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In the horizontal rotating-fluid equation of motion, scale advective acceleration as $U^2/L$ and Coriolis acceleration as $|f|U$, where $f=2\Omega \sin \varphi$. Their ratio is the Rossby number,A small Rossby number means that rotation strongly constrains the flow; with slow evolution, weak friction and a suitable pressure field it permits geostrophic balance. Local time dependence introduces the separate condition $1/(|f|t_{\mathrm{evol}})\ll 1$. A synoptic midlatitude cyclone with $U\sim 10\mathrm{m}{\mathrm{s}}^{-1}$, $L\sim 1000\mathrm{km}$ and $|f|\sim 10^{-4}{\mathrm{s}}^{-1}$ has $\mathrm{Ro}\sim 0.1$. A tornado with $U\sim 50\mathrm{m}{\mathrm{s}}^{-1}$ and $L\sim 1\mathrm{km}$ has $\mathrm{Ro}\sim 500$; its rotational motion does not mean that the Earth's rotation dominates its local dynamics.

The vertical relative vorticity is $\zeta ={\partial }_{x}v-{\partial }_{y}u\sim U/L$, while planetary vorticity is $f$. Thus $|\zeta /f|$ is of the order of the Rossby number, although it is not an identity for arbitrary velocity fields: an irrotational strain can have nonzero $U/L$ but zero $\zeta$. Absolute vorticity is $f+\zeta$.

On large, slowly evolving midlatitude scales, the horizontal pressure gradient force is mainly balanced by the Coriolis force, which is perpendicular to velocity. Consequently the wind is nearly parallel to atmospheric isobars, rather than accelerating directly into a low or out of a high. Curvature adds a gradient wind correction, and surface friction permits the cross-isobar component considered in part (d).

For the real scalar field, the two derivatives entering the Euler-Lagrange field equation areConsequently the equation of motion isThe quartic interaction therefore adds a local cubic term to the free Klein-Gordon equation. Below, write $F=(\square +m^2)\varphi +\lambda {\varphi }^3/6$; thus $F=0$ means that the field is on shell.

First take the trace in four spacetime dimensions:The product rule gives $\square {\varphi }^2=2(\partial \varphi {)}^2+2\varphi \square \varphi$. For the massless equation of motion, $\varphi \square \varphi =-\lambda {\varphi }^4/6$, soBoth coefficients vanish forThe minus sign follows from the order ${\partial }^{\mu }{\partial }^{\nu }-{\eta }^{\mu \nu }\square$ used in the question's stress-energy tensor improvement. Commuting partial derivatives shows that this improvement has identically zero divergence, so it preserves the conservation law.

The precise trace identity, without using the equation of motion, isHence the massless tensor is traceless on shell. It is not traceless for arbitrary off shell configurations: a nonzero constant $\varphi =A$ with $\lambda \ne 0$ has ${\Omega }^{\mu }{}_{\mu }=\lambda A^4/6$. This is a classical result; it does not assert absence of a quantum trace anomaly.

For a Lagrangian density depending on $\varphi$, ${\partial }_t\varphi$, ${\partial }_i\varphi$ and ${\partial }_i{\partial }_j\varphi$, direct variation givesTake $\delta \varphi$ to have compact support, so every boundary term vanishes. One integration by parts for the first-derivative terms and two for the second-derivative term yield the Euler-Lagrange field equation with second derivativesThe plus sign on the last term is essential. For the action in a, it is particularly transparent to vary the Laplacian directly:Thus the classical equation of motion isHere ${\nabla }^4=({\nabla }^2{)}^2$ is the biharmonic operator. Substitution of the plane wave $\varphi =A{e}^{-i\omega t+i\mathbf{p}\cdot \mathbf{x}}$ givesA real classical solution combines complex-conjugate plane waves. Quantization assigns the positive branch to the excitation energy, giving the required dispersion relation. The negative-frequency branch supplies the creation operator part of the field, rather than negative excitation energy. Independently, the Hamiltonian in a gives $\dot{\varphi }=\pi$, $\dot{\pi }=-{\kappa }^2{\nabla }^4\varphi -m^2\varphi$, confirming the same sign.

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.

The transformation $\psi \mapsto \psi +b$ leaves both derivatives unchanged, so the Lagrangian density and action are invariant. For infinitesimal unit field variation, the Noether current for a first-derivative scalar field is the shift current of an elastic rod,Its divergence vanishes by the equation of motion. The charge is the total longitudinal mechanical momentum to this order in the displacement field.

Allow a freely chosen center $x_0$, set $u=k(x-x_0)$ and $z=e^u$. Direct differentiation of the proposed Sine-Gordon kink givesThe angle identity givesSince $k^2/b=a$, these formulas show $\boxed{{\varphi }^{\prime \prime }=a\sin (b\varphi )}$, precisely the static equation of motion.

For $x\to -\infty$ and $x\to +\infty$, the field approaches $0$ and $2\pi /b$ respectively. These are neighboring minima of the scalar potential when $ab>0$, and ${\varphi }^{\prime }$ vanishes exponentially at either end. The energy is therefore finite and the topological charge $Q=b[\varphi (+\infty )-\varphi (-\infty )]/(2\pi )$ is one. Reversing $x-x_0$ gives the corresponding antikink. If $ab<0$, the printed real exponential profile is unavailable; its formal imaginary scale is not a real finite-energy solution of the stated form.

The Lagrangian density is Lorentz invariant, so a scalar-valued static solution can be composed with a Lorentz transformation. Let $X=\gamma (x-vt-x_0)$, with $\gamma =(1-v^2{)}^{-1/2}$ and $|v|<1$. If $\varphi (x,t)=f(X)$, the chain rule givesThe boosted equation of motion is therefore exactly the static one already verified in solution, andis a solution. The center moves as $x=x_0+vt$: $v$ is the physical velocity in units where the limiting speed is one. The kink width contracts from $k^{-1}$ to $(\gamma k{)}^{-1}$, while its topological charge remains unchanged.

For a static finite-energy field, the Hamiltonian density is $\frac{1}{2}({\varphi }^{\prime }{)}^2+V(\varphi )$. Multiply the static equation of motion by ${\varphi }^{\prime }$ and integrate once:At either vacuum endpoint both terms vanish, so $\frac{1}{2}({\varphi }^{\prime }{)}^2=V$. The total energy density is consequently $({\varphi }^{\prime }{)}^2=4k^2{\mathrm{sech}}^2[k(x-x_0)]/b^2$. Its integral givesFor the usual positive $a,b$ convention this isThe coefficient depends on the dimensionless parameter $b$; it is not a universal numerical constant independent of the potential. In two spacetime dimensions, $[\varphi ]=[b]=0$ and $[a]=2$, so the dimensions are consistent. If both $a$ and $b$ are negative while $ab>0$, the real answer is instead $8\sqrt{|a|}/|b{|}^{3/2}$; the literal factor $\sqrt{a}$ would not be real.

An equivalent positive-parameter calculation substitutes $z=b\varphi$ in $E_0={\int }_0^{2\pi /b}\sqrt{2V}d\varphi$ and uses the provided integral of $\sqrt{1-\cos z}$. The following original plot shows that the localized energy density sits where the field passes between the vacua.

Put $\rho ={\phi }^{†}\phi$. The heavy-field equation of motion isThis solution describes the low-energy sector without an independent incoming heavy-particle wave. It is valid when all external momentum invariants and light masses squared are small compared with $M^2$.

After an integration by parts, the terms involving the heavy field are $-\frac{1}{2}\varphi (\square +M^2)\varphi -g\rho \varphi$. Substituting the solution into both terms, rather than the source term alone, givesHence, with the question's sign convention for the local quartic interaction,This is integrating out a field followed by a derivative expansion. The negative quartic means attraction. It also makes the truncated potential unbounded at large fields; minimizing the original potential over $\varphi$ reveals the same negative quartic. Thus neither displayed model, without extra stabilizing interactions, supplies a globally stable nonperturbative vacuum. The matching describes perturbative low-energy scattering about the local vacuum. The simple equation-of-motion matching fixes the leading tree-level coefficient; precision quantum matching requires consistent loop renormalization and the additional effective operators at the requested accuracy.

For a translation-invariant potential, the on-shell energy-momentum tensor obeys ${\partial }_{\mu }{\theta }^{\mu \nu }=0$. The divergence of the dilatation current is thereforeUse the equation of motion $\square \varphi =-V^{\prime }(\varphi )$ to obtain the useful general identityFor $D=4$, $m=0$ and only ${\lambda }_4$ nonzero, every term vanishes. More explicitly, $\varphi \square \varphi =-{\lambda }_4{\varphi }^4/6$ while ${\theta }^{\mu }{}_{\mu }=-(\partial \varphi {)}^2+{\lambda }_4{\varphi }^4/6$; adding the first kinetic term cancels both remaining contributions. Thus ${\partial }_{\mu }{j}^{\mu }=0$, and $\int d^{3}x{j}^0$ is conserved when the boundary flux vanishes. Outside the parameter conditions of solution, the boxed formula displays the classical sources of scale breaking; the same current need not be conserved.