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Put $\kappa =\sqrt{a/b}>0$. Solving the interior Euler-Lagrange equation and imposing both $V(0)=0$ and $\dot{V}(0)=0$ yields $V=C(\cosh \kappa t-1)$, with $\lambda =-2aC$. The distance constraint fixesThe denominator is positive because $\sinh z>z$ for $z>0$. This is the requested solution of the differential equation with the stated initial conditions.

As a minimum-energy problem with free terminal speed, the printed conditions do not admit a smooth minimizer. The formal profile has $\dot{V}(T)=C\kappa \sinh (\kappa T)>0$, violating the natural terminal condition from part (a). Fixing the initial acceleration restricts ${\eta }^{\prime }(0)$, but it does not restrict the independently variable $\eta (T)$ and therefore does not remove that condition.

For clarity, without the extra initial-acceleration restriction the actual unique minimizing profile isIt satisfies $V_{\ast }(0)=0$, ${\dot{V}}_{\ast }(T)=0$, the distance constraint and the interior equation, but ${\dot{V}}_{\ast }(0)>0$. For any perturbation preserving the distance and initial speed, the first variation about $V_{\ast }$ is zero, andwith equality only for $\eta =0$. Its energy is $a{L}^2/[T-\tanh (\kappa T)/\kappa ]$.

This same energy is the unattained infimum under the extra condition $\dot{V}(0)=0$. Indeed multiply $V_{\ast }$ by $1-e^{-t/\epsilon }$ and rescale by a factor tending to one to restore total distance $L$. These smooth nonnegative profiles have both initial values zero and converge to $V_{\ast }$ in value and derivative square integrals as $\epsilon \downarrow 0$. Their energies tend to $E[V_{\ast }]$, but uniqueness of $V_{\ast }$ precludes attaining it with zero initial acceleration. This is nonattainment under an initial derivative constraint.

Here $b>0$ because $a=0$ and the two cost coefficients are not both zero. The interior Euler-Lagrange equation makes $V$ quadratic. Imposing each displayed pair of endpoint data and the distance constraint gives the following formal profiles and their costs:Thus choice (3) is the best strategy, and its true minimum cost isThe qualification from part (b) applies to row (1): its formal quadratic is not a minimizer with free terminal speed, since $\dot{V}(T)\ne 0$. Row (2) is a genuine minimizer for its fixed zero terminal speed, and row (3) is the genuine free-terminal-speed minimizer, satisfying its natural condition $\dot{V}(T)=0$.

A direct lower bound makes the comparison rigorous. For every admissible speed with $V(0)=0$,by integration by parts. The Cauchy-Schwarz inequality gives $L^2\le (T^3/3){\int }_0^T{\dot{V}}^{2}dt$. Equality requires $\dot{V}$ to be proportional to $T-t$, giving exactly row (3). Row (1)'s zero initial acceleration excludes equality; however short initial layers, as in part (b), approach that same infimum. Hence choice (1) has infimum $3b{L}^2/T^3$ but does not attain it, while choice (3) does. This distinction resolves the boundary-condition issue in the printed minimization request.

Work with real-valued functions. The screened sine-Gordon energy is well defined: $0\le 1-\cos u\le u^2/2$, and $fu$ is integrable by the Cauchy-Schwarz inequality. It is coercive, becauseA minimizing sequence is bounded in the Hilbert space $H^1$. Weak sequential compactness of bounded sequences in a reflexive Banach space supplies a weakly convergent subsequence. The squared H1 space norm is weakly lower semicontinuous, the source pairing is weakly continuous, and the nonlinear term is covered by local Sobolev compactness gives lower semicontinuity of a nonnegative integral. The direct method in the calculus of variations therefore gives a minimizer $\varphi$.

Taking its first variation in any $v\in H^1$ givesThus the Euler-Lagrange equation isas a weak solution, equivalently in distributions when tested against smooth compactly supported functions. The derivative of the nonlinear term is justified by $|\sin \varphi |\le |\varphi |$ and the second-order remainder bound $|\cos |\le 1$.

Now $G=f-\sin \varphi \in L^2$, so the supplied elliptic regularity estimate puts $\varphi$ in $H^2$. Applying the Sobolev inequality to $\varphi$ and each first weak derivative gives $\varphi \in W^{1,6}({\mathbb{R}}^3)$. Morrey's inequality and uniformly continuous integrable functions vanish at infinity prove that its continuous representative tends to zero.

The final printed supremum estimate is false in general. The maximum bound for a monotone reaction term involves $h(s)=s+\sin s$, which is odd and strictly increasing: $h^{\prime }=1+\cos s\ge 0$, and its zeros are isolated. If $M=\Vert f{\Vert }_{\infty }<\infty$, a positive maximum $m$ of $\varphi$ satisfies $h(m)\le f\le M$; apply the same argument to $-\varphi$. The valid general estimate isThe bound by $M$ is valid if $M\le \pi$, but $h(s)$ can be smaller than $s$ for larger positive $s$.

For an explicit failure of the source-size bound for the screened sine-Gordon equation, take $a=3\pi /2$, $L=10$, $\varphi (x)=a{e}^{-|x{|}^2/L^2}$ and $f=-\Delta \varphi +\varphi +\sin \varphi$. These are smooth functions in the required spaces. With $q=|x{|}^2/L^2$,Hence $\Vert f{\Vert }_{\infty }\le a-1+6a/L^2<a=\Vert \varphi {\Vert }_{\infty }$. Moreover $V(s)=s^2/2+1-\cos s$ has $V^{\prime \prime }=1+\cos s\ge 0$, so the energy is convex and this critical point is a minimizer. The example therefore satisfies even the minimizing and $C^2$ hypotheses of the printed claim.

Return to the mostly-minus Minkowski metric of Question 3, and write $f={\partial }_a{A}^a$. The field $\xi (x)$ is assumed real and nonzero wherever its inverse occurs. Under the usual Abelian gauge transformation $A_a\mapsto A_a+{\partial }_a\lambda$, with $\xi$ inert, the electromagnetic field tensor is invariant but $f\mapsto f+\square \lambda$. The change in the added density isThis is not generally a total derivative. The action is not invariant under arbitrary gauge transformations. It retains residual Lorenz gauge symmetry for transformations satisfying $\square \lambda =0$, with the same boundary conditions imposed before and after the transformation. The action itself is undefined at $\xi =0$; that value can only be considered as a limiting gauge.

To vary the electromagnetic four-potential, use $\delta F_{ab}={\partial }_a\delta A_b-{\partial }_b\delta A_a$ and $\delta f={\partial }_a\delta A^a$. The integration by parts of both terms giveswith surface terms removed by the variational boundary conditions. HenceFor variable $\xi$, the derivative must act on $1/\xi$ as well as on $f$:Since $\xi$ enters algebraically, its Euler-Lagrange equation isFor a real field this gives $f=0$. Thus a dynamical gauge-fixing parameter imposes a constraint equation in field theory; it is not an ordinary propagating scalar. On configurations satisfying both equations, ${\partial }_a{F}^{ab}=0$, ${\partial }_a{A}^a=0$, and therefore $\square A^b=0$. There is no independent kinetic equation that determines $\xi$.

For the momentum-space equation at a prescribed constant $\xi \ne 0$, use $A_a(x)=\int d^{4}p{e}^{-ip\cdot x}{A}_a(p)/(2\pi {)}^4$. Then ${\partial }_a\mapsto -i{p}_a$, and the linear $A$ equation isEquivalently, $-p^2{A}^a+(1+1/\xi )p^a(p\cdot A)=0$. If the separately varied $\xi$ equation is also imposed on a real classical solution, then $p\cdot A=0$ and its nonzero modes lie on the massless mass shell. For constructing a full-field Green function, however, invert the fixed-background quadratic operator before imposing that on-shell constraint. Holding $\xi$ fixed and integrating over $A$ is a Gaussian functional integral; integrating over the original variable $\xi$ as well is a different constrained problem.

For $p^2\ne 0$, the Lorentzian versions of the transverse projector of a vector field and longitudinal projector of a vector field areThey satisfy $P_T^2=P_T$, $P_L^2=P_L$, $P_T{P}_L=0$ and $P_T+P_L=I$. The mixed-index operator isIts inverse follows by inverting these two scalar eigenvalues. Define $G_{ab}$ by $K^{ac}{G}_{cb}={\delta }_b^a$. The positive-sign covariant gauge-fixing inverse isThis is the algebraic inverse of the kinetic operator. A vacuum photon propagator instead has the factor $i{G}_{ab}$ and a Feynman i-epsilon prescription for its simple and double poles. If the name $G$ is used for the vacuum two-point function, include that factor $i$ consistently in its defining source equation. A retarded Green function would use different boundary conditions at the same poles. The original plus sign corresponds to the usual covariant gauge parameter $\alpha =-\xi$; in particular, $\xi =-1$ gives Feynman gauge and $G_{ab}=-{\eta }_{ab}/p^2$.

Contracting gives the longitudinal gauge propagator contractionThus the contraction is not identically zero as a function of four-momentum for any $\xi \ne 0$, and it is a nonzero vector at every nonzero non-null four-momentum. A particular component may vanish when $p_b=0$. At $p^2=0$, the displayed inverse has poles and must be interpreted using the chosen Green function prescription, rather than as a pointwise finite matrix. In the limiting Landau gauge, $\xi \to 0$, the longitudinal part vanishes. A covariant photon Green function may have a longitudinal component even though physical photon polarizations are transverse.

The statistical Hamiltonian in this question is already in thermal units, as indicated by $e^{-H}$ without an additional ${\beta }_T$. Define $W[h]=\log Z[h]$ and $F[h]=-W[h]$. The mean field and its linear response areThese are functional derivatives of the connected generating functional; the subtraction defines the connected correlation function. In the Landau approximation, neglect loop corrections and evaluate the field integral at a stable saddle $m_h$. It satisfies the Euler-Lagrange equationThis follows by varying the gradient term and integrating by parts, with periodic, decaying, or otherwise appropriate boundary conditions.

The two requested free energy functionals, following the source and Legendre conventions of the question, areIn the scalar-field source Legendre transform on the chosen stable branch, $h$ is chosen to produce $m$, and $\delta {\Gamma }_{\mathrm{L}}/\delta m=h$. Thus the imposed-source Helmholtz free energy and the fixed-order-parameter Gibbs free energy have the appropriate opposite source derivatives. These names are used in the question's magnetic-ensemble convention; the defining sign relation is what fixes the calculation. At leading Landau approximation there is no fluctuation-determinant term in $F_{\mathrm{L}}$.

Differentiate the saddle equation with respect to $h(y)$. The response obeysThereforeEquivalently, the inverse Hessian relation for a connected two-point function states that $G$ is the inverse kernel of ${\delta }^2{\Gamma }_{\mathrm{L}}/\delta m(x)\delta m(y)$. The factor $u_0/2$ follows from twice differentiating the quartic term $u_0{m}^4/4!$; it is not $u_0/6$. This tree-level connected response is obtained by varying the saddle. It does not require replacing the exact connected correlator by a product of the saddle values, which would incorrectly give zero.

For the requested single-momentum formula, assume a homogeneous source and a translationally invariant equilibrium phase, so $m_h(x)=m_0$ and $G(x,y)=G(x-y)$. SetThe Fourier transform with the printed positive sign sends $-{\nabla }^2$ to $q^2$, while the Dirac delta function transforms to one. Hence the Ornstein--Zernike correlation function hasThe inverse convention is $G(x)=\int d^{D}q(2\pi {)}^{-D}{e}^{-iq\cdot x}\tilde{G}(q)$. For general inhomogeneous $h(x)$, $K$ has nonconstant coefficients and $G$ depends separately on its two positions; the displayed momentum-diagonal formula then does not follow. The preceding differential equation still holds in that case.

At zero source and for $u_0>0$, the homogeneous saddle is $m_0=0$ for $r_0>0$, and $m_0^2=-6r_0/u_0$ on either selected stable ordered branch for $r_0<0$. Thus the Landau scalar correlation length isIn the ordered phase the negative bare quadratic coefficient is compensated by the positive curvature at the nonzero saddle. Retaining $m_0=0$ below the transition would instead give an unstable kernel and is not a physical correlation length. With the usual analytic thermal tuning $r_0=a_0(T-T_c)$, $a_0>0$, both branches diverge as $|T-T_c{|}^{-1/2}$, so the correlation-length critical exponent isThe high-temperature amplitude is $\sqrt{2}$ times the low-temperature amplitude for the same $a_0$. This statement is within Landau theory; fluctuations can change critical behavior outside the mean-field regime.

Use the Minkowski metric with signature $(+,-)$, and write $V(\varphi )=\frac{1}{2}(1-{\varphi }^2{)}^2$. The Euler-Lagrange equation isFor the static phi-four kink, ${\varphi }_x=1-{\varphi }^2={\mathrm{sech}}^2(x-a)$ and ${\varphi }_{xx}=-2\varphi (1-{\varphi }^2)=2\varphi ({\varphi }^2-1)$, so the field equation is satisfied. The centre $a$ is arbitrary by translation invariance, and the hyperbolic tangent profile increases monotonically from $-1$ to $+1$, crossing zero at $x=a$. The static kink and its endpoint topological charge are

Both endpoint values are isolated classical vacua, since $V(\varphi )=0$ only at $\varphi =\pm 1$. A continuous finite-energy deformation preserving the vacuum boundary conditions cannot change either endpoint to the other isolated classical vacuum. The topological charge is therefore unchanged: this kink cannot deform into a homogeneous classical vacuum, whose charge is zero. There is also a direct Bogomolny bound in this sector. The square completion for a one-dimensional kink givesThe phi-four kink saturates the bound, so its mass is $4/3$ in these units and it minimizes the energy within its topological sector. Its arbitrary position is a collective coordinate, not an instability. A kink and an antikink together have total charge zero and can annihilate without contradicting the protection of an isolated kink.

For the momentum, the canonical stress-energy tensor of this scalar field isConsequently the physical spatial momentum density and the spatial momentum flux areThe sign of $\mathcal{P}$ makes a right-moving translated kink carry positive momentum. Direct use of the field equation, rather than an assumed static field, yields the scalar-field momentum flux identityThe finite-energy field configuration has $\mathcal{P}\in L^1$ by $|{\varphi }_t{\varphi }_x|\le ({\varphi }_t^2+{\varphi }_x^2)/2$. Integrating the stress-energy conservation law over the left half-line gives the boundary forceUnder the usual vacuum falloff, the stress at the left endpoint is zero. More generally, smooth spatial cutoffs with derivative of order $1/R$ remove the left endpoint using the integrable energy density, so no pointwise limit of every derivative at infinity is needed. The identity expresses the force on the field to the left of $b$: positive force transfers momentum to the right. For well separated solitons, a cut between them measures the interaction force on the left soliton.

Take that cut at $b=0$. The specified symmetric pair has, at the initial time,The printed field profile does not itself specify the initial velocity. If $\psi (x)={\varphi }_t(x,0)$, the exact initial half-line force isFor the intended initially resting pair, or more generally $\psi (0)=0$, put $q=e^{-2c}$ and use $\tanh c=(1-q)/(1+q)$. The at-rest force for a symmetric phi-four pair isThe leading force is attractive, towards the antikink:The antikink feels the opposite force by the symmetry of the resting pair. This is an initial, large-separation interaction calculation, not a claim that the superposed profile is an exact static two-soliton solution. Without the initial-velocity condition, the additional momentum flux above prevents a unique force from being inferred from the printed profile alone.

Set $f=1-2M(v)/r$, so $f_r=2M/r^2$ and $f_v=-2M^{\prime }/r$, where $M^{\prime }=dM/dv$. Dots in this solution mean derivatives with respect to proper time $\tau$. The geodesic Lagrangian becomesThe Euler-Lagrange equation for $r$ gives $\ddot{v}+(f_r/2){\dot{v}}^2-r({\dot{\theta }}^2+{\sin }^2\theta {\dot{\varphi }}^2)=0$. The Euler-Lagrange equation for $v$ givesEliminate $\ddot{v}$ between these equations. The two angular Euler-Lagrange equations follow by differentiating $-2r^2\dot{\theta }$ and $-2r^2{\sin }^2\theta \dot{\varphi }$. Together the four geodesic equations areReading off the symmetric quadratic coefficients gives all nonzero Christoffel symbols of the Vaidya metric:Every unlisted Christoffel symbol vanishes. The lower-index symmetry is that of the Levi-Civita connection. In particular, the time-dependent mass contributes $+M^{\prime }/r$ to ${\Gamma }^r{}_{vv}$; it is absent from ${\Gamma }^v{}_{vv}$. There are eleven independent nonzero entries, or fifteen when both lower-index orders are counted.

Take a smooth variation $u_{\epsilon }=u+\epsilon \varphi$ with $\varphi =0$ on the boundary. Assume enough differentiability for the following operations, for example a smooth bounded open set, $u\in C^2$, and a twice continuously differentiable integrand. Differentiation under the integral givesBy integration by parts, the boundary term vanishes and the first variation isStationarity for every such variation, including every compactly supported smooth one, and the fundamental lemma of the calculus of variations give the Euler-Lagrange equationFor the quadratic integrand, $L_u=2k^{2}u$, $L_{u_x}=2u_x$, and $L_{u_y}=2u_y$. Thuswith the prescribed Dirichlet boundary condition. For real $k$, a stationary function is also the unique minimizer, if it exists: for any nonzero admissible $\varphi$, the linear term vanishes and $I[u+\varphi ]-I[u]={\int }_{\mathcal{D}}(|\nabla \varphi {|}^2+k^2{\varphi }^2)>0$. This last positivity uses the zero boundary values; it also holds at $k=0$.

The principle of stationary action, also called Hamilton's principle, requires $\delta {\int }_{t_1}^{t_2}Ldt=0$ for fixed-endpoint path variations. Here the original PDF has ${\dot{r}}^2$ in the radial kinetic energy; the TeX has lost that dot. With $m>0$ and $r>0$, the Lagrangian isThe Euler-Lagrange equation for each coordinate givesThe two conserved quantities are angular momentum $h=m{r}^2\dot{\varphi }$ and energy $E=\frac{m}{2}{\dot{r}}^2+\frac{m}{2}{r}^2{\dot{\varphi }}^2+km{r}^2$. By Noether's theorem, they arise respectively from rotational invariance and time-translation invariance of the action. The coordinate $\varphi$ is a cyclic coordinate, and $L$ has no explicit time dependence.

The conjugate momentum for each coordinate is $p_r=m\dot{r}$, $p_{\varphi }=m{r}^2\dot{\varphi }$. The Legendre transform $H=p_r\dot{r}+p_{\varphi }\dot{\varphi }-L$ gives the HamiltonianHamilton's equations areSetting $p_{\varphi }=h$ yields $m\ddot{r}=-V_{\mathrm{eff}}^{\prime }(r)$ with the effective potential $V_{\mathrm{eff}}=h^2/(2m{r}^2)+km{r}^2$.

For $k>0$ and $h\ne 0$, this potential tends to infinity both as $r\downarrow 0$ and as $r\to \infty$, and has its unique stationary point atThe effective potential stability criterion proves a stable circular orbit in a quadratic central potential, with $|\dot{\varphi }|=\sqrt{2k}$. For a small radial displacement $\eta$, $\ddot{\eta }+8k\eta =0$ to first order, so radial perturbations oscillate rather than grow. The diagram shows the centrifugal and quadratic contributions and their sum in dimensionless units:

The printed paper only says that $k$ is constant. Its claimed positive-radius stable circular orbit needs the additional attractive-force assumption $k>0$ and nonzero angular momentum. For $k<0$ there is no such minimum; for $k=0,h\ne 0$ the effective potential is strictly decreasing. With $h=0,k>0$, the minimum is at the origin and is not a positive-radius circular orbit. These cases are not represented by the diagram.

Assume the integrand is sufficiently smooth near the positive admissible profiles. Define the second variation as ${\delta }^{2}F[y,\xi ]={d^{2}F[y+\epsilon \xi ]/d{\epsilon }^2|}_{\epsilon =0}$, so it is twice the quadratic Taylor coefficient. Differentiation under the integral givesIntegrating the middle term by parts, with $\xi (\alpha )=\xi (\beta )=0$, yieldsAll coefficients are evaluated along $y$.

For the surface-area integrand $f=2\pi y\sqrt{1+p^2}$, the Beltrami identity for a stationary profile readsEquivalently, its Euler-Lagrange equation is $y{y}^{\prime \prime }=1+y^{\prime 2}$, solved by the positive catenoid profile $y=E\cosh ((x-x_0)/E)$ with $E>0$. Equal boundary radii at $x=\pm L$, for $L>0$, require $x_0=0$, soDirect substitution also verifies the Euler-Lagrange equation. This is a stationary profile whenever a positive $E$ satisfying the boundary relation exists; such an $E$ does not exist for arbitrary ring separations and radii.

On this profile, with $z=x/E$, the coefficients are $f_{yy}=0$, $f_{pp}=2\pi E{\mathrm{sech}}^{2}z$ and $d{f}_{yp}/dx=(2\pi /E){\mathrm{sech}}^{2}z$. Changing variables therefore givesThe printed strict inequality must mean every nonzero admissible variation: the zero function makes the integral zero, so the literal quantifier including it is false.

For completeness, the strict positivity here is sufficient for a strict local minimum in the fixed-endpoint $C^1$ topology, among positive axisymmetric graph profiles. On this finite interval the weight $p(z)={\mathrm{sech}}^{2}z$ has a positive minimum. The regular Sturm-Liouville problem associated with $Q[\xi ]=\int p({\xi }_z^2-{\xi }^2)$ has an attained lowest Dirichlet eigenvalue ${\lambda }_1$ in the Rayleigh quotient. Its first eigenfunction is a nonzero smooth admissible variation; the assumed strict positivity gives ${\lambda }_1>0$. Hence $Q\ge {\lambda }_1\Vert \xi {\Vert }_{L^2}^2$. Combining this with $Q\ge p_{\min }\Vert {\xi }_z{\Vert }_{L^2}^2-p_{\max }\Vert \xi {\Vert }_{L^2}^2$ proves coercivity in the Sobolev space $H_0^1$.

The coefficients of the second variation depend continuously on $(y,y^{\prime })$ while $y>0$. For a sufficiently small $C^1$ perturbation, the quadratic forms along the segment from the stationary profile to the perturbed one retain a uniform positive coercivity bound. Taylor's formula with integral remainder, and the vanishing first variation, then gives $F[y+v]-F[y]\ge c\Vert v{\Vert }_{H^1}^2>0$ for nonzero sufficiently small admissible $v$. The stationary area is therefore a strict local minimum under the stated nonzero-variation condition. This supplies the uniform bound needed in an infinite-dimensional problem, rather than relying on pointwise positivity alone in an arbitrary function space.

For a timelike tangent $v=dx/du$, put $L=\sqrt{-g(v,v)}>0$ and $\mathcal{L}=L^2/2=-g(v,v)/2$, and use the Euler-Lagrange equations. For the quadratic geodesic Lagrangian, they reduce toHere the first equation has been multiplied by an irrelevant minus sign. For the length Lagrangian, its derivatives are $L^{-1}$ times those of $\mathcal{L}$, and differentiating that factor yields insteadThis is the unparametrized geodesic equation, the key to why timelike length and energy have the same geodesic images. Define proper time locally by $d\tau /du=L>0$ and put $w=dx/d\tau =v/L$. The product rule then gives ${\nabla }_{w}w=L^{-2}[{\nabla }_{v}v-(\dot{L}/L)v]=0$ and $g(w,w)=-1$. Conversely an affinely parametrized timelike geodesic has constant $L$, by metric compatibility, and therefore satisfies both equations.

Thus the two variational problems have the same oriented timelike geodesic images, after reparametrization. They do not have exactly the same parametrized solutions with an arbitrary fixed $u$: in Minkowski spacetime, $t(u)=u^2$, $\mathbf{x}=0$, $u>0$, satisfies the length equation but not the quadratic equation. The timelike assumption excludes $L=0$, where this argument and the length derivative would fail.

Let $M_{\mathrm{Pl}}$ be the reduced Planck mass, $M_{\mathrm{Pl}}^{-2}=8\pi G$ in units $\hslash =c=1$. Combining that equation with the Friedmann acceleration equation givesDifferentiate $3M_{\mathrm{Pl}}^2{H}^2=\rho$ to obtain $\dot{\rho }=-3H(\rho +P)=-3H{\dot{\varphi }}^2$. For the homogeneous canonical scalar field, $\dot{\rho }=\dot{\varphi }\ddot{\varphi }+V_{,\varphi }\dot{\varphi }$, soWhere $\dot{\varphi }\ne 0$, division gives the inflaton equation of motion; continuity extends it to isolated turning points:The Hubble friction term describes loss of kinetic energy into expansion.

There is a small logical caveat if the field is frozen on an entire interval: the homogeneous Einstein constraints alone only give the product equation and do not force $V_{,\varphi }=0$. The frozen scalar fields and the Friedmann constraint counterexample makes this explicit. For a canonical scalar, the field equation also follows directly from its inflaton action in an expanding universe, whose homogeneous Lagrangian is $a^3[{\dot{\varphi }}^2/2-V(\varphi )]$. The Euler-Lagrange equation gives $d(a^3\dot{\varphi })/dt+a^3{V}_{,\varphi }=0$, including the frozen case, and yields the same boxed equation without division by $\dot{\varphi }$.

For a regular Lagrangian define the canonical momentum components $p_j=\partial L/\partial {\dot{q}}_j$. Require the velocity Hessian to be invertible locally so that the velocities can be expressed in terms of $(q,p,t)$. The Legendre transform in mechanics definesThe explicit time argument cannot generally be omitted when the given Lagrangian depends on time. Its differential is $dH={\sum }_j{\dot{q}}_{j}d{p}_j-{\sum }_j(\partial L/\partial q_j)d{q}_j-(\partial L/\partial t)dt$, since the $d{\dot{q}}_j$ terms cancel. The Euler-Lagrange equations give $\partial L/\partial q_j={\dot{p}}_j$, and comparison yields Hamilton's equations:For the time-independent stellar gravitational potential below, $H$ is conserved. Its sign convention differs from the binding energy in question 2: here $E=H=v^2/2-\psi$ is the ordinary specific orbital energy.