For the planar good-solvent coil, $R\sim \ell N^{3/4}$. The two-dimensional polymer overlap fraction is thereforeThis is a scaling estimate; the monomer geometry and choice of coil radius set order-one numerical factors.

The polymer overlap concentration is reached when the number of chains per coil area is of order one. If $n_c$ is the chain number per unit interfacial area, then $n_c^{\ast }{R}^2\sim 1$. Equivalently, the monomer area fraction is ${\varphi }^{\ast }\sim N{\ell }^2/R^2$, assuming an occupied monomer area of order ${\ell }^2$. Below this scale coils are mostly separate; above it they interpenetrate.

A three-dimensional self-avoiding walk has swollen size $R\sim b{N}^{\nu }$, with $\nu \simeq 0.59$; the three-dimensional Flory approximation gives the nearby estimate $3/5$. Isotropic projection gives $⟨R_{\mathrm{proj}}^2⟩=(2/3)⟨R^2⟩$, preserving this exponent. For the mean distance rather than its root mean square, the orientation average instead gives $⟨R_{\mathrm{proj}}⟩=(\pi /4)⟨R⟩$. Neither prefactor changes its position between the ideal $N^{1/2}$ walks and the planar swollen $N^{3/4}$ walk.

For an isotropic three-dimensional ideal walk, each Cartesian component contributes $N{b}^2/3$ to the mean square. Keeping two components gives $⟨R_{\mathrm{proj}}^2⟩=2N{b}^2/3$. The planar projection of an isotropic random vector therefore has $R_{\mathrm{proj},\mathrm{rms}}=b\sqrt{2N/3}$, smaller than the unprojected planar ideal walk by $\sqrt{2/3}$. The corresponding two-dimensional Gaussian mean distance has the same ratio.

A planar self-avoiding walk cannot revisit its occupied sites. In the good-solvent scaling regime, the two-dimensional Flory approximation gives $R\sim b{N}^{3/4}$, larger at large $N$ than either ideal walk and the projected three-dimensional self-avoiding walk. This is a scaling statement, with microscopic prefactors dependent on the walk model.

For the planar random walk with independent steps of length $b$, cross terms vanish and $⟨R^2⟩=N{b}^2$. Thus $R_{\mathrm{rms}}=b{N}^{1/2}$. At large $N$, the planar Gaussian end distribution gives $⟨R⟩\simeq b\sqrt{\pi N}/2$.

Kirchhoff's law of thermal radiation equates spectral absorptivity and spectral emissivity at the same wavelength and direction. A surface that is a poor absorber in the thermal infrared is also a poor emitter there; a coating cannot be assigned its solar absorption independently of its thermal emission spectrum.

(i) All-wavelength black paint. It absorbs the sunlight and also radiates maximally at the spacecraft's own temperature. With the isothermal-sphere assumptions of (d), the limiting black-surface distance isThe spacecraft can operate at $R\ge 50r$ in this model. A surface already assumed to be a blackbody is not improved by another black coating; relative to option (ii), however, this paint permits closer operation.

(ii) Black at short wavelengths and grey in the infrared. By Wien's displacement law, changing the black-body radiation temperature from $6000\mathrm{K}$ to $600\mathrm{K}$ shifts the wavelength peak from about $0.5\mu \mathrm{m}$ to about $5\mu \mathrm{m}$. Most solar power is therefore in the highly absorbing short-wavelength band, whereas most spacecraft emission lies beyond $2\mu \mathrm{m}$, where the spectral emissivity is reduced. The coating retains most of the heating but suppresses much of the cooling. Under the intended two-band grey body approximation, option (ii) makes the spacecraft hotter at the same distance and requires a larger minimum distance than black paint.

To express the comparison without inventing a grey absorptivity, let ${\bar{\alpha }}_S$ be radiometric absorptance averaged over the solar spectrum and ${\overline{ϵ}}_T$ be emissivity averaged over the spacecraft's own thermal spectrum. The radiative equilibrium equation becomesAt the failure threshold,For an ideal two-band coating take a constant radiometric absorptance $g<1$ beyond ${\lambda }_c=2\mu \mathrm{m}$, and unity below it. If $f(T)$ is the fraction of the power emitted by a blackbody below ${\lambda }_c$, Planck's law givesSince $f(T_S)>f(600\mathrm{K})$, this model has ${\bar{\alpha }}_S>{\overline{ϵ}}_{600}$ and hence $R_{\min }>50r$. The paper supplies neither a numerical value of $g$ nor a detailed spectrum, so no unique numerical distance for the grey paint follows. If its long-wavelength absorption is not approximately grey, the spectrum must be integrated explicitly rather than inferred from a peak alone. A coating designed to improve close approach would instead reflect much of the solar radiation while retaining strong thermal infrared emission.

Let the spacecraft radius be $a$, and assume that heat redistribution keeps its whole surface at a common temperature $T$. Neglect background radiation, and let $R\gg r,a$. The Sun's luminosity, from the Stefan–Boltzmann law, isAt distance $R$, the incident irradiance is $F_S=L_S/(4\pi R^2)=\sigma T_S^4(r/R{)}^2$. The spacecraft intercepts radiation over its projected disc, area $\pi a^2$, but emits from its whole spherical area $4\pi a^2$. The radiative equilibrium condition is thereforeCancelling the common factors givesThe radius of the spacecraft cancels. The factor $1/\sqrt{2}$ follows from the ratio of absorbing projected area to emitting surface area. Uniform surface temperature is the intended additional idealization behind a single equilibrium surface temperature. A poorly conducting spacecraft would have position-dependent temperatures; for example, a locally reradiating patch facing the Sun would instead have $T_{\mathrm{subsolar}}=T_S\sqrt{r/R}$, and a different maximum-temperature constraint.

The equilibrium photon gas is isotropic, so the energy density per unit solid angle in a narrow angular-frequency interval is $u(T,\omega )d\omega /(4\pi )$. For rays making angle $\theta$ with an outward surface normal, a surface element $dA$ is crossed in time $dt$ by rays initially in a volume $cdtdA\cos \theta$. Only the outward hemisphere contributes. ThusThe hemispherical restriction supplies a factor $1/2$ and the directional projection supplies another $1/2$. A small aperture in a cavity therefore emits this spectral power per unit area. A blackbody at the same temperature emits the same amount: otherwise radiation exchange between it and the aperture would violate thermal equilibrium.

Integrating Planck's law over all positive angular frequencies gives the total emitted heat flux densitySet $x=\hslash \omega /(k_{B}T)$. ThenThe remaining integral is independent of temperature, proving $M\propto T^4$. Its value follows by expanding $(e^x-1{)}^{-1}={\sum }_{m=1}^{\infty }{e}^{-mx}$ and integrating termwise:Consequently the Stefan–Boltzmann law isThis is emitted power alone. A body in a radiation environment also absorbs incoming radiation, which must be included when calculating net radiative cooling.

Each independent cavity mode is a quantum harmonic oscillator of angular frequency $\omega$. Its occupation numbers $n=0,1,\dots$ represent the numbers of photons in that mode. At thermal equilibrium, photons can be absorbed and emitted by the walls, so their number is not conserved and the photon chemical potential is zero. The Boltzmann distribution of the oscillator excitations then givesThe expression supplied in the paper is the thermal excitation energy. The oscillator's temperature-independent zero-point energy is excluded from the emitted thermal radiation.

Putting $ϵ=\hslash \omega$ and $dϵ=\hslash d\omega$ into the mode count from (a) gives the number of modes per unit cavity volume:Multiplying by the mean thermal energy in each mode yields Planck's law in the angular-frequency convention:The two photon polarizations are already included in the density of states; no further factor of two is needed. $u(T,\omega )d\omega$ has units $\mathrm{J}{\mathrm{m}}^{-3}$, whereas $u(T,\omega )$ has units $\mathrm{J}\mathrm{s}{\mathrm{m}}^{-3}$.

A cubic reflecting cavity supports standing waves with positive integer mode indices and wavevector components $k_i=\pi n_i/L$. A cell in mode space has volume $(\pi /L{)}^3$. In the continuum approximation, a spherical shell of radius $k$ and thickness $dk$ has one-eighth of its volume in the positive octant. There are two independent transverse photon polarizations for a generic wavevector. HenceThe photon energy is $ϵ=\hslash \omega =\hslash ck$, so $dk=dϵ/(\hslash c)$. ThereforeThis is the leading bulk density of states, valid when the cavity is large compared with the relevant wavelengths and many modes lie in the counting interval. A finite cavity has a discrete mode spectrum, with boundary corrections beyond this approximation. Modes with vanishing indices require separate boundary counting but do not change the leading $L^3$ term. Here $\hslash$ is the reduced Planck constant, as printed in the original PDF; replacing it by $h$ would introduce an erroneous factor of $(2\pi {)}^3$.

For a diatomic ideal gas in the usual regime with three translational and two rotational quadratic degrees of freedom, but frozen vibrational excitation, the equipartition theorem givesThe Brayton cycle efficiency is thereforeThe constant-pressure expansion has $T_3/T_2=V_3/V_2=2$, while the compression has $T_2/T_1=10^{2/7}=1.93070$. The same relations give $T_4/T_1=2$. Thus, in units of the minimum temperature,The maximum and minimum temperatures are $T_3$ and $T_1$. A Carnot cycle between those extremes hasThe difference is about $25.9$ percentage points. In the Brayton cycle, heat is supplied and rejected over ranges of temperature; the whole input is not supplied at $T_3$, nor is the whole rejection made at $T_1$. This explains the lower efficiency even for mechanically ideal stages. Absolute state temperatures are not supplied, so $\gamma =7/5$ is an explicit standard diatomic-gas approximation rather than a conclusion about all possible rotational and vibrational excitation regimes.

This is the Brayton cycle. Assume fixed gas amount, constant heat capacities, and internally reversible adiabatic processes, as intended by the ideal engine model. Put $r_p=p_h/p_l=p_2/p_1>1$ andCombining $p{V}^{\gamma }=\text{constant}$ with the ideal gas law gives $T\propto p^{(\gamma -1)/\gamma }$ along either reversible ideal-gas adiabat. ThereforeDuring a constant-pressure process the heat supplied equals the enthalpy increase. If $C_p$ is the molar heat capacity at constant pressure, the positive heat input and output areThe insulated branches exchange no heat, so the first law of thermodynamics gives $W=Q_{\mathrm{in}}-Q_{\mathrm{out}}$. HenceAn adiabatic process alone need not be an isentropic process; the relations used here also require internal reversibility. Dissipative compression or expansion would require extra information and would not, in general, obey this efficiency formula.

For the numbered engine, denote the low and high pressures by $p_1=p_4=p_l$ and $p_2=p_3=p_h$. Let $R_g$ denote the molar gas constant. The ideal gas law makes the constant-pressure branches straight rays in a temperature-volume diagram:The high-pressure ray is steeper. For a constant heat capacity ratio $\gamma$, each reversible ideal-gas adiabat obeysConsequently $1\to 2$ curves upward and left during compression; $2\to 3$ goes upward and right along the steeper ray; $3\to 4$ curves downward and right during expansion; and $4\to 1$ goes downward and left along the shallower ray.

The plot is an original calculation for $\gamma =7/5$, $p_h/p_l=10$ and $V_3/V_2=2$; its state directions follow the original paper. The dashed extensions identify the two constant-pressure rays, not additional stages of the cycle.

The complete cycle must be a reversible thermodynamic process: no friction, viscous dissipation, unrestrained expansion, or other source of entropy. Each moving boundary must be arbitrarily close to mechanical equilibrium. Heat transfer to or from a bath must occur across an infinitesimal temperature difference, and the insulated stages must be genuinely adiabatic processes. There must also be no direct heat leak between the heat baths.

For an engine exchanging heat only with these two heat baths, the Second law of thermodynamics givesand thusPractical engines generate entropy through finite temperature differences, friction and other dissipative effects. They therefore fall below the Carnot cycle limit. Approaching reversible heat exchange generally requires vanishingly slow operation for a finite heat-transfer apparatus, limiting useful power.

Because the working substance returns to its original thermodynamic state, its net entropy change is zero. There is no entropy change on either reversible ideal-gas adiabat; on each isothermal process the entropy transfer is the exchanged heat divided by the bath's absolute temperature. HenceThe first law of thermodynamics over a complete cycle gives $W=Q_h-Q_c$. The thermal efficiency is thereforeThe use of kelvins is essential. This result does not depend on the working substance. For the ideal gas illustration, the two reversible ideal-gas adiabats imply $V_C/V_B=V_D/V_A=(T_h/T_c{)}^{1/(\gamma -1)}$, so $\ln (V_B/V_A)=\ln (V_C/V_D)$; the explicit heat expressions in (a) give the same ratio.

Use an ideal gas to illustrate the Carnot cycle, with hot and cold heat baths at $T_h=T_2$ and $T_c=T_1$. The four processes are all reversible thermodynamic processes. Lettered states below belong only to this illustration; the engine later in the question uses numbered states.

Starting at $A$, perform the following four stages:

The clockwise enclosed area in the pressure-volume diagram is the positive net work per cycle, $W=\oint pdV=Q_h-Q_c$.

An incident electromagnetic wave generally produces reflected and transmitted waves because the two media have different electromagnetic response. Consider a flat interface between homogeneous, linear, lossless media with permittivities ${ϵ}_j$ and magnetic permeabilities ${\mu }_j$. Their wave speeds and wave impedances areFor each plane electromagnetic wave, $\mathbf{H}=\hat{\mathbf{k}}\times \mathbf{E}/Z_j$. With no free surface charge or current, the interface conditions for electromagnetic fields require continuous tangential $\mathbf{E},\mathbf{H}$ and normal $\mathbf{D},\mathbf{B}$. Equality of the phase at every point on the boundary gives equal angular frequencies and tangential wavevectors. Hence the reflection angle equals the incidence angle and Snell's law iswhere $n_j=c/v_j$ and each angle is measured from the normal.

At normal incidence use the same tangential axis for all electric amplitudes. The two independent boundary equations areSolving gives the amplitude Fresnel equationsFor nonmagnetic media, these become $r=(n_1-n_2)/(n_1+n_2)$ and $t=2n_1/(n_1+n_2)$. Reflection into a medium of larger refractive index therefore reverses the electric amplitude's sign at normal incidence.

The normal Poynting vector determines the reflected and transmitted fractions of incident power. At normal incidence,It is the power fractions, rather than the squared electric amplitudes alone, that add to unity.

At oblique incidence separate transverse electric polarization ($s$, electric field perpendicular to the incidence plane) from transverse magnetic polarization ($p$, electric field in that plane). For nonmagnetic media, writing $c_i=\cos {\theta }_i$ and $c_t=\cos {\theta }_t$, the Fresnel equations areHere choose a common $s$ direction and, for each ray, the $p$ direction as $\hat{\mathbf{s}}\times \hat{\mathbf{k}}$. This makes the reflected $p$ basis reverse its tangential direction at normal incidence, explaining why the displayed $r_p$ then has the opposite sign to $r_s$. Physical reflectance is independent of this sign convention. For either polarization with propagating transmission,For $p$ polarization, zero reflection occurs at the Brewster angle $\tan {\theta }_B=n_2/n_1$. If $n_1>n_2$, incidence above the critical angle for total internal reflection $\sin {\theta }_c=n_2/n_1$ produces an evanescent field in total internal reflection and $\mathcal{R}=1$ for an isolated lossless interface. A lossy second medium requires complex response coefficients and an absorption term in the energy balance.

Because the magnetic field obeys $\nabla \cdot \mathbf{B}=0$, it can be represented locally as the curl of a magnetic vector potential:This automatically enforces the absence of magnetic monopoles. It is also a global representation throughout ordinary all-space problems without monopole singularities. It is not unique: the gauge transformationleaves $\mathbf{B}$ unchanged because the curl of a gradient vanishes. In a time-dependent problem,makes the same transformation leave both physical fields unchanged and automatically satisfies Faraday's law.

In vacuum magnetostatics, choose the Coulomb gauge $\nabla \cdot \mathbf{A}=0$. Substitution into Ampère's law and the identity $\nabla \times (\nabla \times \mathbf{A})=\nabla (\nabla \cdot \mathbf{A})-{\nabla }^2\mathbf{A}$ givesfor a localized steady electric current density and a potential vanishing at infinity. Taking the curl recovers the Biot-Savart law. The magnetic vector potential converts a constrained vector field problem into componentwise Poisson equations once a gauge is chosen.

For a uniform magnetic field, $\mathbf{A}=\frac{1}{2}\mathbf{B}\times \mathbf{r}$ is a convenient choice. For a localized magnetic dipole moment $\mathbf{m}$, the leading far-field potential isAnother useful relation is Stokes theorem, which relates a closed-path integral to the magnetic flux through a spanning surface:The units of $\mathbf{A}$ are $\mathrm{T}\mathrm{m}$, equivalently $\mathrm{Wb}{\mathrm{m}}^{-1}$.

An electrical insulator has no freely conducting bulk charge in the ideal static description, but an applied electric field can displace bound positive and negative charge relative to one another. The resulting electric dipole moments per unit volume define the polarization density $\mathbf{P}$. Distortion of electronic charge clouds, relative displacement of ions, and partial alignment of permanent molecular electric dipoles can all contribute. Thermal agitation opposes the last mechanism.

For a weak field in a linear isotropic dielectric, the constitutive law isHere ${\chi }_e$ is the electric susceptibility, $ϵ$ is the permittivity, and $\mathbf{D}$ is the electric displacement field. In an anisotropic material, ${\chi }_e$ is a tensor and $\mathbf{P}$ need not be parallel to $\mathbf{E}$. These response laws describe the macroscopic field, not necessarily the microscopic field acting on one molecule.

Spatial variation and surface termination of the polarization density give bound charge and bound current through the bound charge densitieswhere $\hat{\mathbf{n}}$ is outward from the material. Thus uniform polarization density has no bulk bound charge, but generally has opposite bound charges on opposing surfaces. Gauss's law can be rewritten asWith no free surface charge, the normal component of $\mathbf{D}$ is continuous across an interface; the tangential electric field is continuous in electrostatics.

For example, fill a parallel-plate capacitor of plate area $A$ and gap $d$ with a uniform dielectric, neglecting fringing. At fixed free plate charge $Q$, $D=Q/A$, soThe induced bound surface charges oppose the original field, reducing the electric field and potential difference at fixed $Q$. At fixed applied potential difference, the field remains $V/d$ and the battery instead supplies more free charge. The electric displacement field separates these free-charge and bound-charge roles.