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For the charged lepton, $p_{\ell }={\beta }_{\ell }{\gamma }_{\ell }{m}_{\ell }$ and $E_{\ell }={\gamma }_{\ell }{m}_{\ell }$. Their ratio is therefore its speed in natural units. Using part (a),Numerically,The electron is almost ultrarelativistic, while the muon is relatively slow. In ordinary units these are approximately $0.999974c$ and $0.271c$.

There are three free-particle rest-energy terms and five nuclear corrections. In the paper's natural units, their meanings are:

The sign convention is significant: the quantity labeled $E_{\mathrm{bind}}$ in the supplied mass formula is added to the free-particle masses and is normally negative for a bound nucleus. It is the negative of the conventional positive nuclear binding energy $B$:The coefficients must therefore retain their printed signs; using positive $B$ while still adding it to the masses would be inconsistent.

All three processes are weak charged current transformations between neighboring nuclear isobars. Neglecting neutrino masses, and writing $M_a(A,Z)$ for a neutral atomic mass, their Q values are:

These formulae use the paper's natural units. With energies in MeV, beta-plus decay requires an atomic mass-energy difference exceeding $2m_e{c}^2=1.022\mathrm{MeV}$. The extra electron-mass subtraction arises from using neutral atomic rather than bare nuclear masses. Electron capture can occur when the atomic mass-energy difference is positive but below this threshold. For a particular capture shell or excited daughter, the available energy must also cover the electron-hole and nuclear excitation energies, and the small nuclear recoil. Small atomic binding corrections are suppressed in the displayed threshold formulae.

The question's fixed-$A$ family consists of nuclear isobars, rather than isotopes: isotopes have fixed $Z$. The semi-empirical mass formula gives a convex nuclear mass parabola as a function of $Z$ at fixed $A$. For odd $A$, nuclear pairing does not split the family into two branches, so there is typically one beta-stable isobar, at the discrete minimum. Neighboring members can lower their mass by changing $Z$ toward it.

For even $A$, nuclear pairing lowers even-even nuclei and raises odd-odd nuclei, producing two offset nuclear mass parabolas. There are often two beta-stable even-even isobars, separated by an odd-odd member to which neither can energetically decay. The number is not an exact universal rule: some even-$A$ chains have one or more than two such minima against single beta decay, and a few light odd-odd nuclei are stable. Stability against single beta decay also does not guarantee stability against double beta decay.

The baryon hyperfine mass formula is a phenomenological constituent-quark model. The three constituent quark masses include effects of strong-interaction dressing and an approximate spin-independent confinement energy; they are not the small running quark masses in a short-distance Quantum chromodynamics calculation.

The residual correction has the form of a colour-magnetic spin-spin interaction. A nonrelativistic reduction of one-gluon exchange produces a contact contribution proportional toAfter taking its expectation in the common spatial wavefunction and the colour singlet, the colour factor, coupling, and contact density are absorbed into $A^{\prime }$. Summing the three quark pairs gives the supplied formula. Positive $A^{\prime }$ raises aligned-spin configurations relative to lower-spin configurations. In natural units, the spin angular momentum operators are dimensionless, so $A^{\prime }$ has units of energy cubed, as in the given $\mathrm{Ge}{\mathrm{V}}^3$.

Source paper: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/3/Physics%20Papers%202014.pdf?cookie_passthrough=1

Solutions use natural units unless a conversion is displayed. The surviving paper has damaged source illustrations in c and c; the corresponding numerical readings remain explicit TODOs. All eight questions, including all five alternatives in 8, are treated.

Assume coherent production and detection, relativistic neutrinos, two active neutrino flavours, propagation in vacuum, negligible packet separation and no absorption. Invert the given real lepton mixing matrix to obtain $|{\nu }_{\mu }⟩=c|{\nu }_2⟩-s|{\nu }_3⟩$ and $|{\nu }_{\tau }⟩=s|{\nu }_2⟩+c|{\nu }_3⟩$, where $c=\cos \theta$, $s=\sin \theta$. With phases ${\varphi }_a=E_{a}t-{\mathbf{p}}_a\cdot \mathbf{x}$ in natural units,Using the paper's opposite sign convention conjugates this probability amplitude and leaves the probability unchanged. For equal momentum, $E_a\simeq p+m_a^2/(2p)$ and $t\simeq L$, so $\Delta \varphi =\Delta m^{2}L/(2E)$. The equal-energy convention gives the same leading result from the momentum difference; evaluating each component at a different arrival time would incorrectly double the phase.Since $\hslash c=1.97327\times 10^{-7}\mathrm{eV}\mathrm{m}$,This gives the stated factor $1.27$. Neutrino oscillation in this approximation measures $|\Delta m^2|$, not the absolute neutrino mass or its ordering.

Source paper: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/6/Physics%20Papers%202017.pdf?cookie_passthrough=1. The held cover identifies C4 on 14 June 2018; the annual bundle's filename does not determine the year. Natural units $\hslash =c=1$ are used except in the silicon-detector calculations, where SI units and the momentum conversion are stated explicitly. All six questions are solved.

Use dimensionless physical helicity $h=\mathbf{S}\cdot \hat{\mathbf{p}}$, with spin eigenvalues $\pm 1/2$ in natural units. In the Dirac representation,The entries $1$ in the supplied transformation denote two-by-two identities. Since its matrix acts only on the upper/lower spinor blocks, it commutes with the repeated Pauli block. Therefore $u^{-1}=u$ andThe eigenvalues remain the physical spin projection along the direction of motion, $+1/2$ or $-1/2$. A convention omitting the factor one half instead labels them $\pm 1$; this changes no measurement probabilities. A change of gamma-matrix representation cannot change the spin observable.

Source paper: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/9/Physics%20Papers%202020.pdf?cookie_passthrough=1

The source is on bundle pages 34–41. Its cover gives Trinity Term 2021 despite the attachment's 2020 filename. These solutions use natural units $\hslash =c=1$ unless a conversion is written explicitly.

Use the Minkowski metric $(+,-,-,-)$ and choose the incoming electron along $+z$, with the scattered electron in the $xz$ plane. In natural units, neglecting the electron mass, the four-momenta areThe momentum transfer entering the proton is thereforeThe outgoing hadronic system carries $p_X=p_2+q$. The energy transfer is $\nu =E-E^{\prime }$, while $\mathbf{q}$ need not point along the original beam.

Source: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/10/Physics%20Papers%202021.pdf?cookie_passthrough=1

Source bundle pages 31–39. The cover dates this paper to 2022; the attachment filename says 2021. These solutions use natural units $\hslash =c=1$, except where units are restored explicitly.

The lepton mixing matrix is unitary, so each row has unit norm. Explicitly its electron row obeyswhere $c_{ij}=\cos {\theta }_{ij}$ and $s_{ij}=\sin {\theta }_{ij}$. This is also the normalization needed for unit electron neutrino survival at $L=0$.

Use $\mathrm{Re}(1-e^{ix})=1-\cos x=2{\sin }^2(x/2)$ and define ${\Delta }_{ij}=({\varphi }_i-{\varphi }_j)/2=\Delta m_{ij}^{2}L/(4E)$ in natural units. The exact vacuum neutrino oscillation result becomesThe given mass-squared differences satisfy $|\Delta m_{12}^2/\Delta m_{23}^2|\simeq 0.027\ll 1$, so the two fast oscillation frequencies are close: ${\Delta }_{13}={\Delta }_{23}+{\Delta }_{12}$. Approximating their phases as equal combines their coefficients intowhile $4c_{12}^2{s}_{12}^2{c}_{13}^4=c_{13}^4{\sin }^{2}2{\theta }_{12}$. HenceThe complex CP phase cancels from every $|U_{ei}{|}^2$. The phase-replacement approximation is uniformly reliable for $|{\Delta }_{12}|\ll 1$, especially at short reactor baselines; closeness of the mass-squared differences alone does not guarantee close sine values at arbitrarily large $L/E$. At medium baselines the distinct fast frequencies can beat, so the exact three-frequency expression is preferable if their detailed pattern matters.

Source paper: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/4/Physics%20Papers%202015.pdf?cookie_passthrough=1

Solutions use natural units $\hslash =c=1$ and Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$. The d'Alembert operator is $\square ={\partial }_{\mu }{\partial }^{\mu }$. All four questions are covered, including the optional fourth question.

Let brackets denote mass dimension. In natural units, $[x]=-1$ and $[{\partial }_{\mu }]=1$, so a dimensionless action requires $[\mathcal{L}]=D$. The kinetic term fixesThe cubic scalar field theory interaction then requires $[\lambda ]+3[\varphi ]=D$, givingThe coupling constant is dimensionless at this critical dimension of a monomial scalar interaction. For $D<6$ it has positive mass dimension, while for $D>6$ it has negative mass dimension, the usual power counting distinction between super-renormalizable and non-renormalizable cubic interactions.

Source paper: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/5/Physics%20Papers%202016.pdf?cookie_passthrough=1. The cover identifies Hilary Term 2017, 11 January; the bundle's 2016 filename does not determine the examination year. We use natural units $\hslash =c=1$, the mostly-minus Minkowski metric, and Feynman propagators with the $+i0$ prescription.

Source paper: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/9/Physics%20Papers%202020.pdf?cookie_passthrough=1

The source cover identifies Hilary Term 2021 despite the attachment's filename. Use natural units $\hslash =c=1$, the Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$, and $/p={\gamma }^{\mu }{p}_{\mu }$. The three-dimensional scalar field theory in 3 has two spatial dimensions and one time dimension. Results are at the stated order of perturbative quantum field theory.

Source PDF: mmathphys.physics.ox.ac.uk/sites/g/files/vzsiac2991/files/mmathphys/documents/media/a15088w1.pdf

Use natural units $\hslash =c=1$ and the mostly-minus Minkowski metric. Question 3 has two spatial dimensions: its gamma matrices are the specified two-by-two matrices. The printed ${\sigma }^{\mu }$ in its kinetic term is interpreted as those ${\gamma }^{\mu }$, rather than as three unmodified Pauli matrices. All spin sums in question 2 instead use four-component Dirac spinors in four spacetime dimensions.

In $2+1$ spacetime dimensions, the dimensionless action gives $[\mathcal{L}]=3$ and $[{\partial }_{\mu }]=1$. The scalar kinetic term requires $2[\varphi ]+2=3$, whereas the fermion kinetic term requires $2[\psi ]+1=3$. Thus the engineering dimensions of relativistic fields areMatching each interaction to mass dimension three givesHere the brackets mean exponents of mass in natural units. The coefficients $g_1,g_3$ are marginal couplings by classical power counting, and $g_2$ is a relevant coupling.

Source PDF: mmathphys.physics.ox.ac.uk/sites/g/files/vzsiac2991/files/mmathphys/documents/media/a15088h1.pdf

The solutions use natural units $\hslash =c=1$ and the Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$. In question 1, ${\gamma }^5={\gamma }^0{\gamma }^1{\gamma }^2{\gamma }^3$ is exactly the paper's convention: it contains no factor $i$. The qualifications in iii, d and d distinguish the printed expressions from the results of charge conservation, identical final-state symmetry factors and fermionic time ordering.

Use the geodesic radial comoving coordinate, in which the Friedmann-Lemaître-Robertson-Walker metric has the formOn a slice of constant cosmic time, the radial proper-length element is $R(t)\mathrm{d}\chi$. ThereforeIf one starts from a curvature coordinate with radial element $R\mathrm{d}r/\sqrt{1-k{r}^2}$, then $\chi ={\int }_0^r\mathrm{d}{r}^{\prime }/\sqrt{1-k{r}^{\prime 2}}$; the curvature coordinate $r$ itself is not generally the $\chi$ used above.

For a comoving galaxy, $\dot{\chi }=0$, so $\dot{D}=\dot{R}\chi =HD$. The definitions areThis recession law concerns proper distances on common cosmic-time slices. It is not a local special-relativistic relative velocity measurement; a large recession rate is not a signal exceeding the local speed of light.

For a comoving volume $V=V_0{R}^3$, total fluid energy is $E=\rho c^{2}V$. Adiabatic energy conservation gives $\mathrm{d}E+P\mathrm{d}V=0$. Expanding,henceHere $\rho$ is mass-equivalent energy density. Radiation loses energy both by dilution and by pressure work; pressureless matter only dilutes in this description.

In the time-unit convention for the cosmological constant, it represents a uniform vacuum contribution with pressure equal to minus its energy density. There is a units convention in the printed field equations: their term is $\Lambda /3$, so the symbol $\Lambda$ there has units of inverse time squared. If the geometrical constant is instead given in inverse length squared, this symbol equals $c^2{\Lambda }_{\mathrm{geom}}$. The curvature parameter $k$ labels constant-curvature spatial slices; with a conventional scale-factor normalization it is $+1,0,-1$ for positively curved, flat and negatively curved slices. It fixes local curvature sign, not every possible global spatial topology.

For a flat fluid-only model without a cosmological constant and with $P=w\rho c^2$, the two field equations giveThe two printed cases and their scale-factor laws are worked under the literal inline labels below.

Inflation is accelerated expansion, $\ddot{R}>0$, equivalently a decreasing comoving Hubble radius $(RH{)}^{-1}$ on an expanding branch. An approximately constant positive vacuum-like energy can give nearly exponential expansion. Enough inflation can put regions now widely separated inside a common earlier causal patch, suppress $|k|/(RH{)}^2$ and dilute unwanted relics such as monopoles. These are the horizon, flatness and relic-abundance motivations. Reheating is needed to populate the subsequent hot universe. Quantum fluctuations can seed primordial perturbations, but inflation alone is not a universal solution to the initial singularity or a proof that all initial conditions inflate.

Now adopt the printed natural units $c=\hslash =1$. For the Homogeneous massive-scalar equation in an expanding universe,The continuity equation givesWhere $\dot{\varphi }\ne 0$, this implies the expanding-universe Klein-Gordon equationA regular canonical-field solution extends it through isolated turning points by continuity. Purely dividing the conservation equation does not determine a trajectory on an interval where $\dot{\varphi }$ is identically zero; canonical field dynamics, or variation of its action, supplies the equation there. In particular a constant nonzero massive field is not an exact freely evolving solution.

With $\Lambda =k=0$,ThereforeThis inflation condition for a homogeneous massive scalar means a potential-dominated field has sufficiently negative pressure to accelerate expansion. In quadratic chaotic inflation, a large initial field with $V=m^2{\varphi }^2/2$ can enter slow roll, with $3H\dot{\varphi }\simeq -m^2\varphi$ and $H^2\simeq 4\pi G{m}^2{\varphi }^2/3$. The potential slow-roll parameter is ${ϵ}_V=1/(4\pi G{\varphi }^2)$, so a large field relative to the gravitational scale can support many e-folds. The field eventually rolls far enough that this regime ends and oscillation and reheating become relevant. Large potential energy with arbitrary initial kinetic energy does not by itself guarantee an inflationary interval.

Differentiating the flat Friedmann equation and then using the scalar equation,For the expanding branch,This Nonincreasing Hubble rate for a canonical inflaton decreases strictly where the field rolls and can be momentarily stationary at a turning point; the statement “always decreasing” means nonincreasing. This does not conflict with acceleration, because $\ddot{R}/R=H^2+\dot{H}$ remains positive when ${ϵ}_H=-\dot{H}/H^2<1$.

At horizon exit, larger comoving wavelengths leave earlier, when $H$ is larger. For a canonical field in slow roll, with reduced Planck mass ${\overline{M}}_{\mathrm{Pl}}=(8\pi G{)}^{-1/2}$,The decreasing expansion rate gives a red tensor tilt at this order. The scalar curvature spectrum also contains the conversion from field fluctuations to curvature:Here $N=\ln R$ counts elapsed e-folds. Thus the Decreasing Hubble rate does not alone fix scalar tilt. Evolution of ${ϵ}_H$ matters. For the quadratic slow-roll model, ${\eta }_V={ϵ}_V$ and $n_s-1\simeq -6{ϵ}_V+2{\eta }_V=-4{ϵ}_V<0$. If $N_{\mathrm{rem}}$ e-folds remain until the end, leading estimates give $n_s\simeq 1-2/N_{\mathrm{rem}}$ and tensor-to-scalar ratio $r_{\mathrm{TS}}\simeq 8/N_{\mathrm{rem}}$, with small end-of-inflation corrections.

The relevant contemporary comparison is a nearly scale-invariant but red scalar spectrum. The February2015 Planck analysis reported $n_s=0.968\pm 0.006$ for its temperature, low-polarization and lensing combination, and an upper limit $r_{0.002}<0.11$ at95percent confidence. The scalar tilt is qualitatively consistent with a slowly evolving inflaton, but that result does not establish the quadratic potential; its relatively large tensor prediction was already disfavored against lower-tensor alternatives in that analysis.^[ref]