The 2022 paper is omitted because the released copy includes worked solutions. 2022 paper with solutions, PDF pages 448–455

Past papers and worked solutions: tqm.tripos.org/past-papers.html

Worked solutions are publicly provided for 2017–2023 and 2025. Skip these years when choosing papers for new solutions.

For the right-angle geometry of temporal coherence from crossed beams, a physical arm-path delay $\Delta \ell$ gives $I(y)=I_0+\mathrm{Re}\Gamma (\sqrt{2}y/c+\Delta \ell /c)$, with the sign set by the delayed arm. A quarter-cycle optical phase delay requires $\Delta \ell =\lambda /4$; it shifts the fringe pattern by a quarter of its period, $\lambda /(4\sqrt{2})$. A path delay equal to that latter length is a different operation. Replacing the shifted correlation by a phase multiplier additionally requires the delay to be small compared with the envelope optical coherence time.

Let a solid cone of volume $V$ have its planar base in a plane through the origin, and orient its curved surface $C$ outwards. The vector field $\mathbf{F}(\mathbf{x})=\mathbf{x}$ has divergence three. Its flux through the base vanishes because the position vector lies in that plane and is orthogonal to its normal vector. The divergence theorem therefore gives the displayed curved-surface flux. For a circular base of radius $a$ and perpendicular height $b$, this is $\pi a^{2}b$.

A liquid maintained above its equilibrium boiling temperature at the imposed pressure. A barrier to nucleation can delay its conversion into vapor, making this a metastable state.

Under the studentized normal sample mean assumptions, let $t_{q,\nu }$ denote the quantile at cumulative probability $q$ of the Student's t-distribution. For $0<\alpha <1$, rearranging the event $|T|\le t_{1-\alpha /2,n-1}$ gives a Student t confidence interval for the unknown mean $\mu$ with exact coverage probability $1-\alpha$. ^[ref]

Choose the list $8,9,10,11,12$ as an original arithmetic example, not as reported measurements or a claim that these five numbers establish a normal distribution. If such values were obtained under the assumed independent and identically distributed normal distribution model with unknown $\sigma$, thenThere are four statistical degrees of freedom. For a 95% confidence interval, $t_{0.975,4}=2.776445105\dots$, soUsing the standard normal quantile $z_{0.975}=1.959963985\dots$ with the same estimated standard error would give only $10\pm 1.385903824\dots$. That underestimates the correct margin by about 29.4%. Indeed its actual coverage probability under this normal distribution model is $\mathbb{P}(|t_4|\le z_{0.975})\simeq 0.8784$, rather than 0.95.

The observations, the sample mean and the confidence interval all have different meanings. The bar estimates the population mean; it is not an interval intended to contain 95% of individual observations. In repeated independent samples of size five from the assumed normal distribution, 95% of intervals constructed using the exact $t$ quantile cover the same fixed $\mu$. This is not a 95% posterior probability that the fixed statistical parameter lies in this particular realized interval. Such a probability requires a separate Bayesian inference model.

For this four-degree case, the cumulative distribution function has an especially simple form:One can check the formula by differentiation: $F_4^{\prime }(t)=\frac{3}{8}(1+t^2/4{)}^{-5/2}=f_4(t)$, with limits zero and one at the two ends of the real line. Solving the increasing equation $F_4(t)=0.975$ by the bisection method gives the quantile used above.

A release factor is a protein involved in terminating biological translation and releasing the completed polypeptide at a stop codon.^[ref]

Let a closed oriented surface $S$ lose a patch $H$, leaving $S^{\prime }$. Assume a smooth vector field on the relevant surfaces and that its total curl flux through $S$ is zero. The Stokes theorem gives the title formula when the edge is oriented as the boundary of the remaining surface $S^{\prime }$. The boundary of the removed patch $H$, with its original normal, has the opposite direction and hence the opposite circulation.

For a cylindrical side of radius $R$, if $(\nabla \times \mathbf{v})\cdot \mathbf{n}=-\alpha R$ everywhere on the removed patch, the circulation around the boundary of the remaining surface is $\alpha R\mathrm{area}(H)$. This uses actual curved area, rather than its planar projection, and does not depend on the patch shape. For multiple boundary components, the formula concerns the sum with their induced orientations.

For a continuously differentiable vector field on an oriented surface with compatibly oriented boundary, this vector identity follows by dotting both sides with an arbitrary constant vector $a$ and applying Stokes theorem to $\mathbf{F}\times a$. The curl identity $\nabla \times (a\times \mathbf{F})=a\nabla \cdot \mathbf{F}-(a\cdot \nabla )\mathbf{F}$ gives the surface integrand. In components, $[(d\mathbf{S}\times \nabla )\times \mathbf{F}{]}_i=d{S}_j{\partial }_i{F}_j-d{S}_i{\partial }_j{F}_j$, with the derivatives acting on $\mathbf{F}$. For the position vector field $\mathbf{F}=\mathbf{r}$, it equals $-2d\mathbf{S}$; hence the line integral is minus twice the oriented vector area.

The unit square here is the Cartesian product $[0,1]\times [0,1]$ in the Euclidean plane, a square of side length one.

Exchanging two identical bound atoms exchanges their constituent fermions. An atom with mass number $A$ and atomic number $Z$ contains $A$ nucleons and $Z$ electrons, so its exchange sign is $(-1{)}^{A+Z}$ when its internal state is preserved. An odd count gives a composite fermion; an even count gives a composite boson. This is equivalently the integer-versus-half-integer total-spin distinction in the Spin-statistics theorem. For example, neutral lithium-6 has nine fermionic constituents and is a fermion, while neutral lithium-7 has ten and is a boson. Counting nucleons alone instead classifies the nucleus, not the neutral atom. This effective composite-particle description requires energies and separations where the bound atom remains intact.

A detector collects a range of solid angles, not a single direction. If its nonnegative angular response is $w(\Omega )$, defineIntegrating the polarized beta-decay angular distribution gives a count proportional to $G(1+ak)$. For a uniform forward cone with half-angle ${\theta }_0$, put $c_0=\cos {\theta }_0$. Direct integration givesThe opposite cone has mean cosine $-k$. A hemisphere gives $k=1/2$; a narrow cone tends to $k=1$. Hence a spin-reversal cross-ratio for beta decay measures $ak$, and an acceptance correction is needed to obtain $a$.

With unequal angular responses, let the two laboratory mean cosines be $k_{+}$ and $k_{-}$, including their signs. Then the actual expectation iswhich reduces to the simple square only when $k_{-}=-k_{+}$. The cancellation of stable multiplicative detection efficiencies survives, but their angular shapes still matter. An energy-integrated response likewise averages $A(E)P\beta (E)\cos \theta$ with the accepted spectrum; a mean speed cannot in general be substituted independently of acceptance.

Additive backgrounds, unequal polarization magnitudes or detection efficiencies that change on reversal do not have the assumed factorization. For example, a reversal-dependent efficiency cross-ratio multiplies $R$ by ${\eta }_{+}^{+}{\eta }_{-}^{-}/({\eta }_{-}^{+}{\eta }_{+}^{-})$. Background subtraction also creates statistical complications: positive expected signal counts are assumed in this derivation, whereas finite samples with Poisson distributions can contain zero counts. Taking square roots of negative background-subtracted counts is not a valid estimator. These limits explain why reversal and independent preparation checks belong together.

For a diagonalizable matrix with distinct eigenvalues $\lambda ,\mu ,\dots$, this polynomial is the projection operator onto the eigenspace for $\lambda$ along the direct sum of the other eigenspaces. It is one on its own eigenspace and zero on the others. Consequently $P_{\lambda }{P}_{\mu }=0$ for different eigenvalues, $\sum P_{\lambda }=I$ and $A=\sum \lambda P_{\lambda }$. These projections need not be orthogonal projections; that stronger property holds for a normal matrix. This is the polynomial interpolation form of the spectral decomposition.

Use natural units $c=1$ and the Minkowski metric with signs $(+,-,-,-)$. In the ideal centre-of-momentum frame, assume a known initial four-momentum $P=(\sqrt{s},0)$, no unaccounted radiation and exact visible reconstruction. DefineThe missing four-momentum in electron-positron collisions is a subtraction of measured quantities. It is not itself a particle detector hit. With $M_{e\mu }^2=(p_e+p_{\mu }{)}^2$, expansion gives the useful identityFor any collection of future-directed massless invisible four-momenta $q_i=E_i(1,{\hat{\mathbf{n}}}_i)$,Thus a positive missing invariant mass excludes exactly one massless invisible particle under the stated assumptions. It does not exclude one massive invisible particle, identify a neutrino flavor, or determine the number of massless invisible particles beyond that lower bound. Several collinear massless particles can even have zero combined invariant mass. Detector resolution and unaccounted radiation invalidate a literal event-by-event interpretation of these ideal inequalities.

For an original exact example in natural units, choose total energy $6\mathrm{GeV}$ and visible massless four-momenta, in $\mathrm{GeV}$,Two possible invisible massless four-momenta are $(2;-1/2,-1/2,\sqrt{14}/2)$ and $(2;-1/2,-1/2,-\sqrt{14}/2)$. Splitting either into two positive collinear parts gives a three-particle interpretation with identical recoil. Alternatively, one invisible particle of mass $\sqrt{14}\mathrm{GeV}$ has that recoil. This example tests the ambiguity of the subtraction, without asserting any of these alternatives is a tau lepton decay.

In the modern weak charged current interpretation, one allowed assignment isThe electron antineutrino, tau neutrino, muon neutrino and antineutrino of the tau neutrino carry away energy and momentum. There are four undetected particles in this decay assignment. Interchanging which parent decays electronically gives the opposite electric charge assignment. Each parent conserves electric charge and, in this approximation, total lepton number; four-momentum conservation applies separately to production and to each particle decay.

If $B_e$ and $B_{\mu }$ denote the corresponding branching fractions, independent parent decays give an inclusive mixed-flavour fraction $2B_e{B}_{\mu }$. The factor two counts the two assignments to opposite parent charges. For chosen $B_e=B_{\mu }=1/5$, the mixed fraction is $2/25$, before acceptance and detection efficiency. Correlations between the parents' spins can affect angular selection even when the fully inclusive branching fractions factorize.

The neutrinos in that equation are a physical interpretation to test. Neither a deficit in visible energy nor two unusual tracks alone establishes their identities or uniquely counts them. Even a compatible invariant mass and an angular distribution require particle identification, background checks and agreement across further particle decay channels. The direct detection milestone for the tau neutrino was in 2000, separate from the charged tau lepton discovery.

The beam-overlap factor $L$ relating a reaction's cross section $\sigma$ to its ideal production rate $dN/dt$. Integrated collider luminosity $\mathcal{L}=\int Ldt$ has units of inverse area and gives the expected production count $\mathcal{L}\sigma$ when $\sigma$ is constant.

The Homestake experiment, led by Raymond Davis Jr., used electron neutrino capture on chlorine to produce argon subject to radioactive decay. The 1968 paper reported an upper limit, rather than a resolved positive signal.^[ref]

The underground target contained tetrachloroethylene. Circulated helium swept out argon, which was trapped on cooled charcoal, purified, and put into tiny proportional counters. Electron capture back to chlorine supplies the delayed readout. Added stable argon checked extraction recovery; background counting checked spurious pulses. The capture threshold is $0.814\mathrm{MeV}$ and the product's half-life is about 35 days. The chemical extraction retains neither the original neutrino's arrival time nor its direction.^[ref]

The reaction selects electron neutrinos, and its threshold excludes the lowest-energy part of the solar neutrino spectrum. A low count alone cannot distinguish a low source neutrino flux, a changed neutrino flavor mixture, or an underestimated loss in extraction or counting. This is why chemical recovery tests and an independent water Cherenkov neutrino reconstruction are scientifically valuable.