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The cloud absorbs $e\sigma T_g^4$ from below and emits $e\sigma T_c^4$ from each face. It absorbs no stellar radiation. Its net radiative heating per unit area is thereforeFor $e>0$, a cloud with $q^4<1/2$ warms in the absence of other energy transfers; one with $q^4>1/2$ cools. For the thin pressure thickness $\delta p$, its initial temperature tendency is approximately $g{\mathcal{R}}_c/(c_p\delta p)$ if its sensible heat capacity dominates. Latent heat and changing condensate modify that conversion from heating to temperature change, but not the sign of the radiative input.

With $T_g$ fixed, pure radiative equilibrium requiresThis equals the ambient dry-adiabatic process temperature precisely whenFor ordinary dry air, $\kappa \simeq 2/7$, so the ratio is about $0.545$. If $e=0$, the cloud exchanges no thermal radiation and its radiation budget does not select any equilibrium temperature.

At fixed ground temperature, warming the cloud increases its downward radiant flux $e\sigma T_c^4$, giving a positive contribution to the ground's radiative heating. That is the immediate radiative effect. A self-consistent final answer must also remove the convective heat transfer used to pin the original cloud to the dry adiabatic process.

For pure radiative equilibrium of both bodies, with no net nonradiative energy exchange, their two balances areFor $e>0$ this givesThe $e\to 0^{+}$ limit gives the selected limiting cloud temperature; at exactly $e=0$, only $T_g=T_0$ is fixed radiatively. The maximum greenhouse effect at fixed $Q$ occurs at $e=1$:The pinned dry-adiabatic process case instead has maximum factor $1/q=(p_s/p_c{)}^{\kappa }$. It exceeds $2^{1/4}$ for a high, cold cloud with $q<2^{-1/4}$, equals it at the level found in part b, and is smaller for a low, warm cloud.

In particular, for $q^4<1/2$ the final purely radiative ground is cooler than the pinned solution, even though a warmer cloud increases downward radiation at fixed $T_g$. To see why these statements are compatible, let $H$ be the net upward nonradiative energy transfer from the ground to the cloud in the pinned state. Its conservation of energy equation requiresFor the cold cloud, $H<0$: maintaining it on the assumed dry adiabatic process requires removing its net radiative heating and returning energy elsewhere. Setting $H=0$ changes that energy redistribution. It is therefore incorrect to infer the final ground temperature by changing only the downward radiant flux. Indeed the final cloud is warmer than the pinned one for $q^4<1/2$ and $0<e<1$, while the final ground is cooler. At $e=1$ the cloud temperature is $T_0$ in both whole-column equilibria.

Fixed cloud properties are an artificial constraint. Warming a saturated cloud increases the saturation pressure, tending to evaporate its cloud droplets; the released water vapor, changing latent heat budget and altered buoyancy can change its altitude and emissivity. Cooling favours condensation and potentially atmospheric precipitation. A warm perturbation that survives evaporation can rise by convection, but neither a unique vertical displacement nor survival of the cloud follows without moisture supply and an environmental temperature profile. Finally, pure radiative equilibrium of the two bodies does not require the intervening transparent gas to remain on the original dry adiabatic process; imposing that extra constraint is generally inconsistent except at the matching level in part b.

Before condensation, assume a closed parcel with fixed atmospheric mixing ratio, constant specific heat capacities, and negligible irreversible heating. The first law of thermodynamics for a dry adiabatic process isThe parcel remains in approximate pressure equilibrium with a hydrostatic environment. With the standard approximation that parcel and environmental densities are close, $dp/dz\simeq -\rho g$, henceThis is the dry adiabatic lapse rate, about $9.8\mathrm{K}\mathrm{k}{\mathrm{m}}^{-1}$ for air. More precisely the environmental hydrostatic balance gives $dT/dz=-g{\rho }_{\mathrm{env}}/(c_p{\rho }_{\mathrm{parcel}})$; the familiar constant dry adiabatic lapse rate uses their ratio close to one.

Let $e_v$ be the parcel's water vapor partial pressure. Fixed mole fraction before condensation means $e_v/p$ is constant. Its dew point $T_D$ is defined by $E(T_D)=e_v$, where $E$ is the saturation pressure. The Clausius-Clapeyron relation givesThe last approximation uses hydrostatic balance, the ideal gas law, and an environmental temperature close to the parcel's. Rearrangement yieldsFor a modest initial dew-point depression, $T_D/T\simeq 1$ in absolute temperature, soWith $L_v\simeq 2.5\times 10^6\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$, $R_d\simeq 287\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$, and $R_v\simeq 461.5\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$, $H\simeq 1.58\times 10^5\mathrm{m}$. This dew point scale is much longer than the atmospheric pressure scale height.

The exponential function is an approximation, not the exact consequence of a dry adiabatic process. Retaining $T=T_s-{\Gamma }_{d}z$ gives $p/p_s=(T/T_s{)}^{c_p/R_d}$ and the useful comparisonIt also shows explicitly that expansion lowers the dew point even before any water vapor condenses.

At the lifting condensation level, the parcel temperature first equals its dew point. For $z/H\ll 1$, the first-order Taylor expansion gives $T_D\simeq T_{D,s}(1-z/H)$. Equating this with the dry-adiabatic process result givesNear terrestrial surface temperatures, $T_{D,s}/H\simeq 1.7$–$1.9\mathrm{K}\mathrm{k}{\mathrm{m}}^{-1}$, whereas ${\Gamma }_d\simeq 9.8\mathrm{K}\mathrm{k}{\mathrm{m}}^{-1}$. ConsequentlyUse absolute temperature for $T_{D,s}/H$. The temperature difference can be evaluated in kelvins or degrees Celsius, whose increments agree. The coefficient is an approximate near-surface terrestrial value, not independent of composition or ambient conditions for all atmospheres.

A rising parcel cools, lowering the saturation pressure. Once its water vapor supply exceeds that value, atmospheric supersaturation develops. Homogeneous nucleation of pure water is strongly inhibited at the small atmospheric supersaturations typical of clouds; existing aerosol particles provide cloud condensation nuclei. Their dry sizes and composition, together with the parcel's ascent rate, determine how many cloud droplets activate.

Köhler theory combines the Kelvin equation, which raises equilibrium vapor pressure over a curved surface, with Raoult's law, which lowers it for a chemical solution. In a dilute diameter convention, with dry diameter $D_d$ and wet diameter $D$,Here $s=e_v/E(T)-1$ is atmospheric supersaturation, ${\sigma }_w$ is surface tension, and ${\kappa }_h$ measures aerosol hygroscopicity, not the dry-adiabatic process exponent. The maximum of the Köhler curve is the critical supersaturation. Larger or more hygroscopic particles generally have lower thresholds. Cloud-droplet activation requires overcoming that maximum; total aerosol number is not the activated number.

Given an observed aerosol number size distribution $n_j(D_d)=d{N}_j/d\ln D_d$ for composition classes $j$, first assign each class a hygroscopicity and compute its critical supersaturation. If the parcel reaches maximum atmospheric supersaturation $s_{\max }$, the equilibrium-threshold estimate isFor each fitted lognormal distribution, this is its upper-tail integral; it can be evaluated analytically with an error function or numerically from measured bins.

The value $s_{\max }$ must be calculated, not guessed from the aerosol distribution alone. Prescribe parcel upward velocity $w$, initial pressure, temperature and humidity, then integrate expansion cooling, latent heat release, conservation of total water, and diffusional cloud droplet growth. Schematically,where $a_s,b_s>0$, $G$ includes water vapor diffusion and thermal conduction, and $q_l$ is liquid mass fraction. Rising motion produces atmospheric supersaturation; growing particles consume it. Their competition selects the peak and which particles cross their activation barriers. Assume a well-mixed closed parcel, liquid water, soluble spherical particles, specified mixing state, and negligible collision-coalescence and fallout during activation. Slow kinetic growth, weakly soluble particles or entrainment require a more detailed treatment than the threshold integral.

Distinguish the depth-dependent mantle temperature from mantle potential temperature $T_p$: the latter is the temperature a parcel would have after reversible decompression in an adiabatic process to a chosen reference pressure. A convecting mantle is approximately adiabatic through much of its interior, with steep thermal boundary layers near the surface and the core-mantle boundary. It is neither isothermal nor wholly molten.

Early accretion, large impacts and gravitational planetary differentiation supplied heat; metal segregation released gravitational energy and extensive melting could produce a magma ocean. Rapid surface heat loss then crystallized much of the silicate mantle. Subsequent evolution reflects competition between radiogenic heating, heat entering from the core and heat carried out through the upper boundary. In a schematic mantle energy balance,Here $C_m$ is mantle heat capacity, $H_m$ mantle radioactive power and $Q_m$ heat leaving the mantle; continental crustal radioactive production is not included in $H_m$. The convective Urey ratio $H_m/Q_m$ compares internal production with mantle heat loss. A value below one does not alone prove cooling, because core heat also enters; cooling requires $H_m+Q_{\mathrm{cmb}}<Q_m$. Radioactive decay reduces the heating supply, while mantle convection, plate tectonics, melt extraction and changing viscosity control the heat loss. A hotter mantle does not automatically imply proportionally faster plate motion: melting, dehydration and boundary strength can alter that relationship.

The robust long-term picture is an energetic early Earth followed by substantial cooling and a presently mostly solid, convecting mantle. The shape of the solid-state history is less certain. Petrological reconstructions commonly place late-Archean ambient $T_p$ above modern values. One reconstruction infers about $1500$–$1600{}^{\circ }\mathrm{C}$ at $2.5$–$3$ Ga versus about $1350{}^{\circ }\mathrm{C}$ today, and permits warming before that late-Archean maximum after the initial magma-ocean cooling.^[ref] This is a model-dependent ambient-mantle history, not proof that the accreting Earth was cooler than the Archean mantle or that every mantle reservoir followed one curve. An early molten transient, subsequent solid-state heating or cooling, and later secular cooling must be distinguished.

The first energy term is the rate of change of internal plus kinetic energy per unit volume; the second transports that energy with the moving gas; the pressure term is the mechanical work flux across the boundary. There is no heat flux or viscous work in the stated inviscid adiabatic model.

Use mass conservation to replace a conservative derivative of $\rho e$ by $\rho De/Dt$. Dotting the momentum equation with $\mathbf{u}$ gives $\rho D(|\mathbf{u}{|}^2/2)/Dt=-\mathbf{u}\cdot \nabla p$. Subtraction from total energy leavesNow $p=\rho RT$ and $D\rho /Dt=-\rho \nabla \cdot \mathbf{u}$ implyThus $\boxed{DS/Dt=0}$: smooth adiabatic inviscid trajectories are isentropic. Shocks need an entropy inequality rather than this smooth equality.

Adding $\nabla \cdot (k\nabla T)$ models Fourier heat conduction, with heat flux $-k\nabla T$. The same subtraction gives $\rho c_{v}DT/Dt+p\nabla \cdot \mathbf{u}=\nabla \cdot (k\nabla T)$, henceFor a fixed container use impermeability $\mathbf{u}\cdot \mathbf{n}=0$, insulation $k\nabla T\cdot \mathbf{n}=0$, and mass conservation. ThenThe boundary term vanishes and $\nabla (1/T)=-\nabla T/T^2$ supplies the positive sign. For $T>0$, $k\ge 0$, total entropy cannot decrease: this is entropy production by Fourier heat conduction.

Let the bed be $z=0$ and the free surface $z=h(x,t)$. In a long, shallow layer, vertical acceleration is smaller than gravity, so the vertical Euler equation gives hydrostatic pressure $p=p_{\mathrm{atm}}+\rho g(h-z)$. Incompressibility makes the vertical speed smaller than the horizontal speed by the aspect ratio; irrotationality then makes the leading horizontal velocity depth-independent. The horizontal Euler equation becomesIntegrating $u_x+w_z=0$ across the depth, using bed impermeability $w(0)=0$ and the surface condition $w(h)=h_t+u{h}_x$, yieldsCombining these gives the conservative momentum equation $(hu{)}_t+(h{u}^2+g{h}^2/2{)}_x=0$. Integral mass and momentum balance across a stationary thin jump therefore requireThese conservation laws still apply across a jump where the smooth hydrostatic/irrotational derivation does not resolve the internal turbulent structure.

Put $A=h_{-}$, $B=h_{+}$ and $q=A{u}_{-}=B{u}_{+}>0$, orienting the flow from minus to plus. For a nontrivial jump the momentum condition gives $q^2=gAB(A+B)/2$. Its change in mechanical-energy flux isA passive jump cannot create mechanical energy, so the change is nonpositive. Since $q>0$, $\boxed{h_{+}>h_{-}}$ for a genuine jump; equality describes no jump. The upstream flow is supercritical and the downstream flow subcritical, since $F{r}_{-}^2=B(A+B)/(2A^2)>1$ and $F{r}_{+}^2=A(A+B)/(2B^2)<1$.

The gas-dynamic analogue is a compressive normal shock: density and pressure increase and flow changes from supersonic to subsonic. Entropy increases; directed mechanical energy is irreversibly converted into internal energy. Total gas energy, including internal energy, is conserved across an adiabatic shock. This distinguishes that total-energy balance from the reduced shallow-water mechanical-energy loss.

With a thermally insulating boundary, $\delta Q=0$. If the slow piston motion is frictionless and reversible, it is an isentropic process:The first law of thermodynamics gives $dU=-pdV$, so expansion lowers the internal energy and temperature. For constant heat capacity ratio $\gamma$, $p{V}^{\gamma }$ and $T{V}^{\gamma -1}$ are constant combinations of state functions. Insulation alone does not guarantee constant entropy: an irreversible adiabatic process can produce entropy. The conclusion uses the intended reversible, quasistatic piston idealization.

For a reversible adiabatic process of an ideal gas, $\delta Q=0$ and $dU=-pdV$. At fixed composition and constant sample heat capacity at constant volume $C_V$,Mayer's relation gives $n_{\mathrm{mol}}R=C_p-C_V$, so $T{V}^{\gamma -1}$ is constant andThe finite-path power law requires approximately constant heat capacities as well as reversibility. When rotational or vibrational populations change appreciably, use $\int C_V(T)dT/T+n_{\mathrm{mol}}R\ln V=\mathrm{constant}$ instead; the local isentropic pressure exponent remains $\gamma (T)$.

For an intact heteronuclear diatomic gas, quantum excitation progressively activates molecular motion. With rotational and vibrational modes frozen, the three translational degrees of freedom give $C_{V,\mathrm{m}}=3R/2$ and $\gamma =5/3$. Two active rotations add $R$, giving $C_{V,\mathrm{m}}=5R/2$ and $\gamma =7/5$. One active harmonic vibration contributes a further $R$, through kinetic and potential terms, giving $C_{V,\mathrm{m}}=7R/2$ and $\gamma =9/7$.

For molecular moment of inertia $I={\mu }_{\mathrm{red}}{r}_e^2$, the rigid rotor levels are $E_J={\hslash }^{2}J(J+1)/(2I)$, so the rotational excitation scale isThe first gap corresponds to $2{\Theta }_{\mathrm{rot}}$, so the crossover occurs at temperatures of that order. For bond force constant $k_s$ and reduced molecular mass ${\mu }_{\mathrm{red}}$, the vibrational scale isTypically ${\Theta }_{\mathrm{vib}}\gg {\Theta }_{\mathrm{rot}}$. Real gases may condense at the low-temperature end or dissociate and excite electronic states at the high-temperature end; the three plateau values describe the ideal intact-molecule model.

For the piston apparatus, mechanical equilibrium requires $p_{0}A=p_{\mathrm{air}}A+mg$, hence $p_0=p_{\mathrm{air}}+mg/A$. Let $x$ be a small upward displacement, so $V=V_0+Ax$. The local isentropic compressibility givesConsequently the adiabatic gas-spring piston period isThis neglects friction, gas inertia, leakage and dissipation, with internal equilibration much faster than the oscillation and external thermal equilibration much slower. The small oscillation probes $\gamma$ locally; it need not keep $\gamma$ constant over a large temperature interval.

With $p_{\mathrm{air}}=101325\mathrm{Pa}$, the supplied dimensions give $p_0=101351.4\mathrm{Pa}$ and $B=2.20566\times 10^{-5}{\mathrm{s}}^2$. ThusHelium is close to the monatomic ideal gas value $5/3$; NO is close to the diatomic gas value $7/5$, with rotations active and vibration largely frozen. The numbers are approximate experimental inferences, rather than exact rational values.

A thermodynamically reversible process can be reversed without a net change of the system and surroundings. Its entropy change equals reversible heat transfer divided by temperature. A reversible adiabatic process conserves entropy; adiabaticity alone does not forbid internal entropy production in an irreversible evolution.

The Rüchardt experiment infers heat capacity ratio from the period of an oscillation of a piston supported by trapped gas. Use a vertical precision cylinder with a close-fitting movable piston, known moving mass $M$, face area $A$, trapped equilibrium volume $V_0$, and local atmospheric pressure $p_{\mathrm{atm}}$. With negligible static friction, force balance gives the absolute gas pressure $p_0=p_{\mathrm{atm}}+Mg/A$, where $g$ is gravitational acceleration. Displace the piston slightly and release it: compression raises pressure, and expansion lowers it, producing a restoring force. The oscillation is a mechanical probe of thermal response, without separately measuring two heat inputs. ^[ref]

Let $x$ be upward displacement from equilibrium, so $V=V_0+Ax$. For constant heat capacity ratio and a reversible adiabatic process, the net upward force, including the ambient load and weight, isThe linear approximation requires $|Ax|/V_0\ll 1$. Applying Newton's second law gives simple harmonic motion withHere ${\omega }_0$ is the undamped angular frequency and $\mathcal{T}$ the undamped oscillation period, rather than temperature. Record repeated cycles, measure $M,A,V_0,p_{\mathrm{atm}}$ independently, and include all connected trapped volume. An illustrative model with $M=0.020\mathrm{kg}$, $A=2.0\times 10^{-4}{\mathrm{m}}^2$, $V_0=0.010{\mathrm{m}}^3$, $p_{\mathrm{atm}}=100000\mathrm{Pa}$, $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and $\gamma =1.4$ gives $p_0=100981\mathrm{Pa}$ and $\mathcal{T}\simeq 1.182\mathrm{s}$. These stipulated values are not measurements.

In Shearwood and Sloan's implementation, piston-mounted magnets induce an electromotive force in a stationary coil through Faraday's law of induction; an oscilloscope records the signal. They observed faster oscillations at smaller trapped volume values and stronger damping at small cylinder volume values. The signal tracks velocity locally, so its amplitude is not directly piston displacement. Calibration also identified inlet dead space beyond the graduated cylinder volume. ^[ref]

Repeat at smaller initial amplitudes to test the linear approximation, and compare several $V_0$ values. Heat transfer weakens the restoring force toward the isothermal process limit, biasing the inference toward $1$. Check leakage through resting-piston drift and before-and-after trapped volume readings. For viscous damping with envelope $e^{-t/\tau }$, the measured angular frequency obeys ${\omega }_0^2={\omega }_d^2+{\tau }^{-2}$; this correction does not remove thermal leakage. Internal equilibration must be faster than an oscillation period, with heat transfer and leakage slow over a run. Large amplitude or sticking friction require a richer model. ^[ref]

Stellar convective instability occurs when an adiabatic process displaces a fluid element and buoyancy amplifies the displacement. For uniform composition the Schwarzschild criterion gives instability when the actual temperature gradient exceeds the adiabatic temperature gradient; composition gradients instead require the Ledoux criterion.