Lagrangian Mechanics

The Lagrangian of a system is defined as

$L(q_1, \ldots, q_n, \dot{q}_1, \ldots, \dot{q}_n, t) = T - V$

Where $T$ is the kinetic energy and $V$ is the potential energy.

Theorem 3.1 (Euler-Lagrange from D’Alembert). The equations of motion for a holonomic system with ideal constraints are:

$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_j}\right) - \frac{\partial L}{\partial q_j} = 0, \quad j = 1, \ldots, n$

Proof. Start from D’Alembert’s principle with only applied forces (ideal constraints):

$\sum_i (\mathbf{F}_i^{(\mathrm{app})} - m_i\ddot{\mathbf{r}}_i) \cdot \delta\mathbf{r}_i = 0$

Express the virtual displacement in terms of generalised coordinates:

$\delta\mathbf{r}_i = \sum_j \frac{\partial \mathbf{r}_i}{\partial q_j}\delta q_j$

First term (applied forces). For a conservative system, $\mathbf{F}_i^{(\mathrm{app})} = -\nabla_i V$So:

$\sum_i \mathbf{F}_i^{(\mathrm{app})} \cdot \delta\mathbf{r}_i = -\sum_i \nabla_i V \cdot \sum_j \frac{\partial \mathbf{r}_i}{\partial q_j}\delta q_j = -\sum_j \frac{\partial V}{\partial q_j}\delta q_j$

Defining the generalised force $Q_j = \sum_i \mathbf{F}_i \cdot \frac{\partial \mathbf{r}_i}{\partial q_j}$For conservative forces $Q_j = -\partial V/\partial q_j$.

Second term (inertia). Using $\frac{\partial \dot{\mathbf{r}}_i}{\partial \dot{q}_j} = \frac{\partial \mathbf{r}_i}{\partial q_j}$ (which holds when $\mathbf{r}_i = \mathbf{r}_i(q, t)$):

$\sum_i m_i\ddot{\mathbf{r}}_i \cdot \delta\mathbf{r}_i = \sum_i m_i\ddot{\mathbf{r}}_i \cdot \sum_j \frac{\partial \mathbf{r}_i}{\partial q_j}\delta q_j = \sum_j \left[\sum_i m_i\ddot{\mathbf{r}}_i \cdot \frac{\partial \mathbf{r}_i}{\partial q_j}\right]\delta q_j$

Now:

$\sum_i m_i\ddot{\mathbf{r}}_i \cdot \frac{\partial \mathbf{r}_i}{\partial q_j} = \frac{d}{dt}\left(\sum_i m_i\dot{\mathbf{r}}_i \cdot \frac{\partial \mathbf{r}_i}{\partial q_j}\right) - \sum_i m_i\dot{\mathbf{r}}_i \cdot \frac{d}{dt}\frac{\partial \mathbf{r}_i}{\partial q_j}$

Using $\frac{d}{dt}\frac{\partial \mathbf{r}_i}{\partial q_j} = \frac{\partial \dot{\mathbf{r}}_i}{\partial q_j}$ and $\frac{\partial \dot{\mathbf{r}}_i}{\partial \dot{q}_j} = \frac{\partial \mathbf{r}_i}{\partial q_j}$:

$\sum_i m_i\ddot{\mathbf{r}}_i \cdot \frac{\partial \mathbf{r}_i}{\partial q_j} = \frac{d}{dt}\left(\sum_i m_i\dot{\mathbf{r}}_i \cdot \frac{\partial \dot{\mathbf{r}}_i}{\partial \dot{q}_j}\right) - \sum_i m_i\dot{\mathbf{r}}_i \cdot \frac{\partial \dot{\mathbf{r}}_i}{\partial q_j}$

$= \frac{d}{dt}\frac{\partial T}{\partial \dot{q}_j} - \frac{\partial T}{\partial q_j}$

Combining both terms in D’Alembert’s principle:

$\sum_j \left[Q_j - \frac{d}{dt}\frac{\partial T}{\partial \dot{q}_j} + \frac{\partial T}{\partial q_j}\right]\delta q_j = 0$

For conservative forces, $Q_j = -\partial V/\partial q_j$. Since $L = T - V$ and $V$ is independent of $\dot{q}_j$:

$\sum_j \left[\frac{\partial L}{\partial q_j} - \frac{d}{dt}\frac{\partial L}{\partial \dot{q}_j}\right]\delta q_j = 0$

Since the $\delta q_j$ are independent (and we have $n$ degrees of freedom):

$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_j}\right) - \frac{\partial L}{\partial q_j} = 0$

$\blacksquare$

Theorem 3.2 (Hamilton’s Principle). The actual path of a system between times $t_1$ and $t_2$ is The one that makes the action

$S = \int_{t_1}^{t_2} L(q, \dot{q}, t)\, dt$

Stationary.

Theorem 3.3 (Euler-Lagrange Equation from Hamilton’s Principle). The path $q(t)$ that makes $S$ stationary satisfies

$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_j}\right) - \frac{\partial L}{\partial q_j} = 0, \quad j = 1, \ldots, n$

Proof (for one degree of freedom). Consider a variation $q(t) + \epsilon \eta(t)$ where $\eta(t_1) = \eta(t_2) = 0$. The variation of the action:

$\delta S = \int_{t_1}^{t_2} \left(\frac{\partial L}{\partial q}\eta + \frac{\partial L}{\partial \dot{q}}\dot{\eta}\right) dt$

Integrating the second term by parts:

$\delta S = \int_{t_1}^{t_2} \left(\frac{\partial L}{\partial q} - \frac{d}{dt}\frac{\partial L}{\partial \dot{q}}\right) \eta\, dt + \left[\frac{\partial L}{\partial \dot{q}}\eta\right]_{t_1}^{t_2}$

The boundary term vanishes since $\eta(t_1) = \eta(t_2) = 0$. For $\delta S = 0$ for all $\eta$By The fundamental lemma of the calculus of variations:

$\frac{\partial L}{\partial q} - \frac{d}{dt}\frac{\partial L}{\partial \dot{q}} = 0$

$\blacksquare$

Intuition. Hamilton’s principle says nature is “lazy”: out of all possible paths connecting two configurations, the actual path taken is the one that makes the action stationary. This is a profound generalisation of Fermat’s principle of least time in optics.

Problem. Derive the equation of motion for a simple pendulum of length $l$ and mass $m$.

Solution. Take $\theta$ as the generalised coordinate. The position of the bob is $(l\sin\theta, -l\cos\theta)$.

$T = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2) = \frac{1}{2}ml^2\dot{\theta}^2$

$V = -mgl\cos\theta$

$L = T - V = \frac{1}{2}ml^2\dot{\theta}^2 + mgl\cos\theta$

Euler-Lagrange equation:

$\frac{\partial L}{\partial \theta} = -mgl\sin\theta, \quad \frac{\partial L}{\partial \dot{\theta}} = ml^2\dot{\theta}$

$\frac{d}{dt}(ml^2\dot{\theta}) + mgl\sin\theta = 0 \implies \ddot{\theta} + \frac{g}{l}\sin\theta = 0$

For small angles ($\sin\theta \approx \theta$): $\ddot{\theta} + \frac{g}{l}\theta = 0$Giving simple Harmonic motion with $\omega = \sqrt{g/l}$. $\blacksquare$

Problem. Derive the equations of motion for a double pendulum: mass $m_1$ on rod $l_1$Mass $m_2$ on rod $l_2$ attached to $m_1$.

Solution. Generalised coordinates: angles $\theta_1, \theta_2$ from the vertical. Position of $m_1$:

$x_1 = l_1\sin\theta_1, \quad y_1 = -l_1\cos\theta_1$

Position of $m_2$:

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

Velocities:

$\dot{x}_1 = l_1\dot{\theta}_1\cos\theta_1, \quad \dot{y}_1 = l_1\dot{\theta}_1\sin\theta_1$

$\dot{x}_2 = l_1\dot{\theta}_1\cos\theta_1 + l_2\dot{\theta}_2\cos\theta_2, \quad \dot{y}_2 = l_1\dot{\theta}_1\sin\theta_1 + l_2\dot{\theta}_2\sin\theta_2$

Kinetic energy:

$T = \frac{1}{2}m_1(\dot{x}_1^2 + \dot{y}_1^2) + \frac{1}{2}m_2(\dot{x}_2^2 + \dot{y}_2^2)$

$= \frac{1}{2}(m_1 + m_2)l_1^2\dot{\theta}_1^2 + \frac{1}{2}m_2 l_2^2\dot{\theta}_2^2 + m_2 l_1 l_2\dot{\theta}_1\dot{\theta}_2\cos(\theta_1 - \theta_2)$

Potential energy:

$V = -m_1 g l_1\cos\theta_1 - m_2 g(l_1\cos\theta_1 + l_2\cos\theta_2)$

The Euler-Lagrange equations for $\theta_1$ and $\theta_2$ yield two coupled second-order ODEs. For equal masses and lengths ($m_1 = m_2 = m$, $l_1 = l_2 = l$):

$(m + m)l^2\ddot{\theta}_1 + ml^2\ddot{\theta}_2\cos(\theta_1 - \theta_2) + ml^2\dot{\theta}_2^2\sin(\theta_1 - \theta_2) + 2mgl\sin\theta_1 = 0$

$ml^2\ddot{\theta}_2 + ml^2\ddot{\theta}_1\cos(\theta_1 - \theta_2) - ml^2\dot{\theta}_1^2\sin(\theta_1 - \theta_2) + mgl\sin\theta_2 = 0$

$\blacksquare$

Problem. Two masses $m_1$ and $m_2$ ($m_1 > m_2$) are connected by a massless inextensible string over a frictionless pulley. Find the acceleration using the Lagrangian.

Problem. A bead of mass $m$ slides without friction on a circular hoop of radius $R$. The hoop rotates about a vertical diameter with constant angular velocity $\omega$. Find the equilibrium positions and their stability.

Definition. A coordinate $q_j$ is cyclic (or ignorable) if it does not appear explicitly in the Lagrangian: $\partial L / \partial q_j = 0$.

Theorem 3.4. If $q_j$ is cyclic, the conjugate generalised momentum $p_j = \partial L / \partial \dot{q}_j$ is a constant of motion.

Proof. The Euler-Lagrange equation for a cyclic coordinate is:

$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_j}\right) = 0 \implies p_j = \mathrm{const}$

$\blacksquare$

Intuition. Cyclic coordinates correspond to symmetries of the system. Each symmetry gives a conserved quantity --- this is the essence of Noether’s theorem (Section 5).

When holonomic constraints cannot be eliminated by coordinate choice, introduce Lagrange multipliers $\lambda_a$:

$\frac{d}{dt}\frac{\partial L}{\partial \dot{q}_j} - \frac{\partial L}{\partial q_j} = \sum_a \lambda_a \frac{\partial f_a}{\partial q_j}$

The multipliers $\lambda_a$ are proportional to the constraint forces.

Definition. The energy function (also called the Jacobi integral) is:

$h = \sum_j \dot{q}_j \frac{\partial L}{\partial \dot{q}_j} - L$

Theorem 3.5. If $L$ does not depend explicitly on time, then $h$ is conserved. Furthermore, if the transformation $\mathbf{r}_i = \mathbf{r}_i(q)$ does not depend explicitly on time and $V$ is velocity-independent, then $h = T + V$ (the total energy).

Proof. Taking the total time derivative:

$\frac{dh}{dt} = \sum_j \ddot{q}_j \frac{\partial L}{\partial \dot{q}_j} + \sum_j \dot{q}_j \frac{d}{dt}\frac{\partial L}{\partial \dot{q}_j} - \sum_j \frac{\partial L}{\partial q_j}\dot{q}_j - \sum_j \frac{\partial L}{\partial \dot{q}_j}\ddot{q}_j - \frac{\partial L}{\partial t}$

The $\ddot{q}_j$ terms cancel. Using the Euler-Lagrange equation $\frac{d}{dt}\frac{\partial L}{\partial \dot{q}_j} = \frac{\partial L}{\partial q_j}$:

$\frac{dh}{dt} = -\frac{\partial L}{\partial t}$

If $\partial L/\partial t = 0$Then $dh/dt = 0$.

For the second part, when $\mathbf{r}_i = \mathbf{r}_i(q)$ (scleronomic) and $V = V(q)$:

$T = \frac{1}{2}\sum_{i,j,k} m_i \frac{\partial \mathbf{r}_i}{\partial q_j}\frac{\partial \mathbf{r}_i}{\partial q_k}\dot{q}_j\dot{q}_k$

Is a homogeneous quadratic form in $\dot{q}_j$. By Euler’s theorem for homogeneous functions:

$\sum_j \dot{q}_j \frac{\partial T}{\partial \dot{q}_j} = 2T$

Since $\partial L/\partial \dot{q}_j = \partial T/\partial \dot{q}_j$ (as $V$ is velocity-independent):

$h = \sum_j \dot{q}_j \frac{\partial T}{\partial \dot{q}_j} - T + V = 2T - T + V = T + V$

$\blacksquare$

:::caution Common Pitfall The energy function $h$ equals $T + V$ only for natural systems (scleronomic constraints and Velocity-independent potentials). For a bead on a rotating hoop (rheonomic constraint), $h$ is Conserved but $h \neq T + V$. Always check whether the system is natural before identifying $h$ With the total energy.

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