Introduction to Classical Mechanics · Problem 6.33

The Lagrangian method — Morin

Physics — Mechanics · principle-of-stationary-action

Problem

Second-order change * Let xa(t)≡x0(t)+aβ(t)x_a(t) \equiv x_0(t) + a\beta(t). Equation (6.19) gives the first derivative of the action with respect to aa. Show that the second derivative is d2da2S[xa(t)]=∫t1t2(∂2L∂x2β2+2∂2L∂x∂x˙ββ˙+∂2L∂x˙2β˙2)dt.\frac{d^2}{da^2} S[x_a(t)] = \int_{t_1}^{t_2} \left( \frac{\partial^2 L}{\partial x^2}\beta^2 + 2\frac{\partial^2 L}{\partial x \partial \dot{x}}\beta\dot{\beta} + \frac{\partial^2 L}{\partial \dot{x}^2}\dot{\beta}^2 \right) dt.

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