Past exam of the physics A course of the University of Cambridge / 2015 / Paper 2

Original paper: PHYSICS A (2), 28 May 2015; physical PDF pages 106–111.

The photon relation gives $\lambda =hc/E=2.4797\times 10^{-12}\mathrm{m}$. The diffraction-limited angular scale of an aperture of diameter $D$ is thereforeFor a uniformly illuminated circular aperture, the Rayleigh criterion instead gives $1.22\lambda /D=3.03\times 10^{-11}\mathrm{rad}$. Both express the same requested estimate. This is an ideal diffraction limit, conditional on the effective diameter behaving as a coherent imaging aperture; photon energy and diameter alone do not specify the performance of a practical gamma-ray telescope.

At distance $L$, the small-angle approximation gives $\ell =L\theta$. Thus the transverse resolution is approximately $2.5\times 10^{10}\mathrm{m}$, or $3.0\times 10^{10}\mathrm{m}$ with the circular-aperture Rayleigh criterion. This is a transverse separation at the source, rather than a determination of its distance along the line of sight.

Let $T$ be the common tension and ${\mu }_j$ the linear mass densities. The characteristic impedance of a stretched string is $Z_j=T/c_j=\sqrt{T{\mu }_j}$. Continuity of displacement and transverse force at a massless junction gives $1+r=t$ and $Z_1(1-r)=Z_{2}t$. ConsequentlyThe reflected power is $7.18\%$. The minus sign of $r$ represents a phase reversal, not negative power. Both incident and reflected waves travel in the first string, so their power ratio is simply the squared displacement-amplitude ratio. As a check, $\mathcal{T}=(Z_2/Z_1)|t{|}^2=0.92820$ and $\mathcal{R}+\mathcal{T}=1$.

Compare a directly transmitted ray emerging at $A$ with a ray that makes one extra round trip inside the Fabry–Pérot interferometer and emerges at $C$. Both emerge parallel at angle $\theta$ to the plate normal. The extra internal distance is $2d/\cos \theta$, and their emergence points differ laterally by $AC=2d\tan \theta$. The first ray has already travelled $AC\sin \theta$ towards a common outgoing wavefront when the second reaches $C$. The air-gap optical path difference in a Fabry–Pérot interferometer follows because optical path length equals geometric path length in the air-gap approximation:The subtraction is essential: comparing distances only inside the plates would compare different outgoing wavefronts. The corresponding propagation phase difference is $2\pi \Delta /\lambda$. Coating reflection phases can add a separate phase offset; they do not change this geometric optical path difference.

Count two electron spin states in each reciprocal-space cell of volume $(2\pi {)}^3/V$. Filling a Fermi sphere at zero temperature givesThe free electron model then givesThe energy zero is the bottom of the model's conduction band. The supplied carrier density already incorporates the one-electron-per-atom assumption; it must not be multiplied by another atomic count.

Write $a=V_c^{1/3}$ for the primitive cubic lattice spacing. With wavevector convention $e^{i\mathbf{k}\cdot \mathbf{r}}$, reciprocal-lattice primitive vectors have length $2\pi /a$. The first Brillouin zone is the cube $-\pi /a\le k_j\le \pi /a$, henceIt has the same volume as any primitive reciprocal unit cell. A reciprocal coordinate defined in cycles per metre would instead give $1/V_c$; that is a different convention from this $k$-space convention.

Take $m,k,b>0$, and define ${\omega }_0=\sqrt{k/m}$ and $\gamma =b/(2m)$. Substitution of $e^{rt}$ into the unforced equation gives $r^2+2\gamma r+{\omega }_0^2=0$. The roots determine whether the return is oscillatory.

For return after a displacement in this ideal model, critical damping, case (b), is the usual comfort choice: it removes repeated bouncing while avoiding the slow relaxation mode of strong overdamping. For release from rest it gives a monotone return with the fastest asymptotic decay available without oscillation. This criterion does not prove optimal comfort for every road profile or initial velocity; actual comfort also depends on acceleration, road forcing and suspension travel.

For the forced response, set $\Omega =2\pi u/d$ and use $x_{\mathrm{ss}}=\mathrm{Re}(\hat{x}{e}^{i\Omega t})$. ThenThe general forced solution adds the appropriate homogeneous transient from (a), (b) or (c). Initial data were not specified, so its constants cannot be fixed. Positive damping removes this transient at long times. This solution uses the particular driving force in the supplied equation; it does not silently replace it by a different moving-support model.

Maximising $X$ is minimising its squared denominator $D$. Differentiation givesA nonzero displacement resonance exists only if $b^2<2mk$. Here ${\Omega }_r^2=0.175{\mathrm{s}}^{-2}$, and thereforeThe low-frequency limit is $X(0)=A$, whereas $X\to 0$ at large $\Omega$. The parameters give damping ratio $\zeta =b/(2\sqrt{mk})=0.456$, consistent with an underdamped system and the existence of this amplitude maximum.

For $b^2>4mk$, define $\beta =\sqrt{{\gamma }^2-{\omega }_0^2}$. The two characteristic roots are real and negative:This is overdamping: there is no repeated oscillation, and the slow root controls late-time return. A generic initial velocity can still cause a single crossing of equilibrium, so “nonoscillatory” need not mean monotone for all initial data. For large $b$, the slow root approaches $-k/b$.

For $b^2=4mk$, the repeated characteristic root requires a second linearly independent solution $t{e}^{-\gamma t}$:This is critical damping. If $x(0)=x_0$ and $\dot{x}(0)=0$, then $x=x_0(1+\gamma t)e^{-\gamma t}$, demonstrating the monotone return used in the suspension discussion.

For $b^2<4mk$, put ${\omega }_d=\sqrt{{\omega }_0^2-{\gamma }^2}$. Complex-conjugate roots give the real solutionThis is underdamping: oscillation at ${\omega }_d$ persists within a decaying exponential envelope. damping reduces the oscillation frequency below ${\omega }_0$. Constants in all three cases are determined by two initial conditions.

For a monochromatic wave, a constant phase $kx-\omega t$ moves at phase velocity $v_p=\omega /k$. For a narrow wave packet around $k$, expansion $\omega (k+\Delta k)\simeq \omega (k)+(d\omega /dk)\Delta k$ shows that its envelope moves at group velocity $v_g=d\omega /dk$. The envelope interpretation requires a sufficiently narrow spectrum and weak distortion; it is not a general identification with a causal signal front. Since $\omega =k{v}_p$ and $\lambda =2\pi /k$,For the gravity-capillary wave, put $s=\sigma /\rho$ and take $k>0$. Direct differentiation givesThus $v_g=v_p$ when $s{k}_{\ast }^2=g$, orAt this wavelength the gravity and surface-tension restoring terms contribute equally. Also $d(v_p^2)/dk=-g/k^2+s=0$ with positive second derivative: this is the minimum phase speed of a gravity-capillary wave, $v_{p,\min }=(4g\sigma /\rho {)}^{1/4}=0.214\mathrm{m}{\mathrm{s}}^{-1}$. Locally the carrier crests and narrow envelope travel together. The medium has not become nondispersive: $d^2\omega /d{k}^2=k_{\ast }{d}^2{v}_p/d{k}^2\ne 0$ there, so a finite wave packet can still spread. In the gravity-dominated limit $v_g/v_p\to 1/2$; in the capillary-dominated limit it tends to $3/2$.

Take the aperture in $z=0$ and the observation point $P=(0,0,b)$, with $b>0$. Under scalar diffraction theory, the Huygens-Fresnel principle sums complex amplitudes from area elements, rather than summing their intensities. Apart from a common normalization,Here $h$ is the complex aperture transmission, $b$ is the axial distance, and $\theta$ is the angle of the secondary ray to the aperture normal. The obliquity factor $K$ describes the direction dependence of the secondary contribution; one Kirchhoff convention is $K=(1+\cos \theta )/2$ for normal illumination. Other scalar boundary approximations have slightly different angular weights, all tending to unity near the axis.

Put $r^2=x^2+y^2$. The paraxial approximation gives $R\simeq b+r^2/(2b)$ in the phase and $1/R\simeq 1/b$ in the amplitude. Removing common $e^{ikb}/b$ yieldsAssumptions are uniform normally incident plane-wave illumination, coherent monochromatic propagation, aperture dimensions large enough for the scalar description, $r/b\ll 1$, and phase accuracy $k{r}^4/(8b^3)\ll 1$ over the relevant aperture. The last condition justifies dropping the quartic distance term; geometric smallness alone does not guarantee phase accuracy.

For a circular aperture, $h=1$ for $r\le a$ and zero elsewhere. Rotational symmetry and $s=r^2$ give $dxdy=rdrd\phi$, $rdr=ds/2$, andThe constant $\pi$ can be absorbed into the proportionality factor, producing the required one-dimensional integral.

At a focus a distance $f$ away, half-period-zone boundaries obeyWith slowly varying obliquity and illumination, each zone has almost the same area and its integrated phasor has the opposite sign to its neighbour. A zone plate blocks alternate zones, so the remaining phasors add instead of nearly cancelling. Choosing an open central zone, the first three open regions have outer boundaries $n=1,3,5$:These are outer radii, not zone widths; the second open region is the annulus between $r_2$ and $r_3$.

For the specified outer radius, $R^2/(\lambda f)=49$: there are approximately 49 half-period zones, of which 25 odd zones are open. Normalize to the field ${\psi }_0$ of the unobstructed plane wave. The normalized Fresnel aperture integral isFor any odd open annulus this gives $2$, so the whole plate gives $\psi /{\psi }_0\simeq 50$ andThe axial intensity gain of an amplitude zone plate grows as the square of the number of open zones. Treating the number of zones continuously gives $49^2\simeq 2.40\times 10^3$, the same accuracy as an estimate. This large local intensity gain is consistent with conservation of energy because light is concentrated into a small focal spot. It is not a power efficiency; the ideal first-order power efficiency of a binary amplitude zone plate is only $1/{\pi }^2$.^[ref]

“Without the zone plate” has here meant removing the finite plate and its outer stop, giving the unobstructed incident wave. If instead a clear circular aperture of radius $R$ is retained, its axial field is ${\psi }_{\mathrm{clear}}/{\psi }_0=1-e^{i49\pi }=2$, so the ratio to that finite-aperture reference is $2500/4=625$. The two reference experiments differ; neither ratio should be confused with total transmitted power.

A Michelson interferometer divides a source beam at a partially reflecting beam splitter. The two perpendicular arms terminate in mirrors whose reflected beams return to the splitter and overlap at the detector. The recombined fields must have overlapping wavefronts and compatible polarization of waves. Moving one mirror by $\Delta L$ changes the round-trip optical path difference by $2\Delta L$, so $N$ passing fringes determine $\Delta L=N\lambda /2$. Conversely a known mirror displacement measures wavelength. Scanning the delay can determine temporal optical coherence and, by Fourier transform spectroscopy, spectral structure. A compensating plate may equalize glass traversal, particularly for broadband illumination.

For equal real peak field amplitudes, the superposition isTime averaging the square over many optical cycles gives $⟨E^2⟩=2a^2{\cos }^2(\varphi /2)=a^2(1+\cos \varphi )$. With $\varphi =kx$ after absorbing a constant port-dependent phase into the zero of $x$,Here $I_0$ is the sum of the two separate-arm intensities; maximum intensity is $2I_0$ and the ideal minimum is zero. Equal amplitudes give unit interference visibility.

For a stationary source with a flat intensity spectrum, different spectral components add in time-averaged intensity. Keeping their individual two-arm interference terms and defining $\mathrm{sinc}z=\sin z/z$ givesOne must not coherently add unrelated source frequencies as if they had a fixed common phase. The sinc envelope is the Fourier transform of the rectangular spectral intensity.

Numerically, ${\nu }_0=5.00\times 10^{14}\mathrm{Hz}$ and $\Delta \nu =1.00\times 10^{13}\mathrm{Hz}$. The annotated interferogram usesIt is even in $x$, has central value 2 and background 1, and its carrier-fringe period in optical path difference is $c/{\nu }_0=0.600\mu \mathrm{m}$. The first envelope zeros are at $x=\pm c/\Delta \nu =\pm 30.0\mu \mathrm{m}$, corresponding to mirror shifts of $\pm 15.0\mu \mathrm{m}$ from equal paths. Side lobes persist beyond the zeros; a negative sinc reverses fringe phase. The intensity remains between $1-|\mathrm{sinc}|$ and $1+|\mathrm{sinc}|$, not between the signed sinc curves with an assumed fixed ordering.

The free electron model replaces the microscopic ionic potential and interacting electronic motion by independent mobile electrons in a uniform neutralizing background, with approximately parabolic dispersion $ϵ={\hslash }^2{k}^2/(2m^{\ast })$. Its quantum version fills states according to Fermi-Dirac statistics and the Pauli exclusion principle. Scattering is represented by a momentum relaxation time $\tau$, rather than derived from the model's ideal free trajectories. This model explains a Fermi energy much larger than room-temperature thermal energy, an electronic heat capacity proportional to temperature at low temperature, and a Drude electrical conductivity. With appropriate scattering assumptions it also yields the Wiedemann-Franz law. It cannot by itself explain band gaps, detailed band anisotropy, strong correlations, or the microscopic temperature dependence of $\tau$.

The inertial term is mass times the rate of change of mean drift velocity. The electric force is $eE$, taking the printed $e$ as the signed electron charge, and $-\gamma ⟨v⟩=-m^{\ast }⟨v⟩/\tau$ describes loss of directed momentum to scattering. electrons retain large microscopic velocities; this damping is of their mean directed motion. In a steady DC field,Both charge factors are negative for electrons, so the conductivity is positive. If $e$ instead denotes a positive charge magnitude, the force and current equations each require a minus sign and give the same result.

For AC response choose physical fields as real parts of phasors proportional to $e^{-i\omega t}$. Substitution givesThusThe real part decreases from ${\sigma }_0$ to zero. The imaginary part starts at zero, reaches ${\sigma }_0/2$ at $\omega \tau =1$, and falls as $1/\omega$. The real part falls as $1/{\omega }^2$ at high frequency. With an $e^{+i\omega t}$ convention, only the sign of the imaginary part changes. Positive $\mathrm{Re}\sigma$ gives positive cycle-averaged dissipation $⟨jE⟩=\mathrm{Re}\sigma |\hat{E}{|}^2/2$.

The supplied DC conductivity determines the transport time:Inertia matters when $\omega \tau \sim 1$, giving ${\omega }_c\simeq 4.02\times 10^{13}\mathrm{rad}{\mathrm{s}}^{-1}$, or ${\nu }_c={\omega }_c/(2\pi )\simeq 6.40\times 10^{12}\mathrm{Hz}$. Quoting the former number as hertz would introduce an erroneous factor of $2\pi$.

Write the real displacement as $U_n(t)=\mathrm{Re}[u{e}^{i\omega t-iqna}]$, choosing real positive peak amplitude $u$. Nearest-neighbour harmonic springs store energy in their extension $U_n-U_{n+1}$, each spring counted once. Therefore the instantaneous physical energy isThe printed complex displacement is a calculation device: squaring the complex phasor itself would not produce physical energy. The chain sketch in the accompanying figure shows masses at equilibrium spacing $a$, with longitudinal displacements along the chain.

The two spring forces on site $n$ giveSubstituting the normal-mode phasor and cancelling its nonzero amplitude givesThe long-wavelength limit is acoustic, $\omega \simeq a\sqrt{\alpha /M}|q|$, and the first Brillouin zone ends at $|q|=\pi /a$. The maximum angular frequency is ${\omega }_m=2\sqrt{\alpha /M}$.

Time averaging uses $⟨[\mathrm{Re}(z{e}^{i\omega t}){]}^2⟩=|z{|}^2/2$. ConsequentlyThe dispersion relation makes these two contributions equal, givingThis is equality of time-averaged kinetic and potential energies in a harmonic normal mode; the instantaneous energy partition need not be equal.

Use periodic boundary conditions on length $L=Na$ to count modes: $q_j=2\pi j/L$. The density on the signed $q$ axis is $L/(2\pi )$. Using folded wave-number counting for a monatomic chain combines $q$ and $-q$ at positive magnitude and counts both directions, so the convention used by the requested expression isThis is a large-$N$ continuum density, not a second independent branch. It gives $N$ modes after integrating over that half-zone; endpoint counting changes only finite-size corrections.

For positive $q$, $\omega ={\omega }_m\sin (qa/2)$ andThe density of states of a monatomic harmonic chain vanishes outside the band, is constant to leading order near zero, and has an integrable Van Hove singularity at the upper band edge where group velocity vanishes. Normalization checks the directional counting:

A completely filled valence band carries no net current: its occupied crystal-momentum states have velocities that cancel. Near a valence-band maximum, an almost full band is conveniently described by the few missing electrons, or electron holes. Removing an electron of charge $-e_0$, with $e_0>0$, changes the charge relative to the filled band by $+e_0$. This is a quasiparticle description of electronic motion, not a mobile positive ion or a Positron.

For an isotropic parabolic maximum, ${ϵ}_v(\mathbf{k})=E_v-{\hslash }^2{k}^2/(2m_h^{\ast })$. Assign the missing-electron state momentum ${\mathbf{k}}_h=-{\mathbf{k}}_e$ and hole energyIt has positive effective mass $m_h^{\ast }$. Its velocity equals the velocity of the missing electron state, and its positive-charge current replaces the negative of that state's electronic current. Semiclassical motion is thus $\hslash {\dot{\mathbf{k}}}_h=+e_0(\mathbf{E}+{\mathbf{v}}_h\times \mathbf{B})$. With momentum relaxation time ${\tau }_h$, the low-field mobility is ${\mu }_h=e_0{\tau }_h/m_h^{\ast }$ andIf electrons also contribute, $\sigma =e_0(p{\mu }_h+n{\mu }_e)$: the two species' longitudinal conductivities add.

A Hall bar separates carrier concentration from mobility. Apply current density $j_x$, magnetic field $B_z$, and measure transverse electric field $E_y$. For one isotropic hole population, transverse force balance gives $E_y=v_x{B}_z$, soFor thickness $t$, the Hall-voltage magnitude is $|V_H|=IB/(e_{0}pt)$. A longitudinal four-terminal resistance measurement supplies $\sigma$ independently. Use ohmic contacts, known conducting thickness, controlled temperature and low enough current to avoid self-heating. Reversing magnetic field removes even-in-field longitudinal offsets from contact misalignment; reversing current also helps reject thermoelectric offsets. These are central controls, rather than merely taking one transverse voltage reading.^[ref]

The simple inference assumes one carrier type and a known Hall scattering factor. More generally $R_H=r_H/(e_{0}p)$, so $\sigma R_H$ measures Hall mobility rather than automatically the drift mobility. With two carrier populations at weak field,Its sign alone need not identify the more numerous carrier. A Hall experiment also cannot determine $m_h^{\ast }$ and ${\tau }_h$ separately: their ratio sets mobility. cyclotron resonance supplies an independent mass measurement by sweeping frequency or magnetic field through $\omega =e_{0}B/m_c$ and detecting resonant absorption. The measured cyclotron effective mass can depend on band anisotropy and need not equal every curvature mass; the isotropic parabolic model identifies them. Its linewidth provides a scattering-time estimate only with a specified damping model.

semiconductor doping generates holes by adding acceptor impurities. In silicon, a substitutional group-III atom such as boron provides one fewer valence electron than silicon. A shallow acceptor level lies slightly above the valence-band edge: accepting a valence electron makes the dopant negatively ionized and leaves a mobile hole in the valence band. The impurity stays fixed in the lattice. charge neutrality and the nondegenerate equilibrium mass-action law giveFor fully ionized, uncompensated acceptors in the extrinsic regime, $p\simeq N_A$ and $n\simeq n_i^2/N_A$; with fully ionized donor compensation, $p\simeq N_A-N_D$ while holes remain dominant. The Fermi level moves towards the valence band. At low temperature, semiconductor carrier freeze-out suppresses mobile electron holes; at high temperature intrinsic electron-hole generation can dominate. Increasing dopant density also changes scattering, so a conductivity increase alone cannot distinguish increased carrier density from changed mobility. Temperature-dependent Hall and conductivity measurements separate these effects, subject to the carrier and scattering qualifications above.

For a small longitudinal disturbance, let the equilibrium pressure and mass density be $p_0,{\rho }_0$, their perturbations be $p^{\prime },{\rho }^{\prime }$, and the fluid velocity be $v$. Linearizing conservation of mass and the inviscid momentum equation in one dimension givesWhen compression is adiabatic, the equation of state gives $p^{\prime }=c_s^2{\rho }^{\prime }$ with $c_s^2=(\partial p/\partial \rho {)}_s$. Differentiating and eliminating $v$ yieldsfor an ideal gas with heat-capacity ratio $\gamma =C_p/C_v$ and molar mass $M_{\mathrm{mol}}$. Compression raises temperature as well as density, explaining the factor $\gamma$ compared with the isothermal value $\sqrt{p_0/{\rho }_0}$.

A rightgoing plane sound wave has pressure and particle velocity in phase, $p^{\prime }={\rho }_0{c}_{s}v$, and mean intensity $⟨p^{\prime }v⟩={\hat{p}}^2/(2{\rho }_0{c}_s)={\rho }_0{c}_s{\hat{v}}^2/2$ for peak amplitudes. Particle displacement is a quarter-cycle out of phase with particle velocity: particles oscillate locally while the compression pattern transports energy. This continuum linear description requires wavelength much larger than the mean free path, small pressure perturbations and negligible viscous attenuation. Adiabatic propagation requires heat exchange across the compressed region to be slow on the oscillation time; strong thermal coupling would instead favour an isothermal response.

The Q factor measures stored energy relative to loss: $Q=2\pi E/(\text{energy lost per cycle})$, or $Q={\omega }_{0}E/P_{\mathrm{loss}}$ in the weak-loss limit. For viscous damping $m\ddot{x}+b\dot{x}+kx=0$, write $\gamma =b/(2m)$, ${\omega }_0=\sqrt{k/m}$ and ${\omega }_d=\sqrt{{\omega }_0^2-{\gamma }^2}$. The conventional oscillator parameter isIn free ringdown, fit the amplitude envelope $A(t)=A_0{e}^{-\gamma t}$. Its decay time is ${\tau }_A=1/\gamma$, whereas the cycle-averaged energy envelope decays on ${\tau }_E=1/(2\gamma )$; thus $Q={\omega }_0{\tau }_A/2={\omega }_0{\tau }_E$. Equivalently the logarithmic decrement between same-sign peaks is $\delta =\ln (A_n/A_{n+1})=2\pi \gamma /{\omega }_d$. It gives $Q\simeq \pi /\delta$ for weak damping, or the exact parameter relation $Q=\sqrt{1/4+(\pi /\delta {)}^2}$. Fit multiple peaks above the noise floor, measure the damped period, and verify an exponential envelope rather than assuming all amplitude loss is viscous.

For a driven measurement, allow transients to settle and scan frequency at fixed force amplitude. Near resonance and at high $Q$, the response is Lorentzian in displacement power, with half-power points at displacement amplitudes $1/\sqrt{2}$ of the maximum. Their angular-frequency separation gives $Q\simeq {\omega }_0/\Delta \omega$. Halving the amplitude is not a half-power criterion.

The absorbed-power bandwidth of a viscously damped oscillator can be found exactly. Sinceits maximum is at $\Omega ={\omega }_0$, and the half-power frequencies satisfy $|k-m{\Omega }^2|=b\Omega$. HenceFor broad resonances, displacement and absorbed-power peaks differ, so the measured observable must be specified. Frequency-dependent drive, backgrounds or several overlapping modes can otherwise masquerade as a changed Q factor.

For propagation along $z$, a transverse wave has two perpendicular displacement components. With $\varphi =kz-\omega t$, write $u_x=A\cos \varphi$, $u_y=B\cos (\varphi +\delta )$. Their trajectory at a fixed point obeysThe general nondegenerate trajectory is an ellipse: elliptical polarization. For $\delta =0$ or $\pi$, the components remain in phase or antiphase and the trajectory collapses to a line. A single nonzero component is also linearly polarized. Equal amplitudes $A=B$ with $\delta =\pm \pi /2$ give circular polarization. Unequal quadrature amplitudes give an ellipse aligned with the chosen axes, while general phase differences rotate its axes.

For the displayed convention and $\delta =+\pi /2$, at $z=0$ the vector starts at $+x$ and moves towards $+y$ as time increases. Reversing $\delta$ reverses its rotation. Right- and left-handed labels require a stated viewing and phase convention, so rotation is defined explicitly here. A fixed ellipse is a coherent relationship between transverse components; unpolarized light instead has fluctuating orientation and phase relationships on the measurement timescale.

A linear polarizer transmits a linearly polarized beam with intensity proportional to ${\cos }^2\theta$, allowing its direction to be measured by an analyser. circular polarization transmits equal intensity through any linear analyser, but that test alone does not distinguish it from unpolarized light. A quarter-wave plate converts a suitable coherent circular state to linear polarization, allowing the combination to discriminate them. For light the electric field defines the usual polarization; the magnetic field remains perpendicular to it and to propagation. Longitudinal sound in a homogeneous gas lacks the two transverse degrees of freedom needed for these polarization states.