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Let $\ell =\ln ({\theta }_{\max }/{\theta }_{\min })$. The supplied second moment gives the single-scatter root mean square angleFor independent scatters with uniform azimuth, the small angular-deflection vectors have zero mean. Their cross terms average to zero, so their variances add. After $N$ scatters,This is the root mean square magnitude of the two-dimensional angular deflection. Its projection into one specified plane has a variance smaller by a factor of two, hence a root mean square smaller by $\sqrt{2}$.

The magnetic bending angle is $\varphi \simeq |q|BL/p$. For a momentum estimate based on this angle, propagation of uncertainty gives $\sigma (p)/p=\sigma (\varphi )/\varphi$. Comparing the total Multiple Coulomb scattering angle with the bend therefore gives the commonly intended characteristic estimateIts independence of $p$ is the key result for relativistic particles: the scattering angle and magnetic bend both scale as $1/p$.

A precise standard deviation requires specifying how tracks are measured. If $\varphi$ is obtained from independently measured entrance and exit directions and all the material lies between them, the bending-plane projected root mean square is $\mathrm{RMS}(\theta )/\sqrt{2}$. For that estimator the equality isA three-point sagitta fit weights scatter positions differently again. The paper does not specify these measurement details, so it fixes the scaling and characteristic estimate, but not one universal geometric coefficient for all track fits.

Take the initial internal pressure as negligible and the external atmospheric pressure as $p_0=1.01325\times 10^5\mathrm{Pa}$. The window has frontal area $A=\pi (0.400\mathrm{m}{)}^2=0.50265{\mathrm{m}}^2$. Before appreciable air has entered, its inward pressure force is approximately $p_{0}A=5.09\times 10^4\mathrm{N}$. Newton's second law therefore gives the initial collective inward accelerationHere $M=110\mathrm{kg}$ and $g\simeq 9.81\mathrm{m}{\mathrm{s}}^{-2}$. Neglect gravity in this much larger pressure-driven acceleration, together with supports and elastic stresses once the glass has broken. The given thickness $d=0.080\mathrm{m}$ implies a plausible average glass mass density $M/(Ad)\simeq 2.74\times 10^3\mathrm{kg}{\mathrm{m}}^{-3}$. Equivalently, a fragment retaining the original thickness has acceleration $p_0/({\rho }_{g}d)$, independent of its frontal area while it experiences the full pressure difference.

For a vacuum-window implosion energy estimate, the ideal pressure-work scale for filling vessel volume $V=1.5{\mathrm{m}}^3$ is $W\simeq p_{0}V$. Assigning all that pressure-volume work to the glass's kinetic energy givesThe same result follows from retaining acceleration $a_0$ over an effective swept distance $V/A\simeq 2.98\mathrm{m}$, so that $v_{\mathrm{est}}^2=2a_{0}V/A$. This is an ideal upper energy-budget estimate for the common speed, or mass-weighted root mean square speed, of all the glass. It assumes an atmospheric reservoir maintains the driving pressure and that the whole vessel's pressure-work budget can accelerate the glass. In practice, air motion, fragmentation, heating and collisions share that energy, the pressure difference falls, and fragments may hit the vessel after a much shorter path. The unspecified vessel shape and fragment trajectories prevent a unique actual maximum speed; concentrating energy into one small fragment also prevents interpreting this collective estimate as a strict bound on every individual fragment.

Assume a passive linear electrical circuit in sinusoidal steady state, with transients absent and $V_0$ the peak voltage. Write $Z=|Z|e^{i\varphi }=R+iX$. The electric current is $I(t)=I_0\cos (\omega t-\varphi )$, with $I_0=V_0/|Z|$. Average the instantaneous electric power $V(t)I(t)$ over a full period. The double-frequency term averages to zero, leavingThereforeEquivalently, root mean square values give $⟨P⟩=V_{\mathrm{rms}}{I}_{\mathrm{rms}}\cos \varphi$. The reactance exchanges stored electric field or magnetic field energy with the source, whereas the real part produces Joule heating. A purely imaginary complex impedance dissipates no mean electric power under these ideal assumptions.

In vacuum, with possible sources $\rho$ and $\mathbf{J}$, Maxwell's equations areTheir integral forms, with compatible orientations and fixed surfaces, areCharge- and current-free space additionally sets $\rho =0$, $\mathbf{J}=0$. Vacuum constitutive constants alone do not imply that no charges exist.

Take the curl of the source-free Ampere-Maxwell law, use Faraday's law of induction and $\nabla \cdot \mathbf{B}=0$:ConsequentlyFor the specified complex plane wave, the wavevector is $k\hat{\mathbf{z}}$ and $\omega =ck$. Faraday's law of induction fixesThe physical fields are the real parts. They are in phase and their Poynting vector points along $+z$.

The instantaneous electric and magnetic energy densities are $u_E={ϵ}_0{E}^2/2$ and $u_B=B^2/(2{\mu }_0)$. Since $E=cB$, they are equal for this ideal travelling plane wave, and so are their cycle averages: $⟨u_E⟩/⟨u_B⟩=1$. For real peak amplitude $E_0$, each average is ${ϵ}_0{E}_0^2/4$. The Poynting vector and mean intensity satisfyIt represents transported energy per area per time, in $\mathrm{W}{\mathrm{m}}^{-2}$.

For the numerical moonlight estimate, adopt $L_{\odot }=3.828\times 10^{26}\mathrm{W}$ and lunar mean radius $R_M=1.7374\times 10^6\mathrm{m}$. These additional astronomical inputs give solar irradiance at the Moon of ${\mathcal{I}}_{\odot }=L_{\odot }/(4\pi d_{\odot }^2)\simeq 1361\mathrm{W}{\mathrm{m}}^{-2}$.^[ref][ref]

Under the simple isotropic point-reflector approximation, the intercepted power is $\pi R_M^2{\mathcal{I}}_{\odot }$, of which fraction $A=0.12$ is redistributed over $4\pi$ steradians. ThusThese are equivalent sinusoidal peak amplitudes; the corresponding root mean square fields are smaller by $\sqrt{2}$. Actual moonlight is broadband and incoherent.

A total reflected fraction alone does not specify the directional brightness of a full Moon. For a Lambertian sphere at full phase, with the same total reflectance $A$, integrating its radiance gives ${\mathcal{I}}_M=(2A/3){\mathcal{I}}_{\odot }(R_M/d_M{)}^2$, instead giving $E_0\simeq 1.30\mathrm{V}{\mathrm{m}}^{-1}$ and $B_0\simeq 4.32\mathrm{nT}$. If 0.12 is instead the geometric albedo, the full-phase formula is ${\mathcal{I}}_M=0.12{\mathcal{I}}_{\odot }(R_M/d_M{)}^2$, giving $E_0\simeq 1.59\mathrm{V}{\mathrm{m}}^{-1}$. NASA's table itself distinguishes Bond albedo from geometric albedo.^[ref] The numerical estimate therefore needs an explicit reflection convention; the first boxed values use the stated isotropic approximation.

A suitable electrical circuit has $R$ in series with the input and a shunt branch consisting of $C$ in parallel with the lossy inductor $R_L+i\omega L$; take the output voltage across that shunt branch. This is the middle arrangement in the circuit figure. In terms of its complex impedance,At direct current the capacitor is an open circuit and the inductor contributes only $R_L$, so $H(0)=R_L/(R+R_L)=0.01$. ThereforeWhen $R_L$ is neglected, $Z_p=i\omega L/(1-{\omega }^{2}LC)$. At the antiresonance ${\omega }_0^{2}LC=1$ its electrical admittance vanishes, so $|H|$ reaches its maximum value of one. ConsequentlyThis value uses precisely the requested lossless approximation; keeping $R_L$ changes both the peak position and its height. A transfer function peak and one DC measurement do not uniquely identify every possible two-port network, but the displayed arrangement meets both observations to the stipulated approximation. Also, the angle brackets in the source must denote AC wave amplitudes or root mean square values: the literal time averages of sinusoidal voltages are zero, whose ratio would be undefined.

In the high-resolution quantization approximation, a smooth input distribution varies little across one conversion interval, so rounding to the nearest level gives an approximately uniform quantization error $ϵ\in [-\Delta V/2,\Delta V/2]$. ThereforeThe input must stay within the analogue-to-digital converter's range. A large standard deviation alone does not guarantee uniform quantization error: an input concentrated exactly on widely separated conversion levels has zero error. Truncation without a half-step correction also has a different mean and root mean square.

For a rightgoing harmonic travelling wave in the lossless string, its force and transverse-velocity complex amplitudes satisfy $F=Zu$ with real $Z=\sqrt{T\rho }$. Thus the supplied mean-power expression giveswhere $A$ is the displacement complex amplitude and $u=-i\omega A$ for convention $e^{i(kx-\omega t)}$. The same result follows directly from instantaneous string power $P=-T{\Psi }_x{\Psi }_t$ averaged over a cycle. The amplitudes here are peak amplitudes; with root mean square force or velocity the factor $1/2$ is absent. If reflections return to the driver, its net mean input is still $\mathrm{Re}(F{u}^{\ast })/2$, but its input mechanical impedance need not equal the travelling-wave value $Z$.

Interpret “3 mm wire” as a circular wire of diameter $3\mathrm{mm}$, and take the current density as uniform. One conduction electron per atom givesUsing $I=neA|v_d|$, the drift velocity corresponding to a current magnitude of $13\mathrm{A}$ isIt points opposite to conventional electric current. A simple distance scale over the $1/50=0.020\mathrm{s}$ cycle is $\boxed{|v_d|/f=2.71\mu \mathrm{m}}$, but this treats the magnitude as constant. For sinusoidal mains, alternating-current drift excursion gives zero net displacement after a full cycle.

If $13\mathrm{A}$ is the usual root mean square current, $v_{\mathrm{pk}}=\sqrt{2}|v_d|$ and the displacement amplitude and total absolute drift distance per cycle areIf $13\mathrm{A}$ instead denotes peak current, these become $0.431\mu \mathrm{m}$ and $1.72\mu \mathrm{m}$. The paper does not state peak versus RMS; the characteristic drift is the boxed $1.35\times 10^{-4}\mathrm{m}{\mathrm{s}}^{-1}$ on whichever current convention is chosen. It also does not explicitly say diameter; taking $3\mathrm{mm}$ as the radius would reduce all these velocities and drift distances by four. None of these distances is the microscopic path travelled at roughly the Fermi velocity between scattering events; that path cannot be obtained from the supplied current alone.

Use peak phasors with time dependence $e^{-i\omega t}$, real positive electrical conductivity $\sigma$, and absolute permeability ${\mu }_a=\mu {\mu }_0$. Ampère-Maxwell equation and Faraday equations giveso ${ϵ}^{\prime }=ϵ+i\sigma /(\omega {ϵ}_0)$. The sign of the imaginary term follows from the stated time convention. For a transverse plane wave in a homogeneous conductor,In the good-conductor limit, choose the root that decays into $z>0$:The electric phasor and physical field are consequently ${\tilde{E}}_x=E_0{e}^{(i-1)z/\delta }$ and $E_x=\mathrm{Re}[E_0{e}^{i(z/\delta -\omega t)-z/\delta }]$, as required. Tangential boundary matching sets $E_0$; it need not equal the incident amplitude.

Faraday's law fixes the magnetic field, including its phase:For real $E_0$ and $\psi =z/\delta -\omega t$, this means $H_y=E_0{e}^{-z/\delta }(\cos \psi -\sin \psi )/(\omega {\mu }_a\delta )$, with other components zero. The time-averaged Poynting vector isThe energy entering a thin slab exceeds that leaving it by $-d⟨N_z⟩/dz$ per unit volume, givingDirectly, $\mathbf{J}=\sigma \mathbf{E}$ and $\mathbf{J}\cdot \mathbf{E}=J^2/\sigma$ at each time. Therefore $⟨J^2/\sigma ⟩=\sigma |{\tilde{E}}_x{|}^2/2$ is exactly this loss. The factors one half use peak rather than root mean square phasors; amplitude decays over $\delta$, while intensity decays over $\delta /2$.

For a cold collisionless unmagnetized plasma, electron inertia gives $\tilde{\sigma }=i{ϵ}_0{\omega }_p^2/\omega$ and real ${ϵ}^{\prime }=1-{\omega }_p^2/{\omega }^2$. Above the plasma frequency the wave propagates without this resistive attenuation. Below it, a semi-infinite plasma supports an evanescent field with electric and magnetic components in quadrature and zero mean inward energy flux: the incident energy is reflected, not absorbed as Joule heating. electron kinetic energy is reversibly stored, so the nondispersive dielectric field-energy formula alone is incomplete. Collisions introduce a real dissipative electrical conductivity and can produce absorption.^[ref]

For a linear nondispersive medium with real absolute permittivity ${ϵ}_a=ϵ{ϵ}_0$ and permeability ${\mu }_0$, instantaneous field energy density and energy flux arewith local balance ${\partial }_{t}u+\nabla \cdot \mathbf{N}=-\mathbf{J}\cdot \mathbf{E}$. energy stored locally and energy passing through a surface are distinct: integrating the balance relates their rates and the power given to charges.

Between the coaxial conductors, cylindrical Gauss and Ampère laws giveThe inner line charge is $\lambda =2\pi {ϵ}_{a}V/\ln (b/a)$, so the capacitance per length is $C=2\pi {ϵ}_a/\ln (b/a)$. Integrating magnetic energy over the annulus gives $\int B^2(2\pi rdr)/(2{\mu }_0)={\mu }_0{I}^2\ln (b/a)/(4\pi )=L{I}^2/2$, hence $L={\mu }_0\ln (b/a)/(2\pi )$. ThusThese are per-length parameters for ideal conductors, excluding internal magnetic penetration.

Now let the voltage and current amplitudes be $V_0,I_0$ and multiply both transverse fields by $e^{i(kz-\omega t)}$. Their divergences vanish in the annular dielectric. The two nontrivial curl equations reduce toConsequently both $k^2={\omega }^2{\mu }_0{ϵ}_a$ and $V_0=Z_0{I}_0$ are required for a forward wave. Arbitrary independent static $V$ and $I$ cannot acquire the same forward-wave factor and still satisfy Maxwell's equations. This TEM travelling-wave voltage-current relation and the dispersion constraint identify the transverse electromagnetic mode and its speed $v_p=1/\sqrt{LC}$.

For the travelling wave, $⟨N_z⟩=\mathrm{Re}(V_0{I}_0^{\ast })/[4\pi r^2\ln (b/a)]$. Integrating over the dielectric annulus yieldsThis is the familiar circuit power, carried here through the dielectric rather than through the ideal metal. In root mean square notation it is $V_{\mathrm{rms}}{I}_{\mathrm{rms}}$ for the forward lossless wave; the corresponding static fields give $VI$ without a time-average factor. The average line energy is $(C|V_0{|}^2+L|I_0{|}^2)/4=C|V_0{|}^2/2$, whose product with wave speed agrees with the power.

Finite conductor resistance requires a small longitudinal electric field to drive current. Its cross product with the azimuthal magnetic field produces a radial Poynting component into each conductor: inward radially at the inner conductor and outward radially into the outer conductor. This Poynting flux into lossy coaxial conductors explains why forward power decreases along the cable as energy becomes heat in the metal. The mode is then only approximately TEM; strictly zero longitudinal electric field would be inconsistent with resistive current.

For a classical dilute ideal gas in thermal equilibrium, the probability of a single-particle velocity is proportional to the Boltzmann factor for kinetic energy, $e^{-m{v}^2/(2k_{B}T)}$. Velocities with speed between $v$ and $v+dv$ fill a spherical shell of velocity-space volume $4\pi v^{2}dv$. The normalized Maxwell speed distribution is thereforeThe factor $v^2$ counts states, rather than representing an extra Boltzmann energy factor.

Set $\alpha =m/(2k_{B}T)$ and use $I_j={\int }_0^{\infty }{v}^j{e}^{-\alpha v^2}dv$. Normalization cancels in ratios, so the given recurrence givesConsequentlyThe root mean square speed is larger than the mean speed, consistent with the positive variance of the speed distribution.

An electrical transformer transfers alternating electrical power between windings through a shared changing magnetic flux. A core with large magnetic permeability confines most of the flux, so both windings link approximately the same flux $\Phi (t)$. Faraday's law gives induced electromotive forcesFor an ideal electrical transformer, negligible winding resistance and leakage flux imply the terminal-voltage magnitude ratioThe relative phase depends on the winding sense and terminal convention. At a loaded secondary, Lenz's law makes the secondary current oppose the flux change. The primary draws an additional current which approximately cancels the secondary magnetomotive force: $N_p{I}_p\simeq N_s{I}_s$ in magnitude when the magnetizing current is negligible. Consequentlyfor a resistive load, using root mean square values. More generally the ideal device conserves instantaneous power, with the appropriate current signs. A load impedance $Z_L$ appears at the primary as $Z_{\mathrm{in}}=(N_p/N_s{)}^2{Z}_L$, enabling impedance matching and electrical isolation.

For sinusoidal flux, $\Phi ={\Phi }_{\max }\sin \omega t$ gives $V_{\mathrm{rms}}=\omega N{\Phi }_{\max }/\sqrt{2}$. At a given supply voltage, reducing frequency increases the required peak flux and can drive the core into magnetic saturation. Real losses include winding Joule heating, magnetic hysteresis, eddy currents and leakage flux; a laminated core suppresses eddy-current loops. A steady direct current produces no continuing secondary electromotive force, while its small winding resistance can allow a damaging primary current after switching transients.

Detector noise and graphical reading. Noise-equivalent power specifies the incident optical power producing signal-to-noise ratio one, with the electrical bandwidth and averaging convention stated. A spectral NEP has units $\mathrm{W}\mathrm{H}{\mathrm{z}}^{-1/2}$; a band-integrated NEP has units W. Specific detectivity normalizes the latter by detector area and noise bandwidth:Its usual unit, the Jones, is $\mathrm{cm}\mathrm{H}{\mathrm{z}}^{1/2}{\mathrm{W}}^{-1}$. Removing the expected square-root area and bandwidth dependence helps compare different detector sizes and readout bandwidths, provided noise conditions and spectral response are comparable. This does not normalize away every background, temperature or implementation difference.

Reading the native embedded voltage plot gives approximate means $V_C\simeq 0.331\mathrm{V}$, $V_S\simeq 0.406\mathrm{V}$ and $V_H\simeq 0.493\mathrm{V}$. The one-second root mean square scatter is about $0.006\mathrm{V}$; the plotted peak-to-peak excursions are larger and must not be substituted for an root mean square noise. Independent graphical digitization of the blue trace gives means near $0.331,0.406,0.493$ V and scatters about $0.006$–$0.007$ V, consistent with this estimate. These are readings from a drawing, not exact original data.

The supplied Planck approximation gives $B_C\simeq 0.08590$ and $B_H\simeq 0.12836$ in $\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}(\mathrm{c}{\mathrm{m}}^{-1}{)}^{-1}$. Thus the noise-equivalent radiance of one one-second sample isHere the calibrating means are used to determine the response slope; their own uncertainty is included explicitly in the final scene estimate below.

Optical throughput and detectivity. A paraxial $f/F$ cone has half-angle approximately $1/(2F)$ and projected solid angle approximately $\pi /(4F^2)$. With $F=2$ and $A_d=10^{-6}{\mathrm{m}}^2$,Using the exact cone angle $\tan \theta =1/(2F)$ gives projected etendue $A_d\pi {\sin }^2\theta =1.85\times 10^{-7}{\mathrm{m}}^2\mathrm{sr}$, within the intended approximation. The radiance noise corresponds to detector power noiseThe wavenumber bandwidth is $1\mathrm{c}{\mathrm{m}}^{-1}$ and must match the radiance's per-wavenumber unit; it is not an electrical noise bandwidth. For the common order-of-magnitude assignment $\Delta f\sim 1\mathrm{Hz}$ to a one-second sample,Area in the last expression is $1\mathrm{m}{\mathrm{m}}^2=0.01\mathrm{c}{\mathrm{m}}^2$. For a specified one-second boxcar with white noise, the one-sided equivalent noise bandwidth is instead $1/(2t)=0.5\mathrm{Hz}$. That convention gives spectral NEP $\simeq 2.5\times 10^{-10}\mathrm{W}\mathrm{H}{\mathrm{z}}^{-1/2}$ and $D_{\ast }\simeq 3.9\times 10^8$ Jones. The source does not specify the electrical filter, so the two conventions should not be mixed or assigned spurious precision.

Two-point retrieval and its uncertainty. Assume linear, stable response for the three views. The radiometer two-point calibration isWith target radiances treated as known, differentiating givesNotice $b<0$ for an intermediate scene: raising the hot voltage while keeping the scene voltage fixed reduces the inferred scene radiance. Numerically $a\simeq 0.2621$, $b\simeq -0.1214$ in radiance units per volt.

The scene interpolation fraction is about $0.463$, giving $B_S\simeq 0.1056$ andFor independent white noise in each view, each hundred-sample mean has standard error ${\sigma }_{\bar{V}}=0.006/10=0.0006\mathrm{V}$. The radiance error isUsing the given temperature sensitivity,A scatter estimate closer to $0.007$ V changes this to roughly $0.14$ K. These are random standard errors under the averaging assumptions. Target-temperature/emissivity uncertainty, response drift and atmospheric or surface-emissivity corrections are separate: the retrieved quantity is a brightness temperature, not automatically the true sea-surface temperature.

The mean interface height fluctuation is zero; its root mean square roughness is obtained by summing the mode variances. Replacing the two-dimensional wavevector sum by its continuum density gives the thermal roughness of a tension-dominated interface:Take $q_{\min }=2\pi /L$ with $L=\sqrt{A}=10^{-2}\mathrm{m}$, and $q_{\max }=2\pi /d$ with $d=2\times 10^{-10}\mathrm{m}$. At $T=300\mathrm{K}$,The logarithm makes the answer only weakly sensitive to order-one choices of cutoff. Gravity has been omitted as stipulated; the fixed mean height removes the zero wavevector mode.

The Stokes–Einstein relation gives the translational diffusion coefficientat $T=298\mathrm{K}$. For isotropic three-dimensional Brownian motion, the mean squared displacement is $⟨|\Delta \mathbf{r}{|}^2⟩=6Dt$. Thus after one hour the root mean square displacement isThe phrase “on average” needs a convention: the mean displacement vector is zero; the mean displacement magnitude of three-dimensional Brownian motion is $⟨|\Delta \mathbf{r}|⟩=4\sqrt{Dt/\pi }=1.22\mathrm{mm}$, obtained by integrating the three-dimensional Gaussian distribution. These are displacements, not the total microscopic path length. The estimate assumes dilute unconfined water, a spherical hydrodynamic radius and constant temperature.

Availability and total entropy. The reservoir changes by $d{U}_R=-dU$ and $d{V}_R=-dV$. Its particle number and the system's particle number are fixed. The reservoir's first law of thermodynamics therefore givesAdd the system's entropy change. Because $T_R,p_R$ are constant,Thus define the availabilityAn additive constant in $A$ is immaterial; its value above the equilibrium reference is the exergy. The Second law of thermodynamics gives $\Delta S_{\mathrm{tot}}\ge 0$, hence $\Delta A\le 0$. A stable equilibrium maximizes total entropy over the allowed variations and therefore minimizes $A$. The environmental $T_R,p_R$ enter this definition even when the system has not yet reached their values. Because both thermal and mechanical exchanges are allowed, equilibrium here includes mechanical balance; equality of temperatures alone does not ensure the minimum.

Probability and curvature. For a fixed value of a macroscopic coordinate $x$, let $S(x)$ count the compatible system microstates. The product of the system and reservoir multiplicities isAll factors independent of $x$ enter a normalization constant, givingHere $A(x)$ is the constrained or effective availability for this coordinate, including the multiplicity of any eliminated variables. A nonuniform coordinate measure must likewise be retained or incorporated in that effective availability.

At a stable interior minimum, $A^{\prime }(x_0)=0$ and $A^{\prime \prime }(x_0)>0$. Its Taylor expansion givesKeeping the quadratic term yields a normal distribution,This is availability curvature and equilibrium fluctuations. The quoted fluctuation formula holds at quadratic order in the Gaussian fluctuation approximation; it is not an exact identity for an arbitrary nonlinear $A(x)$. Higher terms can shift the mean away from its most probable value $x_0$.

Bubble equilibrium. Keep the number of air molecules fixed and include the surface energy in the system's total internal energy. For constant surface tension $\Gamma$, this contribution is $\Gamma \Omega =\Gamma \alpha V^{2/3}$, with no surface entropy under the stated approximation. Its differential isHenceIndependent thermal and volume variations give $T=T_R$ andThe air pressure is evaluated at $V_0$; since the bubble is closed it generally depends on volume, so this equation is implicit until an air equation of state is supplied. For a sphere, $\alpha =(36\pi {)}^{1/3}$, $\Omega =4\pi R^2$, and the pressure jump becomes $2\Gamma /R$, the Young–Laplace equation for a single gas–liquid interface. There is no second surface as in a thin soap film.

Isothermal volume fluctuations. With the air maintained at $T_R$, eliminate its entropy in favor of its Helmholtz free energy:The identity $(\partial F_{\mathrm{air}}/\partial V{)}_{T,N}=-p_{\mathrm{air}}$ givesUsing the curvature result,These are the volume fluctuations of a closed gas bubble. Stability requires the denominator to be positive. The negative surface contribution reflects the concavity of $V^{2/3}$; gas compression supplies the positive restoring curvature. For ideal air, ${\kappa }_T^{-1}=p_{\mathrm{air}}$, and mechanical equilibrium reduces the denominator toThe units are consistent: $D$ is pressure, so $k_B{T}_R{V}_0/D$ has units ${\mathrm{m}}^6$, the units of a volume variance. Holding $p_{\mathrm{air}}$ externally fixed while varying $V$ would remove the gas-compression term and change this stability problem; that is not the fixed-particle-number situation here.

Numerical size of the fluctuations. The volume conversion is important:Use the specified $\alpha =4.84$, $\Gamma =0.072\mathrm{N}{\mathrm{m}}^{-1}$ and $p_R=10^5\mathrm{Pa}$. ThenWith the Boltzmann constant $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$,Thus the fractional root mean square fluctuation and fractional variance are respectivelyThe absolute root mean square fluctuation is $2.01\times 10^{-19}{\mathrm{m}}^3$. The very small fractional value supports the local Gaussian fluctuation approximation. This estimate neglects exchange of dissolved air, variations of surface tension, shape fluctuations and temperature fluctuations, consistently with the prescribed fixed-number, isothermal bubble model.

A convenient choice for this irregular region is Monte Carlo integration. The region is contained in $[-1,1{]}^2$. For independent uniform points $U_j$ in that square, letThe indicator random variable incorporates the curved boundary directly. This estimator is unbiased; the integrand is bounded on the square, so its variance is finite. The variance of a sum of independent terms gives root mean square error $O(N^{-1/2})$ in big O notation. This is a stochastic error rate. Applying a rectangular Simpson's rule to the indicator-weighted integrand would lose the smoothness required for its fourth-order error bound.