Substitution of the proposed sine and cosine displacements into either linearized equation gives $m{\omega }^2+qB\omega -k=0$. ThusThese are signed angular frequencies, not two positive frequency magnitudes. To first order, with $\epsilon \ll {\rho }_0$,The orbit is therefore a circle of radius $\epsilon$ about $({\rho }_0,0)$ at this order. The positive root runs counterclockwise; the negative root runs clockwise and has the larger magnitude. A positive charge in $B\hat{\mathbf{z}}$ alone circles clockwise. Here the spring permits an additional counterclockwise mode by providing inward force against its magnetic force.

Direct variation of the Lagrangian givesFor a static configuration, $k\rho =qE\cos \theta$ and $qE\rho \sin \theta =0$. Since $qE>0$ and the radius is positive,The magnetic force vanishes at rest and does not shift this equilibrium. In canonical variables it has $p_{\rho }=0$ but $p_{\theta }=qB{\rho }_0^2/2$, rather than zero canonical angular momentum.

The canonical momenta areTaking the Legendre transform givesIts four Hamilton equations, in a polar chart with $\rho >0$, areCanonical angular momentum includes the vector-potential contribution and is not simply mechanical angular momentum.

Choose electric scalar potential $\Phi =-Ex$ and the symmetric gauge $\mathbf{A}=B(-y,x,0)/2$, whose curl is $B\hat{\mathbf{z}}$. Minimal coupling gives $L=m{\dot{\mathbf{r}}}^2/2-k{\mathbf{r}}^2/2+q\mathbf{A}\cdot \dot{\mathbf{r}}-q\Phi$. Since $x=\rho \cos \theta$ and $x\dot{y}-y\dot{x}={\rho }^2\dot{\theta }$,The magnetic term is velocity dependent; it is not an ordinary position-only potential.

The new kinetic matrix is diagonal, givingFor the first mode, $\varphi =0$: the line between the centres stays vertical, both centres translate together, and $\dot{\psi }=2\dot{\theta }$. For the second, $\theta =0$: the outer hoop stays still while the inner hoop rolls within it, with $\dot{\psi }=-\dot{\varphi }$. These coordinate motions are independent only at the retained quadratic order.

Let $\psi$ be the small hoop's absolute spin angle with the same angular sign convention. Equating the two contact-point velocities cancels the large hoop centre's translation. The remaining tangential equality isThis rolling constraint inside a moving circular track adds small-hoop rotational energy $M{a}^2(2\dot{\theta }-\dot{\varphi }{)}^2/2$. At quadratic order its cross term cancels the translational cross term in part a. The potential is unchanged, soStatic friction enforces rolling; ideal no-slip contact dissipates no energy.

The Euler-Lagrange equations from the quadratic Lagrangian areFor a normal mode proportional to $e^{i\omega t}$, let $\lambda ={\omega }^{2}a/g$. Nonzero amplitudes requireHenceThe corresponding amplitude ratios are $\Phi /\Theta =2\lambda /(1-\lambda )$. The lower-frequency mode has $\Phi /\Theta =2(\sqrt{2}-1)$, with the angles in phase; the higher-frequency mode has $\Phi /\Theta =-2(1+\sqrt{2})$, with opposite signs. The small hoop's spin remains zero in both modes.

Measure both angles from the downward vertical, as in the source diagram. With upward $y$, the cylinder centres areThe large hoop has central inertia $4M{a}^2$ and angular velocity $\dot{\theta }$, while the small hoop has no spin in this part. Translation and rotation therefore giveThus $L=T-V$. Counting each angle and velocity as first order, replacing the cosine in the velocity product by one changes only fourth-order terms. Expanding the potential and dropping its constant gives

Assume $b>0$ and temperature-independent $d$ for the local model. The zero-field stationary points obeyFor $d\ne 0$, the favoured ordered sign is opposite to $d$. The nonzero extrema first appear at $a=d^2/(4b)$; their existence is not yet a coexistence condition. To find the equilibrium transition, require both $F^{\prime }(m_{\ast })=0$ and $F(m_{\ast })=F(0)$. Substituting $a=-d{m}_{\ast }-b{m}_{\ast }^2$ into the free energy givesHenceAt coexistence the entire potential factorizes:It has equal minima at zero and $m_{\ast }$ separated by a barrier. Every fixed nonzero cubic coefficient produces a first-order transition in this quartic model, with a finite magnetization jump before the quadratic coefficient reaches zero.

For $d^2/(4b)>a>a_{\ast }$, the ordered minimum is metastable and zero is globally stable. For $a_{\ast }>a>0$, the ordered state is globally stable and zero is metastable. At $a=0$ the zero minimum loses local stability. Thus the spinodals are $a=d^2/(4b)$ and $a=0$, distinct from $a_{\ast }$. For $a<0$, the opposite-sign nonzero state can also be a local minimum, but the cubic term favours the state of sign $-d$.

If $a=a_0(T-T_c)/T_c$, the transition is shifted to $T_{\ast }=T_c[1+2d^2/(9b{a}_0)]$. With $b,d$ constant and $a^{\prime }(T)>0$, the entropy jump from the disordered to ordered phase is $-a^{\prime }(T_{\ast })m_{\ast }^2/2$, giving a latent heat of magnitude $T_{\ast }{a}^{\prime }(T_{\ast })m_{\ast }^2/2$. As $d\to 0$, the jump and shift vanish and the original continuous transition is recovered.

A genuine zero-field uniaxial ferromagnet with exact spin-reversal Z2 symmetry forbids this cubic term. The modified model therefore presumes that symmetry has been changed; it must not be presented as a symmetry-allowed correction for the original ferromagnet.

Differentiate the equation of state at fixed temperature:Above the transition, zero-field equilibrium has $m=0$. Below it the stable pure phases have $m^2=-a/b$. ThusThis is the mean-field magnetic-susceptibility amplitude ratio. Below the transition, take a one-sided response in a selected ordered phase: the globally minimizing magnetization switches discontinuously between opposite phases when $h$ crosses zero. Its two-sided derivative at precisely zero field is therefore not the finite pure-phase magnetic susceptibility used in the ratio.

Equilibrium requires $h=am+b{m}^3$. On the critical isotherm $a=0$, soHere $\delta$ is defined by $h\propto \mathrm{sgn}(m)|m{|}^{\delta }$ at the critical temperature, or equivalently $|m|\propto |h{|}^{1/\delta }$. The positive quartic coefficient selects a stable equilibrium, including the flat but stable zero-field thermodynamic critical point.

The expansion is a local phenomenological description in a small scalar order parameter near a transition. Its coefficients summarize microscopic physics and generally depend smoothly on temperature. At zero field, spin-reversal Z2 symmetry requires even powers of magnetization; the conjugate field supplies the term $-hm$. The additive $F_0$ does not affect minimization with respect to $m$.

For the ordinary continuous-transition quartic model, $b>0$ ensures stability, and the quadratic coefficient changes sign, $a=a_{0}t+O(t^2)$ with $a_0>0$. If $b$ vanishes or is negative, stabilizing higher powers are required and the phase structure can change. Treating one uniform magnetization neglects spatial gradients, fluctuations and domains; the resulting exponents are predictions of mean-field theory.

Use the amplitude placement in the original PDF: $w=[A{\cosh }^2(Bx){]}^{1/(1-n)}$. Introduce the actual peak $C=A^{-1/(n-1)}$ and $p=2/(n-1)$, so $w=C{\mathrm{sech}}^p(Bx)$. Its derivatives satisfyThe static equation is $w^{\prime \prime }=aw-b{w}^n$. Matching the constant and ${\mathrm{sech}}^2$ terms givesChanging the sign of $B$ leaves the profile unchanged. If $A$ is instead defined as an outside prefactor, as in the held TeX transcription, its value is $C$, not the PDF's $A$. This notation discrepancy does not change the physical profile.

The nonconstant positive power-law nonlinear Klein-Gordon pulse is even, smooth, has peak $C$ at zero and decays as $e^{-\sqrt{a}|x|}$. The requested sketches below compare several powers with illustrative $a=b=1$. For odd $n$, its negative is also a solution. A separate constant branch occurs for $B=0$: the positive equilibrium has $C=(a/b{)}^{1/(n-1)}$, or $A=b/a$ in the PDF convention. The zero solution is another equilibrium, obtained as a limiting profile rather than finite positive $A$.

Now apply the Lorentz boost with rapidity $-\beta$ to the static field. It givesFor positive $\beta$, the centre moves to the right at $0<v<1$, with speed one the wave-speed unit. The peak stays $C$ and the width contracts by $1/\cosh \beta$. The shape translates without changing in time. This establishes a travelling solitary pulse; it does not by itself prove stability or elastic multi-pulse scattering.

The Euler-Lagrange field equation is $F^{\prime }(\varphi )-{\partial }_t^2\varphi +{\partial }_x^2\varphi =0$. ThereforePut $c=\cosh \beta$, $s=\sinh \beta$, $X=cx+st$ and $T=ct+sx$. The chain rule gives ${\partial }_t=s{\partial }_X+c{\partial }_T$ and ${\partial }_x=c{\partial }_X+s{\partial }_T$. Cross terms cancel and $c^2-s^2=1$, soThus the boosted scalar field is another solution for any constant rapidity. The nonlinear term in the nonlinear Klein-Gordon equation is unchanged because it is a local function of the scalar itself.

A locally invertible time-independent canonical transformation is canonical whenThe proposed coordinates are locally independent only for $q_1\ne 0$, since their configuration-space Jacobian has determinant $2q_1$. Work on one sign branch of $q_1$.

The mixed brackets first requireThe last bracket condition isExpressing $A,B$ in the $Q$ coordinates turns this into ${\partial }_{Q_2}A={\partial }_{Q_1}B$. Thus their shift one-form is closed and locally is the differential of an arbitrary smooth function $S(Q_1,Q_2)$. The canonical momenta for a squared coordinate therefore have the general local formConversely, any such $S$ satisfies all bracket conditions. This does not give one nonsingular transformation through $q_1=0$, or identify the two sign branches globally.

Choose $S=Q_2^3/3$. Then

In canonical coordinates, the Poisson bracket isIn particular, $\{q_j,g\}={\partial }_{p_j}g$, $\{p_j,g\}=-{\partial }_{q_j}g$, $\{f,q_j\}=-{\partial }_{p_j}f$ and $\{f,p_j\}={\partial }_{q_j}f$. Hence $\{q_i,p_j\}={\delta }_{ij}$ and the coordinate–coordinate and momentum–momentum brackets vanish.

The chain rule along a Hamiltonian trajectory givesThe Poisson bracket of time-dependent conserved quantities follows by defining $D_t={\partial }_t+\{\cdot ,H\}$. The ordinary product rule and Jacobi identity implyTo see the Hamiltonian part explicitly, Jacobi rearranges to $\{\{f,g\},H\}=\{\{f,H\},g\}+\{f,\{g,H\}\}$. Therefore $D_{t}f=D_{t}g=0$ implies $D_t\{f,g\}=0$, even when the conserved functions depend explicitly on time.

Differentiate $\mu \dot{\mathbf{r}}=\mathbf{p}+e^{\ast }\mathbf{B}\times \mathbf{r}$ and substitute the previous momentum equation:For a planar vector, $\mathbf{B}\times \dot{\mathbf{r}}=B(-\dot{y},\dot{x})$, which gives the two component equations with the printed signs.

For circular relative motion of a neutral Coulomb pair of separation radius $R$, $\dot{\mathbf{r}}=\omega \hat{\mathbf{z}}\times \mathbf{r}$ and $\ddot{\mathbf{r}}=-{\omega }^2\mathbf{r}$. ThusSince $(e^{\ast }{)}^2+\mu e^2/M=e^2/4$, the two signed frequencies areOne is positive and the other negative. At $B=0$ they give the two senses of an ordinary Coulomb circular orbit; equal masses remove the term linear in $B$ from their average. With negligible Coulomb attraction and $eB>0$, the branches approach $eB/m_2$ and $-eB/m_1$, the appropriate signed free cyclotron limits.

The actual particle paths follow by integrating the centre-of-mass velocity:Both particles move on circles about the same fixed guiding centre with a constant separation and common signed frequency, with a possible relative phase reversal encoded by a negative coefficient. Their mechanical centre of mass also generally circles. This is consistent with magnetic deflection plus a central mutual force; central interaction alone does not eliminate the external Lorentz forces.

Interpret $\mathbf{P}=0$ as the zero-pseudomomentum sector, rather than assuming the mechanical centre of mass is at rest. Define $\mathbf{u}=\mathbf{p}+e^{\ast }\mathbf{B}\times \mathbf{r}$. Hamilton's equations areFor example, differentiating the relative kinetic term with respect to $\mathbf{r}$ gives $-e^{\ast }\mathbf{B}\times \mathbf{u}/\mu$ because the cross-product matrix is antisymmetric. Hamilton's minus sign then gives the positive cross-product term in $\dot{\mathbf{p}}$. The Coulomb force is attractive. A change to a moving frame need not preserve a pure magnetic external field, so setting $\mathbf{P}=0$ is a restriction on the chosen solution family.

For this neutral Coulomb pair in a uniform magnetic field, the reduced Lagrangian gives the canonical momentaInvert these relations and take the Legendre transform. The velocity-linear terms cancel against their momentum contributions, givingThere is no explicit time dependence, so $dH/dt=0$. There is no dependence on $\mathbf{R}$, so $\dot{\mathbf{P}}=0$. The distinction between kinetic momentum and magnetic pseudomomentum matters: the conserved vector is $\mathbf{P}$, not generally $M\dot{\mathbf{R}}$. The mechanical centre-of-mass momentum $M\dot{\mathbf{R}}=\mathbf{P}+e\mathbf{B}\times \mathbf{r}$ generally varies as the separation changes. A neutral pair can therefore have accelerated centre-of-mass motion in a magnetic field.

Use the magnetic-coupling convention of the supplied Hamiltonian; electromagnetic unit factors are absorbed into the stated $e$ and $B$. Let $\alpha =m_1/M$ and $\beta =m_2/M$, soThe kinetic energy separates into $M{\dot{\mathbf{R}}}^2/2+\mu {\dot{\mathbf{r}}}^2/2$. The symmetric gauge coupling ${\sum }_i{e}_i{\dot{\mathbf{r}}}_i\cdot (\mathbf{B}\times {\mathbf{r}}_i)/2$ becomesThe scalar triple product identity givesDiscarding $-e/2$ times that total derivative, which changes endpoint terms and canonical momenta but not the trajectories, leaves