracket $0.92519$ andabout $173$ times smaller than the unscreened packed-sphere self-energy. This is a specified continuum estimate of packed-state charging free energy; a full packaging free energy also subtracts the screened unpacked reference and includes ion redistribution. It should not be presented as a unique exact packaging work fixed by the problem's few inputs.

The supplied counterion count also permits an independent estimate of the entropic contribution. For $N$ ideal confined ions, $F_{\mathrm{ion}}=-N{k}_{B}T\ln V+\text{constant}$, so compression from an accessible reference volume $V_0$ to $V_f$ costsFor example, approximate the original screening atmosphere as an annulus of thickness ${\lambda }_D$ around a rod. With $a=0.5\mathrm{nm}$, $V_0=\pi [(a+{\lambda }_D{)}^2-a^2]L=4.27\times 10^4\mathrm{n}{\mathrm{m}}^3$. Taking $V_f=V_c-V_D=1.35\times 10^4\mathrm{n}{\mathrm{m}}^3$ gives $\Delta F_{\mathrm{ion}}\simeq 1.91\times 10^5\mathrm{pN}\mathrm{nm}$. This useful order-of-magnitude ion-entropy model is distinct from the screened charging model; the two estimates must not simply be added as independent terms without a consistent ionic free energy functional. The rod cutoff, fixed DNA volume and screening-atmosphere geometry are rough geometric assumptions, not a microscopic reconstruction of the source sketch.

There is a further consistency limit. The nominal ion number density is $n=N/V_c=1.19\mathrm{n}{\mathrm{m}}^{-3}$. For a uniform ideal monovalent screening gas, ${\lambda }_D^2=ϵk_{B}T/(n{e}^2)$ would give ${\lambda }_D\simeq 0.309\mathrm{nm}$ at $300\mathrm{K}$, rather than $1\mathrm{nm}$; added salt would shorten it further. Thus the stated ion inventory and screening length cannot both be interpreted as a self-consistent dilute uniform screening gas with this dielectric. Bound counterions, nonuniform distributions and an effective phenomenological screening length can change that interpretation. Linear screening is itself only an estimate near highly charged DNA.

For ideal freely mobile counterions, their gross osmotic pressure isIf the DNA volume is excluded from the ion-accessible volume, use $V_f=V_c-V_D$ instead, giving $\Pi \simeq 1.23\times 10^7\mathrm{Pa}\simeq 123\mathrm{bar}$. These are ideal-ion estimates; bound counterions contribute less like independent mobile particles, and the net outward osmotic pressure also subtracts the external osmotic contribution.

Stored free energy associated with this pressure, together with bending and interstrand repulsion, can favor initial genome release when a delivery pathway becomes available. This explains a possible mechanical benefit of dense packaging at a conceptual level. It does not imply that internal pressure necessarily supplies all the work of complete genome delivery into a host, whose own environment and interactions also matter.