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The cover establishes Hilary Term 2020 despite the source bundle's 2019 filename. Use $\hslash =c=1$, the Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$, and $/p={\gamma }^{\mu }{p}_{\mu }$ for Feynman slash notation. In question 1, the repeated species index in the Yukawa interaction is summed over $i=1,2$. The spatial mode expansions in question 2 are at a fixed time.

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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.

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Source bundle pages 100–103. The cover establishes Hilary Term 2022 despite the attachment's 2021 filename. Throughout, use $\hslash =c=1$ and the Minkowski metric with signature $(+,-,\dots ,-)$.

Use $\hslash =c=1$, the Minkowski metric $\eta =\mathrm{diag}(1,-1)$, and $p^2=(p^0{)}^2-(p^1{)}^2$. The momentum-space Feynman rules for the real scalar field areAt each interaction vertex impose two-dimensional momentum conservation; before removing the overall conservation factor this is $(2\pi {)}^2{\delta }^{(2)}({\sum }_j{p}_j)$. Integrate each independent loop momentum with $d^{2}k/(2\pi {)}^2$, and divide a Feynman diagram by its symmetry factor. External lines carry the same scalar propagator in an unamputated two-point correlation function; they are removed in an amputated scattering amplitude.

The quadratic action is inverted with the Feynman i-epsilon prescription to obtain the scalar propagator. Expanding the Dyson series gives $-i{g}_4/4!$ per insertion; the $4!$ attachments in the Wick theorem cancel the factorial. Fourier integration over the insertion point supplies the momentum conservation factor. Automorphisms that leave the labelled external attachments fixed account for the remaining symmetry factor. In two spacetime dimensions the mass dimensions are $[\varphi ]=0$ and $[g_4]=2$.

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.

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.

Local equivalence and tides. In Newtonian mechanics, inertial mass $m_{\mathrm{i}}$ measures the response to a force, whereas passive gravitational mass $m_{\mathrm{g}}$ couples to an imposed Newtonian gravitational field. A freely falling body obeysThe observed composition independence of free fall says that this ratio is universal. With a common choice of units it is one. This is the Newtonian content of the weak equivalence principle for test particles; it does not by itself specify all nongravitational physics.

The Einstein equivalence principle further says that local nongravitational experiments in a sufficiently small freely falling laboratory have the form predicted by special relativity, independently of the laboratory's location and motion. In such a local inertial frame, the metric tensor is locally the Minkowski metric and the Christoffel symbols vanish at the selected event. The strong equivalence principle extends the universality claim to local gravitational experiments and bodies with significant gravitational self-energy. None of these statements removes measurable tidal forces across a finite apparatus.

In an ideal small freely falling lift, a released ball initially floats relative to the cabin, because both undergo the same free fall. Two separated balls, however, need not retain their separation: they sample different parts of the gravitational field. For a point source, linearizing $\mathbf{g}=-GM\mathbf{r}/r^3$ across a small separation $\mathbf{\xi }$ gives the tidal tensor predictionA radial separation stretches with coefficient $2GM/r^3$, whereas a transverse separation is compressed with coefficient $-GM/r^3$. Initially parallel velocities therefore develop relative motion even though the individual bodies experience zero proper acceleration.

The relativistic statement is geodesic deviation. With the Riemann curvature tensor conventiona connecting vector between neighboring freely falling worldlines satisfiesHere $D/d\tau$ is the covariant derivative along the reference geodesic, and $U^{\mu }$ its four-velocity. Curvature can be nonzero where the Christoffel symbols have been made zero. In the weak, static, slow-motion limit this equation reduces to ${\ddot{\xi }}^i=-{\partial }_i{\partial }_j\Phi {\xi }^j$, recovering the tidal tensor above. A finite falling lift can detect gravity through these relative accelerations; local equivalence does not imply a globally uniform field.

Weak-field connection and potential. Take $x^0=ct$ and the $+---$ metric signature. There is an index error in the printed formula for the Christoffel symbols: its first differentiated metric component must be $g_{\nu 0}$, not $g_{\mu 0}$. The correctly contracted expression isThe offending term vanishes under the stipulated time independence, so this correction does not change the intended calculation. Write $g_{\mu \nu }={\eta }_{\mu \nu }+h_{\mu \nu }$ and treat both $h$ and its derivatives as first order. To that order $g^{00}=1-h_{00}$, $g^{0i}=O(h)$, and $g^{ij}=-{\delta }^{ij}+O(h)$. A stationary metric tensor givesFor the temporal component the inverse-metric coefficient is already first order, soFor a spatial component the zeroth-order inverse metric contributes:The time component of the approximate geodesic equation says that $dt/d\tau$ is constant to the retained order. In its spatial component, convert the proper time derivatives to coordinate-time derivatives and cancel $(dt/d\tau {)}^2$:Comparison with the Newtonian equation $d^2\mathbf{x}/d{t}^2=-\nabla \Phi$ identifies $h_{00}=2\Phi /c^2+C$. For an isolated source, choose the gravitational potential to tend to zero at infinity and normalize the time coordinate there, giving $C=0$. ThusThe quantity $\Phi$ is the Newtonian gravitational potential, or gravitational potential energy per unit mass, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}={\mathrm{m}}^2{\mathrm{s}}^{-2}$. For a point source, $\Phi =-GM/r$. A static clock has $d\tau =\sqrt{g_{00}}dt\approx (1+\Phi /c^2)dt$, connecting the Newtonian limit with gravitational time dilation.

Frequency of the orbiting source. Treat the gas blob as a test particle on a timelike circular Schwarzschild geodesic, with pressure and other nongravitational forces negligible. Treat our very distant location as infinity, so the observer's own gravitational potential is negligible. These are the assumptions that make a unique answer in terms of $R_S/r$ possible.

Put $f=1-R_S/r$ and use a local static observer at the emission point as an intermediate measuring frame. The remote observer lies on the symmetry axis of the orbit. A ray arriving on that axis has zero conserved axial angular momentum, so its local direction is perpendicular to the emitter's azimuthal velocity. Bending within a meridional plane does not alter this orthogonality. The transverse relativistic Doppler effect therefore giveswhere $v$ is the orbital speed measured by that local static observer. The gravitational redshift between static Schwarzschild observers then givesA gravitational circular orbit has $v^2=GM/r+O((GM/r{)}^2/c^2)$ in the weak-field limit, so $v^2/c^2=R_S/(2r)+O((R_S/r{)}^2)$. Expanding both factors,Solving for the emitted frequency givesThe gravitational part is $R_S/(2r)$ and the transverse kinematic time dilation contributes another $R_S/(4r)$; the absence of line-of-sight motion does not remove this second effect.

As an exact check, the local speed on a timelike circular Schwarzschild geodesic satisfies $v^2/c^2=\mu /(r-2\mu )$, where $\mu =GM/c^2=R_S/2$. Therefore the face-on frequency shift of a circular Schwarzschild source iswhose expansion agrees with the boxed answer. If the gas is supported or accelerated by pressure or magnetic stresses rather than following a geodesic, its orbital speed is not fixed by the mass and radius alone; the appropriate first-order result is instead ${\nu }_{\mathrm{em}}\approx {\nu }_{\mathrm{ob}}[1+GM/(r{c}^2)+v^2/(2c^2)]$, with $v$ supplied independently.

For the supplied plus polarization, the perturbation is traceless, so $h_{\mu \nu }={\varphi }_{\mu \nu }$. At a fixed longitudinal coordinate $z_0$, put $H(t)=h_{\oplus }\cos [\omega (t-z_0)]$. With the $+---$ convention chosen above, the transverse spatial metric measuring positive squared lengths isThe initial coordinate separation is $(\Delta \cos \theta ,\Delta \sin \theta )$ and remains constant in transverse-traceless gauge: $h_{00}=h_{0i}=0$ implies ${\Gamma }^i{}_{00}=0$, so initially stationary particles stay at fixed spatial coordinates. Their measured proper length is insteadThe first expression is understood within the linearized metric; only its first-order expansion is controlled. Separation along the two transverse axes varies oppositely; at $45$ degrees there is no first-order length modulation. This is the angular response of a gravitational-wave separation, despite constant coordinate separation.

If the background is instead chosen as $-+++$ while keeping the same numerical spatial amplitude, the displayed length modulation has the opposite sign. The question does not state a signature here; the convention above agrees with its earlier explicitly specified Minkowski metric. The physical alternating stretch-compress pattern is unchanged by consistently redefining the signed amplitude.

At any event on a massive observer's worldline, choose an orthonormal tetrad with timelike basis vector $e_{(0)}=U/c$. Its components obey $g(e_{(a)},e_{(b)})={\eta }_{ab}=\mathrm{diag}(1,-1,-1,-1)$, so locally $d{s}^2={\eta }_{ab}d{X}^{(a)}d{X}^{(b)}$: the Minkowski metric. A local inertial frame can also remove the Christoffel symbols at that event. Curvature cannot be removed on a finite region, and a frame following an accelerated observer need not remain inertial in its neighbourhood.

In general coordinates, $X^{\mu }(\tau )$ labels the trajectory, while the four-velocity and four-acceleration areCoordinate values $X^{\mu }$ are not themselves a vector under arbitrary nonlinear coordinate changes; their tangent $U$ is. For fixed rest mass $m$, the four-momentum is $P^{\mu }=m{U}^{\mu }$ and the four-force is $f^{\mu }=D{P}^{\mu }/d\tau =m{A}^{\mu }$. With the $+---$ metric signature,where $a$ is the proper acceleration. The middle identity follows by differentiating the constant velocity norm; in the instantaneous rest frame, $U=(c,0)$ and $A=(0,\mathbf{a})$, proving the last identity.

Use the Minkowski metric with signature $(+---)$ and write $U=(U^0,\mathbf{u})$, $V=(V^0,\mathbf{v})$. Future-pointing timelike means $U^0>\Vert \mathbf{u}\Vert$ and $V^0>\Vert \mathbf{v}\Vert$. Thusso $U+V$ is again future-pointing timelike.

Let $a=\sqrt{g(U,U)}>0$ and $b=\sqrt{g(V,V)}>0$. In a Lorentz frame where $U=(a,0,0,0)$, the future-pointing vector $V$ has $V^0\ge b$. Consequently $g(U,V)=a{V}^0\ge ab$, with equality exactly when $V$ has no spatial component in that frame. HenceTaking positive square roots provesEquality holds exactly when $U,V$ are positive scalar multiples of each other. This is the reverse triangle inequality for future timelike vectors, not the Euclidean triangle inequality.

Use coordinates $X=(ct,x,y,z{)}^T$ and Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$. A real $4\times 4$ matrix $L$ is a Lorentz transformation precisely when it preserves the Minkowski interval for every $X$, equivalentlyIt is proper when $\boxed{\det L=+1}$, and orthochronous when $\boxed{L^0{}_0>0}$, equivalently $L^0{}_0\ge 1$. The latter selects the orthochronous Lorentz group, which preserves future-directed timelike four-vectors. Indeed the time-time entry of the preservation identity gives $(L^0{}_0{)}^2-{\sum }_{i=1}^3(L^i{}_0{)}^2=1$, so $|L^0{}_0|\ge 1$. Imposing both conditions gives a proper orthochronous Lorentz transformation.

Now use $(ct,x{)}^T$ and ${\eta }_2=\mathrm{diag}(1,-1)$. WritePreservation of the Minkowski metric says that the first column has squared norm $1$, the second has squared norm $-1$, and the columns are orthogonal for the Minkowski inner product:Orthochronousness gives $a\ge 1$. From the orthogonality equation, $b=de/a$; substituting in the second norm identity gives $e^2=a^2$. Hence $(b,e)=\epsilon (d,a)$ with $\epsilon =\pm 1$. The determinant is then $\epsilon (a^2-d^2)=\epsilon$, so properness forces $\epsilon =1$. Thus $b=d$ and $e=a$.

Set $\beta =d/a$. The first norm identity implies $|\beta |<1$ and $a=(1-{\beta }^2{)}^{-1/2}$. With $u=c\beta$, this givesConversely, the displayed matrix has ${\gamma }^2-(\gamma u/c{)}^2=1$, equal off-diagonal entries, determinant one, and positive time-time entry. Direct multiplication therefore gives $L^T{\eta }_{2}L={\eta }_2$, proving both directions. The plus signs are the Lorentz boost convention used in the paper; reversing the boost direction replaces $u$ by $-u$.

Use the Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$ and coordinates $x^0=ct$. At a point, a smooth path is timelike if its tangent $\dot{x}$ is a timelike vector, here meaning ${\eta }_{ab}{\dot{x}}^a{\dot{x}}^b>0$. It is lightlike if its tangent is a null vector: the same expression is zero and the tangent is nonzero. In these coordinates, a future-directed timelike tangent has spatial speed less than $c$; a lightlike tangent has speed $c$.

Use $u^{\lambda }{u}_{\lambda }=-c^2$ and a constant Minkowski metric. Vary the velocity-dependent square root before imposing the proper-time normalization. Equivalently, vary a reparametrization-invariant worldline action first and then choose proper time. Its canonical covector momentum isThe coordinate derivative at fixed velocity is $\partial \mathcal{L}/\partial x^a=q{u}^{\lambda }{\partial }_a{A}_{\lambda }$. The Euler-Lagrange equation and the chain rule from a therefore giveDefine the antisymmetric field tensor byThe particle equation becomesAntisymmetry ensures $u^a{F}_{a\lambda }{u}^{\lambda }=0$, so the force is orthogonal to the four-velocity and preserves $u^2=-c^2$. The worldline variation before proper-time normalization is essential: substituting $\sqrt{-u^2}=c$ into the Lagrangian before taking its velocity derivative would turn the kinetic term into a constant and incorrectly remove the inertial momentum. The normalization is a property of the proper-time solution, not permission to discard the variational dependence.

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All four numbered questions in the held source are covered. Letters on separately marked paragraphs are inferred; the source's nested letters and Roman numerals are retained. Use the Minkowski metric $(+---)$ and $X^{\mu }=(ct,\mathbf{x})$. Electromagnetic formulas use SI units unless explicitly specialized.

Source paper: Oxford public bundle, pages 67–71. The bundle filename says 2016; the held B2 cover is dated 14 June 2017. Letters identify successive unlettered marked paragraphs; subsidiary source labels are retained. Use $X^{\mu }=(ct,\mathbf{x})$ and the Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$ throughout. Official partial notes are explicitly limited to questions 1 and 2; they do not constitute answers to the whole paper.^[ref]

For a massive particle on a timelike world line, the four-velocity is $U^{\mu }=d{X}^{\mu }/d\tau =\gamma (c,\mathbf{v})$, with $U^2=c^2$. Its four-momentum is $P^{\mu }=m{U}^{\mu }=(E/c,\mathbf{p})$ and its four-force is $F^{\mu }=d{P}^{\mu }/d\tau$. Here $\tau$ is proper time. A massless trajectory does not admit this massive-particle proper-time parametrization.

The asserted classification of arbitrary orthogonal vectors is false. With the chosen Minkowski metric, $D=(0,1,0,0)$ and $G=(0,0,1,0)$ are nonzero, orthogonal and both spacelike vectors. Also $D=(1,1,0,0)$ and $G=2D$ are orthogonal null vectors. The valid statement is that a nonzero vector orthogonal to a timelike vector is spacelike. To prove it, choose the rest frame of the timelike vector, $D=(d,0,0,0)$; then $D\cdot G=d{G}^0=0$ forces $G^0=0$, so $G^2=-|\mathbf{G}{|}^2<0$.

The relativistic energy-momentum relation givesTherefore $\boxed{P^2=0⟹m=0}$. A photon is an example: $E=c|\mathbf{p}|$. A collection of photons need not itself have zero invariant mass.

Use coordinates $X^{\mu }=(ct,\mathbf{x})$ and the mostly-plus Minkowski metric ${\eta }_{\mu \nu }=\mathrm{diag}(-1,1,1,1)$. Along a future-directed timelike world line, proper time satisfiesThe four-velocity and four-acceleration areThe Minkowski norm is fixed: $U^{\mu }{U}_{\mu }=-c^2$. Differentiating with respect to proper time, using the constant Minkowski metric, givesIn the instantaneous rest reference frame, $U^{\mu }=(c,0,0,0)$, so the orthogonality says $A^0=0$ there. This construction uses a timelike world line; a massless particle has no rest-frame proper time parameter.

An invariant has the same value when the same physical event or state is described in different inertial frames. A conserved quantity has the same value as the physical system evolves, subject to its equations and isolation conditions. These are different comparisons. The invariant mass of a heated object can change with time, while an isolated system's conserved energy has different numerical values in boosted frames.

Under a Lorentz transformation $x^{\prime \mu }={\Lambda }^{\mu }{}_{\nu }{x}^{\nu }$, a Lorentz four-vector transforms asA covector transforms by the inverse matrix, $W_{\mu }^{\prime }=({\Lambda }^{-1}{)}^{\nu }{}_{\mu }{W}_{\nu }$. A general tensor has one corresponding factor for each index; for example,Contraction with the Minkowski metric produces Lorentz scalars; it does not by itself make those scalars conserved.

To prove the vanishing component theorem for a Lorentz four-vector, first suppose $Y^0=0$ in every inertial frame. An infinitesimal boost along spatial direction $i$ gives $Y^{\prime 0}=Y^0-\beta Y^i+O({\beta }^2)$. It too vanishes for every $\beta$, so each $Y^i=0$. If instead a specified spatial component vanishes in every frame, spatial rotations first imply that all three spatial components vanish in a chosen frame, and a boost in the specified direction then forces $Y^0=0$. ThereforeA physical consequence is that conservation of energy in every inertial frame implies full four-momentum conservation: apply the result to the difference between outgoing and incoming total four-momenta.

The statement fails for general tensors. A nonzero antisymmetric rank-two tensor has $F^{00}=0$ in every frame, because antisymmetry is preserved. For an off-diagonal component, the nonzero Minkowski metric is another counterexample: ${\eta }^{01}=0$ in all orthonormal Lorentz frames, while $\eta$ is certainly not zero. The transformation law can preserve structural zero components without forcing the entire tensor to vanish.