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Take one-dimensional steady state thermal conduction, constant thermal conductivity $k>0$, and positive volumetric heat generation rate. Write $q_0=q_0^{\prime \prime \prime }$ for brevity, and let $x$ increase away from the illuminated face. Conservation of energy and Fourier's law giveThus $T^{\prime \prime }=-(q_0/k)e^{-\mu x}<0$: the profile is strictly concave and any stationary point is its unique maximum. Two integrations and the equal-surface boundary conditions yield the exponential slab heating solutionIts derivative isDefine $\beta =\mu a$ and $C=(1-e^{-\beta })/\beta$. Setting $T^{\prime }(x_m)=0$ gives the distance from the illuminated surface:For $\beta >0$, $e^{-\beta }<C<1$, so $0<x_m<a$. More specifically $C=e^{-\beta /2}\sinh (\beta /2)/(\beta /2)>e^{-\beta /2}$, hence $x_m<a/2$: the peak is nearer the more strongly heated face.

The apparent singularity at $\mu =0$ is removable. Expansion givesThe uniform-heating limit therefore has its maximum at $a/2$, as symmetry requires.

Write $R_e=R^{+}$ and apply the stated constant-coefficient boundary value problem. A constant particular solution is $-S/(D{B}^2)$. Regularity at the origin removes $\cos (Br)/r$, soThe outer boundary condition gives $A=S{R}_e/[D{B}^2\sin (B{R}_e)]$, henceAt $r=0$, use the limit $\sin (Br)/r\to B$ rather than substitute into the apparent $0/0$. The result satisfies both the differential equation and $\varphi (R_e)=0$ provided $\sin (B{R}_e)\ne 0$. This is the ideal extrapolated-boundary model specified in the question; a separately resolved outer transport layer would need its own matching conditions.

The polarization generated by the relevant third-order nonlinear optical susceptibility is proportional to $E_1{E}_2{E}_3^{\ast }$. Multiplication of its temporal and spatial wave phase factors givesThe two opposing pump arrows in the original diagram mean ${\mathbf{k}}_2=-{\mathbf{k}}_1$. Thus the mixing source points along $-{\mathbf{k}}_3$, and its conjugate signal amplitude drives a backward field. This is counterpropagating-pump phase matching in four-wave mixing.

There is a phase matching qualification to the initial frequency statement. In the stipulated nondispersive isotropic medium, $|{\mathbf{k}}_j|=n{\omega }_j/c$. Exact equality of the backward free-wave vector to $-{\mathbf{k}}_3$ therefore requires ${\omega }_4={\omega }_3$, hence ${\omega }_3={\omega }_P$. For unequal frequencies the backward direction remains the geometry suggested by the source, but the longitudinal phase mismatch has magnitude $2n|{\omega }_P-{\omega }_3|/c$. Efficient generation over a finite length requires that mismatch times $L$ be small; a sinc function factor suppresses conversion otherwise. Nondispersion does not remove the magnitude mismatch. The remainder of the calculation is explicitly degenerate four-wave mixing, so this difficulty disappears.

At the pump frequency, for the strong field $E_1$, self-phase modulation and cross-phase modulation by $E_2$ dominate the weak signal's influence. They change its propagation wave phase through its own intensity and the other pump's intensity. With continuous uniform plane waves, these are nonlinear wave phase shifts; finite pulses acquire temporal chirp, and finite transverse beams can acquire a lens. No universal factor of two should be assigned between the two shifts without specifying their polarization contraction and the field-amplitude convention.

Write $u={\tilde{A}}_3$ and $v={\tilde{A}}_4$. Differentiating the exponential substitutions cancels the terms proportional to $2\kappa$:Consequently $u^{\prime }=i\kappa v^{\ast }$ and $v^{\prime }=-i\kappa u^{\ast }$. Conjugate the latter before eliminating a field: $(v^{\ast }{)}^{\prime }=i\kappa u$, so $u^{\prime \prime }=-{\kappa }^{2}u$. The backward four-wave-mixing boundary problem requires opposite-end boundary conditions: $u(0)=a$, where $a=A_3(0)$, and $v(L)=0$. The transfer matrix isAt $L$, $0=ia\sin \kappa L+b\cos \kappa L$, so $b=-ia\tan \kappa L$. Substitution givesThe printed target expression for $A_4$ has the opposite overall sign. This is present in the original PDF, not just the transcription. With its stated first-order equations, the derivative at the far face must be $v^{\prime }(L)=-i\kappa a^{\ast }/\cos \kappa L$. The boxed result has that derivative; the printed expression has its negative. Reversing the convention for the mixing coefficient could reverse the generated phase, but would also change the coupled equations. The result above satisfies the equations and both boundary conditions together.

At the entrance, $A_4(0)=i{a}^{\ast }\tan \kappa L$, so the returned amplitude is proportional to the complex conjugate of the signal. An ordinary ideal mirror gives $ra$, with a fixed reflection coefficient $r$, rather than $a^{\ast }$. The mixing pumps supply both the extra energy and a wave phase reference. The ideal conjugate reflectivity and signal transmission areThe constancy of $|u{|}^2+|v{|}^2$ along the medium is another direct check. The relation between the incoming signal and outgoing waves allows net output energy to come from the pumps. As $\kappa L\to \pi /2$, the undepleted-pump approximation fails: the predicted fields would extract an appreciable pump fraction. This is the ideal four-wave-mixing oscillation threshold; saturation, pump depletion and noise seeding replace the divergent linearized solution. It is not an unlimited source of reflected power. Amplified phase-conjugated propagation and parametric oscillation are also the effects identified in the primary analysis by Amnon Yariv and David M. Pepper.^[ref]

For the aberrating linear medium, assume its permittivity is real and stationary at this frequency. Taking the complex conjugate of the forward-envelope equation givesThis is the backward-envelope equation with carrier $e^{-i{k}_{3}z}$. Launching the conjugate boundary field therefore permits retracing of the distorted wave. In a transverse lossless phase screen model, let transmission be $t(x)=e^{i\psi (x)}$. A return proportional to $(tU{)}^{\ast }$ acquires $t$ on its second passage and becomes $t(tU{)}^{\ast }=U^{\ast }$. Thus the aberration phase cancels. A mirror instead gives $r{t}^{2}U$, adding the aberration twice; merely reversing the central propagation direction does not reverse the wavefront phase.

The one-dimensional printed equation describes longitudinal variation; genuine transverse aberrations require the corresponding transverse Helmholtz equation or paraxial operator, for which the same real-coefficient conjugation argument applies. Exact reconstruction requires adequate aperture, the correct conjugate amplitude and polarization, uniform pump-product phase, and an unchanged reciprocal lossless medium. Absorption, missing scattered modes or a changed aberrator prevent a universal reconstruction. This is phase conjugation and thin phase-plate cancellation by phase conjugation under specified conditions, rather than reversal of every possible propagation loss.

Use the Backward Euler method in time and the prescribed central finite differences in space. With $j\in \mathbb{Z}$ and $m\ge 0$, the implicit finite difference method for the advection-diffusion equation isEquivalently, for $\nu =\Delta t/\Delta x$ and $\mu =\Delta t/(\Delta x{)}^2$, the new time level solvesThe spatial terms are evaluated at the new time level, making this an implicit scheme. On the whole line there are no finite-endpoint boundary conditions; in the next part its solution is chosen in the l2 sequence space.

The electric potential and magnetic vector potential represent the fields through $\mathbf{B}=\nabla \times \mathbf{A}$ and $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$. They are not unique: a differentiable function $\chi$ giveswithout changing either field. Gauge fixing restricts this freedom to suit the sources, boundaries or symmetry; it does not alter observables.

The Lorenz gauge $\nabla \cdot \mathbf{A}+c^{-2}{\partial }_t\varphi =0$ is Lorentz covariant. Maxwell's equations then reduce to sourced wave equations,Retarded solutions implement causal radiation conditions. This gauge is useful for radiation and relativistic calculations; residual transformations obey the homogeneous wave equation.^[ref]

The Coulomb gauge $\nabla \cdot \mathbf{A}=0$ instead gives ${\nabla }^2\varphi =-\rho /{ϵ}_0$. Its electric potential is an instantaneous Poisson solution, but the accompanying magnetic vector potential makes the full physical electric field causal. It is useful in separating longitudinal electric fields from transverse radiation. In a source-free vacuum region, suitable radiation boundary conditions permit radiation gauge $\varphi =0$, $\nabla \cdot \mathbf{A}=0$, leaving two transverse plane-wave polarizations. These extra conditions are not a gauge choice applicable unchanged to arbitrary nonzero charge distributions.

The magnetic vector potential is defined by $\mathbf{B}=\nabla \times \mathbf{A}$. The gauge transformationleaves both $\mathbf{B}$ and $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$ unchanged. In magnetostatics choose the Coulomb gauge $\nabla \cdot \mathbf{A}=0$. Then Ampère's law and the curl-of-curl identity giveThe free-space Green function identity ${\nabla }^2|\mathbf{r}-{\mathbf{r}}^{\prime }{|}^{-1}=-4\pi {\delta }^{(3)}(\mathbf{r}-{\mathbf{r}}^{\prime })$ therefore givesThe boundary condition selects the homogeneous solution; the usual decay condition assumes suitably localized sources. An infinitely long static wire needs a reference for its logarithmically divergent potential, although its magnetic field is finite away from the wire.

For time-dependent sources use the Lorenz gauge, $\nabla \cdot \mathbf{A}+c^{-2}{\partial }_t\varphi =0$, and the outgoing retarded potentialThe scalar potential has the analogous retarded charge-density integral. The two gauges should not be conflated for a general time-dependent source.^[ref]

For the prescribed uncharged wire, $\rho =0$ and $\varphi =0$ is consistent. Translation invariance makes $\mathbf{A}=A_z(r,t)\hat{\mathbf{z}}$ with $\nabla \cdot \mathbf{A}=0$, so this particular solution satisfies both gauge conditions. With $s=z^{\prime }-z$,It is zero for $t<r/c$. For $t>r/c$, set $L=\sqrt{c^2{t}^2-r^2}$. The causal interval is $-L<s<L$, giving the switched infinite-wire retarded potentialDifferentiation gives the fields and Poynting vector, for $r>0$, $t>r/c$:Here $\mathbf{B}={\mu }_0\mathbf{H}$ and $B_{\varphi }=-{\partial }_r{A}_z$. All three fields vanish before arrival. For positive $I$, $\mathbf{E}$ opposes the imposed current and the radial energy flow after switching a line current points radially outward: $(-\hat{\mathbf{z}})\times \hat{\mathbf{\varphi }}=\hat{\mathbf{r}}$.

For the graph, let $u=ct/r$. Normalized to $E_{\ast }={\mu }_{0}Ic/(2\pi r)$, $H_{\ast }=I/(2\pi r)$ and $N_{\ast }={\mu }_0{I}^{2}c/(4{\pi }^2{r}^2)$, the curves areAt the wavefront, $|E_z|$ and $H_{\varphi }$ diverge as $(u-1{)}^{-1/2}$ and $N_r$ as $(u-1{)}^{-1}$. This abrupt line-current wavefront singularity reflects the ideal instantaneous turn-on of a line current; a finite rise time smooths the front. At late times,The vector potential grows logarithmically with time in this infinite-wire idealization even as the local magnetic field approaches its static limit. The divergent front should not be interpreted as a finite-energy pulse from a real instantaneous switching experiment.

Yes: the same physical resonant cavity and its field remain resonant in every inertial description. A Lorentz transformation changes the mirror worldlines and the field together. The boundary phase matching and the constructive-interference condition are statements about the same reflection events, and the contracted wave phase is a Lorentz scalar. A boosted resonant solution therefore still satisfies the moving-mirror boundary conditions.

It would be incorrect merely to insert a Lorentz-contracted length into the formula for two stationary mirrors while leaving both propagation frequencies unchanged. For example, let the mirrors be separated by $L$ in their rest frame and let ${\omega }_{0}L/c=m\pi$, including the usual equal-mirror reflection-phase convention. For an observer boosted at $+v$ along the cavity axis, the cavity moves at $-v$ and its two travelling components have frequenciesThey need not have one common laboratory frequency. The simultaneous mirror separation is $L^{\prime }=L/\gamma$, while the forward and return flight times are $L^{\prime }/(c+v)$ and $L^{\prime }/(c-v)$. The round-trip interval between events on the same mirror is therefore $2\gamma L/c$.

Along that moving mirror the wave phase rate of either wave component is ${\omega }^{\prime }-k^{\prime }{v}_{\mathrm{mirror}}={\omega }_0/\gamma$. Its advance over the round trip isThus the resonance condition is unchanged even though the observed travelling frequencies, lengths and flight times all change. For a boost in a general direction the beam directions also undergo relativistic aberration; the event-by-event phase-invariance argument still applies. A separate stationary cavity placed in the new frame, or an external laser whose frequency is not transformed with the original setup, is a different physical system.

Write $y=e^{S/\delta }$ and divide the equation by $y$:Balancing the first and third terms gives $\delta =ϵ$. With $S=S_0+ϵS_1+\cdots$, the terms of order ${ϵ}^{-1}$ and one giveFor the slow branch $S_0^{\prime }=0$, choose $S_0=0$ and obtain $S_1=-x^2/2$. For the fast branch $S_0^{\prime }=-1$, choose $S_0=-x$ and obtain $S_1=x^2/2$. Additive integration constants enter the two independent amplitudes. The two-term WKB approximation is consequentlyImposing the boundary conditions on these two approximating modes yieldsThis uses the first two exponent functions $S_0,S_1$ for both branches. The next transport equations, if desired to identify the algebraic error, give $S_2=x-x^3/3$ for the slow mode and $S_2=x+x^3/3$ for the fast mode. Thus each displayed mode has relative amplitude correction $1+O(ϵ)$ uniformly on the fixed interval; no turning point occurs because the two leading exponent roots remain distinct.

Use the Fourier transform convention $\overline{U}(k,y)={\int }_{\mathbb{R}}u(x,y)e^{ikx}dx$, initially in a class where the integral transforms and subsequent limits are defined. Transforming Laplace's equation gives ${\overline{U}}_{yy}-k^2\overline{U}=0$ on real $k$. Decay for $y\to \infty$ selectsThe boundary trace is the sum of its negative-half-line unknown and the given positive-half-line datum. The normal derivative is zero on the negative side. ThereforeFor the $e^{ikx}$ convention, negative support gives a integral transform analytic for $\mathrm{Im}k<0$, and positive support gives one analytic for $\mathrm{Im}k>0$, subject to the appropriate endpoint interpretation. The normal trace in this particular problem requires a finite-part interpretation, detailed below.

For the Wiener-Hopf method, regularize the real-axis kernel byTake the branch cut of $a_{+}$ vertically downward from $-iϵ$ and that of $a_{-}$ vertically upward from $iϵ$. Their product on the real axis is the positive square root $\sqrt{k^2+{ϵ}^2}$. The branch of $H_{+}$ has its cut downward from zero. Dividing the regularized equation by $a_{+}$ givesThe forcing $P$ is analytic in $0<\mathrm{Im}k<ϵ$, between its two cuts. Decompose $P=P_{+}-P_{-}$ by Cauchy integrals on two horizontal lines within that strip. Separation giveswhen the endpoint growth class makes the common entire remainder bounded. For $\mathrm{Im}k<0$ one can take the minus Cauchy integral on $\mathrm{Im}\hat{k}=\delta$, with $0<\delta <ϵ$, obtainingThe integrand is $O(|\hat{k}{|}^{-5/4})$ at infinity, so the Cauchy integral converges there. In the limit choose $\delta =\delta (ϵ)$, for example $\delta =ϵ/2$; a fixed positive $\delta$ cannot remain between zero and $ϵ$. With $\sqrt{k}$ denoting the limiting lower-analytic branch and with the prescribed limiting interpretation,The numerator square root has a cut going upward from $iϵ$; the power ${\hat{k}}^{-3/4}$ has a cut going downward from zero. The contour passes between them. The pole $\hat{k}=k$ is below this contour when $k$ is in the lower integral transform domain. These branch choices are essential, rather than independent principal powers of the factors.

There is an important mathematical limitation in the source's ordinary integral definition of $\bar{g}$. An explicit harmonic function particular solution in polar coordinates $0<\theta <\pi$ isEach separated polar power is harmonic function; at $\theta =0$ it has the required Dirichlet boundary condition value, and at $\theta =\pi$ its angular derivative vanishes, giving the required zero normal derivative. It also decays at infinity. Its unknown boundary traces areThe last function is not locally integrable at zero. Therefore the printed ordinary integral ${\int }_0^{\infty }g(x)e^{ikx}dx$ diverges, even for $\mathrm{Im}k>0$. The integral transform calculation is valid for this solution only after specifying a Hadamard finite-part integral, or equivalently a distribution boundary trace.

For the canonical homogeneous extension of this power, the two transformed traces areThe first is analytic below and the second above. The second formula follows from $\Gamma (3/4)=(-1/4)\Gamma (-1/4)$ and the finite-part continuation of the Laplace transform. Substituting their real-axis boundary values directly verifies the boxed integral transform relation for both signs of $k$, so the qualification can be checked without relying only on formal factorization.

There is also homogeneous corner freedom:satisfies zero Dirichlet boundary condition data on the positive axis, zero Neumann boundary condition data on the negative axis and decay at infinity. Its negative trace has integral transform $C\Gamma (1/2)(ik{)}^{-1/2}$, corresponding to the term $A/\sqrt{k}$. Higher singular modes $r^{-(n+1/2)}\sin ((n+1/2)\theta )$ would also satisfy the homogeneous boundary conditions for integers $n\ge 1$, but are excluded if the negative trace is required to be locally integrable. Thus the one-constant formula belongs to a specified corner-growth class. The original boundary conditions alone, without that class, do not assert uniqueness. These facts form the finite-part normal trace for mixed quarter-power boundary data.

For the left inner coordinate $X=x/\delta$, the equation becomes $Y_{XXX}-Y_X+{\delta }^{2}XY=0$. At leading order the solution that matches a bounded outer expansion is $Y_0=A+D{e}^{-X}$; the growing exponential is excluded. The boundary value gives $A+D=1$. Since $y^{\prime }(0)=1$ is finite, the possible $O({\delta }^{-1})$ derivative must vanish, so $D=0$ and $A=1$.

At the next order, $Y_{1/2}=B+E{e}^{-X}$. Its boundary value is zero, giving $E=-B$, while its physical derivative at zero is $-E=1$. Hence $B=1$, and the left inner expansion is $1+\delta (1-e^{-X})+O({\delta }^2)$.

At the right endpoint put $\xi =(1-x)/\delta$. The equation is $Y_{\xi \xi \xi }-Y_{\xi }=\delta (1-\delta \xi )Y$. Matching the outer expansion at one and imposing $Y(0)=1$ giveAt the next order, $Y_{1/2}^{(3)}-Y_{1/2}^{\prime }=Y_0$. The solution with no growing exponential, the correct matching polynomial, and $Y_{1/2}(0)=0$ isIndeed the operator $D^3-D$ sends $-\sqrt{e}\xi$ to $\sqrt{e}$ and sends $\xi e^{-\xi }$ to $2e^{-\xi }$.

Add the two decaying inner corrections to the outer solution, subtracting their common matching parts. Since $\delta \xi =1-x$, the composite asymptotic expansion isIt reproduces all three boundary conditions to the retained algebraic orders, with exponentially small interactions between endpoints. The right endpoint has an $O(1)$ boundary layer in $y$. The left correction has only amplitude $O(\sqrt{ϵ})$ in $y$, but its derivative is $O(1)$ and is necessary for the derivative boundary condition. Thus the left layer is a weak derivative boundary layer; if “genuine” refers specifically to an order-one change in the function itself, only the right layer is genuine in that sense. Both nonzero corrections belong in the expansion through $O(\sqrt{ϵ})$.

Matching spacetime symmetries to dynamical symmetries is a powerful demand for a coherent interpretation, but an unconditional equality theorem does not follow from logic or from general covariance. A spacetime symmetry preserves the structure declared physically significant: for Minkowski spacetime, the metric tensor and relevant orientations. A dynamical symmetry takes solutions to solutions with the laws' fixed backgrounds held fixed. The appropriate comparison concerns external transformations on the same physical quantities; internal gauge symmetries have no reason to be isometries of spacetime.

One direction becomes a theorem after important premises are supplied. Suppose all fixed background structure is geometrical and the matter equations are constructed covariantly from it. An automorphism of that full structure then takes solutions of those equations to solutions. For example, the Klein-Gordon equation $({\eta }^{ab}{\partial }_a{\partial }_b+m^2)\varphi =0$, in units $c=\hslash =1$, has the Poincaré symmetry of the fixed Minkowski metric. This is an explanation of symmetry inheritance conditional on a coupling prescription; it does not derive that prescription from the bare existence of a metric tensor.

A counterexample exposes the missing premise. On the same Minkowski spacetime, add a fixed timelike vector field $n^a$ and a matter termFor generic nonzero $\alpha$, a boost which preserves $\eta$ but changes $n$ is not a symmetry with that background held fixed. The full background has fewer symmetries than the metric tensor alone. Writing the equations in arbitrary coordinates and transforming $n$ as a vector field gives general covariance, not Lorentz invariance. Conversely, redundant unobservable geometrical structure could make the declared spacetime symmetry group artificially too small. Accidental symmetries of a restricted matter sector could make its dynamical symmetry group too large.

There are deeper motivations for avoiding such mismatches. If all measuring devices obey Lorentz-invariant dynamics, their behaviour supports a common Minkowski metric interpretation; adding an undetectable preferred rest structure sacrifices explanatory economy. If a preferred structure affects some device, then the original metric-only geometry was incomplete as a description of physical background structure. These observations support either revising the geometry or explaining the universal matter symmetry, but they are methodological and interpretive arguments, not an automatic metaphysical identity.^[ref]

Two qualifications matter. Boundary conditions and initial conditions can select a rest frame in an individual solution while the laws remain boost symmetric. Also, a symmetry of the complete theory need not be detectable by transforming only a subsystem relative to its environment. The best principle therefore matches the physically relevant geometry to the external symmetries of the complete laws, after all backgrounds and couplings are included. Calling it merely a bookkeeping preference understates its explanatory value; calling it an unavoidable consequence of covariance overstates its logical force.

Gauss's law states that the outward electric flux through a closed surface is the enclosed electric charge divided by the vacuum permittivity:In electrostatics, the electric field is the negative gradient of the electrostatic potential, $\mathbf{E}=-\nabla V$. Substitution into the differential equation form of Gauss's law gives Poisson's equation ${\nabla }^{2}V=-\rho /{ϵ}_0$. Consequently, in a region containing no charge density,which is Laplace's equation. Charges on the bounding conductors enter through the boundary conditions; they need not vanish for Laplace's equation to hold inside the empty gap.

Neglect fringing field near the edges of the parallel-plate capacitor. Put the plates at $z=0,d$, with prescribed electric potentials $V_A,V_B$. Translational symmetry parallel to a plane makes the electrostatic potential independent of the coordinates parallel to the plates. Thus Laplace's equation reduces to $d^{2}V/d{z}^2=0$. Its general solution is $V=A+Bz$, and the two boundary conditions determine both constants:The electric field is uniform and points toward decreasing electric potential. No plate voltages are specified, so a numerical electric potential cannot be fixed; choosing one plate as the reference only fixes the arbitrary additive constant.

Put $\kappa =e^2/(4\pi {ϵ}_0)$. Substitution of $\psi =R(r)\Phi (\varphi )$ into the time-independent Schrödinger equation, followed by division by $R\Phi$ and multiplication by $r^2$, separates the angular dependence:Thus $\Phi =e^{im\varphi }/\sqrt{2\pi }$ andThe single-valued wavefunction condition, $\Phi (\varphi +2\pi )=\Phi (\varphi )$, requires $m\in \mathbb{Z}$. The regular boundary condition at the origin selects $R\sim r^{|m|}$. Accordingly use $s=|m|$ in the radial ansatz; the printed $x^m$ notation describes the $m\ge 0$ representatives.

For the two-dimensional Coulomb bound state, substitution of $R=x^s{e}^{-x/2}G$ with the stipulated radial and energy scales givesThe coefficient of $x^n$ in the power series isThe regular solution otherwise has an exponentially growing contribution at large $x$, incompatible with being normalizable. Termination at degree $n_r\ge 0$ therefore requiresThe resulting $G$ is proportional to a Generalized Laguerre polynomial.

Define $R_{\mathrm{y}}=\mu {\kappa }^2/(2{\hslash }^2)$, the hydrogenic Rydberg energy for the stipulated mass. The first three distinct two-dimensional energies are $-4R_{\mathrm{y}},-4R_{\mathrm{y}}/9,-4R_{\mathrm{y}}/25$, whereas in three dimensions they are $-R_{\mathrm{y}},-R_{\mathrm{y}}/4,-R_{\mathrm{y}}/9$. Both series approach zero from below, but the planar ground state is four times more tightly bound. Without spin the respective quantum degeneracies are $1,3,5$ and $1,4,9$; if individual degenerate states are counted, the second and third planar states already share the first excited energy. This comparison retains the supplied $1/r$ interaction; it does not replace it by the logarithmic electrostatic potential of a genuinely two-dimensional electromagnetic theory.

For a narrow packet, phase velocity is $u_p=\omega /k$ and group velocity is $u_g=d\omega /dk$. Set $b=\sigma /{\rho }_1$. The uncovered surface hasEquality occurs at $k_{\ast }^2=g/b$, also the minimum of $u_p$. Numerically,The long-wave gravity limit has $u_g/u_p\to 1/2$; the short-wave capillary limit has $u_g/u_p\to 3/2$.

A compact packet subsequently spreads because different wavenumbers have different group speeds. In the gravity regime $\omega \simeq \sqrt{gk}$, so smaller $k$ and longer waves move ahead. Individual crests move twice as fast as the envelope, appearing at its rear and disappearing at its front. In the capillary regime $\omega \simeq \sqrt{b}{k}^{3/2}$, shorter waves move ahead; the envelope overtakes its slower individual crests, which drift backward relative to it. Thus both the wavelength ordering within the spread packet and the crest motion relative to its envelope reverse.

For the sheet define $c=a{\rho }_2/{\rho }_1=8.0\times 10^{-5}\mathrm{m}$. The elastic-cover frequency ratio for water waves isAt small $k$, $F=1-ck/2+O(k^2)$: sheet inertia lowers the frequency, while bending first enters at higher order. At sufficiently large $k$, bending dominates the restoring force and sheet inertia dominates the extra mass, giving $F\sim \sqrt{B/(\sigma c)}{k}^{1/2}$. The curve initially dips below one, reaches a minimum, then rises and crosses one. This asymptote describes the supplied ideal dispersion law; arbitrarily large $k$ eventually violates a thin-sheet continuum model.

For a pinned-square surface displacement mode, $k_{nm}=\pi \sqrt{n^2+m^2}/L$ with $n,m\ge 1$. The fundamental has $k_{11}=74.05{\mathrm{m}}^{-1}$. The uncovered frequency is $4.3729\mathrm{Hz}$ and $F(k_{11})=0.997051$, soThis uses a square two-dimensional surface mode, not the one-dimensional $\pi /L$ estimate. A clamped plate or additional hydraulic boundary conditions would require a different mode calculation.

With periodic boundary conditions, each $k$ state occupies volume $(2\pi {)}^3/V$ and has two spin states. Filling a sphere gives $n=k_F^3/(3{\pi }^2)$ andHere $D$ is density of states per volume and per energy; the extensive $g(E)$ is $VD(E)$. Both spin states are already included.

For divalent magnesium, $n=8.6\times 10^{28}{\mathrm{m}}^{-3}$, givingAt $T\ll T_F$, only electrons within about $k_{B}T$ of the Fermi surface change occupancy. The Sommerfeld heat capacity isEquivalently $C_{e,m}=0.995T\mathrm{mJ}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$, with $T$ in kelvin, so at room temperature it is about $0.299\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$. A temperature was not specified for the estimate; the linear coefficient is the general result.

At sufficiently low temperature, an ordinary nonsuperconducting metal has $C\simeq \gamma T+\beta T^3$: electrons dominate at the lowest temperatures, acoustic phonons produce the cubic contribution, and the lattice approaches $3R$ per mole as temperature exceeds its characteristic Debye scale. Anharmonic expansion can distinguish measured $C_p$ from $C_V$. In the room-temperature regime, $T/T_F\simeq 0.0036$ since $T_F\simeq 8.24\times 10^4\mathrm{K}$, so electron excitation is strongly Pauli-limited. The free-electron estimate is only about $1.2\%$ of the classical $3R$ lattice scale. Calling it negligible means small relative to the lattice contribution, not identically zero or negligible as $T\to 0$. Real band structure and interactions can alter the coefficient.

Stationary states. Inside the infinite square well, the time-independent Schrödinger equation has sinusoidal solutions. The boundary conditions $\varphi (0)=\varphi (a)=0$ select $k_{n}a=n\pi$; normalization then givesEach wavefunction vanishes outside the well. There is no nonzero $n=0$ state.

Immediate energy measurement after expansion. For the one-sided sudden expansion of an infinite square well, take the original wall at $x=0$ as fixed, so the expanded well occupies $0<x<2a$. In the sudden approximation, the initial wavefunction remains $\sqrt{2/a}\sin (\pi x/a)$ on $0<x<a$, extended by zero elsewhere. The new ground state is ${\chi }_1=a^{-1/2}\sin (\pi x/(2a))$. Its probability amplitude isThe Born rule therefore givesThe motion of the walls matters: a symmetric expansion about the old midpoint would instead give $64/(9{\pi }^2)$. The stated width alone does not explicitly identify the final endpoints. The first result uses the natural continuation of the specified left boundary; if both boundaries are moved, the final location must be supplied. For a new well $(b,b+2a)$ containing $(0,a)$, the overlap is $c_1(b)=(4\sqrt{2}/(3\pi ))[\cos (\pi b/(2a))-\sin (\pi b/(2a))]$.