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The transverse stiffness of a stretched spring follows because a spring of length $2l$ develops transverse force $T_0\Delta y/(2l)$ to first order. Its extension changes only at second order, so the leading transverse stiffness comes from its existing tension, not from that extension change. Here $T_0/(2l)=k/2$, givingThe mode vectors are again $(1,1)$ and $(1,-1)$. Treating the springs as unstretched would incorrectly remove this linear restoring force.

For the spin separation at a smooth quantum Hall edge, linearize the confinement energy at a crossing. Compensating a spin energy difference by a potential-energy change givesHere $V^{\prime }$ is interpreted as the specified slope of electron potential energy. If it denotes electrostatic voltage per length instead, the denominator is $e|V^{\prime }|$. The two edges of a neck and the two sides of an antidot have oppositely directed slopes; the formula is a positive local separation on any one edge. For local filling two, only the $n=0$ spin pair is occupied at the neck minimum: $E_{0,+}+U_{\min }<E_F<E_{1,-}+U_{\min }$. Raising both branches with the confinement produces two crossings on each side, as shown in the original neck sketch. This fixed-slope estimate precedes self-consistent electrostatic screening.

An antidot is a depleted potential hill. guiding centres circulate around its closed equipotentials, with chirality opposite to the outer sample edge. A phase-coherent round trip contributes magnetic phase $2\pi BS/{\Phi }_0$, with ${\Phi }_0=h/e$. A localized orbit resonant with the incident energy can transfer an electron between extended channels. If the outgoing channel returns toward the injecting reservoir, the process is resonant reflection and reduces electrical conductance. If coupling instead connects an otherwise blocked incident channel to the far reservoir, it is resonant transmission and adds electrical conductance. Transmission peaks require that forward connectivity; a closed orbit by itself does not guarantee them. Interchannel scattering can change which orbit participates.^[ref]

For the antidot resonance conductance bounds with one active spin-resolved resonance, writeHere ${\Gamma }_L,{\Gamma }_R$ are energy broadenings; the corresponding escape rates are ${\Gamma }_L/\hslash ,{\Gamma }_R/\hslash$. In a reflection-dominated configuration with two otherwise transmitted channels, $G=(e^2/h)(2-\mathcal{L})$: a $2e^2/h$ background can dip to $e^2/h$. In a transmission-dominated configuration with one spectator channel and the second otherwise blocked, $G=(e^2/h)(1+\mathcal{L})$: peaks rise from $e^2/h$ toward $2e^2/h$. These are the two ideal channel-count examples plotted below. More generally, $G=(e^2/h)(N_{\mathrm{open}}-R)$ or $(e^2/h)(N_{\mathrm{dir}}+T)$; a different spectator count changes the integer offset. The constriction filling alone does not specify every tunnel coupling or resonance height. Equal couplings permit unit resonant transfer in the isolated-level model; asymmetry, temperature and dephasing reduce the contrast. Holding the effective orbit area approximately fixed gives $\Delta B\simeq h/(eS)$ for one branch, not a universal charging-corrected period.

In quantum-dot versus antidot resonance topology, a quantum dot is a potential well holding a finite electron island. Its closed guiding-centre circulation is reversed relative to an antidot at the same field. With two barriers in series, current follows source $\to$ dot $\to$ drain; off resonance there need not be a direct conducting path. A gate shifts a level or an addition electrochemical potential into alignment with the reservoirs and produces a electrical conductance peak. For an isolated coherent spin-resolved level, the zero-temperature peak is at most $e^2/h$ and reaches it only for equal couplings; unresolved noninteracting spin degeneracy doubles this bound. In weak sequential tunnelling, peak height depends on rates and temperature rather than being quantized. Charging produces Coulomb blockade peaks with spacing set by addition energy and gate lever arm. An antidot also supports discrete orbit/charging resonances, but can be side-coupled to already open channels and produce dips. Gate voltage changes both orbit area and confinement, whereas a magnetic sweep directly changes enclosed flux.

Coulomb interactions require a self-consistent confinement potential. A gradual decrease in electron density cannot leave arbitrary uncompensated donor charge at no cost. Partly occupied Landau levels have a large charge response: electrostatic screening flattens their effective energy near the Fermi energy over compressible quantum Hall edge strips. Between them, integer-filled incompressible quantum Hall edge strips have a gap, weak density response and most of the potential variation. In an idealized cross-section the effective branch thus alternates between flat, partially occupied portions and rising portions crossing a gap; its carrier density varies oppositely. The bare levels still have Landau level and spin splittings locally, but the common electrostatic potential is no longer an imposed straight slope. exchange interaction can additionally renormalize spin gaps. Widths must follow electrostatics and the density gradient, so the unscreened $E_Z/|V^{\prime }|$ separation is not a quantitative screened-strip width.^[ref]

A modulation-doped GaAs-AlGaAs electron gas is confined on the GaAs side of the interface. The larger-gap AlGaAs supplies a conduction-band barrier; charge transfer from remote donors and the resulting electric field bend the bands into an approximately triangular well. transverse electronic mode wavefunctions are confined normal to the interface but free in its plane. electrons settle into the well until their electrochemical potential agrees with the donor/contact environment. An undoped spacer separates mobile electrons from ionized donors, reducing impurity scattering. Equilibrium has one Fermi level across the structure; the conduction-band edge is not flat. Surface depletion and the cap also affect the profile. The original drawing illustrates one occupied confinement transverse electronic mode; a single spatial electron gas need not in general imply only one occupied vertical transverse electronic mode.

molecular beam epitaxy can grow a clean GaAs buffer/channel, an undoped AlGaAs spacer, a Si-doped AlGaAs supply region and a protective GaAs cap on a suitable substrate. Control composition, interface quality and background impurities during growth; placing dopants directly in the channel defeats the principal mobility advantage. Define a mesa or isolation region, fabricate low-resistance ohmic contacts to the buried gas, and pattern fine metal split gates by electron beam lithography and the lift-off process. Negative gate bias depletes the underlying gas, leaving a narrow channel between the gates. Check contact resistance, leakage, carrier density and mobility; operation with ballistic electron transport requires channel length shorter than the relevant scattering lengths and smooth confinement on the wavelength scale. The lithographic gap does not equal the electrical width because of lateral depletion.

For a long uniform channel supporting ballistic electron transport, transverse confinement creates thresholds $E_n$ with continuous longitudinal propagation above each threshold. One directed spin branch of a transverse electronic mode has states per length $dk/(2\pi )$ and group velocity $v={\hslash }^{-1}dE/dk$. Hence $vdk/(2\pi )=dE/h$: velocity cancels the one-dimensional density of states. Including two unresolved spin states,For low temperature and small bias, a fully open transverse electronic mode has $T_n(E_F)=1$, givingThe assumptions are ideal reservoirs/contacts, negligible reflection and interactions that would invalidate the independent-subband description, and unresolved spin splitting. A resolved spin branch contributes $e^2/h$. The finite two-terminal electrical resistance is a contact/mode-injection effect, despite the absence of dissipative scattering inside the ideal channel.

For voltage partition in a ballistic constriction, a short saddle has strong local electric-field and electrostatic screening variation near its entrances and barrier, rather than the uniform Ohmic voltage gradient of a diffusive wire. Right-moving incident states inherit the source distribution and left-moving states the drain distribution; there is no unique local equilibrium chemical potential inside an ideal ballistic saddle. The electrostatic potential must be determined self-consistently. In a symmetric rigid-barrier approximation, the barrier follows the mean lead potential and half the electrochemical bias lies on each side. This is a stated approximation, not a universal spatial voltage profile.

For a saddle $U=U_0-m^{\ast }{\omega }_x^2{x}^2/2+m^{\ast }{\omega }_y^2{y}^2/2$, transverse electronic mode thresholds are $E_n=U_0+\hslash {\omega }_y(n+1/2)$. The smooth-barrier transmission is $T_n(E)=[1+\exp (-2\pi (E-E_n)/(\hslash {\omega }_x)){]}^{-1}$, approaching a step for a sharp threshold on the bias/thermal scale.^[ref]

Let ${\mu }_{S,D}=E_F\pm eV/2$ and assume no additional bias dependence of $T_n$. At zero temperature,When the source and drain sample $N+1$ and $N$ thresholds, the differential electrical conductance is $(N+1/2)2e^2/h$. These half plateaus arise from the two ends of the bias window, not fractional carriers. The dc ratio $I/V$ need not share their plateau values. Heating, spin splitting and interaction-induced features must be distinguished experimentally.

In bias spectroscopy of one-dimensional subbands, plot differential electrical conductance or its gate derivative against gate bias and source-drain bias. Threshold branches occur where $E_n(V_g)={\mu }_S$ or ${\mu }_D$. When one bias-window edge meets $E_n$ and the other $E_{n+1}$,Thus the intersections of adjacent threshold branches calibrate the one-dimensional spacing directly in energy. A gate lever arm $\alpha =-e^{-1}\partial E_n/\partial V_g$ can then convert gate intervals to energy; it should be measured rather than guessed. Asymmetric bias partition, a moving barrier or a gate-dependent spacing modifies slopes and plateau weights, requiring the corresponding self-consistent bias model.

Because $S_{\mathrm{tot}}^z=N/2-N_f$, the exact finite-size answer is $m_N=1/2-N_f^{\mathrm{gs}}/N$, with $N_f^{\mathrm{gs}}$ determined by the parity-consistent filling above. For the thermodynamic magnetization of the quantum XX chain, the occupied fraction is the length of the negative-energy interval divided by the Brillouin-zone length:ThereforeThe zero-field limit has zero magnetization and the high-field limit has all spins up. The finite-ring curve consists of magnetization steps approaching this continuous limit. Strictly above $h_c$ the fully polarized ground state is unique; exactly at $h_c$ a zero-energy spin-flip state can be degenerate with it. This finite-size qualification does not alter the saturation field or limiting magnetization density.

The Holstein–Primakoff transformation converts a spin algebra into bosonic operators while retaining the finite physical spin Hilbert space. With $[a_n,a_m^{†}]={\delta }_{nm}$, writeThe Holstein–Primakoff occupation constraint is $0\le n_n\le 2S$. Acting on $|n⟩$, the raising matrix element is $\sqrt{n(2S-n+1)}$ and the lowering element is $\sqrt{(n+1)(2S-n)}$. Their squared difference is $2(S-n)$, reproducing $[S^{+},S^{-}]=2S^z$. Operator ordering inside the square roots is essential. For low boson occupation relative to $S$, expand them to obtain $S^{+}\simeq \sqrt{2S}a$ and $S^{-}\simeq \sqrt{2S}{a}^{†}$. Keeping quadratic terms gives the linear spin-wave approximation; higher terms describe magnon interactions. Large $S$ helps control the expansion, but a low-dimensional infrared divergence can invalidate it even at large finite $S$.

For the ferromagnet choose $H_F=-J{\sum }_n{\mathbf{S}}_n\cdot {\mathbf{S}}_{n+1}$, $J>0$, with unit lattice spacing and $\hslash =1$. Expand around the all-up state. A longitudinal bond contributes $-J{S}^2+JS(n_n+n_{n+1})$, and the transverse exchange contributes $-JS(a_n^{†}{a}_{n+1}+a_{n+1}^{†}{a}_n)$. After a Fourier transform,These are ferromagnetic magnons. The polarized vacuum is an exact ground state of the full ferromagnetic Hamiltonian: every bond is in its maximum total-spin sector, minimizing that bond energy. It therefore has no quantum zero-point reduction of its magnetization. The one-magnon dispersion is also exact, although interacting many-magnon states need not be free bosons. Rotating the polarization generates the maximum-total-spin ground-state multiplet. Selecting a definite direction in a thermodynamic symmetry-breaking description breaks $SU(2)$ to $U(1)$; the two transverse spin components form a canonical pair, producing a single quadratic type-B Goldstone boson. A small aligning field gaps it by the Zeeman energy.

At nonzero temperature the one-dimensional ferromagnet is different: its thermal magnon number contains $\int dk[e^{\beta JS{k}^2}-1{]}^{-1}$, with infrared behaviour $\int dkT/k^2$. The divergent magnetization reduction signals the absence of finite-temperature long-range order in an infinite short-range isotropic chain, consistent with the Mermin-Wagner theorem. It does not disallow the polarized zero-temperature ground states just constructed.^[ref]

For the antiferromagnet use $H_{AF}=+J{\sum }_n{\mathbf{S}}_n\cdot {\mathbf{S}}_{n+1}$, still $J>0$, on an even bipartite chain. The classical reference state alternates up and down. Rotate one sublattice through $\pi$ so that both local quantization axes point along their reference spins. In unrotated physical components, to leading order $S_A^z=S-a^{†}a$, $S_B^z=-S+b^{†}b$, $S_A^{+}\simeq \sqrt{2S}a$ and $S_B^{+}\simeq \sqrt{2S}{b}^{†}$. Choosing Fourier phases using the actual site positions givesThe reduced Brillouin zone has $M=N/2$ momenta and length $\pi$. The pair creation and annihilation terms express the fact that an antiparallel reference product state is not an eigenstate. A bosonic Bogoliubov transformation with $u_k=\cosh {\theta }_k$, $v_k=\sinh {\theta }_k$, $\tanh 2{\theta }_k=\cos k$ diagonalizes each two-mode block. With the sign convention $a_k=u_k{\alpha }_k-v_k{\beta }_{-k}^{†}$ and $b_{-k}=u_k{\beta }_{-k}-v_k{\alpha }_k^{†}$, the result isThus the harmonic antiferromagnetic modes have linear, rather than quadratic, long-wave dispersion. In an unfolded Brillouin-zone description the same result is represented by zeros at both zero and the antiferromagnetic wavevector; these are not extra independent degrees of freedom beyond the two-sublattice accounting.

The Bogoliubov vacuum has no $\alpha$ or $\beta$ excitations but contains original spin bosons, with $v_k^2=\frac{1}{2}(|\sin k{|}^{-1}-1)$. Its staggered magnetization is $m_s=S-M^{-1}{\sum }_{k\in \mathrm{RBZ}}{v}_k^2$. This quantum depletion of Néel order diverges logarithmically at small $k$ in one dimension. Linear spin-wave theory therefore does not establish a stable ordered antiferromagnetic ground state for this chain. Its nominal gapless dispersion is a result of expanding around an invalid ordered reference, not a general proof of the exact excitation spectrum. For the nearest-neighbour isotropic chain, integer-spin systems have a Haldane gap, whereas half-integer-spin systems are gapless; the spin-one-half chain in particular has fractional spin excitations rather than a simple dilute gas of ordered-state magnons. This distinction requires physics beyond the harmonic expansion, including the topological term in the continuum theory.^[ref]

The metric uses signature $(+---)$ and a time-valued interval $ds$: a comoving clock has $ds=dt$. Here $t$ is cosmic proper time, $R(t)$ a scale factor with dimensions of length, $c$ the speed of light, $\chi$ a dimensionless comoving radial coordinate, and $\theta ,\varphi$ the usual polar and azimuthal angles. With curvature normalized to $k=+1,0,-1$, $S(\chi )=\sin \chi ,\chi ,\sinh \chi$, respectively. Homogeneity and isotropy apply to each constant-$t$ spatial slice. A comoving object's fixed coordinates do not imply a constant proper separation: spatial lengths scale with $R$.

Radial null propagation obeys $d\chi =cdt/R(t)$ in magnitude. Two neighbouring wave crests between the same comoving emitter and observer have the same radial integral, giving $\delta t_0/R_0=\delta t_e/R_e$. The wavelength redshift is consequentlyPeculiar Doppler shifts are excluded by the comoving assumption. Define angular diameter distance by $d_{\theta }={\ell }_{\perp }/\delta \theta$, where ${\ell }_{\perp }$ is proper transverse size at emission. The transverse metric gives ${\ell }_{\perp }=R_{e}S(\chi )\delta \theta$, henceThis distance is neither the photon travel time times $c$ nor the object's present proper distance.

The comoving particle horizon at $t_1$, with the initial boundary at the Big Bang, is ${\chi }_p(t_1)={\int }_0^{t_1}cdt/R(t)$. Its proper radial distance is $D_p=R(t_1){\chi }_p$; for a power-law scale factor $R=A{t}^p$, $0<p<1$,Thus a matter-only flat Einstein-de Sitter universe, $p=2/3$, has $D_p=3c{t}_1$, and the flat radiation-only case, $p=1/2$, has $D_p=2c{t}_1$. Strictly, Einstein-de Sitter denotes the pressureless matter model; the radiation calculation uses the flat zero-$\Lambda$ radiation analogue requested here.

For the source seen at time $t_1=t_{\mathrm{rec}}$, the radial coordinate is the integral from recombination to today, not the earlier horizon integral. In these flat models,Taking the linear horizon scale in the question to be $D_p$, its small angular extent isThereforeUsing the supplied times gives ${\theta }_{p,m}=0.02684\mathrm{rad}=1.54^{\circ }$ and ${\theta }_{p,r}=0.004244\mathrm{rad}=0.243^{\circ }$. There is a genuine radius/diameter convention to state: the supplied $3c{t}_1$ is a radial horizon distance. If angular diameter means the full diameter of a spherical horizon patch of radius $D_p$, the two answers are instead $2{\theta }_{p,m}=3.08^{\circ }$ and $2{\theta }_{p,r}=0.486^{\circ }$. Both conventions follow from the same distance calculation; a factor of two should not be hidden in the word horizon.

Matter-only evolution is the better of these two approximations because the universe is already matter dominated around recombination and evolves mostly through a subsequent matter era, rather than remaining radiation dominated to the present. Neither is exact: pre-recombination radiation changes the particle horizon, and late-time dark energy changes the distance to the last-scattering surface. No supplied ages have been replaced by later best-fit values.

The cosmic microwave background originates near last scattering and samples many horizon-sized regions. In an always-decelerating hot-Big-Bang model, widely separated last-scattering patches lack overlapping early causal histories, motivating the horizon problem. The measured acoustic angular scale uses the smaller sound horizon, not automatically this light-travel particle horizon; baryon-photon pressure changes the propagation speed. Inflation supplies an earlier accelerated expansion during which the comoving Hubble radius shrinks, allowing a patch with a common earlier causal history to expand to the size of the observable universe. Quantum perturbations can then seed later acoustic and density fluctuations. This does not require local signals to propagate faster than light, and the Hubble radius must not be equated indiscriminately with the particle horizon. CMB temperature and lensing spectra test the resulting perturbation and expansion model.^[ref]

Take $\tau$ as particle proper time, dots as $d/d\tau$, and $f(r)=1-2\mu /r$ with $\mu =GM/c^2$. In the equatorial plane the geodesic Lagrangian isThe cyclic coordinates give $d(c^{2}f\dot{t})/d\tau =0$ and $d(-r^2\dot{\varphi })/d\tau =0$. Set the resulting constants to $c^{2}k$ and $-h$ respectively. The explicit second-order equations areThe polar equation is satisfied by $\theta =\pi /2$, $\dot{\theta }=0$, consistently with spherical symmetry. The proper-time normalization $d{s}^2=c^{2}d{\tau }^2$ supplies the third first integral:The normalization is $2\mathcal{L}=c^2$; it is distinct from choosing an arbitrary affine tangent's norm.

Stationarity and axial symmetry make the covariant time and azimuthal momenta constant. For physical momentum and coordinate $t$ in seconds, $P_t=m{c}^{2}f\dot{t}=m{c}^{2}k$ and $P_{\varphi }=-m{r}^2\dot{\varphi }=-mh$. Thus $m{c}^{2}k$ is the conserved energy normalized at infinity and $mh$ is the signed axial angular momentum. This is Killing energy with positive time signature; a convention with negative timelike norm uses the opposite contraction sign. The supplied covariant-momentum form of the geodesic equation makes these cyclic constants immediate, but its parametrization must be understood: for $p^{\mu }=d{x}^{\mu }/d\lambda$ it is exactly as written. For physical $P^{\mu }=md{x}^{\mu }/d\tau$, proper-time differentiation instead gives $d{P}_{\mu }/d\tau =({\partial }_{\mu }{g}_{\nu \sigma })P^{\nu }{P}^{\sigma }/(2m)$; equivalently, the displayed mass-free form uses $\lambda =\tau /m$. This affine momentum normalization does not change either conservation law.

Multiply the normalization equation by $f$ and substitute the two cyclic integrals. Rearrangement gives the radial energy equationFor radial marginal escape $h=0$ and the future-directed constant is $k=1$, because the speed at infinity vanishes. Choose the outward root:Hence the marginal escape geodesic in Schwarzschild spacetime hasThe constant $a$ has units of length to the power $3/2$ divided by time. This radius is an areal coordinate, not accumulated local proper distance.

A Schwarzschild static observer at exterior radius $r$ uses local radial rulers $d\ell =dr/\sqrt{f}$ and a local clock $dT=\sqrt{f}dt$. Therefore the local radial speed of a Schwarzschild geodesic isFor the marginal trajectory $k=1$ this givesIts equality to $\dot{r}$ is particular to $k=1$; the local definition, rather than that accidental equality, is the required observer measurement. The Schwarzschild-coordinate velocity is instead $dr/dt=f\sqrt{2GM/r}$.

For nonzero asymptotic speed $0\le u<c$, the same energy constant is $k={\gamma }_{\infty }=(1-u^2/c^2{)}^{-1/2}$. Combining the radial equation with the local-ruler relation gives radial Schwarzschild escape with asymptotic speedThe positive root describes the outward particle, at radii on its trajectory with $r\ge r_0$. Independently, a static observer has four-velocity $U=(1/\sqrt{f},0,0,0)$ in these coordinates, giving relative Lorentz factor $g(U,u_{\mathrm{particle}})/c^2=\sqrt{f}\dot{t}=k/\sqrt{f}$ and the same speed. As $r\to \infty$ it tends to $u$; for $u=0$ it gives the preceding result. Its limiting value approaches $c$ as $r\downarrow 2GM/c^2$, but that is a limit through exterior observers: no static timelike observer exists on or inside the horizon.

The Murnaghan–Nakayama rule computes an irreducible character on a permutation with a specified cycle of length $r$ byHere the leg length is one less than the number of occupied rows, and iteration ends at the empty partition with character value one. Conjugating a partition twists its character by the sign character:For a transposition, remove a rim domino and evaluate the remaining character at the identity of $S_3$. The three possible remaining shapes $(3),(2,1),(1^3)$ have degrees $1,2,1$ by the hook formula. For $(5)$ the single removable horizontal domino leaves $(3)$, giving one. For $(4,1)$ the only rim domino leaves $(2,1)$, giving two. For $(3,2)$ the removable bottom-row domino leaves $(3)$, giving one. For $(3,1,1)$, a horizontal domino leaves $(1^3)$ and a vertical domino leaves $(3)$; their contributions are $1-1=0$. The conjugate-partition identity gives the remaining three values with reversed sign.

The Hook-length formula is ${\chi }_{\lambda }(1)=5!/{\prod }_{u\in [\lambda ]}h(u)$. Both requested columns are thereforeAs a check, the squared degrees sum to $1+16+25+36+25+16+1=120=|S_5|$. The transposition column above was obtained from rim-hook removal and conjugation, independently of permutation characters.

Write $\lambda =(a,b)$ with $a\ge b>0$. If the removed corner is in the first row, it must be its last cell, with $a>b$. Remove that entry and output it as $v$; all remaining row and column inequalities persist, and the shape is $(a-1,b)$.

If the corner is the last cell of the second row, remove its value $c$. In the first row find the rightmost entry $d<c$, replace it by $c$, and output $v=d$. Such an entry exists: the first-row cell above the removed corner has smaller value than $c$. The chosen position is therefore at least column $b$.

The new first-row entry exceeds all entries to its left, while every entry to its right is larger than $c$ by the rightmost-choice rule. After deleting the lower corner, no second-row cell remains below the changed position, since that row now has length $b-1$. Thus all column inequalities remain valid too. The resulting shape is $(a,b-1)$, omitting a zero part if $b=1$.The entries remain distinct and increasing along rows and columns, but the missing label is the ejected $v$, not necessarily the original corner value. The resulting tableau is therefore standard on its remaining ordered entry set, the generalized-standard convention in the question.

Track the first row after any prefix of the input permutation. Its $j$th entry is the smallest possible final value of an increasing subsequence of length $j$ in that prefix. This minimum-tail invariant follows inductively from the row-insertion update.

Indeed, when a new value $z$ replaces the first first-row entry larger than it, at position $r$, the previous entry in position $r-1$ is smaller than $z$ when $r>1$. An increasing subsequence ending there can be extended by $z$, so $z$ supplies a new attainable tail of length $r$. It improves the old length-$r$ minimum. The minima for shorter lengths were already smaller than $z$ and remain unchanged; a longer subsequence ending at $z$ is impossible because its previous tail would have to be smaller than $z$, contradicting the choice of $r$. If no first-row entry exceeds $z$, it appends and creates a new longest subsequence. This proves the invariant at every step.

In particular, the number of first-row cells is the maximum increasing-subsequence length. At the final tableau it is ${\lambda }_1$, soThis proves the required nonextension assertion, even without requiring the first-row tuple itself to occur as a subsequence.

That occurrence is not automatic for a general permutation. For example the word $2,4,3,1$ gives $P$ with rows $(1,3),(2),(4)$. Its first-row tuple $(1,3)$ is not a subsequence of the word, because one occurs after three. The first row collects minimum tails for different subsequence lengths; those tails need not come from one common subsequence. If the source's premise is read as an additional hypothesis that the tuple does occur, the preceding maximum-length proof establishes its claimed maximality directly.

The invariant is the first row of RSK as a minimum-tail table.

At a regular point, $f(z_0+\zeta )-f(z_0)=f^{\prime }(z_0)\zeta +O({\zeta }^2)$. Multiplication by a nonzero complex number rotates every tangent by the same angle and multiplies every length by the same positive factor. Oriented angles are preserved. If the first derivative vanishes but the second does not, the leading term is $f^{\prime \prime }(z_0){\zeta }^2/2$: tangent arguments are doubled, so angles are doubled, modulo the identification of rays. This is a local statement; holomorphy alone does not imply global injectivity.

Write $z=x+iy$. ThenFor $-1<x<1$ and $y>0$, the imaginary part is positive. The bottom edge maps monotonically to $[-1,1]$; the left and right vertical edges map to $(-\infty ,-1]$ and $[1,\infty )$ respectively. The inverse below establishes that every point of the open upper half-plane occurs exactly once. At fixed $y=a$, putting $u=\mathrm{Re}w$, $v=\mathrm{Im}w$, one obtainsThis is the upper half of an ellipse, traversed from its left endpoint to its right endpoint.

Choose the branch $s(w)=\sqrt{1-w^2}$ analytic in the upper half-plane and positive on the boundary interval $(-1,1)$. Then the inverse branch into the strip isThe logarithm is chosen continuously so that $-1<\mathrm{Re}z<1$ and $\mathrm{Im}z>0$; for instance $w=iv$, $v>0$, gives $z=(2i/\pi )\mathrm{arsinh}v$. Differentiation gives $dz/dw=2/(\pi \sqrt{1-w^2})$. The only zeros of the original derivative in the closed strip are $z=\pm 1$, with images $w=\pm 1$. The inverse has square-root branch points there, and cannot extend holomorphically across them. At either corner the original map doubles the right angle to a straight angle; a holomorphic inverse with nonzero derivative could not halve it. The open strip interior has no such failure.

Replace the slowly varying interior wave by $C$ and $e^{ik|x-x^{\prime }|}$ by one when both arguments lie within the range. The integral equation givesHenceThe integrated strength has dimensions of energy times length. For $U=-\alpha \delta (x)$, $\gamma =-\alpha$ and this is exactly the result in part c. For a regular weak potential the approximation sums the important ratio $m\gamma /(k{\hslash }^2)$ without requiring it to be small, which is why it remains useful beyond the naive first Born expansion at small momentum.

Let the bunch tail have the higher energy so that it catches the head. For an ultrarelativistic proton, $\delta v/v\simeq (\delta E/E)/({\beta }^2{\gamma }^2)$. A drift of length $S$ then changes the bunch length by approximately $S\delta v/v$. Requiring that to match the initial length givesTaking $\gamma \simeq 62.5/0.938\simeq 66.6$, $\beta \simeq 1$, and the given lengths yields $\Delta E/E\simeq 0.023$, about $2.3\%$. Treating $62.5$ GeV as kinetic rather than total energy changes that by only a few percent.

This is the linear ballistic compression requirement; a real finite-emittance bunch, nonlinear velocity-energy relation, uncorrelated energy spread and collective fields prevent exact zero length. The chirp is a correlated full head-to-tail spread, not automatically an rms energy spread.

The target collision energy requires $175$ GeV per lepton beam. Its initial energy after the half-linac is $62.5$ GeV, so each lepton must gain $112.5$ GeV. Define the transformer ratio as accelerating field experienced by the witness divided by decelerating field experienced by the driver. If the useful stage length exhausts the driver's $62.5$ GeV while delivering the required gain, thenThis matches depletion to acceleration and avoids leaving unused driver energy at the witness target. It is an ideal energy-transfer estimate, assuming suitable loading and longitudinal field profiles; a field ratio alone does not specify a practical wake or its efficiency.

The mean witness gradient must be $112.5\mathrm{GeV}/1.1\mathrm{m}\simeq 102\mathrm{GV}{\mathrm{m}}^{-1}$. To turn this into a density estimate, assume an accelerating wake amplitude of order the cold plasma wave-breaking scale,Eliminating ${\omega }_p$ givesThe associated plasma wavelength is about $31$ micrometres, illustrating why a short driver structure is required. The given length and energy gain do not uniquely fix density without an assumption about wake amplitude relative to the plasma scale. Driver shape, beam loading, relativistic wave-breaking corrections and matching to a Positron witness are substantive additional issues; the estimate is a scale calculation, not a completed accelerator design.

Using the given one-component equipartition estimate, $p^2=M{k}_{B}T$. Substituting into the de Broglie relation givesComparison with a confinement length or interparticle spacing $d$ gives the criterion $T\lesssim h^2/(M{k}_B{d}^2)$. Numerical factors depend on the convention: the statistical thermal de Broglie wavelength is usually $h/\sqrt{2\pi M{k}_{B}T}$, while a three-component root-mean-square momentum introduces a factor $\sqrt{3}$. The question's specified one-component convention is used consistently here. Once degeneracy becomes important, classical equipartition itself ceases to describe the electrons accurately.

The Shockley-Queisser calculation treats a single-junction solar cell with a single band gap, in steady state, with one collected electron-hole pair per absorbed above-gap photon, an ideal absorption edge, thermalized carriers and radiative recombination as the limiting recombination process. The source spectrum, cell temperature, angular acceptance and optical concentration must be specified. Below-gap photons are transmitted; energy above the gap is lost during carrier thermalization; the cell must emit photons by detailed balance, producing a recombination current and an entropy-related voltage loss. Delivering power at finite current also puts the operating voltage below the open-circuit voltage. Nonradiative recombination, resistance and imperfect collection worsen a real cell but are not required for the ideal limit.

Three ways to change those assumptions are tandem solar cells, which divide the spectrum among gaps and reduce both transmission and thermalization losses; hot-carrier solar cells, which aims to collect excess carrier energy before thermalization; and carrier multiplication or spectral conversion, which aims to use a high-energy photon for more than one excitation or combine otherwise unusable sub-gap photons. These concepts have additional practical losses and do not automatically outperform an optimized conventional device. The usual unconcentrated single-junction optimum is approximately one third under a terrestrial solar spectrum, rather than a universal efficiency independent of illumination.

For the geometrical estimate, take the stated horizontal equinox irradiance to set a constant direct-normal reference. The equinox solar-noon elevation is ${\alpha }_e=90^{\circ }-51.75^{\circ }=38.25^{\circ }$, so this reference is $I_n=1/\sin {\alpha }_e\simeq 1.62\mathrm{kW}{\mathrm{m}}^{-2}$. A south-facing panel with tilt $\beta =41.1^{\circ }$ has noon incidence factor $\cos (51.75^{\circ }-\delta -41.1^{\circ })$, where $\delta =\pm 23.44^{\circ }$. Relative to the $1\mathrm{kW}{\mathrm{m}}^{-2}$ plane-of-array rating,These are the results of that idealized geometrical normalization, not credible seasonal production forecasts. Its inferred direct-normal irradiance already exceeds the approximately $1.36\mathrm{kW}{\mathrm{m}}^{-2}$ solar irradiance outside the atmosphere, so the supplied horizontal benchmark cannot describe a purely direct noon beam literally. Real direct and diffuse fractions, atmospheric attenuation at low winter elevation, cell temperature and inverter clipping are unspecified. An alternative assumption of $1\mathrm{kW}{\mathrm{m}}^{-2}$ direct-normal irradiance gives approximately $4.88$ and $4.14$ kW. A physically unique maximum cannot be inferred without an irradiance model.

The stated annual output corresponds to $4100/5=820$ full-rated hours, an average power of $0.468$ kW and a capacity factor of $9.4\%$. Night, cloud, seasonal day length and other losses explain the large difference from an instantaneous sunlit estimate.

For a national scale estimate, take UK area $2.44\times 10^{11}{\mathrm{m}}^2$, population $6.6\times 10^7$, and annual mean solar input $100\mathrm{W}{\mathrm{m}}^{-2}$. Covering five percent with twenty-percent-efficient modules givesThis is comparable with an order-$100$ kWh daily per-person primary-energy budget, and much larger than electricity use alone, of order $10$-$15$ kWh per person per day. It is a gross area-and-flux estimate: access land, shading, seasonal storage, curtailment and conversion between primary energy and electrical end use all affect its interpretation.

With the apsidal origin chosen as specified, $G=0$. For $p_{\varphi }\ne 0$, rearrangement givesLet $u=1/r$. Since $du/d\varphi =-p_r/p_{\varphi }$, the equation is $u^{\prime }=-(u-mk/p_{\varphi }^2)\tan \varphi$. Its solution, continued through points where the intermediate division fails, isAs printed, $e$ has units of inverse length and is not the usual dimensionless eccentricity. Writing $\ell =p_{\varphi }^2/(mk)$ gives the standard form $r=\ell /(1+\epsilon \cos \varphi )$ with $\epsilon =e\ell$. The orbit is an ellipse for $|\epsilon |<1$, parabolic for unit magnitude and hyperbolic for magnitude greater than one on its allowed angular interval. The purely radial case $p_{\varphi }=0$ must be handled separately and is excluded by these divisions.

For monovalent mobile ions, the linearized Debye–Hückel screening length isHere $n_i$ is a number density in ${\mathrm{m}}^{-3}$, not a molarity. Use the question's idealized neutral-water convention, $c_{+}=c_{-}=10^{-7}\mathrm{mol}{\mathrm{L}}^{-1}$, and $T=293.15\mathrm{K}$. Both ion species contribute:For added salt concentration $c_s$ in $\mathrm{mol}{\mathrm{L}}^{-1}$, the ionic strength in this model is $I=c_s+10^{-7}\mathrm{mol}{\mathrm{L}}^{-1}$. With the numerical molarity $I_M=I/(1\mathrm{mol}{\mathrm{L}}^{-1})$,The water contribution is negligible at that salt level. The factor $10^3$ converts molarity into $\mathrm{mol}{\mathrm{m}}^{-3}$; omitting it or counting only one ion species gives a wrong screening length. Exact water autoionization and activities vary with temperature, so the first estimate uses the simplified neutral $10^{-7}\mathrm{M}$ concentrations implied by the question. At the higher salt level, ion-size effects, nonideal activities and changes in permittivity limit the numerical accuracy of this continuum estimate.

For a clean liquid interface at fixed temperature and composition, surface tension is the reversible free-energy cost per added area, $\gamma =(\partial F/\partial A{)}_{T,V,N}$, so $dF=\gamma dA$. Its units are both $\mathrm{J}{\mathrm{m}}^{-2}$ and $\mathrm{N}{\mathrm{m}}^{-1}$. Molecules at an interface have a different environment from those in the bulk; creating more interface changes their interactions and entropy. For constant $\gamma$, the interfacial free energy is $\gamma A$ up to a reference constant. This is not generally identical to the internal surface energy: the interfacial entropy density is $-d\gamma /dT$, giving internal-energy density $\gamma -Td\gamma /dT$.

A line of length $\ell$ in one interface transmits a tangential force $\gamma \ell$, perpendicular to the line. A soap film has two interfaces, giving force $2\gamma \ell$ on a sliding rim. Quasistatic extension by $dx$ creates total area $2\ell dx$, recovering that force from $dF=2\gamma \ell dx$. Uniform positive surface tension favors minimum area at fixed volume, hence an isolated spherical droplet when gravity is negligible. curvature gives the Young–Laplace equation, with outward normal chosen consistently:A spherical liquid drop has $\Delta p=2\gamma /R$; a thin soap bubble with two similar interfaces has approximately $4\gamma /R$. The capillary length $\sqrt{\gamma /(\Delta \rho g)}$ compares gravity with capillarity. surface tension also restores capillary waves, rather than providing a bulk static shear modulus.

Interfacial contamination and added surfactants can change $\gamma$ substantially; a nonuniform value produces tangential Marangoni stress. A rim-force measurement should therefore control temperature, cleanliness and the number of interfaces. Capillary-rise methods additionally need the contact angle, because the vertical component is proportional to $\gamma \cos \theta$. An independent frame-force measurement can distinguish a contact-angle change from a true change of surface tension.