Variational Principle and Lagrange's Equations
Calculus of variations, Euler-Lagrange functional derivatives, Hamilton's principle of stationary action, derivation of equations of motion, undetermined Lagrange multipliers for non-holonomic systems, symmetries, cyclic coordinates, conservation laws, and the double pendulum chaotic dynamical system.
§2.1 Calculus of Variations & The Fundamental Lemma
1. Functionals & Extremal Path Problems
In ordinary calculus, we find points $x^*$ that extremize a function $f(x)$. In the calculus of variations, we seek an entire function or path $y(x)$ that extremizes a functional—an integral of the form:2. Variational Derivation of the Euler-Lagrange Equation
Let $y(x)$ be the true path that renders $J$ stationary. Consider a family of varied paths:§2.2 Hamilton's Principle of Stationary Action
1. Definition of the Action Integral
In 1834, Sir William Rowan Hamilton generalized variational mechanics to dynamics. For a conservative dynamical system with configuration described by coordinates $q(t) = (q_1(t), \dots, q_n(t))$, the action integral $S$ is defined as:2. Statement of Hamilton's Principle
Hamilton's Principle: The actual motion of a holonomic dynamical system from time $t_1$ to $t_2$ follows a trajectory $q(t)$ for which the action integral $S$ is stationary (an extremum, usually a minimum) with respect to arbitrary virtual variations $\delta q_j(t)$ that vanish at the temporal endpoints:
3. Derivation of Lagrange's Equations
Commuting the variation $\delta$ with the time integral:§2.3 Extension to Non-Holonomic Systems & Lagrange Multipliers
1. Constrained Variations with Undetermined Multipliers
When a system is subjected to $m$ non-holonomic constraints in differential form:2. The Method of Lagrange Multipliers
We introduce $m$ time-dependent undetermined multipliers $\lambda_k(t)$. Multiplying each constraint variation by $\lambda_k(t)$, integrating from $t_1$ to $t_2$, and summing yields:§2.4 Symmetries, Cyclic Coordinates & Noether's Theorem
1. Cyclic Coordinates
If the Lagrangian $L(q, \dot{q}, t)$ does not explicitly depend on a specific coordinate $q_k$, so that:2. Immediate Conservation of Conjugate Momentum
Substituting $\frac{\partial L}{\partial q_k} = 0$ into Lagrange's equation:First Integral of Motion: The generalized momentum conjugate to any cyclic coordinate is strictly conserved throughout the motion.
3. Spatial & Temporal Symmetries
- Spatial Translation Invariance: If the Lagrangian is invariant under spatial translation along direction $\hat{n}$, $\sum_i \vec{F}_i \cdot \hat{n} = 0$, the total linear momentum along $\hat{n}$ is conserved.
- Rotational Invariance: If the Lagrangian is invariant under rotation about axis $\hat{u}$, $\frac{\partial L}{\partial \phi} = 0$, the total angular momentum along $\hat{u}$ is conserved.
- Time Translation Invariance & The Jacobi Integral: If the Lagrangian does not depend explicitly on time ($\frac{\partial L}{\partial t} = 0$), the Jacobi energy integral $h$:
$$h(q, \dot{q}) = \sum_{j=1}^n \dot{q}_j \frac{\partial L}{\partial \dot{q}_j} - L = \text{constant}$$is conserved. When transformations to Cartesian coordinates are scleronomic (time-independent), $h = T + V = E$ (total mechanical energy).
§2.5 The Double Pendulum: Nonlinear Dynamics & Chaos
1. Coordinate Parameterization
A planar double pendulum consists of mass $m_1$ connected by a rigid massless rod of length $l_1$ to a fixed pivot, and mass $m_2$ suspended from $m_1$ by a rod of length $l_2$. The generalized coordinates are the deflection angles $\theta_1$ and $\theta_2$ relative to the downward vertical:2. Kinetic and Potential Energy Formulations
Differentiating positions:3. Coupled Nonlinear Equations of Motion
Applying the Euler-Lagrange equations yields:4. Transition to Deterministic Chaos
For small amplitudes ($\theta_1, \theta_2 \ll 1$), these equations linearize into coupled harmonic oscillators yielding two distinct normal modes with real frequencies $\omega_1, \omega_2$. However, at higher energies, the strong nonlinear coupling $\cos(\theta_1 - \theta_2)$ and centrifugal terms induce deterministic chaos:- Extreme sensitivity to initial conditions (positive Lyapunov exponent $\lambda > 0$).
- Two trajectories separated initially by $|\Delta \vec{x}(0)| \sim 10^{-10}$ diverge exponentially: $|\Delta \vec{x}(t)| \sim |\Delta \vec{x}(0)| e^{\lambda t}$.
- Non-repeating, ergodic phase space orbits while conserving total energy $E = T + V$.
Standard University Exam Solved Problems
Find the path connecting two points $A(0,0)$ and $B(x_2, y_2)$ in a uniform downward vertical gravitational field $g$ such that a bead sliding without friction takes the minimum possible transit time.
Orienting the $y$-axis downwards from origin $(0,0)$, conservation of energy gives $v = \sqrt{2gy}$. The transit time functional is $T[y] = \frac{1}{\sqrt{2g}} \int_0^{x_2} \frac{\sqrt{1 + y'^2}}{\sqrt{y}} dx$.
Since the integrand $f(y, y') = \frac{\sqrt{1+y'^2}}{\sqrt{y}}$ does not explicitly depend on $x$, the first integral (Beltrami identity) applies: $y(1 + y'^2) = 2a$, where $a$ is a constant.
Substituting $y = a(1 - \cos \theta)$ yields $dy = a \sin \theta d\theta$, leading directly to $x = a(\theta - \sin \theta)$. The optimal path is a cycloid.
Parametric curve of fastest descent is a cycloid: x = a(θ - sin θ), y = a(1 - cos θ).
A uniform solid cylinder of mass $M$ and radius $R$ rolls down a rough incline of angle $\alpha$ without slipping. Use Lagrange multipliers to simultaneously determine the linear acceleration and the force of static friction.
Treat $x$ (displacement down the incline) and $\theta$ (rotation angle) as initially independent coordinates. For a uniform solid cylinder, moment of inertia is $I = \frac{1}{2}MR^2$.
The rolling condition is $f(x, \theta) = x - R\theta = 0$. The generalized constraint forces are $Q_x = \lambda(1) = \lambda$ and $Q_\theta = \lambda(-R) = -\lambda R$.
Using $\ddot{x} = R \ddot{\theta}$, the rotational equation becomes $\frac{1}{2}M \ddot{x} = -\lambda$. Substituting $\lambda = -\frac{1}{2}M \ddot{x}$ into the $x$-equation gives $M \ddot{x} - M g \sin \alpha = -\frac{1}{2}M \ddot{x} \implies \ddot{x} = \frac{2}{3}g \sin \alpha$.
Acceleration: ẍ = (2/3) g sin α; Frictional constraint force: f_friction = |λ| = (1/3) M g sin α.
Model a linear triatomic molecule (such as CO₂) as three masses on a line connected by identical springs of constant $k$: a central atom of mass $M$ flanked by two equal masses $m$. Find the Lagrangian, normal mode frequencies, and normal coordinates.
Coordinates $x_1, x_2, x_3$ represent linear displacements from equilibrium. Mass matrix is $T_{ij} = \text{diag}(m, M, m)$ and potential matrix $V_{ij}$ has diagonal elements $(k, 2k, k)$ and off-diagonals $-k$.
Expanding the determinant: $(k - m\omega^2)[(k - m\omega^2)(2k - M\omega^2) - 2k^2] = 0$. Factoring yields $\omega^2 [k - m\omega^2][k(2m + M) - m M \omega^2] = 0$.
$\omega_1 = 0$ corresponds to rigid translation of the entire molecule ($x_1 = x_2 = x_3$). $\omega_2 = \sqrt{k/m}$ is the symmetric stretch ($x_1 = -x_3, x_2 = 0$). $\omega_3 = \sqrt{k(2m+M)/(mM)}$ is the asymmetric stretch.
Normal frequencies: ω1 = 0 (rigid translation), ω2 = √(k/m) (symmetric stretch), ω3 = √(k(1 + 2m/M)/m) (asymmetric stretch).