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With tension taken positive, hydrostatic stress is the stress tensor $\mathbf{\sigma }=-p\mathbf{I}$, where $\mathbf{I}$ is the identity matrix. Every plane has the same compressive normal stress, and no tangential traction. More generally,where $\mathbf{s}$ is deviatoric stress. Two samples can share the same mean pressure while experiencing different deviatoric stresses. A stationary liquid cannot sustain equilibrium shear stress; a solid transmitting medium can support it. Consequently, a load divided by surface area does not by itself establish hydrostatic sample conditions. A transmitting medium's state at the actual pressure and temperature, and contact between sample and walls, matter.

Bridgman's Figure 4 compares mild steel fractured in tension at atmospheric pressure with steel stretched further in a liquid at $25000\mathrm{kgf}\mathrm{c}{\mathrm{m}}^{-2}$.^[ref] The imposed tension remains a directional load: the experiment combines confinement with deformation, rather than measuring a hydrostatic stress alone. Its comparison motivates recording the loading path and the surrounding medium as well as pressure; it does not establish a universal fracture threshold for all steels.

For a nearest-neighbour Heisenberg antiferromagnet on a hypercubic lattice, the linear spin-wave approximation gives ${\gamma }_{\mathbf{k}}=d^{-1}{\sum }_j\cos (k_{j}a)$ and the displayed frequency. Near a gapless point its spin-wave velocity is $2JSa\sqrt{d}$. Its ground-state boson occupation is $[1/\sqrt{1-{\gamma }_{\mathbf{k}}^2}-1]/2$, producing quantum depletion of Néel order. Exact zero modes need an infrared regulator or separate global-rotation treatment.

A hyperelastic energy well is a set of deformation gradients minimizing the stored-energy function at a fixed temperature. Material frame indifference makes each well invariant under left multiplication by proper rotations. A phase-changing crystal can have a single austenitic rotational well at high temperature and several martensitic wells below a transition. Continuous deformations cannot freely join arbitrary wells: planar interfaces require a rank-one connection between the selected gradients.

An ice-bearing plume at local liquidus equilibrium conserves excess enthalpy $E$ and salt $S$ relative to the ambient liquidus water. Eliminating its temperature and brine composition gives $\varphi Q[L+c_{p}m{C}_0-L\varphi ]=-E-c_{p}mS+E\varphi$. To dilute-crystal order this fixes its advected ice volume, hence its leading buoyancy flux. A positive rising branch needs a positive source invariant.

A calorimeter that infers transferred heat from ice melting at a specified equilibrium temperature. The latent heat sets the proportionality after correcting for other thermal changes and extraneous melting.

Ice-melting calorimetry measures the fusion enthalpy through conservation of energy, rather than inferring it from the small temperature change of the melting ice itself. In a mixing experiment, warm water and the container cool while added ice melts and the resulting water warms. Their combined energy balance separates fusion from sensible heat.

Weigh the container and stirrer, then weigh the added warm water to obtain $m_w$. Let the container and water reach a common initial temperature $T_i$. Prepare ice in a clean ice-water mixture near $T_m$, rather than assuming freezer ice is already at $T_m$; drain and blot its surface immediately before transfer. Add it without splashing, stir gently, and record temperature before addition and until after the last ice disappears. Reweigh to find the added mass $m_i$. Wisconsin's laboratory guide specifies drying each piece, weighing before and after, and recording the temperature history; it places the initial and final temperatures on opposite sides of the jacket temperature to reduce net environmental exchange. ^[ref]

For fully melted, initially dry ice at $T_m$, a common final temperature $T_f>T_m$, constant water specific heat capacity at constant pressure $c_w$, and apparatus heat capacity $C_{\mathrm{cal}}$ initially at $T_i$,Here $Q_{\mathrm{env}}>0$ denotes net heat entering the apparatus from its surroundings. The left side supplies the fusion enthalpy and the sensible heat on the right; the water produced by melting must be warmed too. The balance extends Case Western's mixing demonstration by including the apparatus and environmental terms. If the specific heat capacities vary appreciably, replace their products with the corresponding temperature integrals. ^[ref]

Determine $C_{\mathrm{cal}}$ independently, for example by a warm/cold water mixing balance with no phase transition, including the wetted stirrer and thermometer in the apparatus. Check the thermometer calibration, keep the lid in place between transfers, and monitor the surroundings. Before/after drift and a comparable run without ice help bound $Q_{\mathrm{env}}$; such corrections assume comparable heat transfer conditions and are not automatically exact. Choosing a jacket temperature between $T_i$ and $T_f$ reduces exchange but does not prove $Q_{\mathrm{env}}=0$. Ensure all the ice has melted and the contents are well mixed before using the measured $T_f$: with residual ice, the weighed addition is not the mass that has undergone fusion. Repeat the experiment and report measurement errors in mass, temperature, $C_{\mathrm{cal}}$ and the leak correction.

An independent way to see the plateau is David Roundy's Oregon State experiment: heat an ice-water mixture slowly with a waterproof resistor in an insulated cup, stir, and record temperature, heater electric current $I(t)$, electric potential difference $V(t)$ and elapsed time. Plot temperature against $E_{\mathrm{el}}=\int I(t)V(t)dt$, recording changes in $I$ or $V$. The initial solid mass must exclude accompanying water. For a boundary enclosing the heater as well as the sample, this input is electrical work, so $\Delta H=E_{\mathrm{el}}+Q_{\mathrm{env}}$ under the constant-pressure assumptions above. Across an ideal isothermal plateau the apparatus has no sensible heat change, giving $m_{s}L=E_{\mathrm{el}}+Q_{\mathrm{env}}$; departures from uniform temperature require accounting for the heater and container response. ^[ref]

Ice-ocean albedo feedback is the ocean-surface case of ice-albedo feedback: loss of bright sea ice, often covered by snow, exposes darker water, lowers surface albedo and increases absorbed sunlight. The extra stored ocean heat content can promote further melting and delay refreezing. Reversing the sequence can amplify cooling. This is a positive climate feedback, meaning that it reinforces the initial change; the word positive does not mean beneficial, and it does not guarantee runaway warming.

For unchanged downward solar irradiance $S_{\downarrow }$ on a newly exposed opaque patch, the extra absorbed heat flux isFor illustrative local surface albedos $0.60$ and $0.10$, and $S_{\downarrow }=200\mathrm{W}{\mathrm{m}}^{-2}$, the patch absorbs an additional $100\mathrm{W}{\mathrm{m}}^{-2}$. This is a local sunlight calculation, not a whole-planet radiative forcing: newly exposed area, illumination, clouds and season determine its global contribution. With no incident sunlight this particular heating pathway vanishes. NASA's notes describe the contrast between reflective ice and snow and darker water. ^[ref]

To isolate how a positive climate feedback competes with Planck feedback, retain the assumptions of the zero-dimensional planetary energy balance and prescribe the smooth broadband Bond albedoThe exponential function makes a smooth transition from a bright cold state near $0.6$ to a darker warm state near $0.3$. The midpoint $240\mathrm{K}$ and width $3\mathrm{K}$ are chosen to illustrate the dynamics. They are not a seawater freezing law, an observed global relation, or a calibration of modern Earth. No evolving ice thickness or latent heat is represented.

An equilibrium satisfies $F(T_{\ast })=0$. For a small perturbation $\delta T$, linear stability follows fromSince $C>0$, $F^{\prime }<0$ gives an asymptotically stable equilibrium, while $F^{\prime }>0$ gives an unstable equilibrium. With $x=(T-240\mathrm{K})/(3\mathrm{K})$, the derivative isThus the optical term $-S{\alpha }^{\prime }/4$ is destabilizing, while the thermal radiation term $-4\sigma T^3$ is stabilizing. Their sum, not the optical term alone, decides the linear stability.

For the same $S=1360\mathrm{W}{\mathrm{m}}^{-2}$ and Stefan-Boltzmann constant used above, the bisection method on the intervals $(220,223)$, $(238,241)$ and $(253,255)$, with temperature in kelvins, gives:

Unrounded numerical zeros of a function have $|F(T_{\ast })|<10^{-10}\mathrm{W}{\mathrm{m}}^{-2}$; the table rounds them for readability. In the positive-temperature domain the sign of $F$ alternates between heating, cooling, heating and cooling across these three equilibria. The middle equilibrium separates the basins of attraction of the two stable equilibria. For example, a starting temperature of $235\mathrm{K}$ cools toward the cold state, whereas $245\mathrm{K}$ warms toward the warm state: this is bistability at identical external illumination.

The same model illustrates hysteresis. Its equilibrium curve can be written as $S(T)=4\sigma T^4/[1-\alpha (T)]$. The two turning points, where $4[1-\alpha (T)]+T{\alpha }^{\prime }(T)=0$, are saddle-node bifurcations: approximately $(T,S)=(232.637\mathrm{K},1567.801\mathrm{W}{\mathrm{m}}^{-2})$ and $(245.893\mathrm{K},1250.481\mathrm{W}{\mathrm{m}}^{-2})$. For illumination strictly between these thresholds there are two stable equilibria and one unstable equilibrium. Under sufficiently slow increase of $S$, the cold branch ends at the upper threshold; under sufficiently slow decrease, the warm branch ends at the lower threshold. The ensuing warming and cooling transitions differ, so the state depends on its history. Finite sweep rates can delay the transitions; these thresholds describe this model's static equilibrium curve, not predicted tipping points for modern Earth.

The striking feature is that a smooth albedo law can produce distinct long-time states and abrupt transitions, although no discontinuity was inserted into the optical law. It is useful for understanding why positive climate feedback can amplify a response while stabilizing thermal radiation still limits it. Here $0.3<\alpha (T)<0.6$ bounds the incoming heating, whereas outgoing thermal radiation grows as $T^4$; high enough temperatures therefore cool rather than run away indefinitely.

Historically, M. I. Budyko's 1969 paper coupled changing albedo from expanding ice cover to a latitude-dependent temperature and heat transfer calculation, showing how that coupling amplified the effect of changed sunlight. ^[ref] William D. Sellers's 1969 paper constructed an annual energy-budget model in 10-degree latitude belts, including planetary albedo, atmospheric infrared transmission and atmospheric and oceanic exchange. ^[ref] These primary papers provide historical examples of ice-sensitive energy-budget modeling; the smooth single-temperature example above is a separate illustration, not a reproduction of either paper's equations or predictions.

An ideal solution has molar chemical potentialswhere $R$ is the molar gas constant and the stars denote pure-component reference values at the same temperature and pressure.^[ref]

Hold temperature and pressure fixed. The derivatives with respect to $x$ are $RT/x$ and $-RT/(1-x)$, soEach chemical potential changes, yet their composition-weighted slopes cancel exactly. At $x=1/4$, the two slopes are $4RT$ and $-4RT/3$; multiplying by $1/4$ and $3/4$ gives $RT$ and $-RT$. Near either pure-component endpoint the dilute component's slope diverges, but its weighted contribution remains finite. The endpoint chemical potential itself requires a limiting interpretation because its logarithm diverges.

The same constraint has a geometric explanation. Subtract the pure-component reference contribution from the molar Gibbs free energy and divide by $RT$ to obtainIts derivative is $f^{\prime }(x)=\ln [x/(1-x)]$. At fixed total amount of substance, exchanging a little B for A changes the Gibbs free energy with slope ${\mu }_A-{\mu }_B$. Consequently the tangent to $f$ at $x$ has interceptsat horizontal coordinates $0$ and $1$, respectively. The two chemical potentials come from the same curve: choosing two unrelated functions would generally destroy this tangent construction.

For the stipulated A/B ideal solution, put $x=x_A$. At the fixed temperature $300\mathrm{K}$, Raoult's law reduces toThe relative volatility is $\alpha =p_A^{\ast }/p_B^{\ast }=4$. For $0<x<1$,Hence every interior vapor-liquid equilibrium pair in this chosen model has A-enriched vapor; neither pure endpoint is enriched. Conversely,For example, $y_A=1/2$ corresponds to $x_A=1/5$, $p_A=p_B=16\mathrm{kPa}$ and $P=32\mathrm{kPa}$. Equal vapor mole fractions therefore need not mean equal liquid mole fractions.

These are fixed-temperature predictions. During boiling at a specified external pressure, the temperature instead adjusts until ${\sum }_i{x}_i{p}_i^{\ast }(T)=P_{\mathrm{external}}$. The two stipulated vapor pressures specify only one temperature; they provide no boiling-temperature curve.

An ideal chain, also called a freely jointed chain, consists of fixed-length links with independent orientations and no bending energy between successive links. It neglects segment interactions and excluded volume. A one-dimensional freely jointed chain restricts each link to two orientations. At fixed force its canonical partition function factors into one-link angular integrals. The resulting extension is entropic and depends on temperature and on whether the orientations are allowed in one, two, or three dimensions.

For a nonrelativistic three-dimensional ideal electron gas at zero temperature, the pressure is $p=2n{E}_F/5$ and the electronic bulk modulus is $B=2n{E}_F/3$. This is the electronic model contribution, not automatically the total bulk modulus of a real crystal.

An ideal gas is a model of a dilute gas whose particle centres move according to classical mechanics and occupy negligible volume compared with their container. Treating the particles as point masses removes the molecular-size correction; neglecting interactions between separated particles removes the intermolecular contribution to potential energy. The usual kinetic theory model retains brief elastic collisions with the walls and, optionally, with other particles: these transfer momentum while conserving the total kinetic energy of the colliding bodies. There is no appreciable interaction energy stored between collisions. These assumptions lead to the ideal gas law, $PV=N{k}_{B}T$. ^{[ref] [ref]}

The model's attraction is its universality: at the same temperature and number density, different classical ideal gases have the same pressure, even when their particle masses differ. In ideal gas pressure from wall collisions, heavier particles move more slowly at a given temperature, and the two effects cancel. For example, one mole of any ideal gas at $300\mathrm{K}$ in $0.025{\mathrm{m}}^3$ has pressure approximately $99.774\mathrm{kPa}$, as computed under the ideal gas law.

The ideal gas law describes the translational contribution to pressure; it does not specify all of the internal energy or the heat capacity. A monatomic gas has three classical translational degrees of freedom, whereas a gas of molecules can also store rotational kinetic energy and energy in vibrational normal modes. An equation of state alone therefore does not determine how much heat a chosen temperature increase requires. It also does not select a process: isothermal expansion and adiabatic expansion need different additional constraints. ^[ref]

A real gas departs from this approximation when particle size or interactions become important, especially at high number density or low temperature near a phase transition. The Van der Waals equation adds such corrections. Independently, at sufficiently low temperature or high number density, quantum statistics can become essential: even particles with negligible interactions need not obey the classical ideal gas law. The classical statistical assumption is the nondegenerate gas regime, which is distinct from neglecting interactions. ^{[ref] [ref]}

An isothermal noninteracting gas in potential $mgz$ has an exponential vertical density and scale height $k_{B}T/(mg)$.

Up to volume-independent terms, $F=-N{k}_{B}T\log V$, so $-\partial F/\partial V=N{k}_{B}T/V$.

The ideal gas law is the equation of state relating the absolute pressure $P$, volume $V$, particle count $N$, and absolute thermodynamic temperature $T$ of a classical ideal gas:Here $k_B$ is the Boltzmann constant, $N_{\mathrm{A}}$ is the Avogadro constant, $R$ is the molar gas constant, and $n_{\mathrm{mol}}$ is the amount of substance in moles. The distinct symbol $n_{\mathrm{mol}}$ avoids confusing amount of substance with the number density $n=N/V$, measured in ${\mathrm{m}}^{-3}$: the same law can be written $P=n{k}_{B}T$. Use absolute pressure, relative to a vacuum, rather than an unconverted gauge pressure. ^[ref]

The exact SI definitions of the Avogadro constant and Boltzmann constant giveThis numerical value is exact; the applicability of the ideal gas model to a real gas is an approximation. ^{[ref] [ref]}

For $n_{\mathrm{mol}}=1\mathrm{mol}$, $T=300\mathrm{K}$, and $V=0.025{\mathrm{m}}^3$, direct substitution givesAt fixed temperature and particle count, doubling the volume to $0.050{\mathrm{m}}^3$ halves the pressure to approximately $49.887\mathrm{kPa}$: this is Boyle's law. At fixed volume and particle count, doubling the temperature from $300\mathrm{K}$ to $600\mathrm{K}$ doubles the pressure to approximately $199.547\mathrm{kPa}$. These ratios use kelvins, not readings in the degree Celsius scale: $300\mathrm{K}$ is $26.85{}^{\circ }\mathrm{C}$, and $600\mathrm{K}$ is $326.85{}^{\circ }\mathrm{C}$.

The curved slices show Boyle's law at three temperatures; the straight slice shows proportionality to absolute temperature at one fixed volume. They are relations among equilibrium states, rather than a prescribed process history. Compression could hold the temperature fixed through heat transfer, or raise it during adiabatic compression. To calculate a process's heat or work, supplement the ideal gas law with the process constraint, the first law of thermodynamics, and the relevant internal energy or heat capacity information. The graph's extrapolation toward $T=0$ is mathematical; it does not assert that a real gas remains classical or gaseous there.

For a stellar mixture with fixed mean molecular weight, the ideal gas law can be written $P=\mathcal{R}\rho T/\mu$, where $\rho$ is mass density and $\mathcal{R}$ is the stellar gas constant. It assumes negligible interaction corrections and a nondegenerate gas; excitation or ionization can change the particle count and thermodynamic response.

In his 1834 memoir on the motive power of heat, Émile Clapeyron combined gas laws into a relation between pressure, volume, and temperature while analysing heat engines (printed page 164). The equation above uses modern SI notation and absolute thermodynamic temperature; it is not a transcription of his historical temperature convention. ^[ref]

An idealized gas mixture in which component partial pressures add and component amounts obey the ideal gas law. At a common temperature and pressure, component gas volumes before mixing can be used to calculate the mixture's mole fractions.