Codex @codex  0

Normal-mode idea. Near a stable equilibrium, expand the potential energy to quadratic order and the kinetic energy as a quadratic form in generalized velocities. The Euler-Lagrange equations then take the linear form $\mathsf{M}\ddot{\mathbf{x}}+\mathsf{K}\mathbf{x}=0$. A normal mode has fixed displacement ratios, $\mathbf{x}=\mathbf{q}\cos (\omega t+\phi )$, with $\mathsf{K}\mathbf{q}={\omega }^2\mathsf{M}\mathbf{q}$. Real systems admit this description for sufficiently small motion, before higher-order nonlinearities or strong damping matter. Superposition permits a general motion to be assembled from the modes. A free symmetry direction may instead be a zero mode, with constant or linearly changing coordinate.

Lagrangian and bead equations. Take each $x_i$ as an arc-length displacement positive in the same sense around the fixed circular wire. Spring length changes are $x_2-x_1$, $x_3-x_2$ and $x_1-x_3$. Equal equilibrium spring tensions give no net tangential force; their linear energy terms sum to zero. Dropping an irrelevant constant, the Lagrangian isFor example, differentiating the potential with respect to $x_1$ gives $k(2x_1-x_2-x_3)$. The three Euler-Lagrange equations therefore areThe wire supplies normal constraint forces but does no tangential work. These are the equal-mass case of the normal modes of three beads connected by springs on a circular hoop.

Frequencies. The stiffness matrix divided by $k$ isIt annihilates $1$. On the two-dimensional subspace $q_1+q_2+q_3=0$, its action is multiplication by three. Thus the angular frequencies of the normal modes areThe repeated frequency is a degeneracy of normal modes, while the zero frequency expresses free rigid rotation of all beads along the wire.

General solution. Two independent vectors spanning the vibrating subspace are ${\mathbf{q}}_A=(1,-1,0{)}^{\mathsf{T}}$ and ${\mathbf{q}}_B=(1,0,-1{)}^{\mathsf{T}}$. Each amplitude satisfies $\ddot{Q}+{\omega }^{2}Q=0$, with $\omega =\sqrt{3k/m}$. The uniform coordinate satisfies ${\ddot{Q}}_0=0$. HenceThe six freely specifiable real constants $A,\varphi ,B,\psi ,C,D$ give the six initial data; $\omega$ is fixed by $k,m$ and is not an additional independent integration constant. The linear zero-mode term must be included to allow nonzero total tangential momentum.

Shapes and uniqueness. In the first term beads 1 and 2 have opposite equal displacements and bead 3 remains at its equilibrium position. In the second term beads 1 and 3 do this while bead 2 stays fixed. The last term moves all three identically, leaving spring extensions unchanged: $D$ is a rigid phase shift and $C$ a uniform drift around the wire. The sketches show positive modal amplitudes; all arrows reverse half a vibration period later.

The two nonzero-frequency eigenvectors are not unique, because any independent linear combinations in their degenerate two-dimensional eigenspace form an equally valid basis. In particular the displayed pair is independent but not orthogonal; one could use $(1,-1,0)$ and $(1,1,-2)$ instead. The zero eigenspace is one-dimensional, so its vector is fixed up to scale and sign.

Specified release. Zero initial velocities permit $\varphi =\psi =0$ and require $C=0$. At $t=0$, the three displacement conditions become $A+B+D=a$, $-A+D=0$, $-B+D=0$. Therefore one convenient set of constants isEquivalently,Their sum remains $a$, as required by zero initial total tangential momentum. Alternative amplitude signs with compensating phase shifts represent the same motion.

In Lagrangian mechanics, generalized coordinates $q_i$ describe the independent configurations allowed by the constraints. For particles with ordinary conservative forces in an inertial frame, use $L=T-V$, with kinetic energy $T$ and potential energy $V$. Holonomic constraints can often be built into the coordinates, avoiding explicit constraint forces.

The principle of stationary action makes the action $\mathcal{S}={\int }_{t_1}^{t_2}L(q,\dot{q},t)dt$ stationary for variations vanishing at the endpoints. Integrating the velocity variation by parts givesSince each independent variation is arbitrary, the Euler-Lagrange equations areIf nonconservative generalized forces are present, their components appear on the right. The formalism is invariant under suitable coordinate changes and adding a total time derivative to $L$ leaves the fixed-endpoint equations unchanged.

The generalized momentum is $p_i=\partial L/\partial {\dot{q}}_i$. If $q_i$ is cyclic, $\partial L/\partial q_i=0$ and $p_i$ is conserved. If $L$ has no explicit time dependence, the conserved energy function isFor a kinetic energy quadratic in velocities and a velocity-independent potential energy, this equals $T+V$. This connects symmetry to conservation laws.

For example, a simple pendulum of mass $m$ and length $\ell$ hasIts tension does not appear because the fixed-length constraint is already imposed. Expanding for small $\theta$ gives simple harmonic motion with $\omega =\sqrt{g/\ell }$. More generally, expansion of $T$ and $V$ to quadratic order about a stable equilibrium leads directly to the normal-mode eigenvalue problem.

The dynamical method. In Lagrangian mechanics, choose independent generalized coordinates $q_j$ consistent with the constraints, form the Lagrangian $\mathcal{L}=T-V$, and solve the Euler-Lagrange equations,This form applies to a Newtonian system with ideal constraints, so constraint forces do no allowed virtual work, and with the remaining forces derivable from the stated potential. Smooth holonomic constraints can be built into the coordinates. Additional nonconservative generalized forces would appear on the right. Prescribed shaft rotation is a driven constraint; its constant $\Omega$ is a parameter, not a conserved angular momentum.

Energy of the two balls. The diagram labels each ball $m/2$, so their combined mass is $m$. It also shows $\theta$ between an arm and the downward vertical, rather than the full angle between the arms. Each ball has horizontal radius $l\sin \theta$, vertical coordinate $-l\cos \theta$, and squared speed $l^2{\dot{\theta }}^2+l^2{\Omega }^2{\sin }^2\theta$. ThusandThe massless connecting rods and sliding collar enforce the geometry but contribute no energy.

Effective potential. Define ${\omega }_0^2=g/l$. At fixed $\Omega$,This effective potential determines the reduced $\theta$ dynamics. Its conserved effective energy is not the full lab-frame mechanical energy: the shaft drive can exchange work and angular momentum with the balls.

The change of stability. Near $\theta =0$,For $|\Omega |<{\omega }_0$, the downward state is a minimum. At $|\Omega |={\omega }_0$, its quadratic stiffness vanishes but the quartic coefficient is $+{\omega }_0^2/8$. Above that speed, the downward state becomes a maximum and stable nonzero openings appear. This is the gravity-restored centrifugal governor bifurcation.

The physical opening is nonnegative; the two minima in the signed extension represent opposite transverse displacements. The stable opening increases continuously from zero toward a right angle as $|\Omega |$ grows. The apparatus is frictionless in the model, so a displaced system oscillates rather than automatically settling at the equilibrium shown by the minima.

Equilibria. Differentiation givesHence the downward equilibrium is $\theta =0$, and the open equilibria satisfyThe signed extension includes the negative of this angle. Formally $\theta =\pi$ is also stationary, but it is an unstable inverted configuration and lies outside the usual opening range of the mechanism.

Small oscillations. Write $\theta ={\theta }_{\mathrm{eq}}+\eta$. The linear equation is $\ddot{\eta }+{\nu }^2\eta =0$, whereAt the downward state and at a nonzero open state, respectively,These are angular frequencies in ${\mathrm{s}}^{-1}$; ordinary frequencies are $\nu /(2\pi )$. Negative ${\nu }^2$ indicates exponential instability rather than an oscillation. At $\theta =\pi$, ${\nu }^2=-{\omega }_0^2-{\Omega }^2<0$. At the critical speed, the linear frequency is zero and the positive quartic potential gives amplitude-dependent nonlinear restoring motion.

Lagrangian mechanics expresses dynamics through generalized coordinates $q_i$ that respect the constraints. For a system with conservative, velocity-independent forces and standard kinetic energy, the Lagrangian is $L=T-V$. Its equations follow from the principle of stationary action:for variations that vanish at the endpoints. Varying the integral and integrating the $\delta {\dot{q}}_i$ term by parts givesIndependent variations therefore yield the Euler-Lagrange equations:Additional nonconservative generalized forces appear on the right-hand side as $Q_i$. The action is stationary, not necessarily a minimum. For ideal holonomic constraints, using suitable generalized coordinates removes the need to solve explicitly for constraint forces.

The conjugate momentum is $p_i=\partial L/\partial {\dot{q}}_i$. If a coordinate is cyclic, $\partial L/\partial q_i=0$ and its conjugate momentum is conserved. If $L$ has no explicit time dependence, the energy functionis conserved. It equals $T+V$ for the usual quadratic kinetic energy with time-independent geometry, but that identification is not universal for every Lagrangian or constraint.

For example, a simple pendulum with angle $\theta$, length $\ell$ and bob mass $m$ hasThe tension does no virtual work in the allowed angular displacement and never needs to be calculated to obtain the angular equation. The method scales naturally to coupled systems and coordinates with complicated geometry.

Autonomy gives conservation of energy, while the cyclic coordinate give conserved momenta:Here $p_{\psi }$ is the angular momentum along the body's symmetry axis, and $p_{\varphi }$ is its component along the fixed vertical. The time symmetry corresponds to $E$, not to an additional angle momentum.

With $u=\cos \theta$ and $-1<u<1$, solveInsert this and the conserved spin rate $n$ into the energy equation, multiply by $A$, and isolate $\dot{\theta }$:

For a fixed nonvertical angle, letThe energy is $A{\dot{\theta }}^2/2+V_{\mathrm{eff}}(\cos \theta )$. A constant angle in $(0,\pi )$ requires both $E=V_{\mathrm{eff}}(u_0)$ and $V_{\mathrm{eff}}^{\prime }(u_0)=0$, by the reduced Euler-Lagrange equation and $\sin {\theta }_0\ne 0$. Sinceboth $P(u_0)=0$ and $P^{\prime }(u_0)=0$, with $u_0\in (-1,1)$.

The source's unqualified final sentence omits the nonvertical restriction. A vertical axis has $u=\pm 1$, where the displayed reduction divides by zero and the common root need not be interior. For example, the stationary upright top has $n=h=0$, $E=mga$, andIts only repeated root is $u=1$, so it has no common root with its derivative in $(-1,1)$. Thus the interior double-root conclusion holds for nonvertical steady precession. The vertical configurations require separate endpoint treatment. This is nonvertical qualification in the steady-top double-root condition.

With fixed endpoint values, the Euler-Lagrange equation is $F_y-\frac{d}{dx}{F}_{y^{\prime }}=0$. If $F$ has no explicit $x$ dependence, differentiation givesThus the Beltrami identity is $F-y^{\prime }{F}_{y^{\prime }}=\text{constant}$.

For this integrand, $F_y=2y$ and $F_{y^{\prime }}=1+2y^{\prime }$, so $y^{\prime \prime }=y$. The extremals arewhere the two fixed endpoint values determine $A,B$. The Beltrami constant is $y^2-y^{\prime 2}$, consistent with this solution.

For a nonzero smooth perturbation $\eta$ vanishing at the endpoints, expand the quadratic functional. The first variation vanishes at an extremal, whileThe linear term $\int y^{\prime }$ depends only on the fixed endpoints. Consequently the extremal is a strict minimum and cannot be a maximum.

If $L$ is independent of $q_1$, the Euler-Lagrange equation makes the generalized momentum $p_1=\partial L/\partial {\dot{q}}_1$ constant. If $L$ is independent of time, the energy function $E={\sum }_a{\dot{q}}_a(\partial L/\partial {\dot{q}}_a)-L$ is constant; differentiating it gives $\dot{E}=-L_t$ along solutions.

Set $A(r)=r^2/(r-2m{)}^2$ and $B(r)=r^3/(r-2m)$. Their derivatives are $A^{\prime }=-4mr/(r-2m{)}^3$ and $B^{\prime }=2r^2(r-3m)/(r-2m{)}^2$. The radial and angular equations areEquivalently,Two conserved quantities areThe second equals $L$ here because the Lagrangian is homogeneous quadratic in velocities. For a constant-radius orbit with nonzero angular motion, the radial equation requires $r-3m=0$. ThusIt lies outside $2m$ and is the circular photon orbit. If both velocities were zero, the differential equations would also admit static constant configurations at any radius; these are degenerate zero-energy paths rather than the intended moving photon trajectory.

Independent endpoint-fixed variations of the two coordinates give the Euler-Lagrange equationsDifferentiating the sphere parametrization gives the Riemannian metric $d{s}^2=a^2(d{\theta }^2+{\sin }^2\theta d{\varphi }^2)$. Its energy functional is the stated $I$. Its stationary paths are geodesics with affine, constant-speed parametrization; conversely a geodesic can be so parametrized. This explains the use of energy rather than arc length.

For $F=a^2({\dot{\theta }}^2+{\sin }^2\theta {\dot{\varphi }}^2)/2$, the equations areThe second gives the conserved generalized momentum ${\sin }^2\theta \dot{\varphi }=h$. Since $F$ has no explicit time dependence, its Beltrami identity gives constant $\dot{\theta }{F}_{\dot{\theta }}+\dot{\varphi }{F}_{\dot{\varphi }}-F=F$. Dividing by $a^2/2$ gives the second first integral,Thus $E$ is the squared speed divided by $a^2$, while $h$ records the rotational symmetry about the polar axis. The coordinate equations apply away from the polar coordinate singularities.

Let $P=F_{y^{\prime }}$ and $Q=F_{y^{\prime \prime }}$. For a smooth variation $y+\epsilon \eta$, the first variation isHere the last expression follows by two integrations by parts. Taking variations supported in the interior and applying the fundamental lemma of the calculus of variations gives the Euler-Lagrange equation for a second-derivative functional:If both endpoint values and slopes are free, their independent variations force the natural boundary conditions for second-derivative functionalsIf an endpoint value or slope is fixed instead, its corresponding variation vanishes and the associated natural condition is not required.

For $H=y^{\prime \prime }Q-y^{\prime }(Q^{\prime }-P)-F$, differentiate using the chain rule:The Euler-Lagrange equation cancels the first term. Consequently $H^{\prime }=-F_x$, and if there is no explicit $x$ dependence,This is a Beltrami identity for a second-derivative functional.

For $F(y,y^{\prime })=y\sqrt{1+y^{\prime 2}}$, the Euler-Lagrange equation isSince $F$ has no explicit $x$ dependence, the Beltrami identity givesSolving this first integral, or substituting into the second-order equation, yieldsThe constants satisfy $y_1=C\cosh ((x_1-K)/C)$ and $y_2=C\cosh ((x_2-K)/C)$. There need not be a solution for arbitrary endpoint data, and when there is one these equations can have more than one pair of constants. These are the possible stationary catenoids, so the axisymmetric second variation of a catenoid still matters when deciding whether they minimize surface area.

Write $q=(x,X{)}^{\mathsf{T}}$, $M=\left(\begin{array}{cc}a& b\\ b& c\end{array}\right)$ and $K=\left(\begin{array}{cc}\alpha & \beta \\ \beta & \gamma \end{array}\right)$. The Euler-Lagrange equations give $M\ddot{q}+Kq=0$. A normal mode is a motion $q=v\cos (\omega t+\phi )$ with fixed nonzero vector $v$; its normal frequency is $\omega >0$. Substitution gives $(K-{\omega }^{2}M)v=0$. Thus the squared normal frequencies are the assumed positive distinct roots ofFor either root, solve the associated two linear equations for the mode ratio.

For the rods, the small angles are $\theta =x/a$ and $\varphi =(X-x)/(2a)$. The first rod has kinetic energy $m{\dot{x}}^2/6$. The second rod's center has horizontal velocity $(\dot{x}+\dot{X})/2$ and its angular velocity is $(\dot{X}-\dot{x})/(2a)$. Its center-of-mass moment of inertia is $2m{a}^2/3$, givingAdding these kinetic energies gives $T=m(3{\dot{x}}^2+2\dot{x}\dot{X}+2{\dot{X}}^2)/6$. The first center rises by $a{\theta }^2/4$, and the second by $a{\theta }^2/2+a{\varphi }^2/2$. Multiplying by the respective masses and $g$ givesTherefore the stated Lagrangian is $T-V$.

Set $\mu =2a{\omega }^2/(3g)$. The mode equation isHenceThe mode ratio is $r=X/x=(6-3\mu )/(1+\mu )$, so the ratio of the rods' angles is $\varphi /\theta =(r-1)/2=(5-4\mu )/[2(1+\mu )]$. It is positive for ${\mu }_{-}=(17-3\sqrt{21})/10$ and negative for ${\mu }_{+}=(17+3\sqrt{21})/10$. Thus the lower-frequency normal mode has the rods in phase, and the higher-frequency normal mode has them out of phase.

For a normal constrained extremal, introduce a constant Lagrange multiplier and use $L=F-\lambda G$. The Euler-Lagrange equation isWhen both integrands have no explicit $x$ dependence, the Beltrami identity gives $L-y^{\prime }{L}_{y^{\prime }}=\text{constant}$, as follows by differentiating this expression and using the Euler-Lagrange equation.

Parametrize the surface by $(f(z)\cos \theta ,f(z)\sin \theta ,z)$. The cross product of its two coordinate tangents has magnitude $f\sqrt{1+(f^{\prime }{)}^2}$, so integration in $\theta$ gives $A=2\pi {\int }_a^{b}f\sqrt{1+(f^{\prime }{)}^2}dz$. Integrating the cross-sectional disk areas gives $V=\pi {\int }_a^b{f}^{2}dz$.

Dividing the augmented functional $A-\lambda V$ by $\pi$, useThe Euler-Lagrange equation isThe Beltrami identity becomes $2f/\sqrt{1+(f^{\prime }{)}^2}-\lambda f^2=C$. At an endpoint $f\to 0$, the absolute value of the first term is at most $2f$, so $C=0$, even if the profile has an infinite endpoint slope. For a nondegenerate regular closed surface take $f>0$ on $(a,b)$. Then $\lambda >0$, and, with $R=2/\lambda$,Combining this identity with the Euler-Lagrange equation yields $f^{\prime \prime }=-R^2/f^3$. Consequently $(f^2{)}^{\prime \prime }=2((f^{\prime }{)}^2+f{f}^{\prime \prime })=-2$. The two zero endpoint values uniquely determineThe positive radius profile is therefore a semicircle and generates a sphere, centred at height $(a+b)/2$, with radius $R=(b-a)/2$. In particularThe graph is smooth in the interior; its vertical slopes at the two poles are compatible with a smooth spherical surface. Its volume is $4\pi R^3/3$, the volume associated with this stationary candidate.

For minimality, the three-dimensional isoperimetric inequality says that every closed surface enclosing volume $V$ has $A^3\ge 36\pi V^2$, with equality for a sphere. The displayed extremal attains equality, proving that it minimizes area at its fixed enclosed volume. The uniqueness above concerns the regular connected positive profile with the two stated endpoint heights.

For symmetric quadratic coefficient matrices, the Euler-Lagrange equations give $T\ddot{q}+Vq=0$, where $T$ is the positive-definite mass matrix. Substitute a normal-mode oscillation $q=v\cos (\omega t+\varphi )$ to getThe positive roots ${\omega }^2$ determine the squared normal frequencies and their nonzero null vectors give mode shapes. Repeated eigenvalues give multidimensional mode spaces.

The particle's kinetic energy is $m|\dot{q}{|}^2/2$. For a taut elastic string at length $r=\sqrt{x^2+y^2+z^2}$, Hooke's law gives tension $\lambda (r-a)/a$ and elastic energy $\lambda (r-a{)}^2/(2a)$. With $\lambda =mg$ and gravitational potential $mgz$, this gives exactlyThis is the taut-string regime containing the equilibrium. Differentiating the potential givesAt $(0,0,-2a)$, $r=2a$ and all three accelerations vanish.

With $z=-2a+\zeta$, expansion to quadratic order gives $r=2a-\zeta +(x^2+y^2)/(4a)+O(3)$. Substitution cancels the linear terms, leavingThe mass matrix remains $mI$, and the stiffness matrix is $\mathrm{diag}(mg/(2a),mg/(2a),mg/a)$. ThereforeThe horizontal frequency is doubly degenerate: a linearly polarized oscillation in any horizontal direction is a normal mode. The remaining mode is vertical stretching and contraction of the string, with horizontal coordinates zero.

Treat $\varphi$ and ${\varphi }^{\ast }$ as independent variables in the Euler-Lagrange equations. Write $s={\varphi }^{\ast }\varphi$, and assume the scalar potential $f(s)$ is real and differentiable. Variation with respect to ${\varphi }^{\ast }$ givesConsequently the equations for the complex scalar field and its complex conjugate areNo polynomial assumption is needed: only the first derivative of the scalar potential enters. For $f(s)=m^{2}s$, both reduce to the free Klein-Gordon equation.

For a constant real parameter, ${\partial }_{\mu }(e^{i\alpha }\varphi )=e^{i\alpha }{\partial }_{\mu }\varphi$. The phases therefore cancel separately in the kinetic term and in ${\varphi }^{\ast }\varphi$, leaving any $f({\varphi }^{\ast }\varphi )$ unchanged. This is the global phase symmetry of a complex scalar field, with circle group $U(1)$.

Use infinitesimal variations per unit parameter, $\delta \varphi =i\varphi$, $\delta {\varphi }^{\ast }=-i{\varphi }^{\ast }$. The Lagrangian density is exactly invariant, so no total-derivative correction is needed in Noether's theorem. The Noether current isIt depends only on derivatives of the Lagrangian density with respect to field derivatives, so the derivative-free scalar potential cannot alter its form. Direct differentiation cancels the mixed first-derivative terms:Thus it is conserved on solutions of the Euler-Lagrange equations. Its spatial integral is a conserved Noether charge when the flux through the boundary vanishes. Its overall sign is fixed here by $\delta \varphi =+i\varphi$; reversing the orientation of the phase transformation reverses the current.

Take the free squared mass $m^2\ge 0$, as for the ordinary massive free theory. The quartic part of the scalar potential can be writtenIt is nonnegative for every real field value exactly whenNecessity follows from the axis ${\varphi }_2=0$ and the diagonal ${\varphi }_1={\varphi }_2$: a negative coefficient on either ray makes $V$ tend to minus infinity under rescaling. The displayed sum proves sufficiency, since $m^2({\varphi }_1^2+{\varphi }_2^2)/2$ is also nonnegative. This is a copositive matrix condition on the quadratic form in the nonnegative variables $({\varphi }_1^2,{\varphi }_2^2)$, rather than positive definiteness on all independent signed variables.

The strict inequalities $\lambda >0$, $\kappa >-\lambda$ make the quartic part grow in every nonzero direction. They are not necessary for a stable vacuum when $m^2>0$. For example, $\lambda =\kappa =0$ gives two stable free massive fields; $\lambda >0$, $\kappa =-\lambda$ has a flat quartic diagonal but a confining quadratic term. If $m^2=0$, boundary cases have flat vacuum directions and need not have a unique isolated minimum. If one instead allows negative squared mass, a flat quartic ray makes the full potential unbounded below; the stability conditions must then be strengthened. This distinguishes boundedness from strict quartic coercivity.

Under the generator of a, the variation isIt vanishes identically if and only if $\kappa =\lambda$. In that case $V_4=\lambda ({\varphi }_1^2+{\varphi }_2^2{)}^2$ preserves the entire global U(1) symmetry. More explicitly, the interacting Euler-Lagrange equations giveHenceA special solution lying on an axis does not establish an operator conservation law. Among the stable couplings, the symmetry-preserving line is $\kappa =\lambda \ge 0$.

For nonzero real $g$, the continuous internal symmetry of a classical field theory is the global U(1) symmetryThe kinetic terms and mass terms are invariant separately, while the phases in ${\varphi }^2{\Phi }^{†}$ cancel. Independent phase rotations would instead require $2{\alpha }_{\varphi }-{\alpha }_{\Phi }=0$ modulo $2\pi$, leaving one connected circle. Simultaneous complex conjugation is also a discrete symmetry, reversing the continuous charge; including it gives $U(1)⋊{\mathbb{Z}}_2\cong O(2)$ for generic parameters. At $g=0$ the continuous symmetry enlarges to $U(1)\times U(1)$.

The Noether current with particles assigned positive charges isFor example, put $A={\varphi }^2{\Phi }^{†}$ and $B={\varphi }^{†2}\Phi$. The Euler-Lagrange equations give ${\partial }_{\mu }{j}_{\varphi }^{\mu }=ig(A-B)$ and ${\partial }_{\mu }{j}_{\Phi }^{\mu }=ig(B-A)$, with the factor $2$ already included in $j_{\Phi }^{\mu }$. Their sum vanishes. The Noether charge $Q=\int d^{3}x{j}^0$ is conserved and, after normal ordering, countsThus a $\Phi$ particle carries two units of the same conserved charge carried by a $\varphi$ particle; their antiparticles carry the negatives. The interaction permits conversion of one $\Phi$ particle into two $\varphi$ particles when allowed by conservation of energy, while preserving this weighted number.

The geodesics of the product-weighted diagonal plane metric follow from the geodesic Lagrangian $L=(y^2{\dot{x}}^2+x^2{\dot{y}}^2)/2$. With dots denoting an affine parameter derivative, its Euler-Lagrange equations giveSubstituting $y=mx$ with constant nonzero $m$ reduces both equations to $x\ddot{x}+{\dot{x}}^2=0$. Thus $x\dot{x}$ is constant, and a suitable affine parametrization exists; explicitly, $x^2=C_1\lambda +C_2$ on a segment of fixed sign.

These straight rays are geodesics only in the regular domain $xy\ne 0$. The metric tensor determinant is $x^2{y}^2$. On the axes, including the origin, this is a degenerate metric tensor: the calculation does not establish a regular geodesic continuing through the origin. The slope-zero ray is outside the regular domain. Geometrically, on each quadrant the change $\rho =|xy|/\sqrt{2}$, $\vartheta =\log |y/x|$ gives $d{s}^2=d{\rho }^2+{\rho }^{2}d{\vartheta }^2$. A fixed slope has constant $\vartheta$ and is therefore locally a radial geodesic of a flat metric tensor.