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The Euler-Lagrange equations areThe coordinate $\varphi$ is a cyclic coordinate, so its conjugate momentumis conserved. Substitution into the first equation givesMultiplying by $2\dot{\theta }$ shows thatThus $h$ is conserved. Since $l^2/{\sin }^2\theta ={\sin }^2\theta {\dot{\varphi }}^2$, $h$ is the squared speed on the unit sphere, equivalently twice the kinetic energy.

The Euler-Lagrange equations are $\ddot{x}+{\omega }^{2}x=0$ and $\ddot{y}+{\omega }^{2}y=0$. They implyWritinggives the conserved value

Use the canonical Kähler potential and normalize superspace integration by $\int d^2\theta {\theta }^2=1$. The superspace Lagrangian density isThe full superspace integration produces a D-term, while the chiral superfield integral produces an F-term. Work in chiral coordinates for the chiral-superfield component expansion, so the omitted spacetime-derivative terms do not enter the extraction of $F$.

Set $\delta \Phi =\sqrt{2}\theta \psi +{\theta }^{2}F$. A Taylor expansion of the superpotential givesHigher terms vanish because there are only two Grassmann variables ${\theta }^{\alpha }$. With $\theta \psi ={\theta }^{\alpha }{\psi }_{\alpha }$ and $\psi \psi ={\psi }^{\alpha }{\psi }_{\alpha }$, their anticommutation gives $(\theta \psi {)}^2=-\frac{1}{2}{\theta }^2\psi \psi$. ThusThe canonical Kähler potential contributes $F^{\ast }F$. Consequently the part of the Lagrangian density containing the auxiliary field isThe fermion terms coming from the superpotential areThe Weyl spinor kinetic term and the complex scalar field kinetic term are canonical because $K={\Phi }^{†}\Phi$.

Treat $F$ and $F^{\ast }$ as independent variables in their algebraic Euler-Lagrange equations. They giveEquivalently,After eliminating the auxiliary field, the Lagrangian density contains $-V$, so the F-term scalar potential isIt is nonnegative for arbitrary complex $m,g,\phi$, since it is a modulus squared. A supersymmetric vacuum satisfies F-flatness, so its zeros are:If $m=0$ and $g\ne 0$ these coincide; if $g=0$ and $m\ne 0$ only the origin remains; if $m=g=0$ every constant $\phi$ is a supersymmetric vacuum.

To compare masses, expand around a supersymmetric vacuum $v$ satisfying $W^{\prime }(v)=0$, writing $\phi =v+\eta$. With $M=W^{\prime \prime }(v)=m+2gv$,Writing $\eta =(A+iB)/\sqrt{2}$ gives the scalar field mass term $-\frac{1}{2}|M{|}^2(A^2+B^2)$; the Weyl spinor has physical mass $|M|$. Therefore the supersymmetric mass degeneracy isIn particular, at $v=0$ the boson and fermion masses are both $|m|$. At the second supersymmetric vacuum $v=-m/g$ the sign of $M$ reverses, but the physical mass is again $|m|$. Equality here concerns fluctuations about a supersymmetric vacuum, not arbitrary backgrounds with $W^{\prime }(v)\ne 0$.

In the convention $V\supset \lambda |\phi {|}^4$ and ${\mathcal{L}}_{\mathrm{Yukawa}}=-y\phi \psi \psi +h.c.$, the same superpotential coefficient gives $\lambda =|g{|}^2$ and $y=g$. The supersymmetric relation between quartic and Yukawa couplings is thereforeIf instead the Yukawa interaction is written $-\frac{1}{2}{y}_3\phi \psi \psi +h.c.$, then $y_3=2g$ and the same physical relation is $\lambda =|y_3{|}^2/4$. The numerical factor depends on the definition of the Yukawa coupling.

The perturbative non-renormalization theorem applies to the local holomorphic superpotential in the Wilsonian effective action, using regularization in quantum field theory that preserves supersymmetry. There are two complementary ways to see why it applies here.

First, in supergraph calculations in perturbative quantum field theory, the algebra of supersymmetric covariant derivatives puts any candidate local correction to the superpotential into a full $d^4\theta$ superspace integration. It therefore corrects a D-term, such as the Kähler potential, rather than a local chiral $d^2\theta$ F-term. Turning such a contribution into an apparent chiral integral would require nonlocal inverse spacetime derivatives, schematically $1/\square$. These are excluded from the local Wilsonian effective action with a fixed nonzero momentum cutoff. This gives the diagrammatic non-renormalization theorem.

Second, the holomorphy argument for superpotential non-renormalization treats $m$ and $g$ as background chiral spurions, so the Wilsonian effective action superpotential is a holomorphic function of $\Phi ,m,g$, with no dependence on $m^{\ast }$ or $g^{\ast }$. Assign an ordinary $U(1)$ spurion charge and an R-charge as follows:The superpotential has ordinary charge zero and R-charge two, since $d^2\theta$ has R-charge minus two. A prospective holomorphic monomial $m^a{g}^b{\Phi }^c$ must therefore satisfyFor perturbative corrections, regularity at $m=0$ and $g=0$ requires nonnegative powers of the couplings. Regularity in $m$ is justified by keeping a nonzero Wilsonian momentum cutoff and retaining $\Phi$ as a light field. Hence only $c=2$ and $c=3$ survive: the existing $m{\Phi }^2$ and $g{\Phi }^3$ structures. The quadratic coefficient is fixed by the free theory at $g=0$, where there are no interaction loops. The cubic coefficient is fixed at first order in $g$, where a connected three-field interaction has only its tree-level Feynman diagram; loop corrections would require further powers of the interaction couplings, which the holomorphy argument for superpotential non-renormalization charge constraints forbid. ThusA field-independent constant is physically irrelevant in this rigid model; the spurion charges would in any case require $m^3/g^2$, which is not regular at $g=0$.

The Kähler potential can still receive wave-function renormalization. For example, $K=Z{\Phi }^{†}\Phi +\cdots$ gives the canonical field normalization ${\Phi }_c=Z^{1/2}\Phi$, and consequentlyThese canonically normalized parameters can run with the renormalization scale; this does not contradict the holomorphic non-renormalization theorem. When massless modes are included in the full quantum effective action, nonlocal terms associated with infrared divergences can also imitate an F-term. Distinguishing that action from the local Wilsonian effective action is part of the theorem's convention.

A collective coordinate describes a position, orientation, or another parameter of a family of static solitons. For a family $\Phi (\mathbf{x};q)$, promote $q$ to a slowly varying $q(t)$ and substitute into the field action. In a scalar theory with unit kinetic coefficient, this gives the collective-coordinate effective LagrangianFor a gauge-theory soliton, one also solves the Gauss law constraint in gauge theory constraint and projects out pure gauge transformations; arbitrary variations of gauge representatives do not define the physical metric. Tangent vectors to an exactly equal-energy family are zero modes in field theory. If $V=E_0$ is constant, the Euler-Lagrange equations of this moduli-space approximation arethe geodesic equations of its Riemannian metric. The approximation neglects radiation and deformation modes; it describes motion sufficiently slow that these omitted degrees of freedom remain unexcited to the required accuracy.

In collective-coordinate quantization, take the wavefunction measure $\sqrt{\det g}{d}^{n}q$ and the minimal scalar Hamiltonian operatorThe Laplace-Beltrami operator supplies coordinate-invariant kinetic energy. Global identifications and statistics must be imposed on the wavefunctions; the classical metric alone does not choose them. Curvature-ordering terms and loop corrections are additional quantum input.

For the phi-four kink, use the normalization and profile of Question 1. Substituting $\varphi (x,t)={\varphi }_K(x-X(t))$ givesThe metric is constant because of translation invariance. This proves the translational dynamics of a phi-four kink: classically the centre moves at constant velocity, and quantum mechanicallyPlane waves label the continuous translational momentum, with no position-dependent potential. Uniform-motion Lorentz invariance upgrades the dispersion to $\sqrt{M^2+p^2}$; the displayed collective Lagrangian is its small-velocity expansion. Small perturbations also include an internal shape mode and continuum radiation, which this single collective coordinate omits. The fluctuation operator of a phi-four kink in this normalization isthe translational eigenfunction has ${\omega }^2=0$, the shape mode has ${\omega }^2=3c^2$, and continuum modes in the spectrum start at ${\omega }^2=4c^2$. Thus the free-coordinate states describe the kink's low translational energies, not its full excitation spectrum or quantum mass correction.

For two Abelian Higgs vortices at critical coupling, the static energy is $2M_v$ and the Abelian Higgs vortex moduli space has four real dimensions. Let $z_1,z_2$ be their positions, $Z=(z_1+z_2)/2$, and $q=z_1-z_2$. The centre of mass decouples; the relative metric is rotationally symmetric and can be writtenAt large separation, $F(\rho )\to M_v/2$, recovering two free particles. Although there is no static separation potential, the nonconstant metric produces velocity-dependent interaction. Coincidence is smooth in the relative coordinate for two identical vortices $w=q^2$, not in the double-valued $q$. Smoothness gives $F(\rho )\sim C{\rho }^2$ for $C>0$ near zero. A head-on geodesic continues through $w=0$ to the opposite real ray, so $q$ changes its line by $\pi /2$: the vortices scatter through a right angle. This geometric argument does not require an explicit formula for $F$.

With ordinary bosonic exchange statistics, relative wavefunctions are single-valued in $w$ and smooth at coincidence. In the separated polar coordinate they obey $\Psi (\rho ,\theta +\pi )=\Psi (\rho ,\theta )$, with even integer angular labels. Their kinetic operator isThe apparent singularity at $\rho =0$ must be resolved with the smooth $w$ coordinate and regularity, rather than arbitrary boundary conditions on a punctured cone. The free centre-of-mass motion and the asymptotically free relative geometry give quantum scattering states; a flat static energy does not imply that the metric is flat or that scattering is absent. This is not a prediction of a discrete family of static two-vortex bound separations. The smooth collision geometry is developed in David Tong's arxiv.org/abs/hep-th/0509216.

For a Skyrmion of baryon number one, the Skyrmion hedgehog ansatz iswhere $\mathbf{\sigma }$ are the Pauli matrices. Include a centre $\mathbf{X}$ and an orientation $A\in SU(2)$ through $U=A{U}_0(\mathbf{x}-\mathbf{X})A^{-1}$. Hedgehog symmetry identifies spatial rotations with opposite internal rotations, so there are three independent orientation coordinates, not six. Since $A$ and $-A$ give the same field, the physical orientation space is $SO(3)$, with SU(2) group as its double cover. Write $A^{-1}\dot{A}=i\mathbf{\omega }\cdot \mathbf{\sigma }/2$. The leading collective Lagrangian has the formwhere $\Lambda >0$ is the rotational moment of inertia obtained by integrating the profile's field kinetic energy.

For the fermionic quantization appropriate to baryons, the Finkelstein-Rubinstein constraints on the double cover impose $\Psi (-A)=-\Psi (A)$. In SU(2) representations, the central element $-1$ acts by $(-1{)}^{2j}$, so $j$ must be half-integer. Left and right group actions supply isospin and spin angular momentum; hedgehog symmetry makes their magnitudes equal. The rotational quantization of a unit Skyrmion therefore givesThe $j=1/2$ level has four spin-isospin states and models the nucleon doublet, proton and neutron, each with two spin states. The $j=3/2$ level has sixteen states and models the Delta baryon quartet, each with four spin states. The rotor predicts a splitting $3{\hslash }^2/(2\Lambda )$. Without the fermionic sign, single-valued functions on $SO(3)$ would instead allow integer $j$, which is a different quantization. High rotor levels can couple to deformation and pion radiation; this semiclassical approximation does not establish that its entire formal tower consists of stable particles.

For an autonomous Lagrangian, the Euler-Lagrange equation usesApply the chain rule to the Hamiltonian along an arbitrary smooth path, without assuming the unforced Euler-Lagrange equation:Consequently the energy balance for an autonomous Lagrangian isWith explicit time dependence, an additional $-{\partial }_{t}L$ enters $dH/dt$. For ideal Gaussian white noise, the identity is understood through smooth-noise regularization or the Stratonovich chain rule. The original PDF correctly differentiates $L$ with respect to $\dot{x}$ in $H$; the supplied TeX's derivative with respect to $x$, its endpoint $t^2$, and its ordinary derivative of $A$ are transcription errors.

Let $h_i=\delta F/\delta p_i$ be the polar molecular field, with the positive functional derivative convention used in the paper. During a pure advective displacement $u_j$, the polar order parameter changes byThe free energy change is $\delta F=\int h_i\delta p_i{d}^{d}r$. Assume periodic boundaries, or boundary conditions that eliminate the surface work, and incompressibility ${\partial }_j{u}_j=0$. Integration by parts in the advective term givesRelabelling the dummy indices in the rotational and flow alignment of a polar order parameter terms givesTherefore $\delta F=\int {\Sigma }_{ij}^p{\partial }_i{u}_j{d}^{d}r$, with the reversible stress of a polar liquid crystalFor a local free-energy density $f(\mathbf{p},\nabla \mathbf{p})$, an explicit choice isIts divergence is exactly $-p_k{\partial }_j{h}_k$, by the Euler-Lagrange equation for $h_k$. A pressure term can be reassigned in an incompressible flow; the chosen representative realizes the printed divergence without that ambiguity. The force density ${\partial }_i{\Sigma }_{ij}^p$ has mechanical power $-\int {\Sigma }_{ij}^p{\partial }_i{v}_j$, the negative of the advective free energy rate, which checks the stress sign.

For the Expanding flat slicing of de Sitter spacetime, take the geodesic LagrangianThe Euler-Lagrange equation for $t$ iswhile each spatial coordinate $x^i$ obeysComparing these equations with the geodesic equation gives the complete list of nonzero Christoffel symbols:

The $x$ Euler-Lagrange equation has the first integralwhere the constant follows from the initial data. Combining this with proper-time normalization gives the Timelike geodesic in the expanding flat slicing of de Sitter spacetime:ConsequentlyIntegrating from $t=0$, where $\tau =0$, givesAs $t\to -\infty$, the first ratio tends to $1$, soThus $t$ tends to $-\infty$ at the finite proper time ${\tau }_{-}$.

In Cartesian coordinates the Lagrangian isIntroduce a Lagrange multiplier $\lambda$ for $x^2+y^2+z^2=c^{-2}$. The Euler-Lagrange equations becomeorDifferentiating the constraint twice givesTaking the scalar product of the vector equation with $\mathbf{x}$ therefore yieldsSubstitution gives

For $c=0$, the equations have the first integralsWriting $X=x+B$ and $Y=y-A$ gives $\dot{X}=-Y$ and $\dot{Y}=X$. Hence every trajectory isThese are helices of unit angular frequency about a line parallel to the $z$ axis, including circles when $C=0$ and a straight line when $R=0$.

Define the canonical momentaWhen the velocity Hessian is nonsingular, these equations can be inverted for ${\dot{q}}_i(q,p,t)$, and the Hamiltonian is the Legendre transformIts differential isbecause the $p_{i}d{\dot{q}}_i$ terms cancel. The Euler-Lagrange equation identifies $\partial \mathcal{L}/\partial q_i={\dot{p}}_i$, so coefficient comparison gives Hamilton's equations

For the symmetric top, the momenta areThusThe Legendre transform gives

Both $\varphi$ and $\psi$ are cyclic coordinates, so $p_{\varphi }$ and $p_{\psi }$ are conserved. The Hamiltonian has no explicit time dependence and is also conserved. Since $H$ is independent of $\varphi ,\psi$,under the Poisson bracket. Hence three independent first integrals in involution are

Let a one-parameter infinitesimal transformation be $\delta q_i=ϵX_i(q,t)$ and suppose its change in the Lagrangian is a total derivative,With canonical momentum $p_i=\partial \mathcal{L}/\partial {\dot{q}}_i$, the chain rule and the Euler-Lagrange equations give, along a motion,Comparison with the assumed total derivative proves Noether theorem:

The given Lagrangian has no explicit time dependence, so its energy function is conserved:It is also invariant under the dilation $(x,y)\mapsto (e^{ϵ}x,e^{ϵ}y)$: the velocities and $y$ all scale by the same factor, while $x/y$ is unchanged. The generator is $(X_x,X_y)=(x,y)$ and $F=0$. SinceNoether's theorem supplies the second first integralThe integrals $E$ and $D$ are independent on an open dense subset of phase space.

The two Euler-Lagrange equations areDifferentiating $\kappa =f-u^{\prime }{f}_{u^{\prime }}-w^{\prime }{f}_{w^{\prime }}$ along an extremal and using these equations gives $d\kappa /dx=f_x$. Hence the Beltrami identity makes $\kappa$ constant when $f$ has no explicit dependence on $x$. If $f$ omits $u$ or $w$, that variable is a cyclic coordinate and the corresponding momentum $f_{u^{\prime }}$ or $f_{w^{\prime }}$ is an additional first integral; continuous symmetries give the analogous conserved quantities through Noether's theorem.

Put $A=1-m/u$. Since $w$ is cyclic,for a constant $\lambda$. The Beltrami integral isTherefore

If $\kappa =-{\lambda }^2$, then $u^{\prime 2}={\lambda }^{2}m/u$ andFor $z>1$, definethe hinted identity gives $F^{\prime }(z)=z^{3/2}/(z-1)$. Hence $w=C\pm mF(u/m)$. Because $F(z)\to +\infty$ as $z\to \infty$, the prescribed limit selects the minus sign:As $u\downarrow m$, the logarithm tends to $-\infty$, so every such solution satisfies $\boxed{w\to +\infty }$.

Write the admissible variation as $y_{\epsilon }=y+\epsilon \eta$, where $\eta (x)\to 0$ at both ends. Differentiating the functional under the integral sign gives the first variationAfter integration by parts, with the boundary term vanishing,Thus the Euler-Lagrange equation isA second differentiation gives the second variation

For unit mass, the Lagrangian in cylindrical coordinates isIntroduce shearing sheet coordinates by $r=r_0+x$ and $\phi ={\Omega }_{0}t+y/r_0$. Evaluate all derivatives of $\Phi$ at $(r_0,0)$. The reference circular orbit obeys ${\Phi }_r=r_0{\Omega }_0^2$, while midplane symmetry gives ${\Phi }_z={\Phi }_{rz}=0$. Its Taylor expansion isExpanding the kinetic energy to the same order givesThe terms linear in $x$ cancel by circular-orbit balance. Discard the constant and the term $r_0{\Omega }_0\dot{y}$ by total-time-derivative invariance of a Lagrangian; these do not change the Euler-Lagrange equations. The particle Lagrangian in a shearing sheet is thereforewith shearing-sheet tidal potentialThe second form follows from ${\Phi }_r=r{\Omega }^2$, the orbital shear parameter $q_0=-d\log \Omega /d\log r$ and the vertical epicyclic frequency ${\Omega }_z^2={\Phi }_{zz}$.

The Euler-Lagrange equations of the particle Lagrangian in a shearing sheet areThe terms coupling $x$ and $y$ are the Coriolis acceleration. The cyclic coordinate $y$ has conserved canonical momentumFor a Newtonian potential of a point mass, $q_0=3/2$ and ${\Omega }_z={\kappa }_r={\Omega }_0$. Set $p_y={\Omega }_0{x}_0/2$ and substitute $\dot{y}=p_y-2{\Omega }_{0}x$ in the radial equation. It becomes $\ddot{x}+{\Omega }_0^2(x-x_0)=0$, a harmonic oscillator equation about the epicyclic guiding center $x_0$. Integration givesThe four real constants in $A,y_0,x_0$ and the two in $B$ account for the six initial position and velocity data.