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Use $c=1$ and the mostly-plus Minkowski metric fixed in part a. The electromagnetic four-potential and its covariant form areDefine the electromagnetic field tensor by $F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }$. Since $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$ and $\mathbf{B}=\nabla \times \mathbf{A}$,Raising both indices changes the signs of its time-space entries. The lower-index signs agree with ^[ref]; the signature and index placement must be kept consistent.

For a passive boost to a frame moving with $v\hat{\mathbf{x}}$, transform the tensor asFor example, $F^{\prime 02}=\gamma F^{02}-\gamma v{F}^{12}=\gamma (E_y-v{B}_z)$ and $F^{\prime 12}=\gamma (B_z-v{E}_y)$. Reading off all entries givesFor an arbitrary boost velocity $\mathbf{v}$, rotational covariance gives the Lorentz transformation of electromagnetic fieldsHere all velocities are in $c=1$ units. Equivalently the parallel components are unchanged and the perpendicular components have the cross-product expressions multiplied by $\gamma$.

Use $\hslash =c=1$ and the mostly-plus Minkowski metric. Then the source's Klein-Gordon equation is $({\partial }_t^2-{\nabla }^2+m^2)\varphi =0$. For a plane wave $\varphi =C{e}^{-i\omega t+i\mathbf{k}\cdot \mathbf{x}}$, substitution givesThus a real propagating wave vector requires $|\omega |\ge m$, or $|E|\ge m{c}^2$ on restoring units. The positive-energy branch has a rest-energy threshold $m{c}^2$. At $|\omega |=m$, $\mathbf{k}=0$; frequencies below the mass threshold give evanescent spatial dependence rather than real-wave-vector propagation. Both signs of $\omega$ occur in the classical equation; a real scalar field combines complex-conjugate modes. Its group velocity is $\mathbf{k}/\omega$ and is subluminal, though the positive-branch phase speed exceeds one when $m>0$.

For a static spherically symmetric scalar field, the equation outside the source isSetting $\psi =r\varphi$ gives ${\psi }^{\prime \prime }-m^2\psi =0$, soDecay at infinity selects $B=0$, the Yukawa potential. It is singular at the origin and describes a point source distributionally; a source-free solution regular at the origin and decaying at infinity must be zero. For $m=0$, $\varphi =A/r+B$, and decay sets the constant $B$ to zero.

Write $u=(a,b{)}^T$. The Pauli matrices giveHenceTherefore $\boxed{U_{\mu }{U}^{\mu }=0}$, with $U^0>0$ for $u\ne 0$: the bilinear is a future null Lorentz four-vector, and $u=0$ gives the zero vector.

For $S=e^{\rho {\sigma }_z/2}=\mathrm{diag}(e^{\rho /2},e^{-\rho /2})$, the transformed components are $a^{\prime }=e^{\rho /2}a$ and $b^{\prime }=e^{-\rho /2}b$. Thus $a^{\prime \ast }{b}^{\prime }=a^{\ast }b$, while the two squared magnitudes acquire factors $e^{\rho }$ and $e^{-\rho }$. Substitution givesThe matrix preserves the Minkowski metric and is an active boost along $+z$ of rapidity $\rho$, with velocity $c\tanh \rho$. Relative to the passive standard-boost convention of part i, its velocity parameter has the opposite sign.

For $S=e^{i\theta {\sigma }_z}$, $a^{\prime }=e^{i\theta }a$ and $b^{\prime }=e^{-i\theta }b$, so $a^{\prime \ast }{b}^{\prime }=e^{-2i\theta }{a}^{\ast }b$. ConsequentlyThis is an active right-handed spatial rotation through $-2\theta$ about $z$. The minus sign follows from the specified positive $i\theta {\sigma }_z$ exponent; the familiar positive-angle spin rotation uses $e^{-i\phi {\sigma }_z/2}$. The doubled angle exhibits the Lorentz spinor double cover: $u$ and $-u$ give the same four-vector.

Use the electromagnetic field tensor convention stated above, and lower indices with the Minkowski metric $g=\mathrm{diag}(-1,1,1,1)$. Lowering a single time index changes its sign. Summing both orders of each antisymmetric index pair gives the Lorentz scalarFor $T^{00}$, $F_j{}^0=F^{j0}=-E_j/c$ and $g^{00}=-1$. Direct substitution in the printed electromagnetic stress-energy tensor yieldsFor a spatial index $i$, $g^{0i}=0$ and $F_j{}^i=F^{ji}=-{ϵ}_{ijk}{B}_k$. HenceThus the full top row isHere $u_{\mathrm{EM}}$ is the electromagnetic energy density and $\mathbf{S}$ the Poynting vector, using ${ϵ}_0{\mu }_0{c}^2=1$. By the symmetry $T^{0i}=T^{i0}$, these components describe both energy flux divided by $c$ and $c$ times the electromagnetic momentum density $\mathbf{g}=\mathbf{S}/c^2$. Keeping the factor $c$ is essential because the time coordinate is $ct$.

For an active clockwise rotation of column vectors in the usual oriented $xy$ plane, takeDirect matrix multiplication givesThus the rotation matrix is an orthogonal matrix, preserving the Euclidean inner product, with determinant $+1$.

For a passive one-dimensional Lorentz boost, use rapidity $\eta$, $\beta =\tanh \eta$:This is not $I$ unless $\eta =0$. Hence a nontrivial Lorentz boost is not an orthogonal matrix for the Euclidean inner product. Instead it preserves the Minkowski metric:by ${\cosh }^2\eta -{\sinh }^2\eta =1$. Thus it is pseudo-orthogonal. The zero-rapidity identity is the explicit exception to the assertion that a Lorentz boost is not orthogonal.

Write the Lorentz transformation as $x^{\prime \mu }={\Lambda }^{\mu }{}_{\nu }{x}^{\nu }$. Acting on a scalar field, the chain rule gives the four-gradientFor the specified Lorentz boost, the inverse is obtained by changing $\beta$ to $-\beta$, soThese are precisely the covector component rules for the lowered coordinates. Raising the index with the Minkowski metric ${\eta }^{\mu \nu }=\mathrm{diag}(-1,1,1,1)$ reverses the time component, yieldingIndeed ${\partial }^{\prime 0}=\gamma ({\partial }^0-\beta {\partial }^1)$ and ${\partial }^{\prime 1}=\gamma ({\partial }^1-\beta {\partial }^0)$, the contravariant four-vector rules. Thus the index on a partial derivative with respect to an upper coordinate is lower.

By four-momentum conservation, $p_{\pi }=p_1+p_2$. With the $(-,+,+,+)$ Minkowski metric, the neutral pion has $p_{\pi }^2=-M_{\pi }^2$, while each photon has $p_i^2=0$ and $|{\mathbf{p}}_i|=E_i$. Taking the Minkowski norm givesThe half-angle identity therefore yieldsThis is a measurement of the diphoton invariant mass using two photon energies and their opening angle, independent of the neutral pion's laboratory velocity.

Use natural units and the Minkowski metric $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$, with $A^{\mu }=(\Phi ,\mathbf{A})$ and $A_{\mu }=(\Phi ,-\mathbf{A})$. Then $F_{0i}=E_i$, $F^{i0}=E_i$, and $F^{ij}=-{ϵ}_{ijk}{B}_k$. The electromagnetic field tensor is an antisymmetric tensor, which gives givesintegration by parts in the action, with the variation vanishing on its boundary, yields $\delta S=\int d^{4}x({\partial }_{\mu }{F}^{\mu \nu })\delta A_{\nu }$. Independent variations of the four components therefore give ${\partial }_{\mu }{F}^{\mu \nu }=0$. The time component is $\nabla \cdot \mathbf{E}=0$; the spatial components are $\nabla \times \mathbf{B}-{\partial }_t\mathbf{E}=0$. These are the two source-free Maxwell equations obtained dynamically. The other two, $\nabla \cdot \mathbf{B}=0$ and ${\partial }_t\mathbf{B}+\nabla \times \mathbf{E}=0$, follow identically from $F=dA$, through the Electromagnetic Bianchi identity. They are not additional Euler-Lagrange field equations.^[ref]

In the same Minkowski metric and electromagnetic units, $T^{\mu \nu }=-F^{\mu }{}_{\lambda }{F}^{\nu \lambda }+\frac{1}{4}{g}^{\mu \nu }{F}_{\alpha \beta }{F}^{\alpha \beta }$. Differentiating givesTo simplify the last two terms, contract ${\partial }_{\mu }{F}_{\nu \lambda }+{\partial }_{\nu }{F}_{\lambda \mu }+{\partial }_{\lambda }{F}_{\mu \nu }=0$ with $F^{\mu \lambda }$. Its first and third terms are equal, and its middle term is $-\frac{1}{2}{\partial }_{\nu }(F_{\alpha \beta }{F}^{\alpha \beta })$. Raising $\nu$ therefore gives $F^{\mu }{}_{\lambda }{\partial }_{\mu }{F}^{\nu \lambda }=\frac{1}{4}{\partial }^{\nu }(F_{\alpha \beta }{F}^{\alpha \beta })$. The cancellation leavesThis is the electromagnetic stress-energy divergence identity in rationalized units. As a sign check, $T^{00}=(E^2+B^2)/2$ is positive and $T^{0i}=(\mathbf{E}\times \mathbf{B}{)}_i$ is the Poynting vector. The integrated four-momentum $P^{\nu }=\int T^{0\nu }{d}^{3}x$ is conserved if the boundary flux vanishes. With matter sources, the electromagnetic field alone instead obeys ${\partial }_{\mu }{T}^{\mu \nu }=-F^{\nu \lambda }{J}_{\lambda }$: energy and momentum transfer to matter restore conservation of the total stress-energy tensor.

Treat $\varphi$ and ${\varphi }^{\ast }$ as independent variables during variation. With Minkowski metric $(+---)$, differentiation with respect to ${\varphi }^{\ast }$ givesThe Euler-Lagrange field equation is therefore $\square \varphi +2ie{A}^{\mu }{\partial }_{\mu }\varphi +ie({\partial }_{\mu }{A}^{\mu })\varphi +m^2\varphi =f$. The static purely temporal potential has ${\partial }_{\mu }{A}^{\mu }={\partial }_{t}V=0$, givingvariation with respect to $\varphi$ gives $(\square -2ie{A}^{\mu }{\partial }_{\mu }+m^2){\varphi }^{\ast }=f(t)$. This is the complex conjugate equation when the driving force is real, as is required for the supplied lagrangian density to be real. For a complex driving amplitude a real source term would instead require conjugate coefficients. The interaction supplied here is linear in the external potential: adding the $e^2{A}_{\mu }{A}^{\mu }|\varphi {|}^2$ term from full minimal coupling would change the problem and its equation of motion.

Choose Minkowski metric $(+,-,-,-)$, so $A_{\mu }=(\varphi /c,-\mathbf{A})$ and $d\tau =dt\sqrt{1-v^2/c^2}$. In laboratory time the relativistic charged-particle action has LagrangianHere $e$ denotes the signed particle charge in the action. The canonical momentum and the position derivative areThe Euler-Lagrange equation therefore givesThe last term is $(\mathbf{v}\times \mathbf{B}{)}_i$, and the first two terms form the electric field. Thus the relativistic Lorentz force equation isIts power relation and acceleration form areThe power relation follows by taking the scalar product with $\mathbf{v}$ and using the relativistic energy-momentum relation. It also shows why the magnetic field alone cannot change the particle's energy.

Use $x^{\mu }=(t,x)$ and Minkowski metric ${\eta }^{\mu \nu }=\mathrm{diag}(1,-1)$, with the wave speed equal to one. Regarding $\varphi$ as a scalar field, the symmetric canonical stress-energy tensor isIt obeys ${\partial }_{\mu }{T}^{\mu \nu }=0$ on the wave equation. For a scalar field the intrinsic spin current vanishes. The total orbital Lorentz current and its integrated angular-momentum tensor areBecause there is only one spatial dimension, its only independent possibly nonzero component iswhere $P=\int pdx$. The corresponding flux is ${\mathcal{J}}^{101}=t\mathcal{E}-xp$. Symmetry and conservation of $T$ giveThus $J^{01}$ is conserved provided the boundary flux vanishes and the integrals exist. For example, localized wave packets on an infinite line have this property. One can also check directly that $d\int x\mathcal{E}dx/dt=P$, so the two terms in $J^{01}$ have cancelling derivatives.

The significance is the boost charge of a one-dimensional scalar field. There is no independent spatial rotation or ordinary spatial angular-momentum component in one dimension. Instead, the action has the mathematical Lorentz invariance of a massless wave field with invariant speed $c_s=1$, and $J^{01}$ generates its boost. Its conservation says that the center of energy moves uniformly. This is a symmetry of the ideal wave-field model, rather than an assertion of fundamental vacuum Lorentz symmetry for a material rod. Fixed boundaries can break it, and a general solution with nonvanishing boundary flux need not conserve the integrated charge.

Use Minkowski metric $g_{\mu \nu }=\mathrm{diag}(-1,1,1,1)$ and units with the speed of light equal to one. Thus $u^{\mu }{u}_{\mu }=-1$, and $h_{\mu \nu }=g_{\mu \nu }+u_{\mu }{u}_{\nu }$ is the spatial projector of a timelike congruence. Keep the medium velocity fixed when varying the electromagnetic four-potential. Since the electromagnetic field tensor is $F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }$, its variation givesBoth tensors are antisymmetric. Integration by parts, with variations vanishing at the boundary, therefore givesThe Euler-Lagrange field equation is ${\partial }_{\mu }(F^{\mu \nu }-M^{\mu \nu })=0$. Renaming the free index and using antisymmetry givesThese are the inhomogeneous Maxwell equations with the bound charge and bound current encoded by the polarization-magnetization tensor.

With the Minkowski metric $(+---)$, the momentum conjugate to the real field vector is $\mathbf{\pi }=\dot{\mathbf{\varphi }}$. The Hamiltonian density isAssume the potential has a global minimum $V_{\min }$ and the boundary conditions permit a constant minimizing field. ThenThe bound is attained byThe homogeneous vacuum minimum criterion proves existence of a homogeneous lowest-energy configuration under those assumptions. On infinite space one subtracts the constant vacuum energy before comparing finite energy differences. This identifies the classical vacuum configuration; it does not assert that the quantum field has no fluctuations.

The claim does not follow from “no derivative dependence in $V$” alone. For example, the allowed functional form $V(\varphi )=-{\varphi }^4$ is unbounded below, so there is no state of lowest energy to identify. Boundary conditions forcing different vacua at opposite ends can instead require an inhomogeneous minimum, such as a domain wall. The positive quartic potential used in the following parts does have homogeneous minima, so these caveats do not obstruct its vacuum calculation.

Source paper: www.whatdotheyknow.com/request/physics_past_exam_papers_2015_16/response/973997/attach/3/Cowell%20Attachment.zip?cookie_passthrough=1

All eight questions are solved. Set $\hslash =c=1$ unless a thermal factor $k_B$ is written explicitly. Relativistic calculations use the Minkowski metric $(+---)$. The unbracketed sums in the first two printed energies are read as sums over all the displayed bond terms, with the common prefactor applying to the entire summand.

Source paper: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/6/Physics%20Papers%202017.pdf?cookie_passthrough=1. The cover identifies Trinity Term 2018; the annual bundle's 2017 filename does not determine the examination year. We use natural units $\hslash =c=1$ for the classical field theory and quantum field theory in four spacetime dimensions and Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$.

The action is dimensionless and $[d^{4}x]=-4$, so $[\mathcal{L}]=4$. A canonical scalar kinetic term then requires $2+2[\varphi ]=4$, and the electromagnetic kinetic term requires $2[F_{\mu \nu }]=4$. ThusThe brackets here denote mass dimension. The Lagrangian has no explicit coordinate dependence, and all indices are contracted as Lorentz scalars. Its continuous spacetime symmetry is the Poincaré group. Noether's theorem gives conserved four-momentum from translations and conserved angular momentum, including the boost generators, from Lorentz transformations. The nonzero mass $M$ and dimensionful couplings prevent a dilation symmetry.

Sign qualification: the printed electromagnetic kinetic term has the opposite healthy relative sign to the scalar kinetic terms. With the stated mostly-minus Minkowski metric, a physical Maxwell Lagrangian is $-F_{\mu \nu }{F}^{\mu \nu }/4$. The printed plus sign instead gives photon Hamiltonian density $-({\mathbf{E}}^2+{\mathbf{B}}^2)/2$; a nonzero purely electric configuration already has negative energy, unbounded below as its amplitude grows. The physical decay widths below use the healthy Maxwell sign with the given interaction normalization. The dimensional and spacetime-symmetry conclusions themselves do not depend on this sign correction.

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

Source bundle pages 48–54; the cover identifies Trinity Term 2022 despite the attachment filename. Solutions use $\hslash =c=1$ except where $\hslash$ is displayed. In Questions 4–6 the Minkowski metric is $\eta =\mathrm{diag}(1,-1,-1,-1)$, and spatial surface terms vanish or cancel under periodic boundary conditions.

Write $d{\Pi }_p=d^{3}p/[(2\pi {)}^{3}2{\omega }_p]$, $p^{\mu }=({\omega }_p,\mathbf{p})$ and ${\omega }_p=\sqrt{{\mathbf{p}}^2+m^2}$. The Fourier expansion in the source is compatible with the oscillator canonical commutation relationsFor the free real scalar field, $\pi =\dot{\varphi }$ andSubstitution of the Fourier expansion produces spatial Dirac delta functions. Terms with two annihilation operators or two creation operators cancel: in $P_0$ their coefficient is $-{\omega }_p^2+{\mathbf{p}}^2+m^2=0$, and in $P_i$ their integral is odd under $\mathbf{p}\mapsto -\mathbf{p}$. The two mixed terms combine, with normal ordering removing the vacuum constant, to giveIn particular, $P_i$ contains the lower component $p_i=-{\mathbf{p}}_i$ in this Minkowski metric convention.

To check commutativity without manipulating a square of a Dirac delta function, define $N_f=\int d{\Pi }_{p}f(p)a_p^{†}{a}_p$. The oscillator algebra gives $[N_f,a_q]=-f(q)a_q$ and $[N_f,a_q^{†}]=f(q)a_q^{†}$. Thus $[N_f,a_q^{†}{a}_q]=0$ and $[N_f,N_g]=0$. Taking $f=p_{\mu }$, $g=p_{\nu }$ provesThese identities hold on the usual dense domain of finite-particle states, or can first be proved in a finite box and then continued to the continuum.

Put $V(\varphi )=m^2{\varphi }^2/2+{\sum }_{n>2}{\lambda }_n{\varphi }^n/n!$. The Euler-Lagrange equation for the scalar field is $\square \varphi +V^{\prime }(\varphi )=0$, or explicitlyThe derivative of ${\varphi }^n/n!$ is ${\varphi }^{n-1}/(n-1)!$; all interaction terms therefore enter with the displayed plus sign for the chosen Minkowski metric.