Physics Classical Mechanics 100% Free Open Access
Chapter 2 • Theory & Derivations

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:
$$J[y] = \int_{x_1}^{x_2} f(y(x), y'(x), x) dx$$
where $y'(x) = dy/dx$, and the endpoints are fixed: $y(x_1) = y_1$ and $y(x_2) = y_2$.

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:
$$Y(x, \alpha) = y(x) + \alpha \eta(x)$$
where $\alpha$ is a continuous parameter, and $\eta(x)$ is an arbitrary differentiable function vanishing at the boundaries: $\eta(x_1) = \eta(x_2) = 0$. The functional becomes a function of $\alpha$:
$$J(\alpha) = \int_{x_1}^{x_2} f(Y(x, \alpha), Y'(x, \alpha), x) dx$$
Stationarity requires $\left.\frac{dJ}{d\alpha}\right|_{\alpha=0} = 0$:
$$\frac{dJ}{d\alpha} = \int_{x_1}^{x_2} \left( \frac{\partial f}{\partial Y} \frac{\partial Y}{\partial \alpha} + \frac{\partial f}{\partial Y'} \frac{\partial Y'}{\partial \alpha} \right) dx = \int_{x_1}^{x_2} \left( \frac{\partial f}{\partial Y} \eta(x) + \frac{\partial f}{\partial Y'} \eta'(x) \right) dx$$
Integrating the second term by parts:
$$\int_{x_1}^{x_2} \frac{\partial f}{\partial Y'} \eta'(x) dx = \left[ \frac{\partial f}{\partial Y'} \eta(x) \right]_{x_1}^{x_2} - \int_{x_1}^{x_2} \frac{d}{dx}\left( \frac{\partial f}{\partial Y'} \right) \eta(x) dx$$
Since $\eta(x_1) = \eta(x_2) = 0$, the boundary term vanishes. Setting $\alpha = 0$:
$$\int_{x_1}^{x_2} \left[ \frac{\partial f}{\partial y} - \frac{d}{dx}\left( \frac{\partial f}{\partial y'} \right) \right] \eta(x) dx = 0$$
By the Fundamental Lemma of the Calculus of Variations, because $\eta(x)$ is arbitrary, the integrand bracket must vanish everywhere:
$$\frac{\partial f}{\partial y} - \frac{d}{dx}\left( \frac{\partial f}{\partial y'} \right) = 0$$

§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:
$$S[q] = \int_{t_1}^{t_2} L(q, \dot{q}, t) dt$$
where $L = T - V$ is the Lagrangian.

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:

$$\delta S = \delta \int_{t_1}^{t_2} L(q_j, \dot{q}_j, t) dt = 0, \quad \delta q_j(t_1) = \delta q_j(t_2) = 0$$

3. Derivation of Lagrange's Equations

Commuting the variation $\delta$ with the time integral:
$$\delta S = \int_{t_1}^{t_2} \sum_{j=1}^n \left( \frac{\partial L}{\partial q_j} \delta q_j + \frac{\partial L}{\partial \dot{q}_j} \delta \dot{q}_j \right) dt$$
Noting that $\delta \dot{q}_j = \delta \left(\frac{dq_j}{dt}\right) = \frac{d}{dt}(\delta q_j)$ and integrating the second term by parts:
$$\int_{t_1}^{t_2} \frac{\partial L}{\partial \dot{q}_j} \frac{d}{dt}(\delta q_j) dt = \left[ \frac{\partial L}{\partial \dot{q}_j} \delta q_j \right]_{t_1}^{t_2} - \int_{t_1}^{t_2} \frac{d}{dt}\left( \frac{\partial L}{\partial \dot{q}_j} \right) \delta q_j dt$$
The boundary term vanishes identically because $\delta q_j(t_1) = \delta q_j(t_2) = 0$. Therefore:
$$\delta S = \int_{t_1}^{t_2} \sum_{j=1}^n \left[ \frac{\partial L}{\partial q_j} - \frac{d}{dt}\left( \frac{\partial L}{\partial \dot{q}_j} \right) \right] \delta q_j dt = 0$$
Since the variations $\delta q_j$ are mutually independent, each coefficient must be zero, delivering the Euler-Lagrange equations of motion:
$$\frac{d}{dt}\left( \frac{\partial L}{\partial \dot{q}_j} \right) - \frac{\partial L}{\partial q_j} = 0, \quad j = 1, 2, \dots, n$$

§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:
$$\sum_{j=1}^n a_{kj} dq_j + a_{kt} dt = 0, \quad k = 1, 2, \dots, m$$
virtual displacements (for which $\delta t = 0$) are constrained by:
$$\sum_{j=1}^n a_{kj} \delta q_j = 0, \quad k = 1, \dots, m$$
Because the variations $\delta q_j$ are no longer independent, we cannot equate each coefficient in $\delta S = 0$ to zero directly.

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:
$$\int_{t_1}^{t_2} \sum_{k=1}^m \lambda_k(t) \left( \sum_{j=1}^n a_{kj} \delta q_j \right) dt = 0$$
Adding this zero sum to Hamilton's action variation $\delta S = 0$:
$$\int_{t_1}^{t_2} \sum_{j=1}^n \left[ \frac{\partial L}{\partial q_j} - \frac{d}{dt}\left( \frac{\partial L}{\partial \dot{q}_j} \right) + \sum_{k=1}^m \lambda_k a_{kj} \right] \delta q_j dt = 0$$
By appropriately choosing the $m$ multipliers $\lambda_k(t)$, the brackets for $m$ of the coordinates vanish; the remaining $n - m$ variations are independent, requiring their brackets to vanish as well. Thus, for all $j = 1, 2, \dots, n$:
$$\frac{d}{dt}\left( \frac{\partial L}{\partial \dot{q}_j} \right) - \frac{\partial L}{\partial q_j} = \sum_{k=1}^m \lambda_k a_{kj} = Q_j^{\text{constraint}}$$
The terms $Q_j^{\text{constraint}} = \sum_{k=1}^m \lambda_k a_{kj}$ represent the generalized forces of constraint. Together with the $m$ constraint equations, this provides $n + m$ equations for $n$ coordinates $q_j(t)$ and $m$ multipliers $\lambda_k(t)$.

§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:
$$\frac{\partial L}{\partial q_k} = 0$$
then $q_k$ is termed an ignorable or cyclic coordinate.

2. Immediate Conservation of Conjugate Momentum

Substituting $\frac{\partial L}{\partial q_k} = 0$ into Lagrange's equation:
$$\frac{d}{dt}\left( \frac{\partial L}{\partial \dot{q}_k} \right) = 0 \implies p_k = \frac{\partial L}{\partial \dot{q}_k} = \text{constant of motion}$$

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:
$$x_1 = l_1 \sin \theta_1, \quad y_1 = -l_1 \cos \theta_1$$
$$x_2 = l_1 \sin \theta_1 + l_2 \sin \theta_2, \quad y_2 = -l_1 \cos \theta_1 - l_2 \cos \theta_2$$

2. Kinetic and Potential Energy Formulations

Differentiating positions:
$$v_1^2 = l_1^2 \dot{\theta}_1^2$$
$$v_2^2 = l_1^2 \dot{\theta}_1^2 + l_2^2 \dot{\theta}_2^2 + 2 l_1 l_2 \dot{\theta}_1 \dot{\theta}_2 \cos(\theta_1 - \theta_2)$$
The total kinetic energy is:
$$T = \frac{1}{2}(m_1 + m_2)l_1^2 \dot{\theta}_1^2 + \frac{1}{2}m_2 l_2^2 \dot{\theta}_2^2 + m_2 l_1 l_2 \dot{\theta}_1 \dot{\theta}_2 \cos(\theta_1 - \theta_2)$$
The gravitational potential energy is:
$$V = -(m_1 + m_2)g l_1 \cos \theta_1 - m_2 g l_2 \cos \theta_2$$

3. Coupled Nonlinear Equations of Motion

Applying the Euler-Lagrange equations yields:
$$(m_1 + m_2)l_1 \ddot{\theta}_1 + m_2 l_2 \ddot{\theta}_2 \cos(\theta_1 - \theta_2) + m_2 l_2 \dot{\theta}_2^2 \sin(\theta_1 - \theta_2) + (m_1 + m_2)g \sin \theta_1 = 0$$
$$m_2 l_2 \ddot{\theta}_2 + m_2 l_1 \ddot{\theta}_1 \cos(\theta_1 - \theta_2) - m_2 l_1 \dot{\theta}_1^2 \sin(\theta_1 - \theta_2) + m_2 g \sin \theta_2 = 0$$

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

Solved Problem Example 2.1: The Brachistochrone Problem via Euler-Lagrange Variational Calculus

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.

Step 1: Set Up the Time Integral Functional
$$v = \sqrt{2gy} \implies dt = \frac{ds}{v} = \frac{\sqrt{1 + y'^2}}{\sqrt{2gy}} dx$$

Orienting the $y$-axis downwards from origin $(0,0)$, conservation of energy gives $v = \sqrt{2gy}$. The transit time functional is $T[y] = \frac{1}{\sqrt{2g}} \int_0^{x_2} \frac{\sqrt{1 + y'^2}}{\sqrt{y}} dx$.

Step 2: Apply Beltrami Identity for Independent Variable x Missing
$$f - y' \frac{\partial f}{\partial y'} = C \implies \frac{\sqrt{1 + y'^2}}{\sqrt{y}} - y' \frac{y'}{\sqrt{y(1 + y'^2)}} = \frac{1}{\sqrt{y(1 + y'^2)}} = \text{const}$$

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.

Step 3: Integrate Using Cycloid Parametrization
$$y' = \sqrt{\frac{2a - y}{y}} \implies x = \int \sqrt{\frac{y}{2a - y}} dy$$

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.

Final Answer & Physical Insight

Parametric curve of fastest descent is a cycloid: x = a(θ - sin θ), y = a(1 - cos θ).

Solved Problem Example 2.2: Lagrange Multipliers: Cylinder Rolling Down an Incline Without Slipping

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.

Step 1: Formulate the Lagrangian with Coordinates x and θ
$$L = T - V = \left( \frac{1}{2}M \dot{x}^2 + \frac{1}{2}I \dot{\theta}^2 \right) - (-M g x \sin \alpha) = \frac{1}{2}M \dot{x}^2 + \frac{1}{4}M R^2 \dot{\theta}^2 + M g x \sin \alpha$$

Treat $x$ (displacement down the incline) and $\theta$ (rotation angle) as initially independent coordinates. For a uniform solid cylinder, moment of inertia is $I = \frac{1}{2}MR^2$.

Step 2: Express the Rolling Constraint in Differential Form
$$dx - R d\theta = 0 \implies a_x = 1, \quad a_\theta = -R$$

The rolling condition is $f(x, \theta) = x - R\theta = 0$. The generalized constraint forces are $Q_x = \lambda(1) = \lambda$ and $Q_\theta = \lambda(-R) = -\lambda R$.

Step 3: Solve Lagrange's Equations with Multiplier λ
$$M \ddot{x} - M g \sin \alpha = \lambda, \quad \frac{1}{2}M R^2 \ddot{\theta} = -\lambda R$$

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 $M \ddot{x} - M g \sin \alpha = -\frac{1}{2}M \ddot{x} \implies \ddot{x} = \frac{2}{3}g \sin \alpha$.

Final Answer & Physical Insight

Acceleration: ẍ = (2/3) g sin α; Frictional constraint force: f_friction = |λ| = (1/3) M g sin α.

Solved Problem Example 2.3: Normal Modes of a Symmetric Linear Triatomic Molecule

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.

Step 1: Set Up Kinetic and Potential Energy Matrices
$$T = \frac{1}{2}m(\dot{x}_1^2 + \dot{x}_3^2) + \frac{1}{2}M \dot{x}_2^2, \quad V = \frac{1}{2}k(x_2 - x_1)^2 + \frac{1}{2}k(x_3 - x_2)^2$$

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 $-k$.

Step 2: Solve the Characteristic Secular Determinant |V - ω² T| = 0
$$\det \begin{pmatrix} k - m\omega^2 & -k & 0 \\ -k & 2k - M\omega^2 & -k \\ 0 & -k & k - m\omega^2 \end{pmatrix} = 0$$

Expanding the determinant: $(k - m\omega^2)[(k - m\omega^2)(2k - M\omega^2) - 2k^2] = 0$. Factoring yields $\omega^2 [k - m\omega^2][k(2m + M) - m M \omega^2] = 0$.

Step 3: Extract the Three Normal Mode Frequencies
$$\omega_1 = 0, \quad \omega_2 = \sqrt{\frac{k}{m}}, \quad \omega_3 = \sqrt{\frac{k}{\mu}}, \quad \mu = \frac{m M}{2m + M}$$

$\omega_1 = 0$ 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.

Final Answer & Physical Insight

Normal frequencies: ω1 = 0 (rigid translation), ω2 = √(k/m) (symmetric stretch), ω3 = √(k(1 + 2m/M)/m) (asymmetric stretch).