Past exam of the physics course of the year 4 of the natural sciences course of the University of Cambridge / 2015 / Paper 1 (General Physics, 3/GPP)

Original public paper: Physical PDF pages 689–699. All questions and alternatives are answered. Idealizations and phase conventions are stated where they affect the answer.

For an infinite cubic potential well of side $L$, the one-particle energy eigenvalues are $E_{n_x{n}_y{n}_z}=E_0(n_x^2+n_y^2+n_z^2)$, where $E_0={\pi }^2{\hslash }^2/(2m{L}^2)$ and each quantum number is a positive integer. The lowest spatial orbital is $(1,1,1)$, with energy $3E_0$ and orbital degeneracy one. The next energy is $6E_0$, with three orbitals given by permutations of $(2,1,1)$.

Spin-zero identical particles are bosons. Both can occupy the lowest orbital, so the two-boson ground state is nondegenerate. The first excited total energy is $9E_0$: one particle occupies the lowest orbital and one an excited orbital. Symmetrizing the two-particle wavefunction gives one state for each of the three excited orbitals, so the first-excited degeneracy is three.

Spin-half identical particles are fermions, so the total state must be antisymmetric. Two particles in the same lowest orbital have a symmetric spatial wavefunction and must form the unique spin singlet state. Thus the fermion ground state is also nondegenerate. At the first excited energy, each choice of excited spatial orbital permits a symmetric spatial state with a singlet state, or an antisymmetric spatial state with a three-dimensional triplet state. ThereforeThe spin multiplicity is essential: counting only the three orbital choices would omit allowed triplet states. The result assumes ordinary spin-statistics, an unsplit spin degeneracy and the hard-wall cube model.

The hard wall imposes $\psi (0)=0$, while the increasing linear potential requires decay as $x\to \infty$. SetThe stationary Schrödinger equation becomes $d^2\psi /d{\xi }^2-\xi \psi =0$. The growing Airy function solution is excluded, leavingThis is the quantum bouncer. For the first four states, $z_n\simeq 2.3381,4.0879,5.5206,6.7867$. Their interior node counts are respectively zero, one, two and three. Each wavefunction starts at zero at the wall, oscillates in the classically allowed interval, and crosses smoothly into an exponentially decaying tail beyond the turning point $x_n=E_n/\beta =z_n\ell$. The turning point is not an extra hard wall: the wavefunction generally remains nonzero there and beyond it.

In the sudden approximation, the nuclear charge changes before the electronic wavefunction has time to evolve. Immediately afterward the state is still the original hydrogenic 1s wavefunction, now expanded in the eigenstates of the new Coulomb Hamiltonian. The final-ground-state probability is the squared overlap, rather than one minus an energy change.

For initial and final charges $Z_1,Z_2$, the normalized 1s overlap isUsing $Z_1=1,Z_2=2$ givesThe remaining probability includes higher bound $ns$ states and the $s$-wave ionization continuum. In this central, instantaneous model the initial orbital has $l=0,m=0$, and angular orthogonality excludes states with $l\ne 0$. The electronic spin is unchanged. Nuclear recoil and interactions with the emitted decay particles are outside this leading sudden-charge-change approximation.

For a homogeneous ideal gas in a two-dimensional box of area $A$, with dispersion $ϵ=p^2/(2m)$ and one internal state, the density of states is $g(ϵ)=Am/(2\pi {\hslash }^2)$. The excited-particle density at temperature $T>0$ isAs $\mu \to 0^{-}$, this diverges. Hence at every fixed positive temperature the excited states can accommodate any finite areal density with $\mu <0$; they never saturate at a finite critical density. There is no finite-temperature Bose-Einstein condensation in the homogeneous two-dimensional ideal gas thermodynamic limit.

The low-energy logarithmic divergence is the reason, not merely the small number of spatial dimensions. A finite box has a low-energy cutoff and a crossover; a trap changes the density of states and can change the conclusion. Condensation at zero temperature and an interacting gas's Berezinskii–Kosterlitz–Thouless transition are separate issues. The proof assumes the uniform quadratic-dispersion model, rather than every conceivable two-dimensional trapping potential.

A spin-half gas has two spin states. Filling the Fermi sphere at zero temperature gives $n=k_F^3/(3{\pi }^2)$, soWith $m_3\simeq 5.008\times 10^{-27}\mathrm{kg}$, the number density is $1.62\times 10^{28}{\mathrm{m}}^{-3}$ and $k_F\simeq 7.82\times 10^9{\mathrm{m}}^{-1}$. ThusThe associated Fermi temperature isFermionic degeneracy matters on cooling to a few kelvin or below in this model, and is pronounced for $T\ll T_F$. Liquid helium-3 is interacting; its quasiparticle effective mass and thermodynamics need not equal the bare-mass free-gas predictions. The requested calculation is the ideal Fermi-gas scale.

For total relativistic energy $E=\gamma m{c}^2$, the speed is $v=c\sqrt{1-(m{c}^2/E{)}^2}$. Both beam energies are enormously greater than the rest energy, so expanding the square root avoids obscuring the small difference:With all energies in GeV, $m{c}^2=0.9383$, $E_1=4000$ and $E_2=6500$, givingThe deficits below the speed of light are about $8.25\mathrm{m}{\mathrm{s}}^{-1}$ and $3.12\mathrm{m}{\mathrm{s}}^{-1}$. The large energy increase therefore produces only a small speed increase; the energy mainly changes the Lorentz factor. These numbers use the quoted beam energies as total energies. Interpreting them as kinetic energies requires adding the rest energy, with only a tiny correction at these scales.

Let $x$ be the cylinder centre's small displacement away from the wall. Rolling without slipping gives rotation angle of magnitude $x/r$. The material point initially at the top moves horizontally by another $x$ relative to the centre to first order, so the spring extension is $2x$, not $x$. Its small vertical displacement affects the extension only at higher order.

The kinetic and potential energies areThe linearized equation is $(3m/2)\ddot{x}+4kx=0$, givingStatic friction enforces the rolling constraint and does no work at the instantaneous contact point. Omitting rotational kinetic energy, or attaching the spring to the centre instead of the top material point, would change the result.

An element at $a(\cos \varphi ,\sin \varphi ,0)$ carries charge $dq={\lambda }_{0}a\sin \varphi d\varphi$. Its field at the origin points opposite its radial position. By Coulomb's law,ConsequentlyFor $q{\lambda }_0>0$ the particle starts along $-y$; for the opposite sign it starts along $+y$. Reflection symmetries make the $y$ axis invariant for motion released there from rest. The ring has zero net charge but a nonzero dipole moment, so zero total charge does not imply zero field at its centre.

Assume independent arrivals at constant rate $r=2000{\mathrm{s}}^{-1}$, so the count follows a Poisson distribution. In $1\mathrm{ms}$ its mean is $\mu =2$. ThereforeOver observation time $T$, the rate estimate is $\hat{r}=N/T$ and its standard deviation is $\sqrt{r/T}$. Thus the fractional one-standard-error uncertainty is $1/\sqrt{rT}$. To make this strictly less than ten per cent requiresThe mean current magnitude is $er\simeq 3.20\times 10^{-16}\mathrm{A}$. The accuracy statement needs a confidence convention: the boxed time uses one standard error, not a guaranteed bound on every realization. A Gaussian 95-percent relative interval would instead require approximately $(1.96/0.1{)}^2$ events, or $T\gtrsim 0.192\mathrm{s}$. Dead time and correlated arrivals would need another count model; the mean rate alone does not establish Poisson statistics.

Normalize the central slit amplitude to one. The two outer amplitudes are $1/2$ each, not $1/\sqrt{2}$, because the specified attenuation concerns amplitude. For centre spacing $d=5\lambda$, adjacent slits have phase difference $\delta =(2\pi d/\lambda )\sin \theta =10\pi \sin \theta$. The far-field amplitude and intensity are thereforewhere $I_0$ is the central slit's intensity alone in the same narrow-slit approximation. Principal maxima have $\sin \theta =m/5$ and height $4I_0$; between them, zeros occur at $\sin \theta =(2m+1)/10$. Only values in $[-1,1]$ are physical. The endpoints are grazing-direction limits of this ideal angular model.

The amplitude weights form a binomial pattern. They remove the secondary maxima that an equal-amplitude three-slit array would have between its principal peaks. No phase shift is needed: changing the coherent weights already changes the interference structure.

For a finite step in a one-dimensional potential and constant mass, the stationary Schrödinger equation requires continuity of the wavefunction and its first derivative. A discontinuous derivative would introduce a delta-function term in its second derivative, unsupported by a merely finite step. A discontinuous wavefunction is still more singular. A delta-function potential has a different derivative jump condition; an infinite wall instead imposes zero wavefunction and does not require matching to an exterior zero derivative.

For a bound state of the half-well, $-V_0<E<0$. The wall at zero excludes a cosine term, and square integrability excludes exponential growth outside. WriteContinuity at $a$ gives $A\sin ka=B$ and $Ak\cos ka=-qB$. HencePut $z=ka$, $w=qa$, and $\nu =a\sqrt{2m{V}_0}/\hslash$. Matching becomes $z\cot z=-w$ with $z^2+w^2={\nu }^2$, $w>0$. The negative branches of $z\cot z$ occur on $((n-1/2)\pi ,n\pi )$. There is one bound-state intersection in the $n$th interval exactly whenAt equality the new level has $q=0$, so its outside tail is constant and not normalizable: it is a threshold state, not a bound state. Thus no bound state exists for $\nu \le \pi /2$, including the strict inequality asked in the question.

Exactly two negative-energy bound states require the second threshold to have been crossed and the third not crossed:The lower endpoint is excluded and the upper endpoint included precisely because a zero-energy threshold state does not count as bound. This half-line finite-well bound-state count differs from an unrestricted symmetric one-dimensional well, which has no hard wall at its centre.

Black-body radiation is the equilibrium electromagnetic radiation associated with a perfect absorber, whose spectrum is fixed by its temperature and the available modes. Photons can be emitted and absorbed, so their equilibrium chemical potential is zero. The model here retains two transverse polarizations in an effectively one-dimensional cavity; the resulting pressure is a one-dimensional mechanical force, not a three-dimensional pressure in pascals.

Standing-wave wave numbers are $k_n=n\pi /L$, with one positive mode index and two polarizations. Since $ϵ=\hslash ck$, the continuum density of states isThe factor two counts polarizations; counting independent positive and negative traveling momenta in addition would double count standing modes. The thermal energy isThus the one-dimensional photon-gas thermodynamics givesReplacing the mode sum by an integral requires the stated temperature to be large compared with the level-spacing scale. This is thermal energy above the vacuum contribution; a separate boundary-dependent vacuum energy is not included.

Differentiating at fixed length gives the heat capacityThe linear temperature dependence reflects the constant one-dimensional mode density, in contrast to the cubic dependence of a three-dimensional photon gas.

Write $b=\pi k_B^2/(3\hslash c)$, so $U=bL{T}^2$. At fixed length, $dS=C_{L}dT/T$, giving $S=2bLT$ with the entropy constant fixed by the unique photon vacuum. The thermal Helmholtz free energy and generalized pressure follow:The entropy constant can also be checked from the exact discrete photon partition function; the continuum approximation should not be extrapolated blindly below its level-spacing regime. In one dimension $P=U/L$, whereas the isotropic three-dimensional relation is $p=U/(3V)$. Both follow from the number of spatial directions carrying momentum flux.

The quasistatic pressure-volume work done on the body is $\delta W=-PdV$. If its volume remains constant throughout the process, every $dV$ vanishes, so the work is zero. Equal initial and final volumes alone would not suffice: an expansion and compression cycle can have nonzero work. Other work modes are excluded by this simple elastic-volume model.

Within the stated small-volume-change approximation, treat $V,{\kappa }_T,\alpha$ in the differential constitutive law as constants. ThenPath (a) first changes temperature at pressure $P_1$, then changes pressure at fixed temperature $T_2$. The two contributions are $-V_1\alpha P_1(T_2-T_1)$ and $V_1{\kappa }_T(P_2^2-P_1^2)/2$. ThereforeWork is path dependent even though the endpoints and approximate total volume change are the same.

Path (b) changes pressure before temperature. Its pressure integral is unchanged, but the thermal expansion occurs at $P_2$:Along the straight path (c), $P=P_1+(P_2-P_1)(T-T_1)/(T_2-T_1)$. Thus ${\int }_{T_1}^{T_2}PdT=(P_1+P_2)(T_2-T_1)/2$, while $\int PdP$ depends only on the pressure endpoints. ConsequentlyThis pressure-temperature path-work identity is a consequence of the linearized volume law and the straight interpolation. It need not hold for strongly varying compressibility or expansion coefficient.

The depicted endpoints have $P_2>P_1$ and $T_2>T_1$. For positive thermal expansion coefficient,Thus path (b) requires the least work on the solid, path (a) the most, and path (c) lies midway between them. Heating at the higher pressure produces more negative work on the system. The ordering would reverse for a negative expansion coefficient; if the product vanishes, the paths have equal work in this approximation.

Three experimental tests probe different consequences of special relativity. The Michelson-Morley experiment compares light round trips along perpendicular interferometer arms while changing the apparatus orientation. Its small observed orientation-dependent fringe shift contradicts the simple stationary-aether prediction and supports orientation-independent round-trip light speed. It does not, alone, exclude every alternative theory.^[ref]

In the muon storage-ring time-dilation experiment, fast circulating muons live longer in laboratory time than muons at rest. Measuring decay counts versus elapsed laboratory time and independently determining the beam Lorentz factor tests ${\tau }_{\mathrm{lab}}=\gamma {\tau }_0$. Circular motion includes acceleration, so this also tests the applicability of accumulated proper time along the stored trajectory.^[ref]

Bertozzi's electron speed-energy experiment measures electron time of flight and deposited kinetic energy independently. An electrostatic accelerator and a linear accelerator increase the energy while the measured speed approaches the light-speed limit. Calorimetry checks that the additional beam energy is real, rather than inferring it only from an assumed voltage-to-speed relation. The result supports relativistic $K=(\gamma -1)m{c}^2$ rather than extending $K=m{v}^2/2$ to arbitrarily large speeds.^[ref]

For the two ships, set launch time to zero and measure Earth time in years. Their Lorentz factors are ${\gamma }_A=5/4$ and ${\gamma }_B=5/3$. Bob's thirtieth birthday occurs after one year of his proper time, so its Earth coordinates areIf Alice sends at Earth time $t_A$, her position is $x_A=3c{t}_A/5$. The left-going light signal obeys $x=x_A-c(t-t_A)$. Intersection with Bob givesAlice's elapsed proper time is thereforeShe sends two months after her twenty-ninth birthday. This compares the emission and reception events; the captains' clocks need only be synchronized at launch, not kept numerically synchronized afterward.

The relative recession speed isThe longitudinal relativistic Doppler effect then givesEquivalently, the left-going photon has Earth frequency $f_0/[{\gamma }_A(1+3/5)]=f_0/2$, and Bob's frame supplies another factor ${\gamma }_B(1-4/5)=1/3$. Both methods give the same redshift.

Newton's law of universal gravitation gives the force $-GmM\hat{\mathbf{r}}/r^2$ on a test mass a distance $r$ from a point source. Gauss's law for gravity statesThe inverse-square point field has the stated flux through a surrounding sphere. Superposition extends it to distributed sources; the divergence theorem relates the integral and local forms. Symmetry is still needed to turn a flux into a pointwise field value.

For the gravitational field of an infinite uniform cylinder, a coaxial cylindrical Gaussian surface has no end-cap flux and constant inward field on its curved surface. The enclosed mass per unit length is $\pi \rho r^2$ inside and $\pi \rho R^2$ outside. HenceIts magnitude rises linearly from zero, reaches $2\pi G\rho R$ at the surface, then decreases as $1/r$. The field is continuous at $R$, while its radial derivative changes because the density has a sharp edge.

Use signed distance $y$ along the diameter hole. A rod displaced by $x$ occupies $y\in [x-R,x+R]$, so its force is $F(x)=\lambda {\int }_{x-R}^{x+R}{g}_y(y)dy$. At equilibrium symmetry makes $F(0)=0$. Differentiating the moving endpoints givesThus the leading restoring force and oscillation frequency areFor an explicit positive-displacement check, put $C=2\pi G\rho$ and integrate the inside and outside pieces for $0<x<2R$:Its first-order term is $-2C\lambda Rx$, agreeing with the endpoint argument. The result is the diameter-rod oscillation in cylindrical gravity; it assumes a sufficiently long source cylinder and a narrow hole whose removed mass can be neglected.

The electric displacement field is $\mathbf{D}={ϵ}_0\mathbf{E}+\mathbf{P}$. A homogeneous linear isotropic dielectric has $\mathbf{P}={ϵ}_0{\chi }_e\mathbf{E}$, so $\mathbf{D}={ϵ}_0{ϵ}_r\mathbf{E}$. At a free-charge-free interface the normal component of D is continuous, while electrostatics requires continuity of tangential E.

At half immersion the conductor centre lies in the flat air-liquid interface. Set $\varphi =C/r$ in both media, with potential zero at infinity. The sphere is an equipotential, and the potential is continuous across the liquid plane. On that plane the radial E field is tangential, so tangential E agrees while both normal D components vanish. Each half-space also satisfies the source-free potential equation. The fields outside the sphere areGauss's law for D on a sphere gives $Q=2\pi {ϵ}_0(1+{ϵ}_r)C$. ThereforeThe half-immersed conducting sphere in a dielectric has radial E lines with equal field strength at equal radius on either side, whereas the D flux density is ${ϵ}_r$ times larger in the liquid. The corresponding free surface charge is larger on the lower hemisphere. The unchanged field direction at this particular interface follows from its tangential geometry, not from a claim that electric fields never refract at dielectric interfaces.

The outward electrostatic pressure is $p_e=\mathbf{E}\cdot \mathbf{D}/2$. Integrating its vertical component over the two hemispheres uses ${\int }_{\mathrm{upper}}\cos \theta dA=\pi a^2$ and the negative of that integral below. ThusThe force is downward for ${ϵ}_r>1$. Initially, buoyancy on the uncharged sphere gives ${\rho }_{\ell }gf(4\pi a^3/3)=Mg$. At half immersion the upward buoyancy is therefore $Mg/(2f)$, leaving a required downward electric force $Mg(1/(2f)-1)$. Equating the two givesThis uses the ideal flat-interface, linear-dielectric model with surface tension and interface deformation neglected. If ${ϵ}_r=1$, there is no electric force of this kind, consistently preventing this mechanism from pulling an initially less-than-half-immersed sphere down to the specified level.

A magnetic dipole is the leading magnetic source moment of a localized steady current distribution. For a small planar loop, $\mathbf{m}=I\mathbf{A}$ with area orientation set by the right-hand rule. A fixed permanent moment in a slowly varying external field has energy $U=-\mathbf{m}\cdot \mathbf{B}$ and force $\mathbf{F}=\nabla (\mathbf{m}\cdot \mathbf{B})$.

For the given purely azimuthal vector potential, the spherical curl givesThese are the magnetic dipole field components outside the localized source.

The conducting shell has $Q=4\pi {ϵ}_{0}aV$ and surface charge density $\sigma =Q/(4\pi a^2)$. A latitude strip of area $2\pi a^2\sin \theta d\theta$ carries current $dI=\sigma \omega a^2\sin \theta d\theta$. Its loop area is $\pi a^2{\sin }^2\theta$, givingIn vector form the magnetic moment of a rotating charged spherical shell isThe positive-z expression in the question takes $\mathbf{\omega }=\omega \hat{\mathbf{z}}$. “Clockwise” needs a viewing direction: this rotation looks clockwise when viewed looking along $+z$ from below, and counterclockwise when viewed from the positive-z side. Looking from the other end reverses the verbal label, not the vector relation.

Since the positive charge and the loop current circulate in the same stated sense, their dipole moments are parallel when judged from a common viewpoint. At $h\gg R$, the loop's axial field magnitude is $B\simeq {\mu }_{0}I{R}^2/(2h^3)$. With the sphere small compared with that variation scale,The magnetic force is attractive, pulling the sphere toward the loop. Reversing only one circulation would reverse that magnetic force. The calculation treats the rotating charge as a prescribed moment and uses the stated separated length scales; it does not add unrelated induction or electric interactions to the dipole model.

Let $R(\theta )$ be the real rotation matrix with columns along the plate's fast and slow axes. Expressing the input in those axes, applying the principal retardance, then rotating back gives the rotated half-wave plate matrixEach fixed quarter-wave plate has $Q=\mathrm{diag}(1,i)$ in the supplied convention. Propagation through the three plates therefore givesThe order matters: this is a product of three propagating plates, not a similarity transformation $QH{Q}^{-1}$. For input Jones vector ${\mathbf{e}}_{+}=(1,1{)}^T/\sqrt{2}$, the Pauli matrix has eigenvalue $+1$, so $J{\mathbf{e}}_{+}=e^{i2\theta }{\mathbf{e}}_{+}$. The final polarization remains at 45 degrees, but its phase depends on the rotating angle.

Take the physical field as the real part of $\mathbf{E}{e}^{-i\omega t}$ and let $\theta =\Omega t$. ThenThus the rotating half-wave plate frequency shift hasThe magnitude is the stated $2\Omega$ shift. Its sign follows the explicit time and rotation conventions above; reversing the physical rotation or using the orthogonal linear input reverses the shift. A time-phasor convention must be changed consistently with the plate matrices, rather than used alone to change a physical redshift into a blueshift.

The physical origin is a rotational Doppler effect and angular-momentum exchange. In the stated convention, the first fixed quarter-wave plate changes the 45-degree input into circular polarization of spin $+\hslash$ per photon. The rotating half-wave plate flips it to spin $-\hslash$, transferring $2\hslash$ to the rotating plate. Maintaining its rotation takes mechanical energy $2\hslash \Omega$ from each photon, agreeing with the redshift. The final fixed plate restores linear polarization; fixed plates can exchange angular momentum with their supports without the rotating element's corresponding mechanical work. The instantaneous Jones treatment assumes slow rotation compared with optical and transit-time scales and ideal retardances over the small shifted bandwidth.