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A state of matter is a broad category of material behavior. A solid resists a sustained change of shape, a liquid flows while remaining difficult to compress, and a gas expands to fill its container. These distinctions are useful first descriptions, but they do not enumerate all thermodynamic phases. Plasma adds collective behavior of charged particles, while liquid crystals show that flow and structural order need not disappear together.

Distinguish three questions: what broad state of matter does the material resemble, which thermodynamic phase is present, and what are its temperature, pressure and other state variables? Two samples at different temperatures can be in the same liquid thermodynamic phase. Conversely, crystal polymorphism gives different solid thermodynamic phases of the same substance, distinguished by their crystal structures. A phase transition between them need not change the broad state-of-matter category.

For an imagined substance with crystalline forms A and B, melting A produces a solid–liquid phase transition, while converting A to B produces a solid–solid phase transition. A sample containing A and B has two thermodynamic phases even though all its material is solid. Cutting one uniform crystal into two pieces does not create a second thermodynamic phase: count distinct equilibrium forms, rather than pieces or grains. This distinction matters when applying the phase rule.

A phase diagram locates stable thermodynamic phases using temperature and pressure. Its liquid–gas phase coexistence curve ends at a liquid-gas critical point, around which a path connects liquid and gas without a first-order phase transition. Broad categories can therefore remain useful without a sharp boundary.^[ref]

stationary convection has a time-independent velocity and temperature field, in the chosen fixed frame. Its onset from a conduction state corresponds to a real growth rate passing through zero. Stability of a resulting steady convection roll is a separate nonlinear question.

A statistical ensemble describes hypothetical copies of the same system under specified constraints, with a probability distribution over their microstates. An ensemble average is an expected value across those copies. The canonical ensemble fixes temperature and particle number while allowing energy exchange; the microcanonical ensemble fixes energy, and the grand canonical ensemble also allows particle exchange. These choices specify different physical preparations. ^[ref]

Statistical mechanics connects microstates to macrostates through probability distributions. Its microcanonical ensemble, canonical ensemble and grand canonical ensemble specify different constraints on exchanges with the surroundings. An ensemble average describes repeated preparations under those constraints; it need not be the value in every individual preparation.

This makes the subject useful whenever tracking every microscopic detail would obscure the question being asked. To predict a heat capacity, we need how the expected value of energy changes with temperature, rather than a record of every unit's state. The partition function packages that calculation into one reusable object.

In his 1902 book, J. Willard Gibbs systematically developed the canonical ensemble. His preface contrasts following one system through time with studying a collection of systems with different microscopic conditions. Chapter IV defines the canonical ensemble through an exponential function of energy and establishes its normalization. This supplied a general strategy for predicting collective measurements by averaging over possible microscopic conditions, with the partition function organizing the calculation. ^{[ref] [ref]}

For steady flow through a fixed control volume with one inlet and one outlet, the first law of thermodynamics and conservation of mass giveHere $\dot{m}$ is the mass flow rate, $\dot{Q}$ incoming heat per unit time, ${\dot{W}}_s$ outgoing non-flow work per unit time, $w$ the flow speed, $z$ elevation and $g$ gravitational acceleration. The specific enthalpy $h=u+p/\rho$ includes both the transported internal energy per unit mass $u$ and the pressure work needed to push volume through the boundary; $\rho$ is mass density. The inlet supplies that pressure work, while the outlet consumes it to displace the downstream fluid.^[ref]

For an insulated stationary throttle, require negligible heat transfer, no shaft or other non-flow work, and negligible changes in bulk kinetic energy and gravitational potential energy. Then $h_2=h_1$. With unchanged composition this is also equality of molar enthalpy. The upstream and downstream states must admit equilibrium pressure, temperature and property values, although the constriction itself is irreversible and can be far from thermodynamic equilibrium. Start-up storage of energy in the walls and fluid, heat leakage, or a large outlet speed invalidates the reduced balance. A nozzle can convert specific enthalpy into bulk kinetic energy even when it is insulated.

The same accounting for a fixed transported amount is $U_2-U_1=p_1{V}_1-p_2{V}_2$, hence $U_1+p_1{V}_1=U_2+p_2{V}_2$. This explains why dropping the two pressure work terms would give the wrong conserved quantity.^[ref]

In one-dimensional steady heat conduction without internal heat sources, conservation of energy gives constant heat flux $q=-k{T}^{\prime }$. Through a layer of thickness $L_j$ and positive thermal conductivity $k_j$, the temperature drop is $q{L}_j/k_j$. If adjoining layers are in perfect thermal contact, temperature and heat flux are continuous at each interface, so summing the layer drops proves the displayed formula. Temperature is piecewise linear and its slope may jump when thermal conductivity changes.

A steady periodic thermal state repeats each drive period $P$ after the startup transient has decayed. It is time dependent, unlike a time-independent steady state. For angular frequency $\omega =2\pi /P$, its single-frequency part has the phasor representation $\theta (x,t)=\mathrm{Re}\{\hat{\theta }(x)e^{i\omega t}\}$. Here $\theta$ is the oscillating temperature about the mean profile. Repeating the fitted amplitudes and wave phases over successive cycles checks that startup drift is small; merely completing one drive cycle is insufficient.

For constant volumetric heat generation rate $q_v$ and thermal conductivity $\kappa$, spherical steady thermal conduction obeys $\kappa r^{-2}(r^2{T}^{\prime }{)}^{\prime }+q_v=0$. Regularity at the centre and a specified surface temperature give the displayed parabolic profile. For a black surface radiating into negligible background radiation, global energy conservation gives ${\sigma }_{\mathrm{SB}}{T}_S^4=q_{v}R/3$. The central excess scales as $R^2$, while the surface temperature scales only as $R^{1/4}$.

The approximately constant strain rate in secondary material creep at specified stress, temperature and material state. A lifetime estimate $t\simeq {\epsilon }_f/\dot{\epsilon }$ assumes that rate remains representative; evolving area, damage and tertiary material creep can shorten it.

The steady-state tracking error iswhen this finite limit exists. A finite steady-state tracking error does not require the reference signal, the plant output, or the control input to become constant: all can grow while their difference settles. If the tracking error grows without bound or keeps oscillating, report that behavior rather than assigning a finite steady-state tracking error. Eventual accuracy and asymptotic stability are different questions.

For the first-order low-pass filter $\tau \dot{y}+y=Ku+d$ introduced in reference signal, assume exact measurement, constant control disturbance $d$, and no actuator saturation or feedback delay. Under proportional control $u=k_p(r-y)$ with $k_p>0$, a constant reference signal $r=R$ givesThe decaying homogeneous solution has rate $(1+K{k}_p)/\tau >0$, so this equilibrium point is approached from any finite initial condition. A nonzero tracking error is needed to maintain the control input that balances the model's tendency to relax toward $y=d$. This differs from the accumulating controlled plant $\dot{x}=u+w$, for which proportional control already tracks a constant reference signal exactly when $w=0$. The presence or absence of a step offset depends on both the controlled plant and the controller.

For a ramp reference signal $r=vt$, the same proportional control givesIts long-time particular solution grows with slope $v/(1+K{k}_p)$ when $v\ne 0$. Thus the controlled plant and homogeneous closed-loop response are stable, but the tracking error has no finite steady-state tracking error. A changing goal requires a different performance test from a fixed destination.

Now use proportional-integral control $u=k_{p}e+k_{i}z$, $\dot{z}=e$, with $k_i>0$. For the same controlled plant, differentiate the error equation on intervals where the inputs are smooth:The characteristic polynomial $\tau {\lambda }^2+(1+K{k}_p)\lambda +K{k}_i$ has both roots of a polynomial in the open left half-plane: real roots have negative sum and positive product, and a nonreal complex conjugate pair has negative real part. This proves decay of the homogeneous error for this model. It does not prove stability of an arbitrary controlled plant controlled by positive gains. At a reference jump, use the original ordinary differential equations to set the new initial conditions, rather than treating the reference as smooth at that jump.

With a constant reference signal and constant control disturbance, proportional-integral control therefore gives $e_{\mathrm{ss}}=0$. At its equilibrium point, $y_{\ast }=R$, $u_{\ast }=(R-d)/K$, and $z_{\ast }=(R-d)/(K{k}_i)$. The stored integral maintains the required command while the current tracking error vanishes. For a ramp reference signal, however, the forcing above is $v$, andThe tracking error settles even though the accumulated integral and the required control input grow. This finite offset is the error needed to keep increasing that command. For $r=a{t}^2/2$ with $a\ne 0$, the forcing is $\tau a+at$ and even this proportional-integral control leaves a tracking error growing with slope $a/(K{k}_i)$. Integral action is therefore a precise mechanism for selected input classes, not a promise of perfect tracking for every reference signal. ^[ref]

Here are completely explicit synthetic examples with $y(0)=z(0)=0$, $d=0$, and no sensor bias. Choose $\tau =2\mathrm{s}$, $K=1{}^{\circ }\mathrm{C}$ per percentage point, $k_p=1$ percentage point per degree, and $k_i=0.5$ percentage points per degree-second. Thus $K{k}_p=1$ is dimensionless, and $K{k}_i=0.5{\mathrm{s}}^{-1}$. In the following formulas, $t$ denotes the numerical time in seconds and $y,e,r$ denote numerical temperature rises in degrees. For the unit-step reference signal $r=1$, proportional control and proportional-integral control respectively giveFor the ramp reference signal $r=0.2t$, the corresponding solutions areThe control inputs here are numerical heater-command percentage points. Substitution into $2\dot{y}+y=u$ and into the controller equations checks these solutions without invoking an unverified limiting rule. The proportional control step offset is $0.5$ degrees; the proportional-integral control step offset is zero. The proportional control ramp error grows; the proportional-integral control ramp offset is $0.4$ degrees, exactly $v/(K{k}_i)$.

The conclusions have specific limits. With constant sensor bias $b$, stable proportional-integral control for a feasible constant reference signal drives the measured tracking error to zero but leaves the physical discrepancy ${\epsilon }_{\mathrm{ss}}=b$. Under proportional control, the measured offset instead becomes $(R-b-d)/(1+K{k}_p)$. If actuator saturation prevents the required control input, zero step offset is not achievable by these formulas. A changing control disturbance contributes $-\dot{d}$ to the error equation, while feedback delay can change stability; neither is covered by the constant-disturbance result. These restrictions are what make steady-state tracking error useful: it identifies the input, measurement convention, and model for which an accuracy claim is actually justified.

The Stefan-Boltzmann constant converts the fourth power of absolute temperature into blackbody radiant exitance:Here $k_B$ is the Boltzmann constant, $h$ the Planck constant and $c$ the speed of light. In the current International System of Units, these three constants have exact defining values, so $\sigma$ is exact as an expression in them and $\pi$. The displayed finite decimal is an approximation, not an exact decimal or a measured uncertainty. The 2022 recommended-constants table lists an ellipsis and an exact uncertainty for $\sigma$.^[ref]

A dimensional check uses $[k_B]=\mathrm{J}{\mathrm{K}}^{-1}$, $[h]=\mathrm{J}\mathrm{s}$ and $[c]=\mathrm{m}{\mathrm{s}}^{-1}$. Consequently $[k_B^4/(h^3{c}^2)]=\mathrm{J}{\mathrm{s}}^{-1}{\mathrm{m}}^{-2}{\mathrm{K}}^{-4}=\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-4}$, exactly the units needed in the Stefan–Boltzmann law.

For an equilibrium photon gas, the energy density is $u=a_{\mathrm{rad}}{T}^4$ with radiation constant $a_{\mathrm{rad}}=4\sigma /c$. The radiation pressure is $p_{\mathrm{rad}}=u/3=4\sigma T^4/(3c)$. These describe the radiation filling an equilibrium cavity; $u$ has units $\mathrm{J}{\mathrm{m}}^{-3}$, whereas the outward radiant exitance through a small opening is $cu/4$. The factor $1/4$ comes from averaging propagation directions and their projected area over a radiation field with isotropy, not from reducing the opening's emissivity.

The Stefan–Boltzmann law gives the total hemispherical radiant exitance of an ideal blackbody at absolute temperature $T$:Here $\sigma$ is the Stefan-Boltzmann constant. Total means integrated over all wavelengths; hemispherical means integrated over all outward directions with the projected area weighting. Thus $M_b$ is emitted power per actual emitting area, rather than directional radiance or spectral radiance. Absolute temperature must be in kelvin; raising a Celsius reading to the fourth power gives the wrong result. A uniform surface of area $A$ emits radiant flux $P_{\mathrm{emit}}=A{M}_b$. A real surface instead has $M=ϵ(T)\sigma T^4$, where $ϵ(T)$ is its total hemispherical emissivity. The older symbol $F$ in the formula above denotes this emitted radiant exitance.^[ref]

For an ideal blackbody, doubling absolute temperature multiplies radiant exitance by $16$. With the approximate Stefan-Boltzmann constant $5.670374419\times 10^{-8}\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-4}$, the radiant exitances at $300\mathrm{K}$ and $600\mathrm{K}$ are approximately $459.300$ and $7348.805\mathrm{W}{\mathrm{m}}^{-2}$. This steep response explains why thermal radiation becomes so consequential at high temperature. It also enables heat transfer through a vacuum, where thermal conduction through a gas and convection are absent.

A short derivation integrates Planck's law in its frequency form. For a blackbody, direction-independent spectral radiance is $B_{\nu }(T)=2h{\nu }^3/[c^2(e^{h\nu /(k_{B}T)}-1)]$, where $h$ is the Planck constant, $k_B$ the Boltzmann constant and $c$ the speed of light. The outward hemisphere contributes $\int \cos \theta d\Omega =\pi$, soExpanding $(e^x-1{)}^{-1}={\sum }_{n=1}^{\infty }{e}^{-nx}$, nonnegative terms allow termwise integration. Repeated integration by parts gives ${\int }_0^{\infty }{x}^3{e}^{-nx}dx=6/n^4$; the Riemann zeta function value $\zeta (4)={\pi }^4/90$ gives the remaining integral as ${\pi }^4/15$. Hence $M_b=\sigma T^4$ with the coefficient in Stefan-Boltzmann constant. The angular integral is essential: omitting it confuses radiance with radiant exitance.

The stellar adiabatic exponents describe pressure and temperature responses to an isentropic density change:

A temperature versus density diagram separates classical, degenerate, nonrelativistic and relativistic stellar-matter limits. The electron degeneracy boundary is $T\sim T_F(\rho )$, with slope $2/3$ at low density and $1/3$ at high density. Thermal relativity is controlled by $k_{B}T\sim m_e{c}^2$, while relativity of degenerate electrons is controlled by $p_F\sim m_{e}c$. Photon pressure and thermal electron-positron pairs add further crossovers; all depend on composition and are gradual, not thermodynamic phase boundaries.

Stellar nuclear fusion converts light nuclei into more tightly bound nuclei and supplies most main-sequence stellar luminosity.

Consider an unnamed model substance with molar latent heat $L_m=40.0\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ and saturation pressure $p_1=10.0\mathrm{kPa}$ at $T_1=300\mathrm{K}$. Stipulate the ideal-vapor Clausius-Clapeyron approximation throughout $280\le T\le 340\mathrm{K}$, constant $L_m$, and $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$. These parameters define an illustrative model, not a measured dataset or a fit to water. At $T_2=330\mathrm{K}$,At the reference state the gas molar volume is $R{T}_1/p_1=0.24942{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$, and the local slope is $dp/dT\simeq 0.5346\mathrm{kPa}{\mathrm{K}}^{-1}$. The tangent estimate $p_2\approx 10.0+30(0.5346)=26.04\mathrm{kPa}$ is much lower than the integrated result: a local derivative should not be extrapolated over this whole interval. By contrast, over a small temperature increment, the fractional sensitivity is $d\ln (p/p_1)/dT=L_m/(R{T}_1^2)\simeq 0.05346{\mathrm{K}}^{-1}$.