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The Euler-Lagrange equations in cylindrical coordinates areThe last two follow because $\varphi$ and $z$ are cyclic coordinates. They give the conserved quantitiesSince the Lagrangian is autonomous, its energy function is conserved as well:The velocity-linear term cancels from this energy, but it still changes the equations of motion. A dimensionless logarithm means $\log (R/R_0)$ for some fixed reference length; replacing the printed $\log R$ by that expression changes the Lagrangian only by a total time derivative and shifts the numerical value of $k$, not the trajectory.

It can describe the motion per unit mass of a charged particle in the azimuthal magnetic field of a long straight current-carrying wire. Choose magnetic vector potentialThen $\mathbf{B}=\nabla \times \mathbf{A}=(mC/(qR))\hat{\mathbf{\varphi }}$ and $(T+q\mathbf{A}\cdot \mathbf{v})/m$ is the given Lagrangian. In terms of a straight-wire current $I$, $B_{\varphi }={\mu }_{0}I/(2\pi R)$, so $C=q{\mu }_{0}I/(2\pi m)$ with signs fixed by charge and current direction.

A static magnetic force does no work, consistent with the conserved kinetic energy. Its radial component is $-q\dot{z}{B}_{\varphi }$ and its axial component is $q\dot{R}{B}_{\varphi }$, reproducing the radial and axial couplings above. The reference length in the logarithm is a gauge convention, not a new physical wire radius in the ideal exterior-field model.

Let $p_i=\partial L/\partial {\dot{q}}_i$ and consider the infinitesimal fixed-time transformation $\delta q_i=ϵK_i(\mathbf{q})$. It induces $\delta {\dot{q}}_i=ϵd{K}_i/dt$. Expanding the Lagrangian givesOn a solution of the Euler-Lagrange equations, $\partial L/\partial q_i={\dot{p}}_i$, so this equals $d({\sum }_i{p}_i{K}_i)/dt$. Exact invariance makes the left side zero, proving the fixed-time form of Noether's theorem:The symmetry direction need not coincide with one original coordinate axis; $\mathbf{K}$ can depend on the generalized coordinates. The conclusion is evaluated on physical trajectories, while the assumed invariance is a property of the Lagrangian under the transformation.

The canonical momentum for the stated symmetric-gauge Lagrangian isSince $\mathbf{B}$ is uniform and time independent, its Euler-Lagrange equation becomesThis is the magnetic Lorentz force, which is unchanged by constant spatial translations. The Lagrangian itself changes byA total time derivative changes the action only by $ϵ[F_{\mathbf{n}}(t_f)-F_{\mathbf{n}}(t_i)]$. For variations with fixed endpoints this boundary contribution has zero variation, so the equations of motion are unchanged. The translation is therefore a symmetry of the dynamics even though this particular gauge makes the Lagrangian only quasi-invariant.

For an infinitesimal continuous symmetry $\delta q_i=\epsilon {\xi }_i(q,t)$, allow the Lagrangian to change by a total derivative, $\delta L=\epsilon dF/dt$. Direct variation givesOn a solution of the Euler-Lagrange equations, the first term vanishes. This proves Noether's theorem in the fixed-time form:Spatial translations yield total momentum, and rotations yield angular momentum. Time translations yield energy when the Lagrangian has no explicit time dependence; this follows either from the extended time-transformation form of the theorem or by differentiating $H={\sum }_i{p}_i{\dot{q}}_i-L$ to obtain $\dot{H}=-\partial L/\partial t$.

The floor contributes gravitational potential $mgz=m{\omega }^2(x_1^2+y_1^2)/2$. Neglecting vertical velocity as specified, for the nonrotating cup orientation $\varphi =-\Omega$ the Lagrangian isFor general $\alpha =\Omega +\varphi$, replace the velocity components by ${\dot{x}}_1-\alpha y_1+R\Omega \sin \varphi t$ and ${\dot{y}}_1+\alpha x_1+R\Omega \cos \varphi t$. The potential remains the same because the paraboloid is axisymmetric.

A constant of motion has zero total derivative on every solution, and may explicitly depend on time. For an infinitesimal transformation $\delta q_i=ϵ{\xi }_i$ with $\delta L=ϵdF/dt$, the on-shell first variation is $\delta L=ϵd({\sum }_i{p}_i{\xi }_i)/dt$. Therefore $\boxed{I=\sum _i{p}_i{\xi }_i-F}$ is conserved. Including a time shift $\delta t=ϵ\tau$ gives the corresponding $I={\sum }_i{p}_i{\xi }_i-H\tau -F$. This proves Noether's theorem in its usual point-transformation form. A symmetry of the action up to a boundary term is sufficient; exact invariance of the displayed Lagrangian is stronger than necessary. Discrete transformations by themselves do not give this continuous-charge construction.

The separability of the printed Lagrangian is important. Every $x_i$ is an independent oscillator of the same frequency, irrespective of mass. Define $X_i=\sqrt{m_i}{x}_i$, $P_i=p_{x_i}/\sqrt{m_i}$. Then the complete constant matrix $Q_{ij}=P_i{P}_j+{\omega }^2{X}_i{X}_j$, and angular quantities $J_{ij}=X_i{P}_j-X_j{P}_i$, include each oscillator energy and the rotational symmetries of this mass-weighted oscillator space. Alternatively its explicitly time-dependent constants are $X_i\cos \omega t-P_i\sin \omega t/\omega$ and $X_i\sin \omega t+P_i\cos \omega t/\omega$.

Translations in all $y_i$ give conserved $P_y=\sum p_{y_i}$. Translation of $z_1,z_2$ together gives $P_{z12}=p_{z_1}+p_{z_2}$; $z_3$ is free and gives $p_{z_3}$. Their center-of-mass positions minus the associated constant velocities times $t$ are boost constants. The decoupled relative $z_1-z_2$ motion conserves $E_{z,\mathrm{rel}}=(p_{z_1}/m_1-p_{z_2}/m_2{)}^2{\mu }_{12}/2+V(z_1-z_2)$, where ${\mu }_{12}=m_1{m}_2/(m_1+m_2)$. In mass-weighted versions, the three free center coordinates also have conserved mutual angular momenta. These families are not all functionally independent.

The $y$ relative subsystem alone is driven through $\cos [\lambda (m_1-m_2)t]$. For generic nonzero $\lambda (m_1-m_2)$ its energy, and hence the total energy, is not conserved. Omitting the mass-weighted oscillator symmetries simply because the physical masses differ would miss genuine symmetries of this particular model.

Vary $S=\int L(q,\dot{q},t)dt$ with fixed endpoints. Integration by parts gives $\delta S=\int {\sum }_i[L_{q_i}-d{L}_{{\dot{q}}_i}/dt]\delta q_{i}dt$, hence the Euler-Lagrange equations. Define $p_i=L_{{\dot{q}}_i}$ and, assuming an invertible velocity-momentum relation, $H=p\cdot \dot{q}-L$. The equivalent action is $\int (p\cdot \dot{q}-H)dt$. Independent variations of $p$ and $q$ yieldBoundary terms vanish for fixed endpoint coordinates; constraining momenta independently at the same endpoints is unnecessary. Singular Lagrangians require constraint analysis instead of this unconstrained Legendre transform.

With zero scalar potential, the nonrelativistic charged-particle Lagrangian in SI units is kinetic energy plus $q\mathbf{A}\cdot \dot{\mathbf{r}}$. ThusThe velocity-dependent magnetic term changes canonical momentum but does not contribute to the mechanical energy.

Define $a=q{B}_0/(2L)>0$. Differentiating the Lagrangian with respect to each generalized velocity givesThe coordinates $y$ and $z$ are cyclic, so $p_y$ and $p_z$ are conserved. The momentum $p_x$ is generally not conserved: its equation is ${\dot{p}}_x=-2ax\dot{z}$. In particular, conserved canonical $p_z$ is not the same as conserved mechanical $m\dot{z}$, which changes as the particle moves across the magnetic-field gradient.

Differentiation givesThe angles $\varphi$ and $\psi$ are cyclic, so their canonical momenta are conserved. The Lagrangian is autonomous, so the total energy is also conserved. The momentum $p_{\theta }$ is not generally constant because gravity and the angular kinetic terms depend on $\theta$.

Use $u^{\lambda }{u}_{\lambda }=-c^2$ and a constant Minkowski metric. Vary the velocity-dependent square root before imposing the proper-time normalization. Equivalently, vary a reparametrization-invariant worldline action first and then choose proper time. Its canonical covector momentum isThe coordinate derivative at fixed velocity is $\partial \mathcal{L}/\partial x^a=q{u}^{\lambda }{\partial }_a{A}_{\lambda }$. The Euler-Lagrange equation and the chain rule from a therefore giveDefine the antisymmetric field tensor byThe particle equation becomesAntisymmetry ensures $u^a{F}_{a\lambda }{u}^{\lambda }=0$, so the force is orthogonal to the four-velocity and preserves $u^2=-c^2$. The worldline variation before proper-time normalization is essential: substituting $\sqrt{-u^2}=c$ into the Lagrangian before taking its velocity derivative would turn the kinetic term into a constant and incorrectly remove the inertial momentum. The normalization is a property of the proper-time solution, not permission to discard the variational dependence.

Differentiate with respect to the generalized velocities:The Lagrangian has no explicit $x$, so ${\dot{p}}_x=\partial L/\partial x=0$. This canonical momentum is also the system's total horizontal mechanical momentum. Horizontal translation symmetry conserves $p_x$ even though the support and bob separately accelerate.

The time-independent Lagrangian conserves $E=T+V$. Substitute $\dot{x}=-m_{2}l\dot{\varphi }\cos \varphi /M$ into the kinetic energy to obtainThis is the angle-dependent effective inertia of a pendulum with a freely sliding support. Solving for the speed givesOn a monotone increasing segment, the travel time is thereforeThe printed oriented integral assumes that increasing-angle branch. For decreasing motion use $|d\varphi |$ or interchange the limits; if a turning point lies between the two events, split the trajectory into monotone segments and add their positive times. The numerator remains positive for positive masses. At an ordinary turning point the inverse-square-root singularity is integrable; reaching the unstable upright separatrix takes infinite time.

The cylindrical entries of $\mathbf{A}$ are physical vector components, so $\mathbf{v}\cdot \mathbf{A}=\mu r^2\dot{\theta }$, not $\mu r\dot{\theta }$. The Lagrangian isIts three equations areFor $r>0$, the angular equation can be expanded as $r\ddot{\theta }+2\dot{r}\dot{\theta }+(2e\mu /m)\dot{r}=0$. Directly computing the fields gives $\mathbf{B}=2\mu \hat{\mathbf{z}}$ and $\mathbf{E}=-2\lambda z\hat{\mathbf{z}}$, agreeing with the equations above.

In Cartesian components, $\partial L/\partial v_i=m{v}_i+e{A}_i$ and $\partial L/\partial x_i=e{v}_j{\partial }_i{A}_j-e{\partial }_i\varphi$. The Euler-Lagrange equation therefore givesThus the minimally coupled particle Lagrangian yields the Lorentz force.

For an arbitrary smooth gauge function $\chi$, take $\mathbf{A}\to \mathbf{A}+\nabla \chi$ and $\varphi \to \varphi -{\partial }_t\chi$. Both fields are unchanged, whileThe action changes only by an endpoint term, so fixed-endpoint variation gives the same motion. Canonical momentum shifts by $e\nabla \chi$, whereas kinetic momentum $m\mathbf{v}$ and the Lorentz force are gauge invariant. This is the relevant gauge invariance of the classical particle dynamics.

Measure both angles from the downward vertical, as in the PDF. With the support as origin and downward coordinate $z$,Thus $|AC{|}^2=2{\ell }^2[1+\cos (\theta -\varphi )]$. The requested expression implicitly uses a zero-rest-length spring: its energy is $\kappa |AC{|}^2/2$. The exact Lagrangian for equal masses isExpand the cosines and count angles and velocities as first order. The cosine correction multiplying $\dot{\theta }\dot{\varphi }$ is fourth order and is dropped. After removing constants,The positive spring term in $L_2$ means the spring lowers its energy when the straight assembly bends. If its natural length were $s_0$, this term would instead be $\kappa \ell (2\ell -s_0)(\theta -\varphi {)}^2/4$. An unspecified ordinary natural length cannot simply be ignored; the displayed target fixes the zero-length idealization.

Removing the hinge mass gives the exact LagrangianThe stated equations again concern its small-angle quadratic approximation. In the sum and difference coordinates this becomesConsequentlyFor unequal squared frequencies, $\nu =0$: the rods remain aligned to first order and the end mass swings like a pendulum of length $2\ell$. This degenerate Lagrangian gives the difference coordinate no quadratic kinetic term, so its equation is an algebraic constraint, not a second oscillator. If ${\omega }_0={\omega }_1$, the quadratic action loses all dependence on $\nu$. Its motion is then undetermined at this order; nonlinear terms or a small hinge mass must resolve the degeneracy. For larger spring stiffness the straight constrained branch is not a minimum of the bending potential, despite the absence of a finite-inertia growing mode in this strictly massless quadratic description.

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.