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I am trying to derive the equations of motion for a Lagrangian which depends on $(q, \dot{q}, \ddot{q}).$ I proceed by the typical route via Hamilton's Principle, $\delta S = 0$ by effecting a variation $\epsilon \eta$ on the path with $\eta$ smooth and vanishing on the endpoints. After some integrating by parts and vanishing of surface terms, I arrive at (to first order in $\epsilon$): $$\delta S = \int\left[\eta\frac{\partial L}{\partial q} - \eta\frac{\mathrm{d}}{\mathrm{d} t}\left(\frac{\partial L}{\partial \dot{q}}\right) + \eta\frac{\mathrm{d}^2}{\mathrm{d}t^2}\left(\frac{\partial L}{\partial \ddot{q}}\right) + \frac{\mathrm{d}^2}{\mathrm{d}t^2}\left(\frac{\partial L}{\partial \ddot{q}} \eta \right)\right]\mathrm{d} t.$$

It is clear to me that either the last term in the integral above should vanish, or else I made an error and it ought not to appear at all. If it is the former case, by what argument does this term vanish?

Qmechanic
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Diffycue
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1 Answers1

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You have to impose that $\eta(t_0)=\eta(t_1)=\dot{\eta}(t_0)=\dot{\eta}(t_1)=0$ where $t_0$ and $t_1$ are the endpoints of the time interval over which you are integrating. Then, the last term is: \begin{equation} \int_{t_0}^{t_1}\frac{d^2}{dt^2} \left(\frac{\partial L}{\partial\ddot{q}}\eta\right)dt = \left[\frac{d}{dt}\left(\frac{\partial L}{\partial\ddot{q}}\eta\right)\right]_{t_0}^{t_1} = \left[\eta\frac{d}{dt}\left(\frac{\partial L}{\partial\ddot{q}}\right)\right]_{t_0}^{t_1}+ \left[\dot{\eta}\frac{\partial L}{\partial\ddot{q}}\right]_{t_0}^{t_1} = 0 \end{equation} The Euler-Lagrange equation is then: \begin{equation} \frac{\partial L}{\partial q} - \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right) + \frac{d^2}{dt^2}\left(\frac{\partial L}{\partial \ddot{q}}\right) = 0 \end{equation} As a justification for the conditions over $\eta$ and its derivative at the endpoints observe that, in general, $\partial L/\partial\ddot{q}$ may depend on $\ddot{q}$, so the equation of motion will be of fourth order. To obtain a solution, four conditions will be needed. In the case of $L$ depending only on $q$ and $\dot{q}$, for a second order equation we needed two conditions: fixing $q(t_0)$ and $q(t_1)$. In the fourth order case, it is reasonable to fix $q(t_0)$, $q(t_1)$, $\dot{q}(t_0)$ and $\dot{q}(t_1)$.

Therefore, as $\delta q=\epsilon\eta$ and $\delta \dot{q}=\epsilon\dot{\eta}$ we have that $\eta(t_0)=\eta(t_1)=\dot{\eta}(t_0)=\dot{\eta}(t_1)=0$.

coconut
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