gen-define-the-lagrangian-formulation-and-give-one-short-worked-ca66

Define the Lagrangian formulation and give one short worked example.

# Lagrangian Formulation ## Definition The **Lagrangian formulation** is a reformulation of classical mechanics based on energy rather than forces. **The Lagrangian** is defined as: $$L = T - V$$ where *T* is kinetic energy and *V* is potential energy. **Lagrange's equations of motion** are: $$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_i}\right) - \frac{\partial L}{\partial q_i} = 0$$ where $q_i$ are generalized coordinates and $\dot{q}_i$ are generalized velocities. ## Worked Example: Simple Pendulum **Setup:** A pendulum of length *l* and mass *m*, using angle θ from vertical as the coordinate. **Step 1 - Kinetic Energy:** $$T = \frac{1}{2}m v^2 = \frac{1}{2}m(l\dot{\theta})^2 = \frac{1}{2}ml^2\dot{\theta}^2$$ **Step 2 - Potential Energy:** $$V = -mgl\cos\theta$$ (taking the pivot as reference) **Step 3 - Lagrangian:** $$L = T - V = \frac{1}{2}ml^2\dot{\theta}^2 + mgl\cos\theta$$ **Step 4 - Apply Lagrange's equation:** $$\frac{\partial L}{\partial \dot{\theta}} = ml^2\dot{\theta} \quad \Rightarrow \quad \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\theta}}\right) = ml^2\ddot{\theta}$$ $$\frac{\partial L}{\partial \theta} = -mgl\sin\theta$$ $$ml^2\ddot{\theta} - (-mgl\sin\theta) = 0$$ **Result:** $$\boxed{\ddot{\theta} + \frac{g}{l}\sin\theta = 0}$$ This is the equation of motion for a simple pendulum.