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For inviscid barotropic flow, conservation gives ${\rho }_t+\nabla \cdot (\rho u)=0$ and $\rho (u_t+u\cdot \nabla u)=-\nabla p$. Linearizing about uniform rest ${\rho }_0,p_0$ giveswith the derivative adiabatic for ordinary sound.

Taking curl gives ${\partial }_t(\nabla \times u)=0$. Therefore the acoustic velocity is irrotational if the initial vorticity is zero. Uniform rest as the background does not alone forbid a superposed stationary vortical perturbation: for example $u=\epsilon (\sin y,0,0)$, ${\rho }^{\prime }=p^{\prime }=0$ solves the linearized equations but has nonzero curl. Thus the printed irrotational conclusion requires the usual restriction to the longitudinal acoustic mode. In that mode set $u=\nabla \varphi$ and absorb a time-only gauge into $\varphi$, so $p^{\prime }=-{\rho }_0{\varphi }_t$. Continuity givesThis separates the sound wave from the stationary vorticity mode in linear acoustics.

Define the quadratic acoustic energy density and acoustic energy flux byMultiplying momentum by $u$ and the pressure form of continuity by $p^{\prime }/({\rho }_0{c}_0^2)$ gives

For any progressive plane wave, not necessarily sinusoidal, let $\varphi =F(\xi )$ with $\xi =\hat{k}\cdot x-c_{0}t$. Then $u=F^{\prime }(\xi )\hat{k}$ and $p^{\prime }={\rho }_0{c}_0{F}^{\prime }(\xi )$. Its kinetic and compressional energy densities both equal ${\rho }_0(F^{\prime }{)}^2/2$, soThe energy is transported at the sound speed in the propagation direction. A superposition of oppositely traveling waves can form a standing wave and need not have this pointwise equality; the statement concerns a single progressive wave.

For a spherical outgoing wave, $\varphi =F(r-c_{0}t)/r$. Its radial velocity and pressure areThereforeThey are generally unequal because of the additional near-field velocity term; their difference is ${\rho }_0[F^2/r^4-2F{F}^{\prime }/r^3]/2$. They agree only to leading far-field order, or for special profile values. This is the near-field energy imbalance in a spherical acoustic wave.

For an incompressible Newtonian fluid the viscous stress is ${\tau }_{ij}=2\mu e_{ij}$, where $e_{ij}=(u_{i,j}+u_{j,i})/2$. Its work converted to heat per unit volume is ${\tau }_{ij}{u}_{i,j}$. The antisymmetric part of the velocity gradient has zero contraction with symmetric $e_{ij}$, soFor potential flow, $u_i={\varphi }_{,i}$ and mixed derivatives commute, giving $e_{ij}={\varphi }_{,ij}$ and $\Phi =2\mu {\sum }_{i,j}{\varphi }_{,ij}^2$.

Use the given linear wave potential with $z$ increasing downward and $\theta =kx-\omega t$. The nonzero strain components areTheir squared contraction is $2{\omega }^2{a}^2{k}^2{e}^{-2kz}$, independent of the phase. Thus the time-averaged dissipation density isDepth integration gives dissipation $2\mu {\omega }^2{a}^{2}k$ per unit horizontal area, or $4\pi \mu {\omega }^2{a}^2$ per wavelength and unit transverse width.

The velocity components are $u_x=-\omega a{e}^{-kz}\cos \theta$ and $u_z=\omega a{e}^{-kz}\sin \theta$. Hence the depth-integrated mean kinetic energy per wavelength isEqual mean potential and kinetic energies give total energy $\mathcal{E}=\pi \rho {\omega }^2{a}^2/k^2$. Dividing the dissipation by that energy yieldsThe amplitude decays at half this exponent. The small-viscosity and deep-water assumptions justify evaluating the leading dissipation in the inviscid wave field, extending the depth to infinity and neglecting higher-order amplitude corrections.

Birkhoff theorem states that every spherically symmetric vacuum solution of Einstein's equations with zero cosmological constant is locally Schwarzschild, with a constant mass parameter. In particular its exterior vacuum geometry is locally static even if the spherical material source moves. Assuming no exterior matter or outgoing matter radiation, the region outside the collapsing star is vacuum with its conserved total mass $M$, so the given Schwarzschild metric is appropriate there. Spherical gravitational radiation does not provide an extra time-dependent vacuum degree of freedom. Matter emission would invalidate the vacuum premise.

Use units $G=c=1$ and signature $(-,+,+,+)$ as in the displayed metric. For a radial timelike geodesic, stationarity gives conserved energy per unit mass $\mathcal{E}=F\dot{t}$. Proper-time normalization giveson the ingoing branch. At the horizon $F\to 0$, so $\dot{r}\to -\mathcal{E}$, with $\mathcal{E}>0$ for an infaller from the exterior. Henceis finite near the horizon. If release is from rest at a finite radius, the integrable square-root behavior at the release point also gives finite elapsed proper time. Thus the observer crosses the horizon in finite proper time; the Schwarzschild coordinate singularity is not a physical barrier.

By contrastso $t\sim -2M\log (r-2M)+\text{constant}$ diverges. Schwarzschild time is the limiting proper time of stationary observers at infinity. For actual signals, an outward radial null ray adds a propagation delay involving $r_{\ast }=r+2M\log (r/(2M)-1)$; signals emitted ever closer to the horizon reach a distant observer ever later and increasingly redshifted. No signal emitted at or inside the event horizon reaches that observer. Thus “seeing crossing only after infinite time” means an asymptotic observational limit, not an eventual finite-time sighting.

There is a sign inconsistency in the printed starting energy: its kinetic term uses $\nabla +i{q}_{s}A/\hslash$, while the requested field equation, current and later zero-covariant-derivative condition use $\nabla -i{q}_{s}A/\hslash$. Both conventions can be derived explicitly. Set $D_{\eta }=\nabla -i\eta q_{s}A/\hslash$ for $\eta =\pm 1$; the intended later equations take $\eta =1$, whereas the literal printed energy takes $\eta =-1$.

The magnetic term is $B^2/(2{\mu }_0)$, because $A_{k,j}(A_{k,j}-A_{j,k})=|\nabla \times A{|}^2$. Vary ${\psi }^{\ast }$ independently with compactly supported variations. Integration by parts in the kinetic term givesIn particular $\eta =1$ gives the requested equation. Its conjugate follows from variation in $\psi$.

For variation in $A$, $\delta D_{\eta }\psi =-i\eta q_s\delta A\psi /\hslash$. The kinetic variation is $-j_{\eta }\cdot \delta A$, whereThe magnetic variation is $(\nabla \times B)\cdot \delta A/{\mu }_0$ after integration by parts. ThereforeFor $\eta =1$ this current equals the two displayed forms in the question. For the literal plus-sign energy, $\eta =-1$, the phase-gradient current term has the opposite sign and the field equation contains $D_{-1}^2$. Thus the requested set follows after correcting that one energy sign, not from silently varying inconsistent expressions. This is the gauge-coupling sign in Ginzburg-Landau superconductivity.

Now use the intended convention and write $\psi =\sqrt{n_s}{e}^{i\theta }$ with $n_s>0$. Dividing $D_1\psi =0$ by $\psi$ separates real and imaginary parts:Thus $n_s$ is constant on each connected region. The remaining field equation requires $\alpha +\beta n_s=0$, so for $T<T_c$,The current is $j=(q_s{n}_s/m_s)(\hslash \nabla \theta -q_{s}A)=0$. Taking the curl locally gives $B=0$, hence also solves the Maxwell equation.

For a closed curve on which the order parameter is nonzero and $D_1\psi =0$, Stokes' theorem givesSingle-valuedness of $\psi$ permits phase change $2\pi N$, so the flux quantum magnitude is $h/|q_s|$ (for Cooper pairs, $h/(2|e|)$). There is an important geometric qualification: if the entire spanning surface lies in the smooth zero-field superconducting region assumed above, its flux and the total boundary winding are zero. Nonzero quantized flux occurs through a hole or vortex core where those assumptions do not hold over the spanning surface. More generally, without $j=0$ along the curve, the exact quantized quantity is the fluxoidThus fluxoid and flux quantization in a superconductor are related but are not interchangeable without the current condition.

The classical one-particle partition function, with phase-space cell $h^3$, isFor the homogeneous trap, rescale $r=V^{1/3}(kT{)}^{1/(2n)}u$. The spatial measure contributes $V(kT{)}^{3/(2n)}$, soHere $I_n=(2m\pi /h^2{)}^{3/2}{\int }_0^{\infty }4\pi u^2{e}^{-u^{2n}}du$, as specified; the last integral is $2\pi \Gamma (3/(2n))/n$.

Using the printed convention $Z_N=Z_1^N$, the free energy isThis convention treats the particles as labeled. For indistinguishable classical particles, $Z_N=Z_1^N/N!$ adds $kT\log N!$ to $F$. It changes the entropy by $-k\log N!$ but leaves the following force, energy and heat capacities unchanged. This normalization qualification is required to distinguish the printed expression from the usual Gibbs-corrected gas free energy.

The generalized force conjugate to the external parameter $V$ is $P=-(\partial F/\partial V{)}_T$, henceIt is the same equation of state as a hard-wall ideal gas, although $V$ now dilates a soft confining potential rather than being the occupied geometric volume.

Writing $L=\log V+a\log (kT)+\log I_n$, thermodynamic differentiation givesThe mean kinetic energy is $3NkT/2$ and the trap potential contributes $3NkT/(2n)$. At fixed $P$, $V=NkT/P$, so ${\partial }_{T}L{|}_P=(a+1)/T$. ConsequentlyThe same result follows from enthalpy $E+PV=(a+1)NkT$. These identities describe the classical ideal gas in a homogeneous soft trap.

In the nondegenerate regime, $\mu$ lies well inside the gap, with $\mu \gg kT$ and $E_g-\mu \gg kT$. Then conduction occupations are approximately $e^{(\mu -E)/(kT)}$, while the hole probability in a valence state is approximately $e^{(E-\mu )/(kT)}$. Using the near-edge density of states and the supplied gamma integral givesThereforeThis assumes the edge approximations are valid over the thermally populated ranges. For an intrinsic semiconductor $n=p$, so each carrier density is proportional to $T^{3/2}{e}^{-E_g/(2kT)}$. Both mobile conduction electrons and holes contribute to conductivity, $\sigma =|e|(n{b}_e+p{b}_h)$ with mobilities $b_e,b_h$. The exponentially varying carrier densities explain strong temperature dependence even when the mobilities vary only algebraically.

Donors supply electrons and acceptors remove electrons to create holes. With complete ionization and negligible minority carriers, the majority densities are $n\simeq N_d$ on the n side and $p\simeq N_a$ on the p side. The corresponding isolated chemical potentials areCarrier diffusion at contact leaves positively charged donors and negatively charged acceptors in a depletion layer. At equilibrium the electrochemical potential is constant; the electron energy shift is $-|e|\varphi$. Hence the n-minus-p electric potential isAll $N$ quantities must have a consistent normalization: if $N_c,N_v$ are state densities per volume, $N_d,N_a$ are dopant densities rather than bare atom counts. The formula neglects incomplete ionization, degeneracy and changes in band structure.

At a common temperature the two equilibrium junction/contact effects do not drive a perpetual current. At different junction temperatures, electrons and holes transport entropy differently, giving different Seebeck coefficients on the two sides. The loop has a thermoelectric electromotive force, oriented as ${\int }_{T_1}^{T_2}(S_n-S_p)dT$, and therefore generally carries current when closed. This is the Seebeck effect: energy is supplied by heat flow from hot to cold. The temperature dependence of the junction barrier is useful intuition, but an exact terminal voltage is not obtained simply by subtracting two equilibrium built-in potentials; it also depends on bulk transport and contact conditions. The given equilibrium density-of-states data do not determine the numerical thermopowers.

Define the interaction picture byDifferentiating and using the Schrödinger equation cancels the $H_0$ terms, leavingIts integral equation, iterated once from the initial $|a⟩$, gives $|{\psi }_I(t)⟩=|a⟩-(i\lambda /\hslash ){\int }_0^t{V}_I(t^{\prime })|a⟩d{t}^{\prime }+O({\lambda }^2)$.

Since $a,b$ are orthogonal energy eigenstates with distinct energies, the zeroth-order transition amplitude is zero, whileReturning to the Schrodinger picture changes this component only by a unit-modulus phase. Squaring the first-order amplitude therefore gives

For exponential switching, put $\Delta =E_b-E_a$. The integral isBecause $\mu >0$, its exponential vanishes at late time, yieldingThis is the leading weak-perturbation result, not an exact probability formula at arbitrary coupling.

Assume the change of variable is locally invertible with ${\xi }^{\prime }>0$; the Liouville-Green choice below requires $q>0$ away from turning points. Write $r=dx/d\xi$ and $w=\sqrt{r}W$. With dots denoting $\xi$ derivatives,The first-derivative terms cancel. Substitution into $w_{xx}=qw$ gives

For ${\xi }^{\prime }=\sqrt{q}$, we have $r=q^{-1/2}$ and $r^{2}q=1$. Expressing the derivative correction in $x$ yieldsIf this correction is negligible over the relevant scale, $W$ is approximately $e^{\pm \xi }$, so the Liouville-Green exponential ansatz isZeros of $q$ are turning points at which this coordinate and approximation need separate treatment.

For the stated Whittaker equation put $\kappa =s(s-1)$. At large positive $x$, $q=1/4+\kappa /x^2>0$, $q^{-1/4}\to \sqrt{2}$, andAfter normalizing the arbitrary leading constants, this suggests the two exponentials times inverse-power series given in the question.

To derive the coefficients, write $w_A=e^{x/2}F(x)$ with $F={\sum }_{n\ge 0}{a}_n{x}^{-n}$ and $a_0=1$. The equation becomes $F^{\prime \prime }+F^{\prime }=\kappa x^{-2}F$. Matching the coefficient of $x^{-n-1}$ givesThusSimilarly $w_B=e^{-x/2}G$ gives $G^{\prime \prime }-G^{\prime }=\kappa x^{-2}G$, so $b_n=(-1{)}^n{a}_n$. These are asymptotic series, not generally convergent series; for integer values making a recurrence factor zero they terminate. The coefficient comparison uses exactly the stated permission to differentiate asymptotically term by term.

For $t>0$, the three-dimensional Kirchhoff formula with zero displacement isThe Strong Huygens principle says that the value depends only on the initial velocity on the sphere of radius $t$, not on points strictly inside it. In particular a compactly supported disturbance leaves no tail at a fixed point after the sphere has passed the support.

For method of descent for the wave equation, extend the two-dimensional initial velocity independently of the third coordinate. Projecting the three-dimensional sphere onto its planar disk gives two sheets, each with area element $tdy/\sqrt{t^2-|y-x{|}^2}$. ThereforeThe local argument can use compact cutoffs in the extra coordinate, so global Schwartz decay there is unnecessary. The weak Huygens principle is finite propagation speed: only data in the closed disk of radius $t$ can influence the point. Unlike the three-dimensional free equation, the full interior contributes and a tail can persist.

For the equation with potential $|x{|}^2$, multiply by $u_t$ and use the product rule:This is the claimed local energy conservation with flux $p=-u_t\nabla u$.

Fix $(t_0,x_0)$ and let $E(t)$ be the integral of $e$ over the shrinking ball $|x-x_0|\le t_0-t$, $0\le t\le t_0-a$ with $a>0$. Its boundary moves inward at speed one. The divergence theorem givesbecause the negative integrand isAll initial data are zero, so $E(0)=0$. Nonnegativity then forces $E(t)=0$. Letting the small terminal ball approach the cone tip proves $u=0$ there; alternatively $u_t=0$ throughout the cone and the initial value fixes $u$. The same argument applied to a difference of solutions whose data agree on the initial ball proves local domain of dependence. Thus the potential equation has propagation speed at most one and satisfies the weak Huygens principle, without requiring finite total energy on all space. It does not assert the strong no-tail property.

Use the minimization convention for Pontryagin maximum principle: $H=c(x,t)+{\lambda }^{T}a(x,u)$, with $\dot{x}=H_{\lambda }$, $\dot{\lambda }=-H_x$, and $u$ minimizing $H$ over its allowed set. The initial state is fixed; since terminal time is fixed and terminal state is free, the transversality condition is $\lambda (T)=\nabla K(x(T))$. No free-terminal-time condition $H(T)=0$ is imposed.

For the advertising problem, ${\lambda }_1(T)={\lambda }_2(T)=0$ and ${\dot{\lambda }}_1=0$, so ${\lambda }_1=0$. Write $z=x_2>0$ and $\lambda ={\lambda }_2$. ThenThe switching function $s=3-\lambda z$ satisfies $\dot{s}=z>0$, while $s(T)=3>0$. Thus there is at most one switch, from $u=1$ to $u=0$, and no singular interval. If $u$ were always zero, then $\lambda (t)=T-t$ and $s(0)=3-z(0)T<0$, contradicting Hamiltonian minimization. Hence a switch occurs at some $0<t_{\ast }<T$.

Before it, $z=z(0)e^{-t}$; after it, $\lambda =T-t$ and $z=z(0)e^{-t_{\ast }}$. Setting $s(t_{\ast })=0$ givesThe left side increases and the right side decreases, so the switch is unique.

Sufficiency can also be proved directly. For any control with total effort $a={\int }_0^{T}udt$, the cumulative effort by time $t$ is at most $\min (t,a)$. Since $z(t)=z(0)e^{-{\int }_0^{t}u}$, the best arrangement for that fixed effort is to advertise immediately at full strength for time $a$. Its cost isIts derivative is $3-z(0)(T-a)e^{-a}$ and its second derivative is $z(0)(T-a+1)e^{-a}>0$ on $[0,T]$. The unique minimizer is exactly the preceding $t_{\ast }$. This verifies a global optimum, not only a necessary extremal.