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For $(u,s),(v,t)$ in the second-order cone, the Cauchy-Schwarz inequality gives $u\cdot v+st\ge -\Vert u\Vert \Vert v\Vert +st\ge 0$. Thus the cone is contained in its dual cone. Conversely, if $s<\Vert u\Vert$ and $u\ne 0$, pairing $(u,s)$ with $(-u/\Vert u\Vert ,1)$ is negative. If $u=0$ and $s<0$, use $(0,1)$. Every point outside the cone therefore lies outside its dual, proving that it is a self-dual cone.

No cyclic heat engine exchanging heat only with two fixed thermal reservoirs at positive absolute temperatures $T_h>T_c$ can exceed the thermal efficiency of an engine whose cycle is a reversible thermodynamic process between them. All such reversible engines have the same thermal efficiency, regardless of their working substance. Let $Q_h>0$ be heat absorbed from the hot thermal reservoir and $Q_c>0$ the magnitude rejected to the cold one. ThenThe work convention is the one in the first law of thermodynamics: $W>0$ is net output by the engine. Use absolute thermodynamic temperatures, such as values in kelvin; ratios of readings in degree Celsius are invalid because their zero is shifted.

For $T_h=600\mathrm{K}$, $T_c=300\mathrm{K}$ and $Q_h=1000\mathrm{J}$, the maximum thermal efficiency is $50\%$. A reversible engine delivers $W=500\mathrm{J}$ and rejects $Q_c=500\mathrm{J}$. The thermal reservoirs have entropy changes $-1000/600$ and $+500/300\mathrm{J}{\mathrm{K}}^{-1}$, which cancel. The engine's own entropy change is zero over its thermodynamic cycle.

A short derivation now becomes transparent: the Second law of thermodynamics gives $-Q_h/T_h+Q_c/T_c\ge 0$, and the first law of thermodynamics gives $W=Q_h-Q_c$. Together they yield the bound and, more precisely,At fixed $Q_h,T_h,T_c$, every extra unit of total entropy production reduces output by $T_c$ units of energy. ^[ref]

Fixed thermal reservoirs are essential. Finite bodies cool and warm, so their initial temperature ratio cannot be used throughout an entire extraction process. As a computed contrast, the bodies with equal constant heat capacity in the entropy example would reach $T_f=\sqrt{600\times 300}=424.26\mathrm{K}$ if a reversible engine extracted work until their temperatures matched. Zero total entropy change gives this geometric mean; the first law of thermodynamics then gives $W_{\max }=C(600+300-2T_f)=5147.19\mathrm{J}$. Direct contact reaches $450\mathrm{K}$ and extracts no work.

Carnot's 1824 book asked whether the useful output of machines driven by heat was fundamentally limited and whether changing the working substance could improve that limit. Its reversible comparisons are the historical setting of Carnot's theorem. Carnot used the then-current caloric picture, treating heat as conserved in its passage through an engine; the modern balance $Q_h-Q_c=W$ should not be read back into that text.^[ref]

Rotate an ellipse with positive semiaxes $a,b$ about an axis a distance $r>a$ from its centre. For the parametrized surface $\sigma (u,v)=((r+a\sin u)\cos v,(r+a\sin u)\sin v,b\cos u)$, put $R=r+a\sin u$ and $D^2=a^2{\cos }^{2}u+b^2{\sin }^{2}u$. An outward unit normal is $N=(b\sin u\cos v,b\sin u\sin v,a\cos u)/D$. Direct differentiation gives the first fundamental form and second fundamental form, with the convention $I{I}_{ij}=⟨{\sigma }_{ij},N⟩$:Reversing the unit normal reverses $II$, but the Gaussian curvature $K=a{b}^2\sin u/(R{D}^4)$ is unchanged.

Differentiate the orthogonality of the unit normal to the two independent coordinate tangent vectors. If the second fundamental form vanishes, each derivative of the normal is orthogonal to both tangent vectors; differentiating its unit length also makes it orthogonal to the normal. Thus both normal derivatives vanish. On a connected parameter domain the normal is constant, and differentiating the position's scalar product with that constant normal proves the surface lies in a plane.

At every point its second fundamental form is a scalar multiple of its induced metric. Equivalently its shape operator is scalar, so all principal normal curvatures agree. For a three-dimensional hypersurface, $K=3\lambda$ and $K_{ab}{K}^{ab}=3{\lambda }^2$. The sign of its intrinsic scalar in a vacuum Lorentzian ambient space then follows from the Gauss equation for a nonnull hypersurface.

For an additive time series $X_t=T_t+S_t+{\epsilon }_t$ with period-$m$ seasonal component and ${\sum }_{j=1}^m{S}_j=0$, a centred average spanning one complete period removes the seasonal component while estimating the slowly varying trend. When $m$ is even, average two adjacent length-$m$ averages to centre the filter on an observation; the endpoints receive half the usual weights. Estimate each seasonal phase by averaging the corresponding detrended residuals across cycles, then subtract the mean of those phase averages to enforce the zero-sum convention. Multiplicative seasonality instead uses ratios and a mean-one convention. Repeated averaging is filter composition and induces serial dependence even from white noise.

Steric sea level change results from changes of seawater mass density at fixed ocean mass. To first order, thermal expansion gives $\Delta {\eta }_{\mathrm{thermal}}=\int \alpha \Delta Tdz$; changes in salinity give a halosteric contribution. The reference column, pressure and equation of state must be specified.

Eustatic sea level change is a change in global sea level associated with changes of ocean water amount or ocean-basin capacity. For contemporary climate budgets the relevant contribution is principally transfer of water between land and ocean, including melting grounded ice and changes in terrestrial water storage. It is distinct from steric sea level change.

With positive freshwater runoff $R$, net sea ice melt $I$ and precipitation $P$, and evaporation loss $E$, the ocean water-mass budget is $\dot{M}=R+I+P-E$. With salt-free ice and fixed total dissolved salt mass, mean salinity satisfies $\dot{S}/S=-\dot{M}/M$. Gross regional ice melt must not be substituted for net global melt without accounting for ice formation elsewhere and the averaging period.

Sea ice anchored to a coast, grounded feature or other fixed support is fast ice. It can be smooth or deformed, young or old. Currents can transport an oil spill under sea ice beneath it even though the ice itself remains stationary.

A melt pond is liquid water collected in a depression on melting ice. Its lower albedo than snow-covered ice can strengthen ice-albedo feedback. Pond drainage and connected pores can redistribute an oil spill under sea ice during warming.

The Odden ice tongue is an intermittent extension of young sea ice into the Greenland Sea. Repeated formation and export of frazil ice and pancake ice can produce brine rejection without a steadily thickening local ice cover, increasing the salinity of the water left behind.

The sea-ice crystal fabric describes grain shapes and crystallographic orientations. Calm-water congelation ice tends to develop vertical columns with predominantly horizontal crystal c-axes, whereas accumulated frazil ice produces a granular texture with much less systematic orientation.

// Target: geophysics.bigb

Nilas is a thin, flexible, relatively continuous initial sea-ice sheet. Further freezing at its underside in quiet water produces congelation ice.

// Target: geophysics.bigb

Snow ice forms when flooded snow or a seawater-snow mixture freezes. Snow loading of a thin floe can depress the ice surface below water level and initiate flooding; wave overwash is another route.

// Target: geophysics.bigb

A sea-ice thickness distribution assigns fractional horizontal area to ice thickness intervals: $g(h)dh$ is the fraction between $h$ and $h+dh$. Include open water as a mass at $h=0$ when normalizing over the entire ocean area. The first moment is ice volume per ocean area. With growth rate $f$ and velocity $v$, its conservative evolution is ${\partial }_{t}g+\nabla \cdot (vg)+{\partial }_h(fg)=\psi$, with mechanical redistribution of sea ice represented by $\psi$.

// Target: geophysics.bigb

A sea-ice lead is a fracture or elongated opening in a sea-ice cover, containing open water or young thin ice. Divergence and shear create leads, while freezing and convergence close them.

// Target: geophysics.bigb

The horizontal sea-ice drift momentum balance per unit area includes air and water drag forces, the Coriolis force, depth-integrated internal ice stress and the sea-surface pressure gradient. Here $m$ is ice-plus-snow mass per area and $\Sigma$ is a depth-integrated stress tensor.

// Target: geophysics.bigb

The tortoise coordinate is $r_{\ast }=r+(r_h/2)\log |(r-r_h)/(r+r_h)|$. Set $u=t-r_{\ast }$ and $v=t+r_{\ast }$, then $U=-e^{-u/r_h}$ and $V=e^{v/r_h}$ in the right exterior. The extended radial metric is $-r_h^2(r+r_h{)}^2{e}^{-2r/r_h}dUdV/r^2$, regular at $r=r_h$. The horizons are $UV=0$, while $r=0$ gives $UV=1$ and is a spacelike curvature singularity. Compactification by $\arctan U,\arctan V$ displays two exteriors, a future black hole and a past white hole.

A freely falling source emits optical pulses at proper times $\delta +nP$, for integers $n\ge 0$. If its horizon-crossing proper time is finite, only pulses beginning before $s_H$ can escape. Their number is the displayed ceiling, for $0\le \delta <P<s_H$. Infinite Schwarzschild time at crossing does not give infinitely many emitted pulses. The pulse phase $\delta$ fixes both the count and the redshift of the last pulse. If a finite pulse straddles the horizon, its exterior portion has an arbitrarily late, increasingly redshifted received tail.