Lagrange's equations of motion

Remarks

The proof starts from d'Alembert's form of the principle of virtual work, $\sum_k (F_k - m_k \ddot r_k) \cdot \delta r_k = 0$, which is where the hypothesis on ideal constraints enters. Two identities follow from $\dot r_k = \sum_j \frac{\partial r_k}{\partial q_j}\dot q_j + \frac{\partial r_k}{\partial t}$: first $\frac{\partial \dot r_k}{\partial \dot q_j} = \frac{\partial r_k}{\partial q_j}$, and second $\frac{d}{dt}\frac{\partial r_k}{\partial q_j} = \frac{\partial \dot r_k}{\partial q_j}$. With them, $\sum_k m_k \ddot r_k \cdot \frac{\partial r_k}{\partial q_j} = \frac{d}{dt}\frac{\partial T}{\partial \dot q_j} - \frac{\partial T}{\partial q_j}$, and since the $\delta q_j$ are independent, the equations follow. The second form is the system of Euler–Lagrange equations of $L$. Without the ideal-constraint hypothesis, for example with sliding friction at a contact, the constraint forces contribute to $Q_j$ and do not drop out.

Example: a simple pendulum of length $\ell$ and mass $m$, with angle $\theta$ from the downward vertical, has $T = \tfrac12 m\ell^2\dot\theta^2$ and $V = -mg\ell\cos\theta$. The equation is $m\ell^2\ddot\theta + mg\ell\sin\theta = 0$, and the tension of the rod never appears.

Going deeper

For a student

These are the equations of the second kind; the equations of the first kind keep Cartesian coordinates and add Lagrange multipliers for the constraint forces. For conservative forces they are the Euler–Lagrange equations of $L = T - V$, which links them with Hamilton's principle and the stationary-action principle. Through the Legendre transform they become Hamilton's equations.