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The action of a path is the time integral of its Lagrangian. The principle of stationary action requires its first variation to vanish for every admissible fixed-endpoint variation.

If the spatially integrated Lagrangian changes by $dB[\Phi ]/dt$, where $B$ depends on fields and their spatial derivatives but not velocities, the canonical momentum components change by ${\pi }_a\mapsto {\pi }_a+\delta B/\delta {\Phi }_a$. This follows by varying $dB/dt$ and using spatial integration by parts. The canonical one-form changes by the exact variation $\delta B$, so the symplectic form and canonical commutation relations are preserved. For the two Feynman gauge Maxwell densities, $B=\int A_0{\partial }_i{A}_i{d}^{3}x$ gives shifts $\Delta {\pi }^0={\partial }_i{A}_i$ and $\Delta {\pi }^i=-{\partial }_i{A}_0$. Boundary flux must vanish for this spatial functional calculation.

For the linear-in-potential charged-scalar Lagrangian, $\pi ={\dot{\varphi }}^{\ast }-ie{A}_0{\varphi }^{\ast }$ and ${\pi }^{\ast }=\dot{\varphi }+ie{A}_0\varphi$. With $A_{\mu }=(A_0,-\mathbf{A})$, the Hamiltonian density isThe first product includes $+e^2{A}_0^2|\varphi {|}^2$ after the Legendre transform; that term can arise in canonical variables even though the original Lagrangian has no interaction quadratic in $A$.

For a massive test particle with electric charge $e$, choose the electromagnetic four-potential $A_t=-Q/r$ and proper time $\tau$. The cyclic coordinates $t,\varphi$ in the equatorial Lagrangian give the conserved quantities $E=mF\dot{t}+eQ/r$ and $l=r^2\dot{\varphi }$. Here $l$ is angular momentum per unit mass. Substituting these quantities into the four-velocity normalization $g(U,U)=-1$ givesThe energy includes the gauge-dependent electrostatic term; the displayed convention fixes that gauge by taking the potential to vanish at infinity.

Let a planar Lagrangian have the form $L=a(r){\dot{r}}^2+b(r){\dot{\theta }}^2+c(r)$, with smooth coefficients and $b(r)>0$. Its cyclic coordinate $\theta$ has constant conjugate momentum $p_{\theta }=2b(r)\dot{\theta }$. For a circular orbit $r=R$, the radial Euler-Lagrange equation reduces to $b^{\prime }(R){\dot{\theta }}^2+c^{\prime }(R)=0$. If $b^{\prime }(R)\ne 0$, eliminating $\dot{\theta }$ gives the displayed condition. A real orbit requires $-c^{\prime }(R)/b^{\prime }(R)\ge 0$; the radial kinetic coefficient $a$ does not enter this constant-radius condition, although it affects radial perturbations. If $b^{\prime }(R)=0$, the original condition must be used rather than divided by zero.

A compound Atwood machine replaces one hanging mass by another pulley carrying two masses. Let the upper mechanical string carry mass $M$ opposite the moving pulley, and let the lower mechanical string carry masses $a,b$ at upward heights $x,y$. The length constraints on the mechanical strings give the height of $M$ as $-(x+y)/2$ up to a constant. Thus its Lagrangian isThis follows by adding the three kinetic energies and subtracting their gravitational potential energies. Constant coordinate shifts annihilating the linear potential energy give conserved generalized momentum components by Noether's theorem.

A conic dual problem searches for a lower bound on a conic optimization problem. A feasible point exhibits an attainable cost; a feasible dual variable explains why every allowed decision costs at least a specified amount. If the two costs agree, the pair certifies a global minimum without relying on how the points were found.

Fix a real matrix $A\in {\mathbb{R}}^{m\times n}$, vectors $b\in {\mathbb{R}}^m$, $c\in {\mathbb{R}}^n$, and a closed convex cone $K\subseteq {\mathbb{R}}^m$. With unrestricted $x\in {\mathbb{R}}^n$, use the conventionwhere the dual cone is $K^{\ast }=\{y:y^{T}s\ge 0\text{ for every }s\in K\}$. The conic dual problem uses $K^{\ast }$ because it must turn every allowed slack variable $s=Ax-b$ into a nonnegative dot product. Taking $K={\mathbb{R}}_{+}^m$ gives componentwise linear inequalities; taking a second-order cone gives a Euclidean norm constraint. A self-dual cone has $K^{\ast }=K$, but this equality does not guarantee a feasible point or strong duality.

The Lagrangian supplies a short derivation. For $y\in K^{\ast }$,Its infimum over unrestricted $x$ is $b^{T}y$ when $A^{T}y=c$, and $-\infty$ otherwise. Maximizing the resulting Lagrange dual function gives the stated conic dual problem. The minus sign matches the convention $Ax-b\in K$.

A constituent quark mass is an effective mass parameter in a low-energy constituent-quark model. It accounts approximately for strong-interaction dressing and some spin-independent bound-state dynamics; it is not identical to a renormalized quark mass in the Standard Model Lagrangian.

A generalized coordinate absent from the Lagrangian is cyclic. Its conjugate generalized momentum is conserved by the Euler-Lagrange equations.

For a uniform solid cylinder of mass $M_c$ and a frictionlessly sliding wedge of mass $M_b$, take the horizontal wedge position $X$ and uphill contact coordinate $s$. Contact and rolling without slipping leave two degrees of freedom. Up to a constant, the Lagrangian is $L=\frac{1}{2}(M_b+M_c){\dot{X}}^2+M_c\cos \alpha \dot{X}\dot{s}+\frac{3}{4}{M}_c{\dot{s}}^2-M_{c}gs\sin \alpha$.

The cyclic coordinate $X$ gives $(M_b+M_c)\ddot{X}+M_c\cos \alpha \ddot{s}=0$. The other Euler-Lagrange equation is $\cos \alpha \ddot{X}+(3/2)\ddot{s}=-g\sin \alpha$. Elimination proves the displayed acceleration, with $\ddot{X}=-M_c\cos \alpha \ddot{s}/(M_b+M_c)$. For $M_b\gg M_c$, the cylinder approaches the fixed-incline acceleration $-(2/3)g\sin \alpha$.

A Lagrangian is degenerate when its Hessian with respect to velocities is singular. The momentum–velocity relation then cannot be inverted for every velocity. Some Euler–Lagrange equations can become algebraic constraints rather than independent second-order equations. A massless limit can produce this behaviour; disappearance of an equation at quadratic order does not establish an unconstrained mode of the exact nonlinear model.

The University of Toronto's institutional demonstration “Chaotic Behaviour (Double Pendulum)”, activity Me-020, uses a metal stand with four leveling feet and a pendulum arm carrying a white dot. Preparation includes leveling the stand and correcting any wobble. The instructions call for holding the arm vertically upward and releasing it; the readout is the visible motion of the dot. A further comparison inserts a metal dowel in a groove on the release arm and repeats the release at the same location. Keeping the stand steady and repeating the release position provide practical controls. The page supplies no numerical angle trace or calibrated timing, and does not explain the dowel's mechanical effect. It therefore documents a qualitative release demonstration, without quantitatively testing the equal-point mass model. Its extended arms also have moments of inertia absent from massless rods. Irregular visible motion alone does not establish sensitive dependence on initial conditions.^[ref]

A separate comparison with normal modes of a double pendulum would require small downward releases and measurement of both angles over time. The actual arm masses, centres of mass and moments of inertia would first have to enter the Lagrangian. The upward release in this demonstration explores a different regime from small oscillations about the downward stable equilibrium.

Write $\Delta ={\theta }_1-{\theta }_2$. Applying the Euler-Lagrange equations to the double pendulum Lagrangian, then dividing by $m{\ell }^2$, givesFor example, the upper generalized momentum is $\partial L/\partial {\dot{\theta }}_1=m{\ell }^2(2{\dot{\theta }}_1+\cos \Delta {\dot{\theta }}_2)$. Its time derivative contains $-m{\ell }^2\sin \Delta ({\dot{\theta }}_1-{\dot{\theta }}_2){\dot{\theta }}_2$. Subtracting $\partial L/\partial {\theta }_1=-m{\ell }^2\sin \Delta {\dot{\theta }}_1{\dot{\theta }}_2-2mg\ell \sin {\theta }_1$ cancels the mixed product and leaves the positive $\sin \Delta {\dot{\theta }}_2^2$ term in the first equation. This cancellation is a useful check on the sign convention.

The matrix multiplying the two angular accelerations isIt is invertible for every configuration, so specified angles and angular velocities determine both angular accelerations. For the same reason $T>0$ whenever the angular velocities are not both zero. The mass cancels from the motion equations, while the timescale is set by $\sqrt{\ell /g}$.

As an exact constructed release, put ${\theta }_1(0)=\alpha$, ${\theta }_2(0)=0$, and both angular velocities zero. Solving the two equations at release yieldsFor $0<\alpha <\pi /2$, the upper rod initially moves back toward vertical, while the lower rod initially moves away from it. The lower pivot's acceleration makes this possible even though the lower rod starts vertical. The small-angle approximation gives ${\ddot{\theta }}_1(0)\simeq -2g\alpha /\ell$ and ${\ddot{\theta }}_2(0)\simeq 2g\alpha /\ell$.

The conserved mechanical energy is $E=T+V$, because the Lagrangian has no explicit time dependence. Directly, $dE/dt$ is $m{\ell }^2$ times the first equation multiplied by ${\dot{\theta }}_1$ plus the second multiplied by ${\dot{\theta }}_2$, so the equations imply $dE/dt=0$. This also supplies a check when computing a trajectory.

With equal rod lengths $\ell$ and absolute downward-vertical angles $\theta ,\varphi$, the endpoint separation is $s^2=2{\ell }^2[1+\cos (\theta -\varphi )]$. A zero-rest-length spring between the fixed support and the final mass contributes $-\kappa {\ell }^2[1+\cos (\theta -\varphi )]$ to the Lagrangian. It favours bending away from maximum extension and can reduce the gravitational bending stiffness.

A dual variable is a multiplier associated with a primal constraint in a dual linear program. In a minimization linear program with equality constraints $Ax=b$ and nonnegative primal variables, the Lagrangian $c^{T}x+y^T(b-Ax)$ yields the dual objective $b^{T}y$ and inequalities $A^{T}y\le c$; the equality-constraint multipliers $y$ are unrestricted in sign. For a transportation problem, row and column dual variables give potentials $u_i,v_j$ satisfying $u_i+v_j\le c_{ij}$. Complementary slackness makes occupied routes attain equality.

A transformation taking solutions of a specified set of dynamical equations into solutions, with the theory's fixed background data held fixed. For a Lagrangian, invariance of the action up to a boundary term is a useful sufficient criterion. Symmetry of the equations does not imply that every individual solution has that symmetry: initial data and boundary conditions can select a direction or state of motion.

For an autonomous Lagrangian $L(x,\dot{x})$, define its action $A=\int Ldt$ and Hamiltonian $H=\dot{x}{L}_{\dot{x}}-L$. The chain rule givesThus the identity holds along forced trajectories as well as solutions of the homogeneous Euler-Lagrange equation. Explicit time dependence adds $-{\partial }_{t}L$; stochastic work requires a consistent regularization or the Stratonovich chain rule.

For constant symmetric mass and stiffness matrices, $E=\frac{1}{2}{\dot{q}}^{\mathsf{T}}M\dot{q}+\frac{1}{2}{q}^{\mathsf{T}}Kq$ satisfies $\dot{E}={\dot{q}}^{\mathsf{T}}(M\ddot{q}+Kq)$. Every equation $M\ddot{q}+Kq=G\dot{q}$ with antisymmetric $G$ conserves this same mechanical energy, because ${\dot{q}}^{\mathsf{T}}G\dot{q}=0$. Thus a scalar conservation law alone cannot select the dynamics of several degrees of freedom. The specified conservative-force or Lagrangian structure supplies the missing information.

Two identical planar simple pendulums of length $l$ and bob mass $m$, with pivots separated horizontally by $r>0$, are coupled by a straight spring of natural length $r$ and spring constant $k$. In the ideal model without contact restrictions, their configuration space is the torus $S^1\times S^1$. Measure both angles from the downward vertical towards the same horizontal direction. Their bob separation isThe exact Lagrangian is the expression in the title. To derive it, use bob positions $(l\sin {\theta }_1,-l\cos {\theta }_1)$ and $(r+l\sin {\theta }_2,-l\cos {\theta }_2)$. Kinetic energy from Cartesian position vectors gives $T=m{l}^2({\dot{\theta }}_1^2+{\dot{\theta }}_2^2)/2$. Gravitational potential energy is $-mgl(\cos {\theta }_1+\cos {\theta }_2)$ and spring potential energy is $k(d-r{)}^2/2$; subtracting these from $T$ proves the formula. Near the downward equilibrium, $d-r=l({\theta }_2-{\theta }_1)+O(|\mathbf{\theta }{|}^3)$ for fixed $r$, giving the familiar normal modes of two spring-coupled equal pendulums.

For a uniform slender ladder of length $2l$ sliding while touching a smooth wall and floor, use its angle $\theta$ from the vertical. Its centre of mass is $(l\sin \theta ,l\cos \theta )$ and its central moment of inertia is $m{l}^2/3$. Hence the Lagrangian is $L=2m{l}^2{\dot{\theta }}^2/3-mgl\cos \theta$, and its Euler-Lagrange equation is $\ddot{\theta }=3g\sin \theta /(4l)$. The contacts are physically possible only while their normal forces are nonnegative. If released from rest at ${\theta }_0$, energy conservation gives ${\dot{\theta }}^2=3g(\cos {\theta }_0-\cos \theta )/(2l)$, and the wall force becomes $3mg\sin \theta (3\cos \theta -2\cos {\theta }_0)/4$. Contact is lost at $\cos \theta =2\cos {\theta }_0/3$, after which the constrained equation no longer describes the motion.