Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2012 (2S98, 2S99)

Source paper: www-thphys.physics.ox.ac.uk/people/AndreiStarinets/short_options_exam_2012.pdf

Unlettered requests are numbered in source order; explicit letters and Roman parts are retained.

The Laurent series isIts principal part contains infinitely many nonzero negative powers, so zero is an essential singularity. The first three displayed terms are not an asymptotic truncation as $z\to 0$; they are the beginning of the convergent Laurent series in powers of $1/z$.

The coefficient of $z^{-1}$ in the Laurent series is $-1$. The residue theorem therefore gives, for the standard positive orientation,The residue theorem applies to an isolated essential singularity as well as to a pole.

Parametrize the positively oriented unit circle by $z=e^{i\theta }$. Then $e^{-1/z}=e^{-\cos \theta +i\sin \theta }$ and $dz=i{e}^{i\theta }d\theta$, soEquating the real part and imaginary part with $-2\pi i$ gives

The Joukowski map $f(z)=z+z^{-1}$ is holomorphic on $\mathbb{C}\setminus \{0\}$. A holomorphic function is locally conformal where its derivative is nonzero. Here $f^{\prime }(z)=1-z^{-2}$ vanishes at $z=\pm 1$. Thereforewhere the exclusion of zero is because the function is undefined there. Holomorphy alone does not supply conformality at the two critical points.

Put $z=1+\zeta$. Expanding the holomorphic function near its critical point givesFor a small ray $\zeta =r{e}^{i\theta }$, the leading image displacement is $r^2{e}^{2i\theta }$. Thus oriented ray angles are doubled modulo $2\pi$ at $z=1$. This is angle multiplication at a holomorphic critical point of multiplicity two; an unoriented angle between full tangent lines must also be interpreted modulo $\pi$. The map is not locally one-to-one there.

For $z=e^{i\theta }$ the image is $w=2\cos \theta$, a real number. Therefore each semicircle maps onto the real interval $[-2,2]$, with endpoints included for closed semicircles. Under positive traversal of the full unit circle, the upper half runs from $2$ to $-2$ and the lower half returns from $-2$ to $2$. Every interior point has two unit-circle preimages. The Joukowski circle-to-ellipse mapping degenerates to a doubled line segment at this radius.

For a radius $\rho >0$, write $z=\rho e^{i\theta }$. ThenFor $\rho \ne 1$ its image is the ellipseIts horizontal semiaxis is $\rho +{\rho }^{-1}$, its vertical semiaxis is $|\rho -{\rho }^{-1}|$, and its foci are $\pm 2$ because the squared semiaxes differ by $4$. Reciprocal radii give the same ellipse, traversed in opposite directions as $\theta$ increases.

The square-root branch points are zero and infinity. At zero the local behavior is $z^{-1/2}$; in the coordinate $u=1/z$ near infinity it is $u^{1/2}$. In either case one circuit reverses the sign and two restore it. Thus the ramification index is $2$: these are two-sheeted algebraic branch points, sometimes described as branch order one when order means ramification index minus one. The decay to zero at infinity does not remove the branching.

TODO: the original closed-contour drawing is absent from the source. The following independent Bromwich contour deformation establishes the required discontinuity relation without asserting the missing drawing's exact geometry. Orient the vertical line upward, as required by the positive imaginary answer in the next part, and use the principal square root, $-\pi <\arg z<\pi$.

Close a truncated vertical line to the left with a rectangle, slit it along the negative real axis, and exclude a small circle about zero. The integrand is holomorphic inside that slit domain, so the boundary contour integral is zero. Horizontal boundary integrals are $O(R^{-1/2})$: integrating $e^x$ along $-R\le x\le 1$ gives a bounded factor while $|z{|}^{-1/2}\le R^{-1/2}$. The left boundary contributes $O(e^{-R}{R}^{1/2})$, and the small circle contributes $O(\sqrt{\epsilon })$. All vanish in the indicated limits.

At $z=-x$ with $x>0$, the upper and lower bank values are respectively $-i{e}^{-x}/\sqrt{x}$ and $+i{e}^{-x}/\sqrt{x}$. The boundary follows the upper bank toward zero and the lower bank away from zero. Their combined contribution is $-2i{\int }_0^{\infty }{e}^{-x}{x}^{-1/2}dx$. HenceThe vertically improper integral converges conditionally; these estimates also justify the contour limit. Reversing the vertical orientation reverses its sign.

The Gamma function identity $\Gamma (1/2)=\sqrt{\pi }$ gives the Bromwich integral of an inverse square root directly:Equivalently, put $x=u^2$ in the real integral. It becomes $2{\int }_0^{\infty }{e}^{-u^2}du=\sqrt{\pi }$ by the Gaussian integral.

For two bodies of masses $M_1,M_2$, the relative Kepler orbit has semimajor axis $a$ and orbital period $P$ satisfying Kepler's third lawHere $a$ is the semimajor axis of the separation vector, not either star's individual barycentric orbit. In conventional stellar units the same relation is approximately $(P/\mathrm{yr}{)}^2=(a/\mathrm{AU}{)}^3/[(M_1+M_2)/M_{\odot }]$, with year, astronomical unit and solar mass as the units.

Equal radial velocity amplitudes imply equal stellar masses. Under the circular, edge-on eclipsing binary interpretation, each star moves at $K=100\mathrm{km}{\mathrm{s}}^{-1}$, and the relative speed is $2K$. ThusUsing $P=5.45$ days givesMore generally, spectroscopic binary amplitudes determine $a\sin i=P(K_1+K_2)\sqrt{1-e^2}/(2\pi )$ and $(M_1+M_2){\sin }^{3}i=P(K_1+K_2{)}^3(1-e^2{)}^{3/2}/(2\pi G)$. Thus these numerical masses and separation use $e=0$, $\sin i=1$; eclipses alone imply high inclination rather than its exact value.

TODO: the source light curve, its numerical contact phases and the meanings of the four plotted time labels are absent, so unique numerical stellar radii cannot be recovered. The central eclipse contact-radius relations explain what the missing measurements would determine.

If $t_1,\dots ,t_4$ denote the usual first through fourth contacts of a central eclipse, define $T_{14}=t_4-t_1$ and $T_{23}=t_3-t_2$, and let $R_L\ge R_S$ be the two radii. For a circular edge-on orbit,Taking half their sum and difference gives the radii. For short eclipses the familiar approximation is $R_L\simeq v_{\mathrm{rel}}(T_{14}+T_{23})/4$ and $R_S\simeq v_{\mathrm{rel}}(T_{14}-T_{23})/4$, with $v_{\mathrm{rel}}=2K$. At first and fourth contact the discs touch externally; at second and third the smaller disc is just internally tangent to the larger. Noncentral eclipses require the projected impact parameter as well.

Build the star by adding spherical shells. With $m(r)$ the enclosed mass, its gravitational binding energy isFor a nondegenerate extended star, $r(m)<R_{\odot }$ on the interior, soThis is a gravitational binding-energy lower bound for a spherical star. The Kelvin-Helmholtz cooling time is of orderThe displayed energy bound gives $|\Omega |/L_{\odot }>1.6\times 10^7$ years; it supplies a lower bound on that reservoir-to-luminosity ratio. A stellar structure model sets its coefficient. For an ideal monatomic, gas-supported star the virial theorem puts about half the released gravitational energy into heat, leaving half available for radiation; a uniform-density model would give about $9\times 10^6$ years. These contraction estimates are all of order $10^7$ years, much shorter than the nuclear lifetime.

Use the supplied atomic masses consistently. Their mass defect per four hydrogen atoms is $4(1.0078)-4.0026=0.0286$ atomic mass units. The conversion efficiency isThe atomic-mass bookkeeping includes the associated positron-annihilation energy; an exact photon-heating budget also subtracts energy carried away by neutrinos. With the stated ten-percent fuel fraction and constant present solar luminosity, the main sequence lifetime estimate isThe much longer timescale explains why stellar nuclear fusion can power the Sun throughout its long main-sequence history.

A Hertzsprung-Russell diagram plots stellar luminosity upward and effective temperature increasing toward the left. The main sequence runs from hot luminous stars at upper left to cool faint stars at lower right, with the Sun near $T_{\mathrm{eff}}\simeq 5800\mathrm{K}$, $L=L_{\odot }$.

In the evolution of a solar-mass star, core hydrogen fusion initially supplies the luminosity. As core hydrogen is exhausted, hydrogen-shell fusion around a contracting helium core carries the star through the subgiant and red giant phases toward higher luminosity and lower surface temperature. Helium ignition is followed by core helium fusion with a hydrogen-burning shell, placing a solar-mass star in the red clump. The later Asymptotic giant branch has hydrogen- and helium-burning shells around a carbon-oxygen core. Envelope ejection exposes a hot central star of a planetary nebula; residual shell burning ends and the resulting white dwarf radiates stored thermal energy as it cools.

Thermal interstellar dust produces a bright galactic foreground concentrated toward the Milky Way plane, with additional cloud structure. Its emission spectrum differs from the approximately $2.73\mathrm{K}$ blackbody spectrum of the cosmic microwave background, so comparisons across frequencies help separate it. In the sketch below this is the bright equatorial band.

Observer motion produces the cosmic microwave background dipole: to first order in $\beta =v/c$,The direction of motion is hotter and the opposite direction cooler. Its full-sky pattern has one broad positive hemisphere and one broad negative hemisphere; it is not concentrated in the galactic plane.

After subtracting the mean, the motion-induced dipole and galactic foregrounds, intrinsic Cosmic microwave background anisotropy appears as angular hot and cold patches, typically at the $10^{-5}$ fractional scale. These spatial temperature fluctuations encode early density and velocity structure. The displayed patch pattern is illustrative, with no claim to reproduce an observed sky map.

Measure the absolute spectral radiance over a range of frequencies and fit the Planck spectrumFor example, the COBE FIRAS blackbody spectrum experiment used a differential Michelson interferometer to compare the sky with an internal blackbody; an external calibrated blackbody could cover the sky entrance. Cold bolometers measured the difference spectrum, and foreground separation distinguished Milky Way emission from the nearly isotropic background. The fitted background temperature is approximately $\boxed{T_0=2.73\mathrm{K}}$.^[ref] The result relies on an absolute spectral comparison, rather than merely measuring a map's hot-to-cold contrast.

The stated one-sided temperature contrast is $\beta =0.00105$, so the Doppler dipole givesThe motion is toward the hotter region. The contrast already gives one side relative to the mean; the hot-to-cold difference would be twice as large.

Adiabatic expansion preserves the blackbody form while each photon's energy redshifts as $a^{-1}$. Thus the cosmic microwave background temperature obeys $T(z)=T_0(1+z)$, giving

Convert both inputs to energy density. The matter contribution, including rest energy, is $u_{M,0}={\rho }_{M,0}{c}^2=2.25\times 10^{-10}\mathrm{J}{\mathrm{m}}^{-3}$. The given photon value is $u_{\gamma ,0}=5\times 10^5(1.602\times 10^{-19})=8.01\times 10^{-14}\mathrm{J}{\mathrm{m}}^{-3}$. HenceMatter energy density scales as $(1+z{)}^3$, whereas the cosmic microwave background scales as $(1+z{)}^4$. The photon-only matter-radiation equality condition is thereforeThis uses the supplied CMB photon density. Additional relativistic species change equality for total radiation.

Black-body radiation and the photoelectric effect exposed failures of classical continuous-wave energy exchange. Classical equipartition predicted an unbounded short-wavelength thermal spectrum, the ultraviolet catastrophe. Quantized oscillator energies $E_n=nh\nu$ give the mean excitation energy $h\nu /(e^{h\nu /(k_{B}T)}-1)$ and suppress high-frequency modes, producing Planck's law. In the modern description, thermal radiation consists of photons with energy $h\nu$ and Bose occupation numbers.

For the photoelectric effect, increasing intensity below a threshold frequency does not eject electrons, whereas the maximum electron kinetic energy grows with frequency above threshold. Absorption of one photon transfers $h\nu$ to one electron, giving $\boxed{K_{\max }=h\nu -W}$ and threshold ${\nu }_0=W/h$, where $W$ is the work function. Intensity primarily changes the photon arrival rate and hence the photocurrent in the ordinary single-photon regime. Thus changing energy per quantum and changing the number of quanta have distinct consequences. Historically the thermal derivation first quantized matter's energy exchange; it should not be described as an already complete derivation of modern photon theory.

For a photon, $p_{\gamma }=E/c$ and ${\lambda }_{\gamma }=hc/E$. The neutron is nonrelativistic here because $18.8\mathrm{keV}\ll m_n{c}^2$, so $p_n=\sqrt{2m_{n}E}$ and ${\lambda }_n=h/p_n$. Using the supplied common energy givesEqual kinetic energies do not imply equal momenta: the massive particle obeys $E=p^2/(2m_n)$, while the massless photon obeys $E=pc$. The neutron momentum is about $316$ times larger, so its de Broglie wavelength is about $316$ times smaller.

To give the neutron the photon wavelength, set its momentum to $E/c$, not its kinetic energy to $E$ again. ThenIts kinetic energy would then be about $0.188\mathrm{eV}$, connecting this comparison with the next calculation.

For a transverse periodic row of scatterers of spacing $d$, normal incidence makes their incident phases equal. Scattering at angle $\theta$ from that normal creates an outgoing path difference $d\sin \theta$ between neighbours. Constructive interference requires a phase difference $2\pi n$, givingUsing this printed relation as the intended transverse diffraction grating model, the nonrelativistic neutron wavelength at $0.188\mathrm{eV}$ is $\lambda =6.5964\times 10^{-11}\mathrm{m}$. Hence the first-order data giveGeometry qualification: this is not a derivation from reflection at crystal planes parallel to the surface, as the source prose says. The ordinary Bragg's law for planes of normal separation $d$ is $2d\sin {\theta }_B=n\lambda$, where ${\theta }_B$ is the incident glancing angle to those planes. At normal incidence on parallel planes, ${\theta }_B=\pi /2$ and specular reflected rays leave along the normal, so their angle from the normal is zero; they cannot obey the printed positive-order relation $d\sin 0=n\lambda$ for $\lambda >0$. This is a concrete counterexample to identifying the two geometries. The numerical spacing above is therefore conditional on the printed grating-like relation, rather than a forced assertion that the stated Bragg geometry proves it.

For nonrelativistic neutrons, $\lambda \propto E_K^{-1/2}$, so first-order uncertainty propagation gives $|\Delta \lambda |/\lambda =\frac{1}{2}|\Delta E_K|/E_K=0.05$. ThusAt fixed $d$, differentiating the stipulated first-order relation $d\sin \theta =\lambda$ gives $\Delta \theta =\tan \theta \Delta \lambda /\lambda$. ConsequentlyFor photons, $\lambda \propto E^{-1}$, so the same fractional energy uncertainty gives $\Delta {\lambda }_{\gamma }/{\lambda }_{\gamma }=0.10$. Their $18.8\mathrm{keV}$ wavelength almost equals the low-energy neutron wavelength, so the conditional central angle is $0.116973\mathrm{rad}$ andabout twice the neutron angular spread. This difference comes from the different energy-momentum relations, not from substantially different central wavelengths. The small-angle approximation is adequate at $0.117\mathrm{rad}$: replacing $\sin \theta$ by $\theta$ changes it by about ${\theta }^2/6=0.23\%$, and replacing $\tan \theta$ by $\theta$ changes the sensitivity by about $0.46\%$. The numerical angular results retain the preceding geometry qualification; a different physical scattering geometry would require its own angle convention.

In the Franck-Hertz experiment, electrons emitted from a heated cathode are accelerated through a dilute atomic vapour toward a collector. A small retarding potential before the collector rejects electrons that have lost too much kinetic energy. Below the first excitation threshold, mainly elastic collisions occur and the collector current rises as the accelerating voltage increases. At threshold an inelastic collision can transfer a definite excitation energy to an atom, causing a current drop because the slowed electron cannot overcome the retarding potential.

Raising the voltage further permits acceleration again, and repeated losses produce successive current minima. For mercury their spacing is about $4.9\mathrm{V}$, corresponding to an atomic excitation of about $4.9\mathrm{eV}$ and an associated ultraviolet photon near $254\mathrm{nm}$. Contact potentials shift the absolute voltages, so the spacing is more informative than the first minimum's position alone. Discrete energy losses support discrete atomic energy levels; a continuously variable excitation energy would not produce the same recurring fixed loss scale. The experiment establishes quantized excitations without requiring the classical electron orbit of the Bohr model to be literally correct.

A classical orbiting electron accelerates and therefore radiates according to the Larmor formula. Its mechanical energy becomes more negative, its radius decreases, and the orbit spirals inward rather than remaining a stable atom. The Bohr model replaces this prediction with postulated nonradiating stationary states and quantized orbital angular momentum, together with photon emission or absorption only between states, $h\nu =|E_n-E_m|$.

For a heavy nucleus, $m_{e}vr=n\hslash$. Since the de Broglie wavelength is $h/(m_{e}v)$, this impliesThe electronic wave fits an integer number of wavelengths around the orbit. This is a standing-wave consistency interpretation of the quantization rule, not a classical mechanism that prevents an accelerating charge from radiating. For finite nuclear mass, the two-body relative coordinate uses reduced mass $\mu =m_{e}M/(m_e+M)$; the next derivation retains that correction. In the limit $M\gg m_e$, it reduces to the fixed-nucleus electron description above.

Let $r$ be the electron-nucleus separation and use the reduced mass $\mu =m_{e}M/(m_e+M)$. Coulomb attraction supplies the relative centripetal force, and Bohr quantization fixes the relative angular momentum:Eliminating $v$ givesThe total energy is $E=\mu v^2/2-Z{e}^2/(4\pi {ϵ}_{0}r)=-Z{e}^2/(8\pi {ϵ}_{0}r)$, henceHere $a_0$ is the ordinary Bohr radius and $R_{\infty }$ the infinite-nuclear-mass Rydberg constant. The electron's radius about the centre of mass is $M{r}_n/(M+m_e)$; for a heavy nucleus this distinction and the reduced-mass correction are small, but the supplied nuclear mass permits them to be stated precisely.

The emitted photon carries the decrease in atomic energy, so $hc/\lambda =E_n-E_m$ for $n>m$. Using the preceding energy levels givesThe lower state's energy is more negative, making the right-hand side positive. This is the hydrogenic Rydberg formula, with the finite-nuclear-mass factor included in $R_M$.