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An isothermal compression reduces the volume of a thermodynamic system at fixed temperature. For a fixed-composition ideal gas compressed reversibly from $V_i$ to $V_f<V_i$, isothermal ideal-gas work givesThe surroundings supply positive work $W_{\mathrm{on}}$, and the gas rejects the same amount of heat to the heat bath. Compressing $1\mathrm{mol}$ at $300\mathrm{K}$ from $20$ to $10\mathrm{L}$ requires approximately $1.729\mathrm{kJ}$ of work input and rejects $1.729\mathrm{kJ}$ of heat. If that heat transfer cannot keep up, the temperature rises: the process departs from isothermal compression toward adiabatic compression. The finite-rate isothermal piston model explains the required small temperature excess.

The isoplanatic angle is the characteristic angular separation over which a single-direction atmospheric correction remains useful. It depends on the altitude distribution of atmospheric turbulence as well as on wavelength.

For two groups carrying angular momenta $L_a,L_b$ with mutual inclination $\Delta I$, resolving total angular momentum parallel and perpendicular to the second group's normal gives its orbital inclination relative to the invariable plane:For a positive denominator, $\tan I_b=\sin \Delta I/(L_b/L_a+\cos \Delta I)$. The larger-angular-momentum group lies closer to the invariable plane.

A compactness proof of the interior Schauder estimate can magnify a pair of points where the Hölder seminorm of the Hessian matrix is nearly attained. If the proposed estimate fails, normalize that seminorm to one while lower norms and forcing tend to zero. The separation $r_j$ of the chosen pair tends to zero. Subtract the quadratic Taylor polynomial and divide by $r_j^{2+\alpha }$ after rescaling space by $r_j$.

The resulting functions have uniformly bounded Hessian Hölder seminorms on expanding domains and vanishing Taylor data at zero. Compact embedding of Hölder spaces yields an entire harmonic function whose second derivatives have finite global $\alpha$-Hölder seminorms, with $0<\alpha <1$. The polynomial-growth Liouville theorem for harmonic functions makes those derivatives constant, contradicting the normalized nonzero oscillation. The Simon absorption lemma removes the residual small multiple of the larger-ball seminorm.

The bounded-forcing norm alone cannot control the Hölder seminorm of the Hessian matrix. On a unit Euclidean ball take $u_m=m^{-2}\sin (m{x}_1)$ and $f_m=\Delta u_m=-\sin (m{x}_1)$. The derivative supremum norm $|u_m{|}_2$ and $\Vert f_m{\Vert }_{\infty }$ stay bounded, but the Hessian matrix difference between zero and $\pi /(2m)e_1$ gives $[D^2{u}_m{]}_{\alpha }\ge (2m/\pi {)}^{\alpha }$ on $B_{1/2}$.

For $-{\epsilon }^2{y}^{\prime \prime }+2(x-x_{\ast })y^{\prime }+\cdots =f$ on an interval straddling $x_{\ast }$, the bounded leading interior solution cannot connect unequal constants. A common bulk value is instead selected by an inner solvability condition, and width-${\epsilon }^2$ endpoint layers enforce the two boundary values. At endpoints $x=0,2$ with $x_{\ast }=1$, their leading decaying factors are $e^{-2x/{\epsilon }^2}$ and $e^{-2(2-x)/{\epsilon }^2}$. The interior layer still corrects derivative mismatches at smaller amplitude.

The Fatou lemma makes every time value of a nonnegative discrete-time local martingale integrable, assuming the usual integrable initial value. The integrable discrete-time result then makes it a true martingale.

The integer multiplication map on the circle is an ergodic transformation for Lebesgue measure. In the Fourier basis, an invariant function has coefficients $c_j=0$ when $K$ does not divide $j$, and $c_j=c_{j/K}$ otherwise. Every nonzero index eventually reduces to the first case, so only the constant coefficient remains.

If the horizontal magnetic field perturbations vanish at $z=\pm H$, their allowed vertical wavenumbers are $k_m=m\pi /(2H)$, $m=1,2,\dots$. The lowest wavenumber is unstable exactly whenIf that magnetic mode is stable, all higher magnetic modes are stable as well. A weak field admits longer unstable wavelengths that fit inside the slab; sufficiently strong magnetic tension stabilizes every allowed wavelength.

For $x\ge 1$, the inverse hyperbolic cosine on its nonnegative real branch is $\mathrm{arcosh}x=\log (x+\sqrt{x^2-1})$. It is the real cosine member of the inverse hyperbolic functions.

A giant planet can also remain inflated when power is deposited sufficiently deep to alter its internal energy budget. Candidate sources include tidal heating, Joule heating by wind-driven currents, and dissipation of circulation-driven mechanical energy. The depth and efficiency of deposition matter, as discussed in models of hot-Jupiter heating.

Delayed cooling preserves a young giant planet's large radius by reducing the escape of internal energy. Enhanced atmospheric opacity, an irradiation-maintained outer radiative blanket, and composition gradients that inhibit convection can slow Kelvin-Helmholtz contraction. These mechanisms retain existing heat rather than supplying a new deep heat source.

In the steady state of the horizontally invariant magnetized shearing-sheet equations, let $B_x(\pm H)=\pm B_{+}$ and $B_y(\pm H)=0$. For $q>0$ and $B_z\ne 0$, the symmetric equilibrium hasThis follows by combining azimuthal induction with radial magnetic tension balance to obtain $B_x^{\prime \prime }+K^2{B}_x=0$. Nonzero $B_{+}$ requires $\sin (KH)\ne 0$; even homogeneous additions at $\cos (KH)=0$ are excluded by imposing midplane symmetry.

The hook length of cell $(i,j)$ in the Young diagram of a partition of an integer $\lambda$ is the number of cells in its Young-diagram hook:Here ${\lambda }^{\prime }$ is the conjugate partition. There is one hook length for every cell, so the multiset of hook lengths has cardinality $|\lambda |$.

For $\lambda =(3,2)$ and ${\lambda }^{\prime }=(2,2,1)$, compute $h_{1,1}=(3-1)+(2-1)+1=4$ and $h_{1,2}=(3-2)+(2-1)+1=3$. Continuing cell by cell gives the hook graph of a partitionor the row array $[[4,3,1],[2,1]]$. Its hook product is $4\cdot 3\cdot 1\cdot 2\cdot 1=24$. A cell has hook length one exactly when it is a Removable node of a Young diagram: it has no cell to its right or below. Here these are $(1,3)$ and $(2,2)$. Transposing the Young diagram gives the array $[[4,2],[3,1],[1]]$ with the same hook product; this explains geometrically why conjugate shapes have equal counts in the hook-length formula.

For a code with product state inputs to $n$ uses of a memoryless quantum channel, Subadditivity of Von Neumann entropy bounds its output Holevo quantity by $n{\chi }^{\ast }(\Lambda )$. The Holevo bound and Fano's inequality then give, for asymptotic rate $R$, average and hence maximum error at least $1-{\chi }^{\ast }(\Lambda )/R$ in the limit inferior. In particular, no code above that rate can have vanishing maximum error, even with a collective decoder measurement.

A channel's Holevo capacity is additive with another channel when their joint capacity equals the sum of the individual capacities. If this holds with every other channel, the channel's tensor powers have capacity $n{\chi }^{\ast }(\Lambda )$ and regularization does not increase its unassisted classical capacity of a quantum channel. Additivity is a special property, not a general rule for quantum channels.

A finite-dimensional irreducible representation of a complex semisimple Lie algebra is minuscule if its weights form one Weyl group orbit. Every weight multiplicity is then one, because multiplicities are invariant under the Weyl group and the highest weight has multiplicity one.

If a dominant integral weight $\lambda$ lies in the root lattice, its finite-dimensional irreducible highest-weight representation contains zero as a weight. First $\lambda$ is a nonnegative integral combination of simple roots: a negative part would have negative inner product with $\lambda$, contradicting dominance. For any nonzero weight $\mu =\sum m_i{\alpha }_i$ with $m_i\ge 0$, the identity $(\mu ,\mu )=\sum m_i(\mu ,{\alpha }_i)>0$ supplies an index with $m_i>0$ and $⟨\mu ,{\alpha }_i^{\vee }⟩>0$. By Injectivity of sl2 lowering above weight zero, applying the corresponding lowering operator produces a nonzero vector of weight $\mu -{\alpha }_i$. Repetition reaches zero.

For $m_H>2m_Z$, the Higgs boson coupling to Z bosons gives $\mathcal{M}=(2m_Z^2/v){ϵ}_1^{\ast }\cdot {ϵ}_2^{\ast }$. The massive vector polarization sum gives $\sum |\mathcal{M}{|}^2=(m_H^4/v^2)(1-4x+12x^2)$, with $x=m_Z^2/m_H^2$. The two-body decay phase space and identical final-state symmetry factor $1/2$ giveThe contraction uses $k_1\cdot k_2=(m_H^2-2m_Z^2)/2$. The on-shell rate vanishes at threshold; below threshold an off-shell calculation is required.

For a finite-index subgroup $L\le {\mathbb{Z}}^2$, its projection to the first coordinate is $a\mathbb{Z}$ and its intersection with the second axis is $\{0\}\times d\mathbb{Z}$, with $a,d>0$. Choose $(a,b)\in L$, reducing $b$ modulo $d$. Together with $(0,d)$ it is a basis of this row lattice: subtracting a multiple of $(a,b)$ from any vector leaves a vector on the second axis. These parameters are unique, and reduction of the two coordinates shows $[{\mathbb{Z}}^2:L]=ad$. Consequently an integral matrix of positive determinant $n$ has a unique representative of this form under left multiplication by $S{L}_2(\mathbb{Z})$, with $ad=n$. This proves the lattice facts underlying determinant-n matrix representatives for Hecke operators.