2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – coconut Dec 7 '16 at 19:51
  • $\begingroup$ @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. $\endgroup$ – Diffycue Dec 7 '16 at 19:53
  • $\begingroup$ @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? $\endgroup$ – Jahan Claes Dec 7 '16 at 19:57
  • $\begingroup$ 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? $\endgroup$ – Diffycue Dec 7 '16 at 20:08
  • $\begingroup$ 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. $\endgroup$ – coconut Dec 7 '16 at 20:16
1
$\begingroup$

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 $\eta(t_0)=\eta(t_1)=\dot{\eta}(t_0)=\dot{\eta}(t_1)=0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.