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A useful plot must keep three different times separate: source emission $t_e$, atomic absorption or re-emission $t$, and return reception $t_R$. For the cloud centroid,Let $q=f_0/f<1$. Resonant excitation requires $q{D}_{\mathrm{in}}(t)=1$. Since $D_{\mathrm{in}}$ decreases from $\gamma (1+\beta )$ to $\gamma (1-\beta )$, its zero-width resonance first becomes possible atAt equality the line centre is reached only at infinite approach distance, where illumination vanishes. A finite source start time further limits how closely that ideal threshold is approached. For $\beta >{\beta }_{\min }$ the line-centre event satisfiesThere is a second useful critical speed,for which excitation is centred on closest approach, $t_{\ast }=0$. For ${\beta }_{\min }<\beta <{\beta }_{\perp }$ it occurs on approach; for $\beta >{\beta }_{\perp }$ it occurs after closest approach. Even a receding ultra-relativistic atom can see the incident light blueshifted because $\gamma$ is large.

A photon emitted toward the origin near the atomic rest-frame frequency $f$ has return Doppler factorThe sign differs from that in $D_{\mathrm{in}}$ because the returning ray travels in the opposite direction. The returning line centre decreases from $\gamma (1+\beta )f$ on early approach to $\gamma (1-\beta )f$ on late recession, and is $f/\gamma$ at closest-approach emission. The origin receives that closest-approach radiation at $t_R=L/c$, not at zero. Prompt re-emission at exact resonance would giveIt is above $f_0$ below ${\beta }_{\perp }$, equal at ${\beta }_{\perp }$, and below $f_0$ above it. A finite fluorescence lifetime shifts emission later along the orbit and therefore usually toward a lower return frequency.

The following plots adopt a compact, optically thin cloud, weak excitation, isotropic incoherent spontaneous emission in the atomic rest frame, and negligible velocity dispersion. They show an intensity envelope as the observable amplitude; its rms electric-field amplitude is proportional to the square root of this intensity. An incoherent fluorescence signal has no fixed optical phase. The four rows include a nonrelativistic speed, both critical speeds, and an ultra-relativistic speed. All rows share one intensity normalization, so off-resonant signals remain visibly weak.

A model for fluorescence from a uniformly moving cloud generating the plotted envelopes iswhere $\mathcal{L}(\Delta )=[1+4{\Delta }^2/{\Gamma }^2{]}^{-1}$ is a Lorentz line shape of full width $\Gamma$. The factor $1-\beta s$ is the incident photon-flux correction per source-frame time. The population is an exponential memory of earlier absorption events, not an instantaneous resonance indicator. For a detector of fixed small area at the origin, its received bolometric intensity scales asTwo powers of $D_{\mathrm{out}}$ come from aberration of light, one from photon energy, and one from arrival-rate compression. Finite lifetime is dilated to $\gamma \tau$ in the source frame; over an interval with slowly changing geometry its observed decay time is $\tau /D_{\mathrm{out}}$. Thus approach can compress and amplify a pulse, while recession stretches and redshifts it. Illumination and collection each supply an inverse-square geometric factor in the prompt weak-excitation limit. Strong forward beaming can even make a detuned approaching precursor brighter than the later line-centre feature, as in the ultra-relativistic row; the intensity maximum need not occur at exact resonance.

Absolute amplitude and a unique detailed spectral line profile cannot be inferred from $f,\tau ,v$ alone: cloud size, atom number, source power, optical depth, angular emission pattern and saturation were not specified. Finite cloud size convolves these point-centroid curves with additional delays. The plots describe incoherent, frequency-redistributed emission near $f$; a coherent Rayleigh scattering component would instead retain the driving frequency in the atom frame and add a different spectral component. These qualifications do not change the kinematic resonance thresholds or return-time relations.

An invariant has the same value when the same physical event or state is described in different inertial frames. A conserved quantity has the same value as the physical system evolves, subject to its equations and isolation conditions. These are different comparisons. The invariant mass of a heated object can change with time, while an isolated system's conserved energy has different numerical values in boosted frames.

Under a Lorentz transformation $x^{\prime \mu }={\Lambda }^{\mu }{}_{\nu }{x}^{\nu }$, a Lorentz four-vector transforms asA covector transforms by the inverse matrix, $W_{\mu }^{\prime }=({\Lambda }^{-1}{)}^{\nu }{}_{\mu }{W}_{\nu }$. A general tensor has one corresponding factor for each index; for example,Contraction with the Minkowski metric produces Lorentz scalars; it does not by itself make those scalars conserved.

Use $\hslash =c=1$ and the mostly-plus Minkowski metric. Then the source's Klein-Gordon equation is $({\partial }_t^2-{\nabla }^2+m^2)\varphi =0$. For a plane wave $\varphi =C{e}^{-i\omega t+i\mathbf{k}\cdot \mathbf{x}}$, substitution givesThus a real propagating wave vector requires $|\omega |\ge m$, or $|E|\ge m{c}^2$ on restoring units. The positive-energy branch has a rest-energy threshold $m{c}^2$. At $|\omega |=m$, $\mathbf{k}=0$; frequencies below the mass threshold give evanescent spatial dependence rather than real-wave-vector propagation. Both signs of $\omega$ occur in the classical equation; a real scalar field combines complex-conjugate modes. Its group velocity is $\mathbf{k}/\omega$ and is subluminal, though the positive-branch phase speed exceeds one when $m>0$.

For a static spherically symmetric scalar field, the equation outside the source isSetting $\psi =r\varphi$ gives ${\psi }^{\prime \prime }-m^2\psi =0$, soDecay at infinity selects $B=0$, the Yukawa potential. It is singular at the origin and describes a point source distributionally; a source-free solution regular at the origin and decaying at infinity must be zero. For $m=0$, $\varphi =A/r+B$, and decay sets the constant $B$ to zero.

Write $\Gamma =(1-v^2/c^2{)}^{-1/2}$ for the Lorentz factor of the inertial frame $S^{\prime }$, distinguishing it from the particles' changing Lorentz factors. Apply the Lorentz transformationto each world line in a. Both particles initially have velocity $-v$ in $S^{\prime }$. The switching events have coordinatesThus the Positron starts accelerating earlier in $S^{\prime }$. Its velocity first becomes less negative, vanishes when its laboratory velocity reaches $v$, and then becomes positive. The electron continues at $-v$ until $t^{\prime }=0$, then accelerates toward $-c$. The relativistic velocity-addition formula makes this precise:The electric field parallel to the Lorentz boost has unchanged magnitude, but its switching surface is $t^{\prime }+v{x}^{\prime }/c^2=0$, rather than $t^{\prime }=0$ everywhere.

In the laboratory, the total three-momentum is zero for all $t$: the two Lorentz forces cancel at equal laboratory times. In $S^{\prime }$, each active force is still $qE$, since this is one-dimensional motion with a parallel electric field. Define $t_p^{\prime }=-\Gamma vd/c^2$. For $v>0$ the total three-momentum, evaluated at equal $t^{\prime }$, isThis follows by integrating $d{p}_p^{\prime }/d{t}^{\prime }=F$ from its switching event and $d{p}_e^{\prime }/d{t}^{\prime }=-F$ from its later switching event. During the interval between them only the Positron feels the prescribed electric field, so momentum conservation for the particle pair fails. Once both forces act, the total three-momentum is constant again.

The equal-time momentum of an externally driven particle pair illustrates a distinction. The sum of particle four-momenta at a fixed pair of events is always a four-vector. However, a fresh sum at equal time in each inertial frame generally uses different events. Here transforming an equal-$t$ laboratory sum would give $-2\Gamma v\sqrt{m^2{c}^4+F^2{c}^2{t}^2}/c^2$, which varies with $t$; this differs from the equal-$t^{\prime }$ answer above. For example, with $c=m=F=d=1$ and $v=0.6$, the late-time equal-$t^{\prime }$ sum is $-0.75$, whereas transforming the equal-$t=0$ sum gives $-1.5$. This is a concrete failure of treating the pair's equal-time sums as components of a single frame-independent four-vector.

A closed system's total four-momentum, including the electromagnetic field and the apparatus supplying it, is conserved and independent of the choice of spacelike integration surface under the usual isolated-system boundary conditions. The particles alone exchange energy and momentum with an external system, so neither that independence nor universal particle-only momentum conservation is guaranteed.

With the reversed electric field, reflection about $x=d/2$ exchanges the electron and Positron. They therefore meet there with equal and opposite momenta. Each travels $d/2$, and the work-energy theorem gives its kinetic energy:Their combined kinetic energy is $1.000\mathrm{MeV}$, while each total relativistic energy is $\mathcal{E}=0.511+0.500=1.011\mathrm{MeV}$. The laboratory is the center-of-momentum frame of the colliding pair.

Local four-momentum conservation in the electron-positron annihilation excludes one real photon: the incoming total four-momentum is timelike, with zero spatial component and positive energy, while a single photon's four-momentum is null and has $|\mathbf{p}|=\mathcal{E}/c>0$. Two photons can balance their momenta by travelling in opposite directions, soThe equal photon energies follow from opposite equal momenta, not merely from sharing the incoming energy arbitrarily. Here $hc\simeq 1.23984\mathrm{MeV}\mathrm{pm}$. This is the free local annihilation idealization implicit in the question; significant momentum transfer to the external apparatus would change the kinematic system being considered.

The laboratory remains the center-of-momentum frame. At the relativistic collision mass threshold, the two final W bosons can have arbitrarily small opposite momenta. The available relativistic energy is $2m_e{c}^2+eEd$, soEach incoming particle must acquire about $80.378489\mathrm{GeV}$ of kinetic energy, travelling about $16.08\mathrm{km}$. Equality is the ideal kinematic threshold; a nonzero production rate needs energy above it. This estimate neglects acceleration losses and treats the W bosons with the supplied nominal rest mass.

A linear particle accelerator avoids repeatedly bending high-energy electrons and Positrons. This greatly reduces the bending synchrotron radiation loss, which at fixed bending radius grows approximately as the fourth power of the relativistic energy for a given particle species. Its useful collision energy can therefore be increased without the severe circular electron collider radiation penalty. The price is long structures or high accelerating gradients: each bunch normally passes once, requiring substantial beam power, fresh bunches, and very tight focusing for high luminosity. A circular particle accelerator reuses its accelerating structures and circulates bunches through many collisions, but needs a large ring and substantial radiofrequency power to replace synchrotron radiation losses. Heavier particles such as protons radiate far less at a given energy and radius, although a proton's constituent structure makes the hard-collision energy less sharply specified than in an electron-positron annihilation experiment.

Let $z^{\mu }(\tau )$ be the point electric charge's world line, with proper time $\tau$ and four-velocity $U^{\mu }=d{z}^{\mu }/d\tau =\gamma (c,\mathbf{u})$. For the field event $x^{\mu }$, choose the retarded source event: $R^{\mu }=x^{\mu }-z^{\mu }({\tau }_{\mathrm{ret}})$ is future-directed and null. Write $R^{\mu }=(R,R\mathbf{n})$, where $R=|\mathbf{x}-\mathbf{z}(t_{\mathrm{ret}})|=c(t-t_{\mathrm{ret}})$. In the instantaneous rest inertial frame of that source event,Consequently the covariant expression matching this Coulomb potential isThe denominator is a Lorentz scalar, so the right-hand side is a four-vector with the required instantaneous-rest components. This describes the retarded potential in Lorenz gauge, with no added free electromagnetic wave. Lorentz covariance alone does not establish the boundary conditions or exclude additional terms depending on four-acceleration. The physical input is the retarded point-source solution of Maxwell equations: the potential at the retarded event depends on the instantaneous four-current, while its derivatives generate the acceleration field. Equivalently, evaluating the retarded source integral gives a Jacobian $1/(1-\mathbf{n}\cdot \mathbf{\beta })$ from the light-cone constraint; this fixes that input for arbitrary smooth subluminal motion.^[ref]

In any inertial frame, put $\mathbf{\beta }=\mathbf{u}/c$ and $\kappa =1-\mathbf{n}\cdot \mathbf{\beta }$. Then $-R\cdot U=\gamma cR\kappa$, and the Liénard–Wiechert potential isDifferentiate using $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$ and $\mathbf{B}=\nabla \times \mathbf{A}$, including the position and time dependence of the retarded time. For example, $\partial t_{\mathrm{ret}}/\partial t=1/\kappa$ and $\nabla t_{\mathrm{ret}}=-\mathbf{n}/(c\kappa )$. The resulting Liénard-Wiechert field isHere the dot denotes a derivative with respect to source coordinate time. The first term is the velocity or non-radiative field: it falls as $R^{-2}$ and exists even without acceleration. The second is the acceleration or radiation field, falling as $R^{-1}$ and transverse to $\mathbf{n}$. Its Poynting flux falls as $R^{-2}$, so its outward power integrated over a large sphere stays finite as its radius tends to infinity. The first term's squared-field flux instead vanishes in that limit. This large-distance distinction defines radiation; a moving near field can carry local energy without constituting radiation escaping to infinity.

For uniform circular motion, the transverse Lorentz force supplies the change in direction of three-momentum:The magnetic rigidity relation gives $pc=0.299792458B[\mathrm{T}]r[\mathrm{m}]\mathrm{GeV}=2.99792458\mathrm{GeV}$. Using $m_e{c}^2=0.511\mathrm{MeV}$,The coordinate centripetal acceleration is $a=u^2/r$. For transverse acceleration, the magnitude of the four-acceleration, or proper acceleration, is $a_0={\gamma }^{2}a$. The supplied Larmor formula, in its proper acceleration form, therefore givesThe emitted power here is also the laboratory total emitted power: in the instantaneous rest inertial frame the dipole radiation pattern has zero net radiated three-momentum, so both emitted energy and elapsed coordinate time gain the same Lorentz factor on transformation. This does not assert that a directional detector flux is invariant.

Numerically $\mathcal{T}\simeq 1.048\times 10^{-5}\mathrm{s}$, $P\simeq 2.18\times 10^{-10}\mathrm{W}$ per electron, andThe synchrotron's radiofrequency accelerating system must replace this lost energy to keep $r$ and magnetic rigidity fixed over many revolutions. The fraction lost per turn is about $4.77\times 10^{-6}$, so a fixed-orbit calculation over one turn is self-consistent.^[ref]

Use the electromagnetic field tensor convention stated above, and lower indices with the Minkowski metric $g=\mathrm{diag}(-1,1,1,1)$. Lowering a single time index changes its sign. Summing both orders of each antisymmetric index pair gives the Lorentz scalarFor $T^{00}$, $F_j{}^0=F^{j0}=-E_j/c$ and $g^{00}=-1$. Direct substitution in the printed electromagnetic stress-energy tensor yieldsFor a spatial index $i$, $g^{0i}=0$ and $F_j{}^i=F^{ji}=-{ϵ}_{ijk}{B}_k$. HenceThus the full top row isHere $u_{\mathrm{EM}}$ is the electromagnetic energy density and $\mathbf{S}$ the Poynting vector, using ${ϵ}_0{\mu }_0{c}^2=1$. By the symmetry $T^{0i}=T^{i0}$, these components describe both energy flux divided by $c$ and $c$ times the electromagnetic momentum density $\mathbf{g}=\mathbf{S}/c^2$. Keeping the factor $c$ is essential because the time coordinate is $ct$.

Let starred quantities denote the capacitor's rest inertial frame, $u_0={ϵ}_0{E}^2/2$, $\beta =v/c$, and $\gamma =(1-{\beta }^2{)}^{-1/2}$. Work in the uniform interplate region of an ideal parallel-plate capacitor, neglecting its fringing field of a capacitor. To transform from that rest inertial frame to a laboratory in which the capacitor moves along $+x$, use the positive-sign Lorentz transformationFor a purely electric field, the spatial block of the electromagnetic stress-energy tensor isThis is the negative of the Maxwell stress tensor: it has $-u_0$ along the electric field and $+u_0$ perpendicular to it. The stress contribution to energy transport in a moving capacitor is essential to the Lorentz transformation of the energy density and flux.

Plates parallel to $xy$. Choose ${\mathbf{E}}_{\ast }=E\hat{\mathbf{z}}$ and ${\mathbf{B}}_{\ast }=0$. The electromagnetic field tensor and electromagnetic stress-energy tensor areThe transformed nonzero independent electromagnetic field tensor components are $F^{03}=\gamma E/c$ and $F^{13}=\gamma \beta E/c$. Thus $\mathbf{E}=\gamma E\hat{\mathbf{z}}$ and $\mathbf{B}=-\gamma vE\hat{\mathbf{y}}/c^2$, consistent with $\mathbf{B}=\mathbf{v}\times \mathbf{E}/c^2$. Transforming the electromagnetic stress-energy tensor givesIn particular, the laboratory Poynting vector is $\mathbf{S}=2u_0{\gamma }^{2}v\hat{\mathbf{x}}$, and the electromagnetic energy density is $u=u_0{\gamma }^2(1+{\beta }^2)$. The ratio $S_x/u=2v/(1+{\beta }^2)$ is subluminal for $v<c$, but differs from the capacitor's velocity. The difference from a simple convective flux isThe extra term is $v{T}_{\ast }^{xx}$, the rest-frame stress multiplied by the velocity. This shows why treating stored field energy as an unstressed parcel moving rigidly with the plates misses part of the energy transport.

Plates parallel to $yz$. Now ${\mathbf{E}}_{\ast }=E\hat{\mathbf{x}}$. The rest tensors areThe electric field is parallel to the Lorentz boost, so $F^{01}=E/c$ remains unchanged and $\mathbf{B}=0$. The cancellation between $T_{\ast }^{00}$ and $T_{\ast }^{11}$ givesThere is no Poynting flux in the ideal interior in this orientation, although the region containing the electric field moves. The longitudinal negative stress contribution cancels the convective term: $S_x={\gamma }^{2}v(u_0+T_{\ast }^{xx})=0$. Field energy is removed at the receding boundary and supplied at the advancing boundary by work on the charged plates; it need not flow through the interior to accompany the moving region. At edges, omitted fringing fields provide additional local structure.

For a finite idealized interplate rest volume $V_0$, length contraction gives $V=V_0/\gamma$. The field-only integrated laboratory energies would consequently be $U_{xy}=\gamma (1+{\beta }^2)U_0$ and $U_{yz}=U_0/\gamma$, with $U_0=u_0{V}_0$. Their different transformation laws are not paradoxical: the field alone is not a closed system. Include the plates and the supports holding them apart, whose mechanical stresses balance the Maxwell stress tensor. The total closed-system four-momentum then has the ordinary Lorentz transformation law. This is the same distinction between a subsystem and a conserved whole that appears in b.

By four-momentum conservation, $p_{\pi }=p_1+p_2$. With the $(-,+,+,+)$ Minkowski metric, the neutral pion has $p_{\pi }^2=-M_{\pi }^2$, while each photon has $p_i^2=0$ and $|{\mathbf{p}}_i|=E_i$. Taking the Minkowski norm givesThe half-angle identity therefore yieldsThis is a measurement of the diphoton invariant mass using two photon energies and their opening angle, independent of the neutral pion's laboratory velocity.

In the center-of-momentum frame, four-momentum conservation gives two opposite photon momenta, each of energy $M_{\pi }/2$. Let $z=\cos {\theta }_{\ast }$, with the axis along the Lorentz boost from that frame to the laboratory. Reversing the axis specified in the source changes $z$ to $-z$ and has no effect on the following product. The Lorentz transformation of four-momentum givesWriting $s=\sin (A/2)$, b now impliesThe isotropy assumption is $dN=a|dz|$ on $-1\le z\le 1$. Each interior opening angle has two inverse branches, $z$ and $-z$, both of which must enter the change of variables for probability densities. The positive branch has derivativeConsequently the diphoton opening-angle distribution isIts support is $2\arcsin (1/\gamma )<A<\pi$ for $0<\beta <1$, and its integral is $2a$, the integral of the original angular probability density before normalization. Thus the normalized probability density is $(2a{)}^{-1}dN/dA$. Omitting one inverse branch would incorrectly halve the total count.

Metric-sign qualification. With this paper's $(-,+,+,+)$ metric signature, b givesThe minus sign in the printed Larmor formula would therefore make a positive radiative loss negative. A concrete counterexample is an instant of rest with nonzero acceleration: $F^{\alpha }=(0,m\mathbf{a})$, so the printed expression is $-q^2{a}^2/(6\pi {ϵ}_0{c}^3)$. That sign is appropriate for the opposite metric signature. With the convention already fixed in question 1, positive emitted power instead isAlternatively the printed negative quantity can be interpreted as the particle's signed radiative energy change, rather than the positive loss magnitude.

For fixed rest mass, resolve the acceleration parallel and perpendicular to the velocity, as in b. This givesWriting $C_q=q^2/(6\pi {ϵ}_0{c}^3)$, the Liénard radiated-power formula becomes $P_{\mathrm{rad}}=C_q({\gamma }^6{a}_{\parallel }^2+{\gamma }^4{a}_{\perp }^2)$. For uniform circular motion, the acceleration is perpendicular to the velocity; for straight-line motion it is parallel. Comparing the same particle at equal relativistic energy and equal laboratory acceleration magnitude $a$ therefore givesThis particular comparison makes the linear loss larger. The familiar larger loss from bending applies to a different comparison: at equal applied laboratory force magnitude $F$, $a_{\perp }=F/(m\gamma )$ and $a_{\parallel }=F/(m{\gamma }^3)$, giving $P_{\mathrm{circular}}=C_q{\gamma }^2{F}^2/m^2$ and $P_{\mathrm{linear}}=C_q{F}^2/m^2$. Stating which quantity is held fixed is essential.

Both the muon and electron have the same electric charge magnitude. At equal kinetic energy $K$ per particle, their Lorentz factors are $\gamma =1+K/(m{c}^2)$, so the heavier muon has the smaller Lorentz factor. At equal total relativistic energy the same ordering follows from $\gamma =E/(m{c}^2)$, provided that energy is large enough to create either particle.

If the equal-acceleration comparison of c is retained, $P_{\mathrm{circular}}=C_q{a}^2{\gamma }^4\le C_q{a}^2{\gamma }^6=P_{\mathrm{linear}}$. Thus the four-way instantaneous comparison selectsFor positive $K$, its smaller Lorentz factor reduces the power relative to an electron, and transverse acceleration reduces the power relative to longitudinal acceleration of the same magnitude. This conclusion follows from the constraint actually stated in c.

For the customary engineering comparison at equal applied force, the result changes. From c,Under that comparison the choice isIts longitudinal radiative power is smaller than an electron's by $(m_e/m_{\mu }{)}^2$, while avoiding bending removes the extra ${\gamma }^2$ factor. At a fixed bend radius $R$, the circular loss is $C_q{\gamma }^4{v}^4/R^2$, giving the familiar $m^{-4}$ suppression at fixed ultrarelativistic energy and $R$.

An energy budget alone does not specify a unique optimization. The force, acceleration, bend radius, acceleration time, and length are otherwise unrestricted. For example, in a linear accelerator a constant small force $F$ gives radiative power proportional to $F^2$, and reaching a specified final three-momentum takes time proportional to $F^{-1}$; the total radiated energy therefore tends to zero as $F\to 0$. One cannot infer a unique accelerator choice without an additional constraint. The standard practical preference is the second boxed answer, with its stated comparison. Muon decay introduces a separate lifetime constraint if slow acceleration is contemplated, but it was not part of the specified radiative-loss calculation.

The first energy term is the rate of change of internal plus kinetic energy per unit volume; the second transports that energy with the moving gas; the pressure term is the mechanical work flux across the boundary. There is no heat flux or viscous work in the stated inviscid adiabatic model.

Use mass conservation to replace a conservative derivative of $\rho e$ by $\rho De/Dt$. Dotting the momentum equation with $\mathbf{u}$ gives $\rho D(|\mathbf{u}{|}^2/2)/Dt=-\mathbf{u}\cdot \nabla p$. Subtraction from total energy leavesNow $p=\rho RT$ and $D\rho /Dt=-\rho \nabla \cdot \mathbf{u}$ implyThus $\boxed{DS/Dt=0}$: smooth adiabatic inviscid trajectories are isentropic. Shocks need an entropy inequality rather than this smooth equality.

Adding $\nabla \cdot (k\nabla T)$ models Fourier heat conduction, with heat flux $-k\nabla T$. The same subtraction gives $\rho c_{v}DT/Dt+p\nabla \cdot \mathbf{u}=\nabla \cdot (k\nabla T)$, henceFor a fixed container use impermeability $\mathbf{u}\cdot \mathbf{n}=0$, insulation $k\nabla T\cdot \mathbf{n}=0$, and mass conservation. ThenThe boundary term vanishes and $\nabla (1/T)=-\nabla T/T^2$ supplies the positive sign. For $T>0$, $k\ge 0$, total entropy cannot decrease: this is entropy production by Fourier heat conduction.

Rigid walls require zero normal velocity, so ${\varphi }_x=0$ at $x=0,L$ and ${\varphi }_y=0$ at $y=0,b$. Separation of variables gives spatial modes $\cos (m\pi x/L)\cos (n\pi y/b)$, for nonnegative integers $m,n$. The wave equation therefore givesThe $(0,0)$ spatially constant mode has zero frequency and is not an ordinary positive-frequency acoustic oscillation.

A spatially uniform boundary forcing excites the $n=0$ transverse mode. Put $\varphi =X(x)\cos \omega t$. Then $X^{\prime \prime }+(\omega /c_0{)}^{2}X=0$ and $X^{\prime }(L)=0$, hence $X=C\cos [\omega (L-x)/c_0]$. The condition $X^{\prime }(0)=a$ fixes $C$ and givesArbitrary free oscillations can be superposed. The denominator vanishes at longitudinal resonances. Near them the lossless steady harmonic response ceases to be a uniformly small-amplitude approximation; damping from viscosity, heat transfer or wall losses regularizes the response, and large amplitudes can require nonlinear equations of motion or a more accurate moving-boundary model. At exact resonance a switched-on ideal forcing produces secular growth rather than a bounded harmonic particular solution.

For the semi-infinite duct expand $F(y)={\sum }_{n\ge 0}{F}_n\cos (n\pi y/b)$. Each axial mode hasFor a propagating mode a particular solution is ${\varphi }_n=(F_n/k_n)\cos (n\pi y/b)\cos (k_{n}x-\omega t)$; differentiating at zero gives the prescribed $F_n\cos (n\pi y/b)\sin \omega t$. A nonzero higher transverse coefficient propagates when $|\omega |>n\pi c_0/b$.

The outgoing radiation condition selects $c_0^2{k}_n/\omega >0$, its positive axial group velocity. Thus $\boxed{\omega k_n>0}$. This sign is selected by the absence of waves incident from infinity, not by the differential equation alone. If $F$ has a nonzero mean, its $n=0$ component can propagate even below every higher-mode cutoff.

For $F=\cos (\pi y/b)$ and $0<|\omega |<\pi c_0/b$, put $\kappa =\sqrt{(\pi /b{)}^2-{\omega }^2/c_0^2}$. The decaying solution isIt is evanescent and carries no mean energy to infinity. These mode and sign conditions are summarized by the radiation condition for a waveguide mode.

Let the bed be $z=0$ and the free surface $z=h(x,t)$. In a long, shallow layer, vertical acceleration is smaller than gravity, so the vertical Euler equation gives hydrostatic pressure $p=p_{\mathrm{atm}}+\rho g(h-z)$. Incompressibility makes the vertical speed smaller than the horizontal speed by the aspect ratio; irrotationality then makes the leading horizontal velocity depth-independent. The horizontal Euler equation becomesIntegrating $u_x+w_z=0$ across the depth, using bed impermeability $w(0)=0$ and the surface condition $w(h)=h_t+u{h}_x$, yieldsCombining these gives the conservative momentum equation $(hu{)}_t+(h{u}^2+g{h}^2/2{)}_x=0$. Integral mass and momentum balance across a stationary thin jump therefore requireThese conservation laws still apply across a jump where the smooth hydrostatic/irrotational derivation does not resolve the internal turbulent structure.

Put $A=h_{-}$, $B=h_{+}$ and $q=A{u}_{-}=B{u}_{+}>0$, orienting the flow from minus to plus. For a nontrivial jump the momentum condition gives $q^2=gAB(A+B)/2$. Its change in mechanical-energy flux isA passive jump cannot create mechanical energy, so the change is nonpositive. Since $q>0$, $\boxed{h_{+}>h_{-}}$ for a genuine jump; equality describes no jump. The upstream flow is supercritical and the downstream flow subcritical, since $F{r}_{-}^2=B(A+B)/(2A^2)>1$ and $F{r}_{+}^2=A(A+B)/(2B^2)<1$.

The gas-dynamic analogue is a compressive normal shock: density and pressure increase and flow changes from supersonic to subsonic. Entropy increases; directed mechanical energy is irreversibly converted into internal energy. Total gas energy, including internal energy, is conserved across an adiabatic shock. This distinguishes that total-energy balance from the reduced shallow-water mechanical-energy loss.

Insert the harmonic variables into the forward map and reduce the squared trigonometric functions. This givesIndeed $p=m\dot{x}$ at this order. The second harmonic and mean displacement of a weakly cubic oscillator show asymmetry of the motion: for positive $\lambda$ the mean shifts towards negative $x$. There is no first-order frequency shift, although the waveform is no longer purely sinusoidal.