Lagrangian Mechanics | Physics

sources:

• text: “Goldstein, H., Poole, C., & Safko, J. (2002). Classical Mechanics (3rd ed.). Addison Wesley.”
• text: “Landau, L. D., & Lifshitz, E. M. (1976). Mechanics (3rd ed.). Butterworth-Heinemann.”

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):

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

$\blacksquare$

flowchart TD
    A[3_Lagrangian Mechanics] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

The Lagrangian formulation is like giving nature a GPS route planner. Instead of tracking every force individually, you describe the landscape of kinetic and potential energy, and nature automatically picks the path that makes the action stationary. Think of a ball rolling down a hill: you do not need to compute each force component if you know the shape of the terrain. The Lagrangian L = T - V encodes this terrain, and the Euler-Lagrange equations are nature’s optimization algorithm. Cyclic coordinates are like flat stretches of the landscape where nothing changes in a particular direction, so the corresponding momentum stays constant. The whole framework is coordinate-independent, meaning you can describe the same physical system using any convenient set of variables without changing the physics.

• Generalised Coordinates and Constraints: Generalised coordinates and constraints provide the foundation for formulating the Lagrangian.

• Hamiltonian Mechanics: The Hamiltonian formalism extends Lagrangian mechanics to phase space and provides energy conservation.

• Noether’s Theorem: Noether’s theorem connects symmetries of the Lagrangian to conserved quantities.

• Calculus

• Linear Algebra

Mistake 1: Including constraint forces in the Lagrangian The Lagrangian formulation eliminates constraint forces (like normal forces or tension in a string) by choosing appropriate generalised coordinates. A common error is adding the constraint force back into the Euler-Lagrange equations. Constraint forces do no work in holonomic systems and should not appear in $L = T - V$.

Mistake 2: Assuming $L = T - V$ always gives the correct equations The Lagrangian $L = T - V$ is valid when the potential $V$ is velocity-independent and the coordinate transformation is scleronomic (time-independent). For rheonomic constraints or velocity-dependent potentials, the correct Lagrangian may differ. Always verify the conditions under which $L = T - V$ applies.

Mistake 3: Forgetting that the action is stationary, not necessarily minimised Hamilton’s principle states the action is stationary ($\delta S = 0$), not necessarily minimised. The actual path can be a saddle point or even a local maximum of the action. The stationarity condition is what yields the Euler-Lagrange equations, and the second variation determines whether the path is a minimum.