Let $B_1,B_2$ be real square invertible matrices whose columns are lattice bases. They generate the same Euclidean lattice precisely when^[ref]

If $B_2=B_{1}U$, multiplication by $U$ merely permutes the integer coordinate vectors, so the two generated sets agree. Conversely, equality of the generated sets means every column of either basis is an integer linear combination of the other. Hence $B_2=B_{1}U$ and $B_1=B_{2}V$ for integer matrices $U,V$. Substitution followed by multiplication by $B_1^{-1}$ gives $UV=I$, so the integer inverse criterion for a unimodular matrix applies.

Equal real linear spans are a weaker condition: any two real square invertible matrices span the same ambient vector space. Even equal covolumes do not force the same Euclidean lattice. The requirement that the change matrix and its inverse matrix have integer entries provides the missing arithmetic information.

For a square integer matrix $U$ with nonzero determinant, the inverse matrix has integer entries exactly when $\det U=\pm 1$. The adjugate matrix formula proves one direction:Every entry of $\mathrm{adj}(U)$ is an integer cofactor, so division by $1$ or $-1$ leaves integer entries. Conversely, if both $U$ and $U^{-1}$ are integer matrices, their determinants are integers with product one. The only possibilities are that both are one or both are minus one. This explains why a general integer matrix with nonzero determinant need not have an integer inverse matrix.

Multiple edges are distinct edges with the same pair of endpoint vertices in an undirected graph. A multigraph permits them; a simple graph does not. In an orientation of a graph of a multigraph, each such edge receives its own direction, so two of them can supply opposite directed edges between the same vertices.

For a specified optical transition, the on-resonance saturation parameter is the light intensity divided by the saturation intensity. The independent-beam Doppler cooling force model assumes weak total excitation and defines $s$ separately for each beam.

For a convergent series with sum $S$, the error from keeping only the first $N$ terms is the remaining sum $R_N={\sum }_{n>N}{a}_n$. This truncation error exists even with exact arithmetic. The zeta series truncation error gives explicit bounds for positive reciprocal-power terms.

A sequence $(x_n)$ in a topological space converges to $x$ if every open neighbourhood of $x$ contains all sufficiently late terms:For a metric space, this is the usual limit of a sequence condition expressed through open balls. For a Hausdorff space, distinct proposed limits have disjoint open neighbourhoods, so a sequence cannot eventually stay in both and the limit is unique. The sequence convergence in the cofinite topology demonstrates what can change when that separation property is absent.^[ref]

The category of topological manifolds has topological manifolds without boundary as its objects of a category and all continuous maps between them as its categorical morphisms. Allowing every dimension of a manifold lets maps connect spaces of different dimensions. Identity functions are continuous, and function composition preserves continuity, so the category laws are inherited from functions. Its isomorphisms are exactly homeomorphisms.

In coordinate charts $(U,\phi )$ on $M$ and $(V,\psi )$ on $N$, a continuous map $f:M\to N$ is represented locally by $\psi \circ f\circ {\phi }^{-1}$ on $\phi (U\cap f^{-1}(V))$. Changing coordinate charts changes this expression through topological transition maps, while $f$ itself stays fixed. This viewpoint separates the underlying continuous map from the coordinates used to describe it; restricting to smooth manifolds and smooth maps gives the corresponding smooth category.

For $n\ge 1$, a topological $n$-manifold with boundary is a Hausdorff second-countable space locally homeomorphic to relatively open sets in the half-space ${\mathbb{H}}^n$. The openness here is in the subspace topology of ${\mathbb{H}}^n$, not necessarily in ${\mathbb{R}}^n$. Its intrinsic boundary $\partial M$ consists of points sent to $x_n=0$; the remaining points have ordinary coordinate charts with images open in Euclidean space. The intrinsic boundary is independent of the chosen chart.^[ref]

The closed interval $[0,1]$ illustrates the distinction: its interior points have open interval charts, while $0$ and $1$ have charts modeled on $[0,\epsilon )\subseteq {\mathbb{H}}^1$, using $x$ and $1-x$. Its intrinsic boundary is $\{0,1\}$. The circle has empty intrinsic boundary even though its boundary of a set as a subset of the Euclidean plane is the whole circle. A topological manifold without boundary is therefore a special case with $\partial M=\varnothing$, and a smooth atlas on the half-space charts gives a manifold with boundary in the smooth sense.

Take the circle $S^1=\{(x,y)\in {\mathbb{R}}^2:x^2+y^2=1\}$ with its subspace topology. Write $W=(-1,0)$ for its west point and $E=(1,0)$ for its east point. Two coordinate charts obtained by stereographic projection areBoth take values in the entire real line. Geometrically, join $W$ to $(x,y)$ and intersect the resulting line with $x=0$: the height is $t$. Projecting from $E$ instead gives height $u$. Each projection omits its own starting point, but the other coordinate chart includes that point. Hence $U\cup V=S^1$.^[ref]

To verify the homeomorphisms, solve the displayed equations for $(x,y)$. The inverse functions areEach formula lies on the circle, avoids its excluded point, and is continuous for every real argument. Substitution into the projection formulas returns $t$ or $u$; substitution in the other direction returns $(x,y)$. Thus these are continuous bijections with continuous inverse functions, as required for coordinate charts. Since the Euclidean plane is Hausdorff and second countable, the subset $S^1$ with its subspace topology is too. This checks all three conditions for a topological manifold, rather than just drawing a smooth-looking curve.

On the overlap $U\cap V$, both $x=\pm 1$ are excluded, so $y\ne 0$. Using $y^2=1-x^2$, the topological transition map isFor example, $(x,y)=(3/5,4/5)$ has $t=1/2$ and $u=2$. The two connected components of the overlap, the upper and lower arcs, correspond to the positive and negative parts of the punctured real line. The values $t=0$ and $u=0$ represent $E$ and $W$, respectively, and belong to only one chart each. The reciprocal coordinate change and its inverse function are smooth functions away from zero, so this topological atlas also supplies a smooth atlas.

Why can neither chart be enlarged to a single coordinate chart on the whole circle? The circle is a closed set and a bounded set in the Euclidean plane, hence compact by the Heine-Borel theorem. A global coordinate chart would give a homeomorphism to a nonempty open set $O\subseteq \mathbb{R}$. Its image would be compact and therefore closed in the real line. By connectedness of the real line, this forces $O=\mathbb{R}$, which is not compact. This is the one-dimensional instance of no compact manifold has a single Euclidean chart. Locally the circle is indistinguishable from the real line; globally compactness obstructs a single chart.

A topological atlas is a family $\mathcal{A}=\{(U_i,{\phi }_i)\}$ of coordinate charts whose domains form an open cover of a topological manifold. Each point must belong to at least one chart domain; no individual domain needs to cover the entire topological manifold. On overlaps, topological transition maps describe the same points in two coordinate systems.

A thermodynamic process changes a thermodynamic system between states. A quasistatic process supplies equilibrium intermediate states and therefore a path on equilibrium coordinates. An irreversible thermodynamic process can have equilibrium endpoints while lacking a uniform intermediate temperature or pressure. Endpoint values still determine changes in a state function; computing heat and work requires the actual interaction history.

An equilibrium thermodynamic state specifies sufficient independent macroscopic properties, together with material identity and constraints, to determine the other equilibrium properties. For a uniform fixed-composition ideal gas, its amount of substance, temperature and volume determine its pressure through the ideal gas law. These properties specify a state, without specifying a thermodynamic process that prepared it.^[ref]

The International Temperature Scale of 1990 is a practical scale for reproducible temperature measurements. Its assigned reference points and interpolation procedures approximate thermodynamic temperature. A calibrated standard platinum resistance thermometer transfers this scale over specified intervals; it does not define the modern kelvin.^[ref]

Polarization-gradient cooling uses spatial variation in electromagnetic polarization together with optical pumping between internal quantum states to cool moving chemical atoms.

Jean Dalibard and Claude Cohen-Tannoudji distinguish cooling by changing elliptical polarization from cooling by rotating linear polarization. The hill picture below concerns the former.^[ref]

The UCSB demonstration uses an aluminum cylinder turned by a crank, with a wrapped rope supporting a hanging mass. An embedded electrical thermometer connects through sliding contacts to an electrical resistance meter; a supplied electrical resistance–temperature table or fitted equation converts the readout. A counter records turns. Record the initial temperature, crank steadily, then stop and convert the minimum recorded electrical resistance to the final temperature.^[ref]

The proposed control is to start below room temperature and finish comparably above it, approximately balancing incoming and outgoing environmental heat. Neglected accessory heat capacities and rough temperature estimates limit accuracy. The mechanical input and thermal response test an energy balance; warming alone does not establish quantitative equivalence.^[ref]

This is a modern companion to Joule's paddle-wheel experiment, with a different apparatus. Choose the cylinder as the closed system. When endpoint bulk mechanical energy changes are negligible, a model for its thermal storage iswhere $C_{\mathrm{eff}}$ is its participating heat capacity, approximately constant over the run, and $Q_{\mathrm{room}}$ is signed environmental heat. The cylinder is not an isolated system, since mechanical work enters. The temperature rise estimates a change in internal energy, rather than an amount of externally supplied heat. This boundary distinction connects a measured thermal response to the same stored quantity used in two reversible ideal-gas paths between the same states.

Take a closed system consisting of one mole of a classical monatomic gas, idealized as an ideal gas with constant heat capacity at constant volume $C_V=3nR/2$. Stipulate amount of substance $n=1\mathrm{mol}$ and molar gas constant $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ for this calculation. Set initial temperature $T_0=300\mathrm{K}$ and volume $V_0=0.010{\mathrm{m}}^3$, and define $E_0=nR{T}_0=2.4942\mathrm{kJ}$ and $p_0=E_0/V_0=249.42\mathrm{kPa}$. The specified equilibrium endpoints areThus $B$ has volume $0.030{\mathrm{m}}^3$, temperature $600\mathrm{K}$ and pressure $166.28\mathrm{kPa}$. The fixed material inventory and ideal gas law make these complete endpoint specifications for this model. All numbers below are calculated values, rather than measurements of a particular gas.

Use a frictionless piston, ignore bulk mechanical energy changes, and allow only pressure-volume work. Each step is the limit of a reversible thermodynamic process: adjust loads by infinitesimal pressure differences, and supply heat through infinitesimal temperature differences. Heating with changing temperature requires a succession of matching thermal reservoirs; contact with a single hotter heat bath would give an irreversible finite difference. On a fixed-volume leg the piston is locked. These assumptions justify using the actual equilibrium path's pressure inAn endpoint chord is not the path; substituting a guessed straight line would compute a different thermodynamic process.^[ref]

The upper route heats at fixed volume from $A$ to $C=(V_0,2T_0,2p_0)$, then undergoes isothermal expansion at $2T_0$ to $B$. The lower route undergoes isothermal expansion at $T_0$ from $A$ to $D=(3V_0,T_0,p_0/3)$, then heats at fixed volume to $B$. The tuples use the same $(V,T,p)$ ordering as the endpoints. Vertical legs do zero expansion work, and the two curved legs have different pressures at every shared intermediate volume.

The internal energy of an ideal gas gives the same endpoint difference for both routes:On either fixed-volume heating leg, $W_{\mathrm{by}}=0$ and $Q=3E_0/2$. On each isothermal expansion, $\Delta U=0$, so the first law of thermodynamics gives $Q=W_{\mathrm{by}}$. The ideal gas law and the integral of $1/V$ then giveHere $\ln$ is the natural logarithm. Rounded to $0.001\mathrm{kJ}$, the upper route has $(Q,W_{\mathrm{by}},\Delta U)=(9.222,5.480,3.741)\mathrm{kJ}$ and the lower route has $(6.481,2.740,3.741)\mathrm{kJ}$. Subtracting separately rounded entries can differ by $0.001\mathrm{kJ}$; the unrounded balances are exact in the stipulated model.

The shaded area between the curves explains the entire difference:Following the upper route forward and the lower route backward makes a thermodynamic cycle with this positive net work and matching net heat. Its internal energy change is zero. The two open routes themselves are reversible, so their different work outputs should not be mistaken for a loss caused by entropy production: they receive different heat transfers at different temperatures.

Two other state functions provide independent endpoint checks. With enthalpy $H=5nRT/2$ in this model,To obtain the entropy change along either route, integrate $C_{V}dT/T$ on its heating leg, giving $C_V\ln 2$, and $nRdV/V$ on its isothermal expansion, giving $nR\ln 3$. Reordering the legs changes their heat totals but leaves the sum of their entropy changes unchanged. In particular, neither route's total $Q$ divided by its final temperature equals $\Delta S$, and neither total $Q$ equals $\Delta H$: both routes include an isothermal expansion, rather than a single fixed-pressure heating step.