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Use Minkowski spacetime with $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ and units $\hslash =c=1$. The transformation rotates the two components in their internal plane. In particular,Thus both terms in the Lagrangian density are invariant: this is an internal rotation symmetry of two real scalar fields, rather than a transformation of spacetime. The equal masses are essential to this symmetry.

The Noether theorem gives the Noether currentIndeed, the cross terms in its divergence cancel, and the two Klein-Gordon equations implyFor fields decaying sufficiently at spatial infinity, there is no outward current flux. The conserved Noether charge is therefore

In canonical quantization, the canonical momenta are ${\pi }_k={\dot{\varphi }}_k$, with equal-time canonical commutation relationsConsequently $Q=\int d^{3}x({\pi }_1{\varphi }_2-{\pi }_2{\varphi }_1)$. Operators of different components commute, so this expression is Hermitian without an ordering correction. Write ${\int }_{\mathbf{p}}=\int d^{3}p/(2\pi {)}^3$, $E_{\mathbf{p}}=\sqrt{{\mathbf{p}}^2+m^2}$ and $a_{k\mathbf{p}}=a_{\mathbf{p}}^k$. Substituting the oscillator expansions and using the Dirac delta function from the spatial integral givesThe terms with two annihilation operators cancel after $\mathbf{p}\mapsto -\mathbf{p}$; the terms with two creation operators cancel in the same way. In the remaining terms useThere is no commutator constant between the distinct components. The result is already in normal ordering:

For the charged oscillator basis of a scalar doublet, computeIt follows that $[Q,b_{\pm \mathbf{p}}^{†}]=\pm b_{\pm \mathbf{p}}^{†}$. The Fock vacuum satisfies $Q|0⟩=0$, so $b_{+\mathbf{p}}^{†}|0⟩$ has charge $+1$, and $b_{-\mathbf{p}}^{†}|0⟩$ has charge $-1$. To obtain a normalized one-particle state, replace the sharp momentum by ${\int }_{\mathbf{p}}f(\mathbf{p})b_{\pm \mathbf{p}}^{†}|0⟩$, where ${\int }_{\mathbf{p}}|f(\mathbf{p}){|}^2=1$. The charge is unchanged. Equivalently,The generator convention is $\delta {\varphi }_k=i\alpha [Q,{\varphi }_k]$, which reproduces the original rotation signs.

For stability of a two-scalar quartic potential, the relevant potential energy density isAlong either axis boundedness requires $\lambda \ge 0$; along ${\varphi }_1={\varphi }_2$ it requires $\lambda +\mu \ge 0$. Conversely the displayed sum is nonnegative whenever both conditions hold. With $m^2>0$ the origin is then the unique global minimum, including the equality cases. Hence the stable-vacuum conditions areStrict positivity of the quartic term away from the origin would instead require $\lambda >0$ and $\mu >-\lambda$. That stronger condition is unnecessary for a positive mass term. If $m=0$, the same non-strict conditions give boundedness, but the equality cases have flat directions and do not give an isolated classical vacuum.

Finally, the interaction changes the variation of the potential energy byThe interacting Euler-Lagrange equations consequently giveThus the same Noether charge is conserved for all field configurations precisely when $\mu =\lambda$. Then $V_4=\lambda ({\varphi }_1^2+{\varphi }_2^2{)}^2$ preserves the internal rotation symmetry of two real scalar fields; imposing stability additionally requires $\lambda \ge 0$.

Work in units $\hslash =1$ and with the particle mass absorbed into the stated kinetic term. Divide the interval into $N$ pieces of length $\Delta =T/N$. For a suitable self-adjoint Hamiltonian operator, the Trotter product formula gives the regulated limitInsert $1=\int d{q}_j|q_j⟩⟨q_j|$ between neighboring factors. Because $V(\hat{q})$ is diagonal in the position operator basis, the given free-particle propagator yieldsThe exponent is the time-sliced action. Its continuum notation isThe measure is defined by the displayed finite-dimensional limits and their normalization, not by a translation-invariant flat measure on an infinite-dimensional space. The oscillatory limit needs a damping/analytic-continuation prescription. This is the time-sliced configuration-space path integral derived from the operator kernel.

For a quadratic potential energy, write $q=q_c+\eta$, where the classical path satisfies ${\ddot{q}}_c+V^{\prime }(q_c)=0$ and the endpoint conditions, and $\eta (0)=\eta (T)=0$. Expansion is exact:Integration by parts makes the linear term equal to $-[{\int }_0^T({\ddot{q}}_c+V^{\prime }(q_c))\eta dt]$, since the endpoint term vanishes. It is zero by the Euler-Lagrange equation. Also $V^{\prime \prime }$ is constant, independent of the endpoints. Thus classical-path factorization of a quadratic path integral is exact:Only the classical action carries the endpoint dependence.

Let $D={\partial }_t^2+V^{\prime \prime }$ on homogeneous Dirichlet boundary conditions. Expand $\eta$ in its normalized eigenfunctions, with eigenvalues ${\lambda }_n$. The fluctuation integral becomes a product of regulated integrals $\int d{a}_n\exp (-i{\lambda }_n{a}_n^2/2)$. Each is proportional to ${\lambda }_n^{-1/2}$, with a prescribed complex phase. Relative to $D_0={\partial }_t^2$, the normalization is thereforeThe subscript denotes the Dirichlet operator domain, and the square-root branch is continued from the free kernel with its convergence prescription. Equivalently one can use determinants of $-D$ and $-D_0$. For $V^{\prime \prime }={\omega }^2$, the Dirichlet oscillator determinant ratio isThis also checks the free limit. At conjugate times, the fluctuation operator has a zero eigenvalue and the ordinary finite-prefactor expression fails; the kernel is obtained by distributional continuation. Away from these times the classical boundary problem is unique and the stated factorization applies directly.

For the infinite-time oscillator with an external source, use vacuum boundary conditions and the Feynman i-epsilon prescription. For a smooth source of compact support, defineThe imaginary term damps the real-field integral. Integration by parts gives $S_{ϵ}[q]=\frac{1}{2}q{A}_{ϵ}q+Jq$, with the pairings understood as time integrals. Completing the square givesTranslate the integration variable. The source-independent Gaussian functional integral is $N$, soOn Fourier modes $e^{i\omega t}$, $A_{ϵ}$ has eigenvalue ${\omega }^2-m^2+iϵ$, henceFor $m>0$, closing the contour above for $t>0$ and below for $t<0$ gives $G(t)=-i{e}^{-im|t|}/(2m)$. The physical time-ordered oscillator quantum field theory propagator is $iG$, not $G$ itself. The shifted stationary path is $q_c=-GJ$ and obeys ${\ddot{q}}_c+m^2{q}_c=J$ with the same vacuum prescription. A retarded inverse would define a different boundary problem. Dividing by the zero-source integral removes $N$ and gives the vacuum source functional of a quantum oscillator. At zero mass an additional infrared prescription is needed for the inverse on constant modes.

For a variation of the covariant metric tensor, $\delta g^{\alpha \beta }=-g^{\alpha \mu }{g}^{\beta \nu }\delta g_{\mu \nu }$ and $\delta \sqrt{-g}=\frac{1}{2}\sqrt{-g}{g}^{\mu \nu }\delta g_{\mu \nu }$. Writing $\mathcal{L}=-\frac{1}{2}{g}^{\alpha \beta }{\partial }_{\alpha }\varphi {\partial }_{\beta }\varphi -V$, one obtainsThe specified metric-variation convention therefore givesFor $\varphi =\varphi (t)$, its invariant gradient square is $-{\dot{\varphi }}^2$. Hence $T^{00}={\dot{\varphi }}^2-({\dot{\varphi }}^2/2-V)={\dot{\varphi }}^2/2+V$, while $T^{ij}=a^{-2}{\delta }^{ij}({\dot{\varphi }}^2/2-V)$ and $T^{0i}=0$. Lowering one index makes $T^{\mu }{}_{\nu }=\mathrm{diag}(-\rho ,P,P,P)$, withPotential energy has negative pressure, whereas scalar kinetic energy contributes positive pressure. This sign distinction is what permits accelerated expansion.

Dot Newton's second law with $\dot{\mathbf{r}}$. The magnetic cross-product term is perpendicular to this velocity, soThusorThe two halves are the kinetic energy and the isotropic spring's potential energy. The expression in the question is twice their sum. The Lorentz force changes the direction of motion but supplies no power, because it is perpendicular to velocity; this explains why the magnetic field does not appear in the conserved mechanical energy. The sign corresponds to charge $-e$.

Let $r=|{\mathbf{r}}_1-{\mathbf{r}}_2|$ and $\hat{\mathbf{r}}=({\mathbf{r}}_1-{\mathbf{r}}_2)/r$. A differentiable central potential energy gives the internal forcesThere are no external forces. Differentiating the total kinetic energy and using Newton's second law therefore givesSince $\dot{r}=\hat{\mathbf{r}}\cdot ({\mathbf{v}}_1-{\mathbf{v}}_2)$, the last expression is $-U^{\prime }(r)\dot{r}=-dU/dt$. ThusThis conserves total mechanical energy, including the kinetic energy of centre-of-mass motion, not merely the relative kinetic energy.

The relevant parameters are $a,h,g$ and the ring mass $m$, with dimensions length, length, length/time${}^2$ and mass. The wire is fixed and smooth, so there is no additional friction parameter. Because $m$ is the only parameter carrying a mass dimension, it cannot enter a dimensionless group governing the period. Using $a$ and $g$ as the length and time scales, the independent dimensionless variables are $T\sqrt{g/a}$ and $h/a$. Dimensional analysis therefore givesThe physical mass cancellation also follows from the energy equation in part (ii), since kinetic and gravitational potential energy are both proportional to $m$.

Place the four fixed charges at $(sa,ta)$ with $s,t\in \{-1,1\}$, and write the moving particle's position as $(x,y)$. Introduce its mass $m>0$, which is needed for an oscillation frequency but is not named in the question. Let $C=Qq/(4\pi {ϵ}_0)>0$. By Coulomb's law and superposition, the moving particle's potential energy isThe electrostatic potential itself is $V/q$. Put $r^2=x^2+y^2$ and ${\delta }_{st}=-(sx+ty)/a+r^2/(2a^2)$. Near the centre, a Taylor expansion givesOn summing over the four corners, linear terms cancel. To quadratic order, $\sum {\delta }_{st}=2r^2/a^2$ and $\sum {\delta }_{st}^2=4r^2/a^2+O(r^4/a^4)$, since $\sum st=0$. Reflection symmetry removes all odd-degree terms. HencewhereThus the centre has zero planar force and a positive-definite planar Hessian matrix $\kappa I_2$. It is a strict local minimum of the constrained potential energy, so it is a stable equilibrium. More explicitly, conservation of the total energy confines sufficiently small-energy motions within a small neighbourhood of this isolated local minimum.

Keeping only the quadratic potential, the two coordinates obey independent small oscillations:Both planar modes have the same angular frequency, and the ordinary frequency isThe planar constraint is essential to this constrained electrostatic equilibrium. Along the perpendicular axis, $V(0,0,z)=4C/(2a^2+z^2{)}^{1/2}=V(0,0)-C{z}^2/(\sqrt{2}{a}^3)+O(z^4)$, so the same stationary point is unstable to unconstrained normal displacements. That also checks the cancellation of the three-dimensional Hessian trace required by Laplace's equation away from charges; it does not contradict the positive-definite restriction to the given plane.

To distinguish the moment quantity from the gravitational constant, write it as $\mathcal{G}={\sum }_i{\mathbf{p}}_i\cdot {\mathbf{r}}_i$. Newton's equations give ${\dot{\mathbf{p}}}_i={\mathbf{F}}_i$ and ${\dot{\mathbf{r}}}_i={\mathbf{p}}_i/m_i$. Differentiating provesAverage from $0$ to $\tau$, divide by $\tau$, and apply the supplied long-time boundary condition. The result isThe PDF's intermediate formula omits this minus sign. It cannot follow from the preceding derivative identity: an attractive inverse-square circular orbit already has $\mathbf{F}\cdot \mathbf{r}<0$ and $T>0$, contradicting the printed positive-sign version. The later homogeneous-potential relation is consistent with the corrected sign.

Interpret $W$ as the total pair potential energy, with each pair counted once, for example $W={\sum }_{i<j}C|{\mathbf{r}}_i-{\mathbf{r}}_j{|}^{p+1}$. The printed pair sum leaves one particle index unspecified; an ordered-pair convention with the corresponding factor absorbed into $C$ has the same degree of homogeneity. Under a common dilation of all positions,Differentiating with respect to $\lambda$ at one is the Euler theorem for homogeneous functions. Since ${\mathbf{F}}_i=-{\nabla }_{i}W$, it givesSubstitution into the correctly signed averaged identity proves the virial theorem for a homogeneous pair potential:For $p\ne -1$, the conserved total energy can be evaluated using these averages:Use the kinetic temperature convention $⟨T⟩=\kappa \Theta$, where $\kappa =3N{k}_B/2>0$ for three-dimensional thermalized particles, or the same positive coefficient as a kinetic-temperature definition. At fixed particle number and interaction parameters,Thus the equilibrium heat capacity is negative precisely whensubject to existence of the bounded equilibrium sequence assumed by the virial theorem. At $p=-3$ the virial energy is zero and this derivative vanishes. At $p=-1$ the written power-law potential is constant, its force vanishes, and the bounded-motion virial condition excludes nonzero kinetic energy; division by $p+1$ is not permissible. The Newtonian case $p=-2$ recovers $E_{\mathrm{tot}}=-⟨T⟩$.

Write $\mu =G{M}_{\odot }$, and interpret the specified radial force field as acceleration, or force per unit test-particle mass. Since it is a central force, the specific angular momentum $\mathbf{h}=\mathbf{r}\times \mathbf{v}$ is conserved: $\dot{\mathbf{h}}=\mathbf{r}\times \mathbf{F}=0$. For $h=|\mathbf{h}|\ne 0$, motion is confined to the fixed plane perpendicular to $\mathbf{h}$. Choose the orientation of the polar angle so that $h=r^2\dot{\theta }>0$.

To derive the Binet equation here, let $u=1/r$ and differentiate with respect to $\theta$. Then $\dot{r}=-h{u}^{\prime }$ and $\ddot{r}-r{\dot{\theta }}^2=-h^2{u}^2(u^{\prime \prime }+u)$. Equating this to the radial acceleration givesPut $K=h^2+C$ and $\alpha =\sqrt{K/h^2}$. For $K>0$, the solution isThe radial potential energy per unit mass is $V(r)=-\mu /r+C/(2r^2)$, so the conserved specific orbital energy isSubstitute the displayed $u$ and $\dot{r}=-h{u}^{\prime }$ to obtain $\mathcal{E}={\mu }^2(e^2-1)/(2K)$. Therefore defineFor regular bound motion, $K>0$ and $-{\mu }^2/(2K)\le \mathcal{E}<0$, giving $a>0$ and $0\le e<1$. Since $K/\mu =a(1-e^2)$, choose the angular origin at a pericentre, so ${\theta }_0=0$, and obtainThe turning radii are $r_{\mathrm{p}}=a(1-e)$ and $r_{\mathrm{a}}=a(1+e)$. Thus $a$ is the arithmetic mean of the turning radii; when the orbit precesses it should not be interpreted as the semimajor axis of a fixed ellipse. If $K\le 0$, the centrifugal barrier needed for a regular two-turning-point bound orbit is absent, and this bound-orbit parametrization does not apply.

For $\alpha =1$, one has $C=0$ and the orbit is a closed Kepler orbit: an ellipse of eccentricity $e$ with the Sun at a focus. In addition to $\mathcal{E}$ and the three components of $\mathbf{h}$, the Laplace-Runge-Lenz vectoris conserved. Indeed,so $\dot{\mathbf{A}}=0$. It points toward the fixed pericentre. The constraints $\mathbf{A}\cdot \mathbf{h}=0$ and $A^2={\mu }^2+2\mathcal{E}{h}^2$ leave five generically independent isolating integrals. One may takechoosing a component $A_j$ whose differential is independent of the other four on the local region under consideration. The extra integral fixes the apsidal direction in the already fixed orbital plane.

For $\alpha \ne 1$, successive pericentres are separated by $2\pi /\alpha$ in polar angle. The generic bound orbit is a rosette in that same plane. Its perihelion advance per radial cycle isFor $\alpha <1$ this is prograde; for $\alpha >1$ it is retrograde. If $\alpha$ is rational, the orbit closes after a finite number of radial cycles; if it is irrational and $0<e<1$, its successive radial phases densely fill the accessible annulus. A generic independent set of single-valued isolating integrals isThe apsidal direction is no longer fixed. For irrational frequency ratio, any continuous integral of motion must be constant on the whole two-dimensional orbital torus, because one orbit is dense there; there is no further independent continuous integral specifying a fixed apsidal direction. Exceptional closed resonant orbits do not provide a fifth global integral on an open region of phase space. Circular and radial orbits are degenerate cases of these generic counts.

The plotted precession is enlarged for visibility. Both examples have $e=0.5$ and the same turning radii; the second has $\alpha =\sqrt{0.9}$.

For the precession rate, it is useful to derive the radial period rather than assume an unperturbed angular period. Set $r=a(1-e\cos \chi )$ in the radial energy equation. On a complete radial cycle,Hence $n=2\pi /P_r=\sqrt{\mu /a^3}$. Also $K=\mu a(1-e^2)$ and $h^2=K-C$, so ${\alpha }^{-1}=\sqrt{1-C/K}$. For $|C|\ll K$, inverse-cube perturbation of Kepler apsidal motion givesThere is a dimensional issue with the printed parameter: $C$ has units $L^4/T^2$ and $\mu$ has units $L^3/T^2$, so $\eta =aC/\mu$ has units $L^2$. Retaining that definition literally, the requested expression isA dimensionless perturbation coefficient would instead be $\tilde{\eta }=C/(\mu a)=\eta /a^2$. In that convention the rate is $-n\tilde{\eta }/[2(1-e^2)]$. These two conventions must not be conflated.

Use the stated residual precession and period. The advance in one radial cycle isWith $e=0.206$, the small-perturbation estimate becomesThe negative sign is required: an additional attractive inverse-cube acceleration advances perihelion, whereas the positive-$C$ term would make it regress. If a length-unit estimate of the printed $\eta$ is desired, $P_r=0.24\mathrm{yr}$ and the solar value $\mu \simeq 1.3271\times 10^{20}{\mathrm{m}}^3{\mathrm{s}}^{-2}$ give $a\simeq 5.78\times 10^{10}\mathrm{m}\simeq 0.386\mathrm{au}$ through the radial-period law. ThusThe dimensionless value is the estimate obtainable directly from the supplied period, eccentricity and advance; the dimensional value additionally uses the solar gravitational parameter to set the length scale.

Self-adjointness gives $⟨\mathbf{\xi },F\mathbf{\xi }⟩=⟨F\mathbf{\xi },\mathbf{\xi }⟩=⟨\mathbf{\xi },F\mathbf{\xi }{⟩}^{\ast }$, soThe integrated expression from part (a) also writes it explicitly asThis is the quadratic change in potential energy associated with the perturbation, including pressure, magnetic and imposed-gravity restoring effects. Its sign need not be positive term by term.

The perturbation kinetic energy is $T=\frac{1}{2}⟨\dot{\mathbf{\xi }},\dot{\mathbf{\xi }}⟩$, and the total quadratic perturbation energy isSince the equilibrium and $F$ are time independent, differentiating and using the self-adjoint operator identity givesThus $\boxed{dE/dt=0}$ along the linearized motion. This conserved kinetic-plus-restoring energy is the basis of the magnetohydrodynamic energy principle.

In the Boussinesq approximation, put $\gamma =K{W}^2\sqrt{g/{\rho }_0}$. The buoyancy-driven turbulent density diffusion equation gives, wherever $q={\bar{\rho }}_z>0$,The coefficient outside the bracket is positive, so the local density gradient increases precisely whenAt a point where $q=0$, use the original degenerate diffusion equation rather than dividing by $\sqrt{q}$.

For the proposed self-similar solution, $\bar{\rho }={\rho }_0[1+F(t)G(z)]$ and $F(0)=1$. Substitution into ${\bar{\rho }}_t={\partial }_z{F}_{\rho }$ givesAn odd power profile $G(z)=C\mathrm{sgn}(z)|z{|}^p$ has right-hand spatial power $(3p-5)/2$. Matching it to $p$ requires $p=5$. Thus choose $G=C{z}^5$ with $C>0$, an odd, increasing profile. Since $G^{\prime }=5C{z}^4$, its flux is proportional to $z^6$ on both sides of the midplane. The resulting quintic density-profile similarity in an unstable tube satisfiesIntegration gives $F^{-1/2}=1-At/2$. ConsequentlyAlthough mixing transports density downwards, the local density gradient can grow because spatial variations of the flux redistribute density. In this example the downward flux increases towards each end, driving the formal amplification of the odd profile.

This self-similar solution deliberately ignores the end conditions: its nonzero end flux cannot satisfy impermeable ends. Moreover, the Boussinesq approximation requires $FC|z{|}^5\ll 1$, which fails before the formal divergence. A closed tube has only finite available potential energy, so it cannot sustain indefinitely growing turbulent kinetic energy or the assumed instantaneous flux law. End effects, changing eddy statistics and decay of the unstable stratification must eventually replace this local model. The conflicting endpoint labels in the source do not enter the calculation because end conditions have been expressly excluded.

At infinity the potential energy tends to zero, so the conserved mechanical energy is $E=\frac{1}{2}{v}_{\infty }^2$. The impact parameter $b$ is the perpendicular distance from the origin to the incoming asymptotic line. Conservation of angular momentum gives $|h|=b{v}_{\infty }$.

For $b>0$, a turning point of the radial motion satisfiesOnly the positive root is physical, givingFor $Q=0$ this reduces to the straight-line closest distance $b$.

For nonzero $Q$, put $\eta =b{v}_{\infty }^2/|Q|\ll 1$. If $Q<0$, the numerator is $|Q|(\sqrt{1+{\eta }^2}+1)$, soIf $Q>0$, rationalize the numerator instead:Thus the requested leading distances areRepulsion gives a finite barrier even for zero impact parameter; attraction allows much closer passage. At $b=0$, the attractive formula has the limiting closest distance zero, corresponding to collision with the singular centre rather than a positive turning point.

Let $h_0=h(0)$ and $S=\pi R^2$. The closed tank has constant mass, soTherefore $({\rho }_l-{\rho }_u)h=({\rho }_l-{\rho }_u(0))h_0$. Under the Boussinesq approximation, define $g^{\prime }=g({\rho }_l-{\rho }_u)/{\rho }_l$ and obtainThe gravitational potential energy is especially easy to evaluate with the stipulated downward depth coordinate: the gravitational potential per unit mass is $-gz$. HenceThis proves the potential energy of two-layer entrainment relationThe positive sign expresses the lifting and homogenization of dense entrained fluid within the upper layer. The fixed thin interface is neglected in this leading two-layer energy integral.