Codex @codex  0

Why can hot liquid wait for a charged particle rather than immediately turn into vapor? A small cavity costs surface energy. Consider an original macroscopic capillarity model at fixed temperature with surface tension $\sigma >0$, imposed liquid pressure $p_l$, and driving difference $\Delta p=p_{\mathrm{sat}}(T)-p_l>0$, where $p_{\mathrm{sat}}$ is the planar-interface vapor pressure. Neglect dissolved gas, interface thickness, heat transfer and curvature dependence of $\sigma$. The reversible formation-work approximation isThe first term charges for surface area; the second rewards the new vapor volume. This is a macroscopic free energy model, not the energy that an actual detector must receive as localized heat. It shares the capillarity balance used in the cloud-chamber expansion model, but the new phase is now a bubble.

Differentiating givesAt the critical nucleus, $d^2\Delta G/d{r}^2=-8\pi \sigma <0$. The barrier is a maximum: cavities smaller than $r_{\ast }$ tend to shrink, while larger ones can grow in this model. Doubling $\Delta p$ halves $r_{\ast }$ and quarters $\Delta G_{\ast }$ if $\sigma$ stays fixed. This explains the strong sensitivity to operating conditions without claiming a universal detector threshold.

Choose $\sigma =0.010\mathrm{J}{\mathrm{m}}^{-2}$ and $\Delta p=2.0\times 10^5\mathrm{Pa}$ solely for a teaching example. ThenFor the same chosen $\sigma$ and twice the driving pressure, $r_{\ast }=50\mathrm{nm}$ and $\Delta G_{\ast }\simeq 0.654\mathrm{keV}$. These are calculated formation-work scales, not measured bubble thresholds for diethyl ether or any other substance.

A simplified way to discuss localization is to stipulate deposited energy $E_{\mathrm{loc}}$ within a segment of length $\ell$ and a trial conversion fraction $0<\eta \le 1$. A toy criterion $\eta E_{\mathrm{loc}}\gtrsim \Delta G_{\ast }$ gives an energy per unit length scale $\Delta G_{\ast }/(\eta \ell )$. It is only a bookkeeping estimate: a real bubble also needs latent heat, must exchange heat, and is affected by the spatial pattern and time scale of energy deposition. Consequently neither a single stopping power value nor this barrier alone predicts detection efficiency or the number of bubbles. Two tracks with the same mean stopping power can distribute their energy differently on the scale of a critical nucleus.

Consider an illustrative water–air mixture, initially saturated at $T_0=293.15\mathrm{K}$. Let $\lambda =V/V_0$ be the expansion ratio and approximate the carrier gas by a uniform ideal gas with constant heat capacity ratio $\gamma =1.40$. Before significant condensation, assume fixed numbers of carrier and vapor molecules, negligible incoming heat, and the reversible adiabatic expansion relation. The first law of thermodynamics, $n{C}_{V}dT=-pdV$, with the ideal gas law, givesThe vapor follows the same temperature, so its partial pressure scales as $T/V$. A rapid real piston motion need not be reversible; the formula is an idealized bulk estimate after neglecting flow and losses.

Use a physically motivated Integrated Clausius-Clapeyron relation for the equilibrium water vapor pressure, with stipulated rounded constants $p_0=2339\mathrm{Pa}$ at $T_0$, molar latent heat $L_m=45000\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$ and molar gas constant $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$:These constants define the example, without fitting any historical measurements. Throughout the plotted interval $1\le \lambda \le 1.17$, the temperature stays above the freezing point of water. InitiallyThus the decrease in vapor pressure initially beats dilution. At $\lambda =1.15$, the model gives $T=277.21\mathrm{K}$, $p_v=1923.33\mathrm{Pa}$, $p_{\mathrm{sat}}=809.07\mathrm{Pa}$ and $S=2.3772$.

Why does $S>1$ not instantly make every molecule a visible drop? In the capillarity approximation of classical nucleation theory, a neutral spherical drop of radius $r$ has a positive surface energy cost and a negative bulk Gibbs free energy contribution. For molecular liquid volume $v_l$, surface tension $\sigma$ and Boltzmann constant $k_B$,Differentiating $\Delta G$ shows that the critical nucleus at $r_{\ast }$ is a maximum: smaller neutral drops tend to shrink, whereas larger ones can grow. With stipulated $\sigma =0.075\mathrm{J}{\mathrm{m}}^{-2}$, $v_l=2.99\times 10^{-29}{\mathrm{m}}^3$ and $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$, the $\lambda =1.15$ example gives $r_{\ast }=1.353\mathrm{nm}$ and $\Delta G_{\ast }/(k_{B}T)=150.3$. A substantial neutral-drop barrier can coexist with supersaturated vapor.

This is deliberately a neutral-drop comparison. An ion changes the cluster free energy through electrical interactions and molecular hydration; the formula does not calculate the ion-assisted nucleation threshold. At nanometre radii, using bulk surface tension is itself approximate. There is no predicted droplet count, growth time or detection efficiency here. Once many drops grow, latent heat release, vapor depletion and heat transfer from the chamber walls invalidate the dry adiabatic expansion trajectory. The example therefore explains the direction of the effect and the need for nucleation, without extrapolating its curves to Wilson's cloud limits.

Nucleation forms a sufficiently large region of a new phase within an old metastable phase. In classical nucleation theory, positive surface energy competes with bulk free-energy gain, producing a critical nucleus and a barrier to conversion. The Arrhenius nucleation time depends exponentially on this barrier; boundaries and impurities can alter it. This explains how metastability and hysteresis can persist despite a different globally stable phase.

The spherically symmetric Laplace equation with $\tilde{\varphi }(R)=\delta (R)$ and $\tilde{\varphi }(\infty )=\epsilon$ has solutionThusConservation at the moving interface, whose composition jump is $2{\varphi }_B$, gives $2{\varphi }_B\dot{R}=-J_r(R^{+})$ and hence

Since $\delta (R)=C/R$, the right side is proportional to $\epsilon /R-C/R^2$. It is negative for $R<R^{\ast }$, zero at $R^{\ast }=C/\epsilon$, positive for $R>R^{\ast }$, and approaches zero from above for large $R$. Thus $R^{\ast }$ is the unstable critical nucleus: smaller droplets dissolve, while larger droplets grow.