# Variation of the lagrangian

1. Aug 27, 2011

### Identity

Given an action:

$$S = \int L(q,\dot{q},t) \,dt$$

The variation is:

$$\delta S = \int \left(\frac{\partial L}{\partial q}\delta q+\frac{\partial L}{\partial \dot{q}}\delta\dot{q}\right)\,dt$$

I'm guessing this is some type of chain rule, but I haven't been able to derive it... how is it justified?

2. Aug 27, 2011

### WannabeNewton

Let $x_{l}(t)$ be the path of least action. Consider the path $x_{l}(t) + \zeta (t)$ where $\zeta (t) << 1$. So, $$S(x_{l}(t) + \zeta (t)) = \int_{t_{i}}^{t_{f}}L(x_{l}(t) + \zeta (t); \dot{x_{l}}(t) + \dot{\zeta }(t))dt$$ After a taylor expansion to first order in $\boldsymbol{\zeta }$, $$S(x_{l}(t) + \zeta (t)) = \int_{t_{i}}^{t_{f}}[L(x_{l}(t), \dot{x_{l}}(t)) + \frac{\partial L}{\partial x(t)}\zeta (t) + \frac{\partial L}{\partial \dot{x}(t)}\dot{\zeta }(t)]dt$$ (with the expansion terms along the path of least action) so the variation is given by $$\delta S = \int_{t_{i}}^{t_{f}}[ \frac{\partial L}{\partial x(t)}\zeta (t) + \frac{\partial L}{\partial \dot{x}(t)}\dot{\zeta }(t)]dt$$

3. Aug 27, 2011

### Identity

Ah, thanks for that and the fast reply :)