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The photon leaves the front rocket at $(t,x)=(0,a+L)$ and moves toward decreasing $x$, so its laboratory path is $x_{\gamma }(t)=a+L-ct$. Intersect it with the rear hyperbola, keeping the future root. Squaring and cancelling $c^2{t}^2$ givesThe rear rocket's speed there isAt emission the front rocket is instantaneously at rest in S, so the photon laboratory frequency is ${\omega }_H$. At reception the rear rocket moves in $+x$ while the photon travels in $-x$. The Doppler factor for this head-on reception isThe receiver sees a blueshift. It is essential to use the front velocity at emission and the rear velocity at reception; comparing the two velocities at one laboratory time would describe the wrong pair of measurements. The result also agrees with the Rindler light-transfer frequency ratio for static accelerated observers at radii $a$ and $a+L$. In the weak-acceleration limit $L/a\ll 1$, the shift is $1+L/a$, and the arrival time approaches $L/c$, providing useful checks.

Let $t$ label the reception event in the source reference frame. Its distance from the source and the relevant retarded time areThe received phase is $2\pi f_0{t}_e$. Because the observer's proper time obeys $d\tau =dt/\gamma$, its measured frequency isEquivalently, write $\beta =v/c$ and $s(t)=vt/r(t)$. The incoming Doppler factor is $D_{\mathrm{in}}=\gamma (1-\beta s)$, obtained also by contracting the photon four-momentum with the observer's four-velocity. The frequency decreases monotonically, withIn particular, the instant of closest approach is not the zero-shift point: a moving receiver looking perpendicular to its motion sees a factor $\gamma$, not $1/\gamma$. The latter factor applies to a different moving-emitter geometry. The graph uses source-frame reception time, with observer proper time $\tau =t/\gamma$.

An excited-state population lifetime $\tau$ gives a natural linewidth whose Lorentzian full width in ordinary frequency iswhen no extra dephasing broadens it. Additional thermal motion, collisions or inhomogeneity may increase the effective width. Therefore excitation is appreciable over $|D_{\mathrm{in}}{f}_0-f|\lesssim \Gamma /2$, rather than only at one event. A Lorentz line shape has nonzero wings, so ${\beta }_{\min }$ is a line-centre threshold, not a mathematically sharp cutoff of all signal. Below it, only a weak detuned wing may be excited; at it, finite-width illumination produces a broad weak precursor rather than a delta-function burst at infinity.

For a narrow line with a finite resonance time, a useful local excitation duration isNear the far-approach threshold this linear estimate becomes large and the full line shape and inverse-square illumination must be retained. After re-emission, the received full width is approximately $D_{\mathrm{out}}\Gamma$ at a fixed emission event. Lifetime memory and the changing return Doppler factor additionally produce a chirp and a redder tail.

The plots in part i use $\Gamma /f=0.02$ and $c\tau /L=0.4$, consistently corresponding to $fL/c=1/[2\pi (0.4)(0.02)]$ if the width is entirely natural. They are illustrative dimensionless parameters, not an asserted measurement for a specified atomic species.

Decreasing $q=f_0/f$ increases the blueshift required for excitation. Both characteristic speeds move toward $c$:If $q$ is near one, modest approach speeds can excite the line. If $q\ll 1$, only strongly relativistic motion reaches the line centre. At a fixed qualifying speed, decreasing $q$ moves the resonance toward earlier approach, since a larger $D_{\mathrm{in}}$ is required. It can also move the event outside the illumination window set by $t_e\ge -T$.

At fixed $q$, increasing $\beta$ moves the resonance from the far approaching tail, through closest approach at ${\beta }_{\perp }$, and eventually to the receding side. The received pulse frequency cannot be estimated simply by applying the same blueshift twice: the inward and outward Doppler factors are different, and decay may occur well after excitation. The critical-speed rows of part i make these changes explicit.

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.

Hamilton's equations give $\dot{X}=P/m+O({\lambda }^2)$ and $\dot{P}=-m{\omega }^{2}X+O({\lambda }^2)$. Hence, with constants $A\ge 0$ and $\delta$ fixed by transformed initial data,An equivalent form uses $X(0)\cos \omega t+[P(0)/(m\omega )]\sin \omega t$ for $X(t)$. The indicated errors are a fixed-time perturbative statement; a higher-order frequency correction can accumulate phase on longer times.

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.

The small parameter is $|\lambda |A/(m{\omega }^2)\ll 1$. A purely cubic perturbation is unbounded in one direction, so these formulas describe small local oscillations around the stable origin, not globally confined motion at arbitrary energy. Amplitude $A$ labels the transformed orbit and must be inferred using the inverse map if physical initial position and momentum are supplied.

For the specified parameter relation,Thus there are $|n|$ axial oscillations per transverse revolution, and the nondegenerate three-dimensional trajectory closes after $2\pi /{\Omega }_c$. Since $\lambda >0$, the integer $n=0$ is excluded; changing its sign changes nothing. More generally commensurate cyclotron and axial oscillations close for any rational frequency ratio, whereas irrational ratios give quasiperiodic motion when both components are present. An integer ratio is sufficient, not the only closure condition. The same frequency argument also applies to transverse cyclotron circles centred away from the chosen cylindrical origin.

Put $G=g/\ell$ and $s=\kappa /m$. Applying the Euler-Lagrange equations to the quadratic Lagrangian givesA nonzero motion satisfying the specified equal-and-opposite initial displacements can have one frequency only if $(1,-1)$ is a normal-mode vector. Its acceleration terms are proportional to $(1,0)$, while the restoring terms are proportional to $(2G-2s,2s-G)$. HenceandThe rods swing oppositely. The hinge mass moves sideways, while the final mass has no first-order displacement: the two horizontal contributions cancel and its vertical change is quadratic in angle. In the corresponding geometric ansatz it moves vertically at twice the small-oscillation frequency to leading nonzero order, but the displayed single-frequency solution is a linearized result. The trivial case $\xi =0$ imposes no parameter condition.

Removing the hinge mass gives the exact LagrangianThe stated equations again concern its small-angle quadratic approximation. In the sum and difference coordinates this becomesConsequentlyFor unequal squared frequencies, $\nu =0$: the rods remain aligned to first order and the end mass swings like a pendulum of length $2\ell$. This degenerate Lagrangian gives the difference coordinate no quadratic kinetic term, so its equation is an algebraic constraint, not a second oscillator. If ${\omega }_0={\omega }_1$, the quadratic action loses all dependence on $\nu$. Its motion is then undetermined at this order; nonlinear terms or a small hinge mass must resolve the degeneracy. For larger spring stiffness the straight constrained branch is not a minimum of the bending potential, despite the absence of a finite-inertia growing mode in this strictly massless quadratic description.

Differentiate $\mu \dot{\mathbf{r}}=\mathbf{p}+e^{\ast }\mathbf{B}\times \mathbf{r}$ and substitute the previous momentum equation:For a planar vector, $\mathbf{B}\times \dot{\mathbf{r}}=B(-\dot{y},\dot{x})$, which gives the two component equations with the printed signs.

For circular relative motion of a neutral Coulomb pair of separation radius $R$, $\dot{\mathbf{r}}=\omega \hat{\mathbf{z}}\times \mathbf{r}$ and $\ddot{\mathbf{r}}=-{\omega }^2\mathbf{r}$. ThusSince $(e^{\ast }{)}^2+\mu e^2/M=e^2/4$, the two signed frequencies areOne is positive and the other negative. At $B=0$ they give the two senses of an ordinary Coulomb circular orbit; equal masses remove the term linear in $B$ from their average. With negligible Coulomb attraction and $eB>0$, the branches approach $eB/m_2$ and $-eB/m_1$, the appropriate signed free cyclotron limits.

The actual particle paths follow by integrating the centre-of-mass velocity:Both particles move on circles about the same fixed guiding centre with a constant separation and common signed frequency, with a possible relative phase reversal encoded by a negative coefficient. Their mechanical centre of mass also generally circles. This is consistent with magnetic deflection plus a central mutual force; central interaction alone does not eliminate the external Lorentz forces.

Substitution of the proposed sine and cosine displacements into either linearized equation gives $m{\omega }^2+qB\omega -k=0$. ThusThese are signed angular frequencies, not two positive frequency magnitudes. To first order, with $\epsilon \ll {\rho }_0$,The orbit is therefore a circle of radius $\epsilon$ about $({\rho }_0,0)$ at this order. The positive root runs counterclockwise; the negative root runs clockwise and has the larger magnitude. A positive charge in $B\hat{\mathbf{z}}$ alone circles clockwise. Here the spring permits an additional counterclockwise mode by providing inward force against its magnetic force.

Define the positive focusing frequencyIn the paraxial approximation $w\simeq v$, traversal time is $T\simeq l/v$. The exit values areAcross the exit fringe, $p_{\theta }=0$ and $A_{\theta }\to 0$ remove the azimuthal velocity. The radial velocity remains continuous in the thin-fringe model. Outside, the transverse trajectory is therefore straight along a fixed azimuth, not a continuing helix. Taking the outgoing axial speed also as $v$, the distance to the axis isIt is independent of ${\rho }_0$, so all equal-speed paraxial rays focus to a common point. The condition $0<\omega l/v<\pi /2$ ensures the first focus lies beyond the exit. For a weak lens, $\boxed{f\simeq 4({\gamma }_{v}mv{)}^2/(q^2{B}^{2}l)}$.

The relativistic solenoid magnetic lens focuses either charge sign, while its rotation reverses with $qB$. Keeping the exact radius-dependent $w$, or the exit axial-speed change, produces radius-dependent corrections and spoils an exact common focus. Thus the requested focus is appropriately a paraxial result, with the end fields included even when their axial-speed changes are neglected.

A retarded Green function obeys $G(\tau )=0$ for $\tau <0$ andFor a bounded or suitably tempered response represented by the real-axis Fourier integral, analyticity in the upper frequency half-plane is required, with suitable behaviour at infinity. For negative $\tau$, $e^{-i\omega \tau }$ decays on a large upper semicircle. If that half-plane has no singularities, contour closure gives zero response before the source. Lower-half-plane poles give factors decaying for positive time. Real poles are approached from below for a retarded prescription; at $\epsilon =0$, for example, $G_R(\tau )=\Theta (\tau )\sin \tau$.

The causality versus stability of a Green function distinction matters: an unstable initial-value problem can still have a strictly causal response that grows exponentially. Such a response has no ordinary real-axis Fourier transform. Its inverse Laplace transform must run on a frequency contour above its growth poles. Thus “upper poles violate causality” is correct only with the stable/tempered real-axis Fourier assumptions, not as an unconditional theorem about differential equations.

A constant electrostatic energy offset is removed locally by the quantum phase substitution $\psi =e^{-i{V}_{0}t/\hslash }{\psi }_0$, since $(i\hslash {\partial }_t-V_0)\psi =e^{-i{V}_{0}t/\hslash }i\hslash {\partial }_t{\psi }_0$. Therefore the two frequency branches becomeThe normalization uses the kinetic frequency ${\omega }_k$, while the canonical energy eigenvalues are $E=V_0\pm \hslash {\omega }_k$. The opposite-frequency branch is not obtained by replacing ${\omega }_k$ by the same shifted positive frequency in both terms. A globally constant offset produces no electric field, whereas different offsets on opposite sides of a step cannot be removed by a single smooth global phase.

First trace the state explicitly specified in (c). orthogonality in the $y$ oscillator removes terms with different occupation numbers:For a thermal oscillator of frequency $\omega$, the ratio of successive Boltzmann weights is $e^{-\hslash \omega /(k_{B}T)}$. Consequently the conditional answer for that two-mode squeezed state isAt $\theta =0$ this is interpreted by the limit $T_{\mathrm{sq}}=0$. The reduced state is mixed for nonzero squeezing although the joint state is pure; its effective thermal statistics arise from entanglement, not from heating by a bath.

The ground-state identification must still be qualified. For the true coordinate-coupled ground state, direct normal-mode averages giveTheir ratio is $m^2{\omega }_{+}{\omega }_{-}$, rather than $m^2{\omega }^2$. Therefore this reduced state is not thermal with respect to the original bare frequency $\omega$ for $k\ne 0$. It does have an exact thermal representation for an effective oscillator of mass $m$ and frequencyThe corresponding occupation basis is that of $\nu$, not that of the original $a$. This follows by matching the Gaussian variances to $⟨x^2⟩=\hslash \coth [\hslash \nu /(2k_{B}T)]/(2m\nu )$ and its momentum counterpart. A direct reduced-Gaussian eigenfunction calculation gives the same effective frequency and geometric eigenvalues.^[ref]

A Klein-Gordon solution is second order in time: both $\Psi$ and ${\partial }_t\Psi$ are needed as initial data. Its free modes have both signs of frequency,This differs from the first-order Schrödinger evolution of a fixed one-particle wavefunction.

Moreover, $|\Psi {|}^2$ does not generally obey the required probability continuity equation. The conserved Klein-Gordon density is proportional toup to a positive conventional constant. A pure positive-frequency mode gives $+2\omega |\Psi {|}^2$ and a negative-frequency mode gives $-2\omega |\Psi {|}^2$. This density is not positive definite and cannot be a universal position probability density.

For a free field, one can choose a positive-frequency subspace with a positive integrated inner product. Time-dependent backgrounds generally mix it with the negative-frequency subspace, so simply deleting the latter is not dynamically consistent. In quantum field theory the complex Klein-Gordon field instead has particle and antiparticle sectors; its conserved current is interpreted as charge current, and creation or annihilation of pairs need not fit a fixed-particle-number wavefunction. These are limitations of the naive single-particle interpretation, not a failure of the relativistic field equation.

Work in temporal gauge, with zero scalar potential, and use the charge sign exactly as in the given covariant spatial derivative. A spatially uniform vector potential acts on a Fourier mode asThere are no spatial derivatives of $\mathbf{A}$. Substitution into the field equation and multiplication by $c^2$ therefore givesThe canonical wavevector $\mathbf{k}$ is conserved by spatial translation symmetry; the kinetic momentum is $\hslash \mathbf{k}-e\mathbf{A}(t)$. The absence of a term involving $\dot{\mathbf{A}}$ in this mode equation follows from temporal gauge and the spatial minimal-coupling operator. The electric field still affects evolution through the changing oscillator frequency.

For this mode, abbreviateThe electric pulse is $\mathbf{E}(t)=-{\mathbf{A}}_0\delta (t)$. In temporal gauge the mode equation has a finite step in its frequency squared, with no delta multiplying the mode. Thus both ${\psi }_{\mathbf{k}}$ and its first time derivative are continuous at zero.

Writing the positive-frequency incident amplitude as $a_{\mathbf{k}}^{\mathrm{in}}$, the matched solution isIts initial value is $a_{\mathbf{k}}^{\mathrm{in}}/\sqrt{2\omega }$ and its initial derivative is $-i\omega a_{\mathbf{k}}^{\mathrm{in}}/\sqrt{2\omega }$, as required. In the normalized outgoing-frequency basis it isThe negative-frequency contribution appears even though only positive frequency was incident. These are temporal-mode matching coefficients, so their conserved difference is hyperbolic rather than the sum-one rule of ordinary spatial reflection and transmission probabilities.

The Euler-Lagrange equation for the real scalar field givesThus the conjugate momentum and Hamiltonian density areThe spatial Fourier transform reduces the Klein-Gordon equation to ${\ddot{\varphi }}_{\mathbf{p}}+{\omega }_{\mathbf{p}}^2{\varphi }_{\mathbf{p}}=0$, with positive frequency ${\omega }_{\mathbf{p}}=\sqrt{{\mathbf{p}}^2+m^2}$. Hence a general real solution is the mode expansion of a free fieldThe two coefficients are complex conjugates because the field is real. Each pair of initial data $\varphi (0,\mathbf{x}),\pi (0,\mathbf{x})$ fixes these coefficients. The normalization anticipates canonical quantization and the relativistic normalization of one-particle states.