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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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  • It's a total derivative, so when you integrate it you get $\frac{d}{dt}\left(\eta\frac{\partial L}{\partial \ddot{q}}\right)$, which vanishes if $\eta=\dot{\eta}=0$ at the endpoints. – coconut Dec 07 '16 at 19:51
  • @AccidentalFourierTransform, I understand that "standard argument" here to be that of eq. (2.8) in Landau; there he mandates that it be a total derivative of a function $f = f(q,t),$ which is not the case here. – Diffycue Dec 07 '16 at 19:53
  • @Diffycue But it is! It's the total derivative of $f(q,t)=\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial \ddot q}\eta$. Here $\eta$ is a function of $t$, and the derivative of the Lagrangian is a function of $q$ and it's derivatives. What part don't you like? – Jahan Claes Dec 07 '16 at 19:57
  • Thank you all for responding.

    @JahanClaes The part I do not like is that $q,$ $\dot{q}$, $\ddot{q}$ are to be treated as independent coordinates in the variational formalism; so then I do not believe it is the case that $\frac{\mathrm{d}}{\mathrm{d} t} \left[\frac{\partial L}{\partial \ddot{q}} \eta\right]$ is only a function of $q$ and $t,$ since it is not the case that we can write $\dot{q}$ or $\ddot{q}$ as functions of $q$ and $t$ before effecting the variation.

    Is the root of my confusion clearer?

     – Diffycue Dec 07 '16 at 20:08
  • If you are dealing with a lagrangian that depends on $q$, $\dot{q}$ and $\ddot{q}$ then I would say that the usual vanishing of total derivatives isn't true unless you impose the extra condition that the derivative of $q$ vanishes on the endpoints. – coconut Dec 07 '16 at 20:16
  • @coconut, I grant that the term will vanish if we specify that $\dot{\eta} = 0$ at the endpoints. But the approach Landau takes is as follows: "Let the system occupy, at the instants $t_1$ and $t_2$, positions defined by two sets of values of the co-ordinates, $q^{(1)}$ and $q^{(2)}.$" This is what justifies setting the variation on the path to $0$ at the endpoints; what is the justification for setting its time derivative to $0$ here? (edit: I wrote this comment in response to your first comment, but I don't think the question is answered by your second comment so I'll leave this up) – Diffycue Dec 07 '16 at 20:22
  • @Diffycue I think he did answer your question. You have to assume $\dot\eta=0$ at the endpoints to get the right answer. Landau didn't do this because he was not considering a Lagrangian that depended on $\ddot q$. Further, Landau's equation 8 explicitly only holds for a Lagrangian that depends on $q,\dot q$. – Jahan Claes Dec 07 '16 at 20:46
  • @Diffycue When you have a $\ddot{q}$-dependent lagrangian, the equation of motion will in general be of order four (instead of two). That means that instead two conditions (the positions at endpoints) we need four. We can do this by fixing both position and speed at endpoints, which means that $\eta=\dot{\eta}=0$ there – coconut Dec 07 '16 at 21:12
  • I have posted an answer summarizing what I have said here – coconut Dec 07 '16 at 21:33
  • Related: http://physics.stackexchange.com/q/109518/2451 , http://physics.stackexchange.com/q/119750/2451 , http://physics.stackexchange.com/q/169419/2451 , and links therein. – Qmechanic Dec 07 '16 at 22:25

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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