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The absence of magnetic monopoles is expressed by $\nabla \cdot \mathbf{B}=0$. In a regular simply connected region, a magnetic vector potential may be introduced so thatIts units are $\mathrm{T}\mathrm{m}=\mathrm{Wb}{\mathrm{m}}^{-1}$. It is not unique: adding a gradient leaves the magnetic field unchanged because $\nabla \times \nabla \chi =0$. In time-dependent electromagnetism,is a gauge transformation preserving both physical fields. Stokes theorem gives ${\oint }_C\mathbf{A}\cdot d\mathbf{\ell }={\int }_S\mathbf{B}\cdot d\mathbf{S}={\Phi }_B$ for a contour bounding an appropriate surface.

For localized steady currents, choose the Coulomb gauge $\nabla \cdot \mathbf{A}=0$. Substitution into Ampère's law and the curl-of-curl identity givesBoundary conditions are needed in addition to the gauge choice. For a uniform magnetic field one convenient choice is $\mathbf{A}=\frac{1}{2}\mathbf{B}\times \mathbf{r}$. For a magnetic dipole, $\mathbf{A}={\mu }_0\mathbf{m}\times \mathbf{r}/(4\pi r^3)$ outside its source. The magnetic vector potential simplifies field calculations and appears directly in the Lagrangian of a charged particle, $L=\frac{1}{2}m{v}^2+q\mathbf{v}\cdot \mathbf{A}-q\varphi$. In multiply connected regions, a field-free region can retain nontrivial circulation of $\mathbf{A}$, relevant to the Aharonov-Bohm effect; a local gauge argument cannot remove every global circulation.

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.

Definition and equations. A normal mode is a motion in which every component of a linear system has the same time dependence with fixed amplitude ratios and phases. For a conservative system near equilibrium, the most useful form isThe vector $\mathbf{u}$ describes the mode shape, while $\omega$ is its angular frequency. Linear combinations of modes describe arbitrary sufficiently small motions when the modes form a complete basis.

Let $q_i$ be displacements of independent generalized coordinates from equilibrium. Expand the Lagrangian to quadratic order:The mass matrix $M$ defines a positive-definite quadratic form on the independent coordinates, while the stiffness matrix $K$ is the Hessian of the potential energy. Substituting a mode gives the generalized eigenvalue problem for small oscillationsA practical calculation identifies the equilibrium, chooses independent coordinates, computes the quadratic kinetic and potential energies, solves for the eigenvalues and then obtains their eigenvectors. One can equivalently diagonalize the real symmetric matrix $M^{-1/2}K{M}^{-1/2}$. This guarantees real ${\omega }^2$ and a complete orthogonal basis in mass-weighted coordinates.

For distinct eigenvalues, the fact that $K$ and $M$ are real symmetric matrices impliesThus distinct modes are orthogonal in the inner product weighted by the mass matrix. Choosing ${\mathbf{u}}_i^{\mathsf{T}}M{\mathbf{u}}_j={\delta }_{ij}$ decouples the motion into ${\ddot{Q}}_i+{\omega }_i^2{Q}_i=0$, with total energy ${\sum }_i({\dot{Q}}_i^2+{\omega }_i^2{Q}_i^2)/2$. A positive-definite matrix $K$ gives positive eigenvalues and oscillatory stability. Zero modes can describe rigid translations or rotations without restoring force; a negative eigenvalue gives exponential instability rather than an oscillatory mode.

Symmetry. A symmetry preserving both quadratic energies maps modes into modes of the same frequency. Coordinates adapted to symmetry divide the eigenvalue problem into smaller independent blocks. For two identical subsystems, exchanging them gives symmetric and antisymmetric modes. The degeneracy of normal modes means that independent modes share one frequency. Higher-dimensional group representations of a symmetry can enforce such degeneracies; symmetry does not require every pair of modes to be degenerate, and accidental degeneracies can also occur. Within a degenerate eigenspace, orthogonal linear combinations are equally valid mode shapes.

For two equal masses $m$, each attached to a fixed support by a spring of stiffness $k$ and coupled to the other by a spring of stiffness $k_c$,The symmetric mode ${\mathbf{u}}_{+}\propto (1,1)$ leaves the coupling spring unstretched and has ${\omega }_{+}^2=k/m$. The antisymmetric mode ${\mathbf{u}}_{-}\propto (1,-1)$ stretches it and has ${\omega }_{-}^2=(k+2k_c)/m$. Superposing the two gives beats when their frequencies are close.

Continuous systems. The same idea replaces matrices by differential operators and mode vectors by functions satisfying boundary conditions. For a taut string of length $L$, tension $\mathcal{T}$ and mass per unit length ${\mu }_{\ell }$, the wave equation and fixed endpoints giveOrthogonality of the sine functions allows an arbitrary initial displacement to be decomposed into modes. Changing the endpoints changes the spectrum: boundary conditions are part of the physical mode problem.

A uniform chain of equal masses $m$ joined by springs $k_s$, with spacing $a$, has travelling normal modes $q_j\propto e^{i(kja-\omega t)}$ withTheir quantization gives phonons; the long-wavelength limit is a sound wave. Rectangular electromagnetic cavity modes satisfy Maxwell's equations and conductor boundary conditions, with $\omega =c|\mathbf{k}|$ in vacuum. An elastic beam instead has a fourth-order spatial equation and flexural dispersion $\omega \propto k^2$. These examples show how restoring forces and geometry determine the frequency spectrum.

Real systems. Small-amplitude motion is often approximately linear even when the full equations are nonlinear. Large amplitudes produce frequency shifts, mode coupling and transfer of energy between modes. Friction broadens resonances and makes free oscillations decay. A damping matrix proportional to $M$ or $K$ can retain the undamped mode shapes; general damping need not, and the appropriate eigenfrequencies and modes can be complex. External periodic forcing predominantly excites modes near resonance, with amplitude set by coupling to the forcing and dissipation. Imperfections split ideal degeneracies, while radiation or coupling to a surrounding fluid can remove energy from the chosen subsystem. Normal modes remain useful when linearization, the boundary conditions and the treatment of damping accurately describe the motion being measured.

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.

Inertia is a tensor. For a rigid body, distances between material points are fixed. Write each position relative to the centre of mass as $\mathbf{r}$, so its rotational velocity is $\mathbf{\omega }\times \mathbf{r}$. Integrating $\mathbf{r}\times (\mathbf{\omega }\times \mathbf{r})$ gives angular momentumThe rotational kinetic energy isThe inertia tensor is real and symmetric, so orthogonal principal axes diagonalize it. Angular momentum is parallel to angular velocity only along a principal axis, or within a subspace of equal principal moments. For translation plus rotation about the centre of mass, $T=M{v}_{\mathrm{CM}}^2/2+T_{\mathrm{rot}}$. Moving the reference to a parallel axis adds $M(d^{2}1-\mathbf{d}{\mathbf{d}}^{\mathsf{T}})$ to the tensor, the tensor form of the parallel-axis theorem.

Euler's equations. The external torque about the centre of mass obeys $\mathbf{\tau }=(d\mathbf{L}/dt{)}_{\mathrm{space}}$. In body axes, the transport rule givesThe principal moments are constant in that frame, henceThese are Euler's equations. With zero torque, spatial angular momentum and kinetic energy are conserved, although the body components generally vary. For distinct moments, small perturbations about rotation around the largest or smallest principal moment oscillate, whereas those about the intermediate moment grow. This is the intermediate-axis instability.

A lamina. Choose the lamina's plane as $z=0$. Then $I_{xz}=I_{yz}=0$, so the normal is a principal axis, andThis perpendicular-axis theorem applies to a planar mass distribution; it is not a general identity for a three-dimensional body. Rotate the in-plane axes to diagonalize the remaining symmetric two-by-two block. For a uniform rectangular lamina with sides $a,b$ and mass $M$, its central principal moments are $I_x=M{b}^2/12$, $I_y=M{a}^2/12$ and $I_z=M(a^2+b^2)/12$. If $a>b$, then $I_x<I_y<I_z$: torque-free rotation about the long in-plane axis or the normal is stable, while that about the short in-plane axis is unstable. Coincident moments in a square require the degenerate, rather than strict intermediate-axis, case. Under a constraint to rotate only about the normal, the motion reduces to the scalar equation ${\tau }_z=I_z{\dot{\omega }}_z$.

A symmetric top without torque. Let $I_1=I_2=I_{\perp }$ and $I_3=I_{\parallel }$. Euler's equations reduce toThus ${\omega }_1+i{\omega }_2$ rotates at signed body-frame frequency $\nu$, while the axial component stays fixed. The symmetry axis has constant angle to fixed spatial angular momentum. Indeed $\mathbf{L}=I_{\perp }\mathbf{\omega }+(I_{\parallel }-I_{\perp }){\omega }_3{\mathbf{e}}_3$, soIt precesses in space about $\mathbf{L}$ with angular speed $L/I_{\perp }$. This spatial precession and the body-frame frequency $\nu$ describe different motions and should not be identified.

A heavy top. If a symmetric top is pivoted a distance $\ell$ from its centre of mass, gravity gives torque ${\mathbf{r}}_{\mathrm{CM}}\times M\mathbf{g}$. Here the relevant moments are about the fixed pivot. The axial angular momentum and the vertical angular-momentum component are conserved, because gravity supplies neither an axial torque nor a torque about the vertical. Using tilt $\theta$, precession angle $\varphi$ and spin angle $\psi$, its Lagrangian isEliminating the two conserved cyclic momenta leaves one-dimensional motion in an effective tilt potential; bounded oscillation in tilt is nutation. Steady precession at $\dot{\varphi }=\Omega$ and fixed tilt satisfiesFor a rapidly spinning top on the slow branch, the first term is small andGravity then changes the direction of angular momentum rather than simply tipping the top straight down. The approximation fails near conditions where nutation or the fast-precession branch matters; the full equations retain those possibilities.

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.

The Lagrangian is $L={\sum }_i{m}_i|{\dot{q}}_i{|}^2/2-f(r)$. Away from a collision, its equations of motion areSumming forces proves conservation of total linear momentum. The total mechanical energy $E={\sum }_i{m}_i|{\dot{q}}_i{|}^2/2+f(r)$ is also conserved: its derivative is $F\cdot ({\dot{q}}_1-{\dot{q}}_2)+f^{\prime }(r)\dot{r}=0$, because $\dot{r}=(q_1-q_2)\cdot ({\dot{q}}_1-{\dot{q}}_2)/r$.

For the first proposed angular momentum,which need not vanish. Thus particle one's angular momentum about the fixed origin is not generally conserved. Its force is central relative to particle two, not necessarily relative to that origin.

For small oscillations about equilibrium, put $M_{ij}=T_{ij}(0)$ and $K_{ij}={\partial }_i{\partial }_{j}V(0)$. The potential has no linear term at equilibrium; to quadratic order the Lagrangian is $L_2={\dot{x}}^{\mathsf{T}}M\dot{x}/2-x^{\mathsf{T}}Kx/2$. Its equations are $M\ddot{x}+Kx=0$. A normal mode $x=v{e}^{i\omega t}$ satisfies the generalized eigenvalue problem $(K-{\omega }^{2}M)v=0$. For a positive-definite mass matrix, diagonalization $M^{-1/2}K{M}^{-1/2}$ gives the squared frequencies and independent normal coordinates. A strictly positive Hessian gives positive frequencies; zero eigenvalues require separate higher-order stability analysis.

Here the spring extensions are $x_1$, $x_2-x_1$, and $-x_2$. ThusThe stiffness matrix is $k\left(\begin{array}{cc}2& -1\\ -1& 2\end{array}\right)$ and $M=mI$. Its eigenvectors $(1,1)$ and $(1,-1)$ have stiffness eigenvalues $k$ and $3k$, givingIn the first mode the masses move together with equal displacements and the middle spring keeps its length. In the second they move oppositely with equal amplitudes, stretching and compressing the middle spring twice as much as either end spring.

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.

The lower mass has coordinates $(x+l\sin \theta ,-l\cos \theta )$. Its squared speed is ${\dot{x}}^2+2l\cos \theta \dot{x}\dot{\theta }+l^2{\dot{\theta }}^2$. Adding the bead's kinetic energy gives the kinetic part of the supplied Lagrangian. Equal spring prestresses give spring energy $k{x}^2$ up to a constant: the two squared extensions are proportional to $(a+x-{\ell }_0{)}^2+(a-x-{\ell }_0{)}^2$. The gravitational potential is $-m_{2}gl\cos \theta$. Thus $L=T-V$ has exactly the stated form, with constant energy offsets discarded.

Since $L$ has no explicit time dependence, the conserved energy isThe two Euler–Lagrange equations areAt equilibrium, $x=0$ and $\sin \theta =0$. There are two geometrically distinct positions, $\boxed{(x,\theta )=(0,0),(0,\pi )}$ modulo $2\pi$. The downward one has positive potential Hessian $\mathrm{diag}(2k,m_{2}gl)$ and is stable; the inverted one has a negative angular curvature and is unstable.

Keeping only quadratic terms about the downward equilibrium givesFor unit parameters, the mass and stiffness matrices are $M=\left(\begin{array}{cc}2& 1\\ 1& 1\end{array}\right)$ and $K=\left(\begin{array}{cc}2& 0\\ 0& 1\end{array}\right)$. The equation $\det (K-{\omega }^{2}M)=0$ is ${\omega }^4-4{\omega }^2+2=0$. Therefore the normal angular frequencies are

By reflection symmetry, $G$ is the midpoint of the cylinder axis. With position $r$ measured from $G$, define $J_{ij}^G=\int (|r{|}^2{\delta }_{ij}-r_i{r}_j)dm$. Symmetry makes the off-diagonal entries zero. Uniform density gives $⟨x^2⟩=⟨y^2⟩=a^2/4$ and $⟨z^2⟩=b^2/3$, by integrating over a disk and the interval $[-b,b]$. HenceFor parallel axes based at $P$, the tensor form of the parallel-axis theorem is $J^P=J^G+M(|d{|}^{2}I-d{d}^T)$, where $d=\overrightarrow{PG}$. This follows by substituting $r+d$ in the defining integral and using $\int rdm=0$. Here $d$ is an axial vector of length $b$, soThe Euler angles $\psi ,\varphi$ are cyclic coordinates. Their conserved conjugate momenta are $p_{\psi }=C(\dot{\psi }+\dot{\varphi }\cos \theta )=Cn$ and $p_{\varphi }=A\dot{\varphi }{\sin }^2\theta +Cn\cos \theta =h$. The Lagrangian has no explicit time dependence, so the conserved energy is kinetic plus gravitational potential, namely the expression given with constant $E$.

Eliminate the angular velocities using $\dot{\varphi }=(h-Cn\cos \theta )/(A{\sin }^2\theta )$ and $\dot{\psi }+\dot{\varphi }\cos \theta =n$. The energy equation becomesThusThe elimination is valid away from the coordinate poles; the multiplied equation extends by continuity to any pole reached by a regular motion.

Take a uniform nearest-neighbour harmonic chain with displacement $u_j$, conjugate momentum $p_j$, mass $m_0$, spring stiffness $\kappa$, lattice spacing $a$ and periodic length $\ell =Na$. These specify a standard model for the requested notes, not additional numerical exam data. Its Lagrangian and Hamiltonian areVariation gives $m_0{\ddot{u}}_j=\kappa (u_{j+1}-2u_j+u_{j-1})$. For a mode $e^{i(kja-\omega t)}$, periodicity requires $k=2\pi n/\ell$ andThe lattice makes $k$ periodic modulo $2\pi /a$, so choose one Brillouin zone. At long wavelength ${\omega }_k\simeq v|k|$, where $v=a\sqrt{\kappa /m_0}$ is the sound speed. The continuum limit of a harmonic chain expands $u_{j+1}-u_j\simeq a{\partial }_{x}u$ and replaces ${\sum }_j$ by $a^{-1}\int dx$. Defining mass density $\rho =m_0/a$ and one-dimensional elastic coefficient $K=\kappa a$ givesThe momentum density is $\Pi =\rho {\partial }_{t}u$, and the Hamiltonian density is ${\Pi }^2/(2\rho )+K({\partial }_{x}u{)}^2/2$. Translational invariance of $u$ makes total momentum conserved. The continuum approximation describes acoustic wavelengths large compared with $a$; it is not a justification for extending the lattice dispersion and zero-point sums to arbitrarily large wavevector.^[ref]

canonical quantization promotes the coordinates and momenta to operators with $[u_j,p_l]=i\hslash {\delta }_{jl}$ and the other equal-time commutators zero. In the continuum, $[u(x),\Pi (x^{\prime })]=i\hslash {\delta }_{\mathrm{per}}(x-x^{\prime })$. The nonzero normal modes can be represented by independent phonon operators obeying $[b_k,b_q^{†}]={\delta }_{kq}$. A useful lattice expansion iswhere $X=N^{-1}{\sum }_j{u}_j$ is the centre-of-mass displacement. The corresponding momentum isIncluding the centre-of-mass term restores the full site commutator: the nonzero-mode sum alone gives $i\hslash ({\delta }_{jl}-1/N)$. Substitution diagonalizes the Hamiltonian,A phonon is one quantum of a collective normal mode, carrying energy $\hslash {\omega }_k$ and lattice momentum $\hslash k$ modulo a reciprocal-lattice vector. Reality of $u_j$ relates its Fourier coordinates by $u_k^{†}=u_{-k}$; it does not remove the independent oppositely travelling phonon modes. Using one complete Brillouin zone avoids double-counting real standing-wave coordinates.^[ref]

The free centre of mass has no restoring force and must not be treated as an oscillator of zero frequency. Fixing it, weakly pinning it, or specifying a finite centre-of-mass volume defines the corresponding zero-mode factor. With a fixed centre of mass, the internal thermal partition function and occupations areThe internal energy is ${\sum }_{k\ne 0}\hslash {\omega }_k(⟨n_k⟩+1/2)$; its high-temperature heat capacity approaches $(N-1)k_B$. For a large one-dimensional chain at temperatures below the lattice cutoff but above its finite-size mode spacing,This one-dimensional acoustic phonon heat capacity shows that quantization turns classical normal waves into bosonic excitations with a linear low-temperature heat capacity in one dimension. The Euclidean field action $S_E=\frac{1}{2}{\int }_0^{\beta \hslash }d\tau dx[\rho ({\partial }_{\tau }u{)}^2+K({\partial }_{x}u{)}^2]$ is Gaussian, with periodic imaginary-time boundary conditions; its Fourier modes reproduce these oscillators. Absolute displacement correlations require the zero-mode convention, and the lattice regulates short-distance fluctuations.

Keep the electromagnetic units of the supplied Lagrangian, including its factor $c$, and regard $e$ as the signed coupling used there. Set $a_B=e{\Phi }_B/(2\pi c)$. The canonical momentum is $p_{\varphi }=I\dot{\varphi }+a_B$, so the Legendre transform givesSingle-valued periodic wavefunctions have normalized eigenstates $e^{in\varphi }/\sqrt{2\pi }$ with integer $n$. Defining dimensionless flux $f=e{\Phi }_B/(hc)$ gives $E_n={\hslash }^2(n-f{)}^2/(2I)$, and the trace is thereforeThe flux-threaded quantum rotor spectrum is periodic under $f\mapsto f+1$ after relabelling $n$, corresponding to flux period $hc/|e|$ in these units. A local gauge change can remove the magnetic vector potential from the differential operator but then imposes a twisted boundary condition with phase $e^{-2\pi if}$; it cannot remove the flux dependence from both the Hamiltonian and boundary condition simultaneously. In SI conventions the flux coupling has no $c$, so one must not mix the two unit systems.

Differentiate the Lagrangian with respect to the generalized velocities:The Legendre transform cancels the vector-potential energy term, giving $H=p\cdot \dot{x}-L=m|\dot{x}{|}^2/2+V$. Eliminating velocity with $m\dot{x}=p-eA$ therefore yieldsFor the given gauge, set ${\pi }_1=p_1+eB{x}_2/2$, ${\pi }_2=p_2-eB{x}_1/2$, ${\pi }_3=p_3$. Hamilton's equations give ${\dot{x}}_j={\pi }_j/m$, ${\dot{p}}_1=(eB/2){\dot{x}}_2$, ${\dot{p}}_2=-(eB/2){\dot{x}}_1$, ${\dot{p}}_3=0$. Hence the conserved quantities areThe two transverse combinations are components of magnetic pseudomomentum. The printed conserved combinations omit the electric charge factor $e$. With the stated vector potential and general electric charge they need the factors displayed above; as printed they hold only if $e=1$ or if electric charge is already absorbed into the field convention.