I'm almost embarrased to ask this question because I thought I was by now very confident with classical mechanics.

Someone has stated that given a mechanical system with a Lagrangian $L$ s.t. $\frac{\partial L}{\partial t}=0$ where the kinetic energy $T$ is NOT homogeneous quadratic in the generalised velocities, one cannot infer that the total energy $E=T+V$ is NOT conserved.

However, I think this is already enough to show that $\dot{E} \neq 0$.

Let's assume that $L$ looks as follows: $L = \frac{1}{2}(\dot{q}^2 + 2 \dot{q}f(q))-V(q)$.

Then, after plugging in the equation of motion $\ddot{q} = -V^{\prime}$, I obtain \begin{equation} \dot{E} = \dot{q} [\dot{q}f^{\prime}(q) - V^{\prime}(q)f(q)]. \end{equation}

I can't see how one can make this vanish.

To me it is clear that scleronomic constrains imply that $T$ is homogeneous quadratic in $\dot{q}$. Then, one has of course energy conservation. But, does $T$ being not homogeneous quadratic in $\dot{q}$ also imply that there is a rheonomic constraint? (Because then it is also physically clear, why $E$ is not conserved.)

I'd be grateful for answers!

  • $\begingroup$ Is the Euler-Lagrange equation of motion correct? There should be another term $2\dot{f}(q)$ or am I missing something? $\endgroup$
    – yuggib
    Jul 25 '14 at 22:43
  • $\begingroup$ Related: physics.stackexchange.com/q/11905/2451 , physics.stackexchange.com/q/37725/2451 , and links therein. $\endgroup$
    – Qmechanic
    Jul 26 '14 at 0:03
  • $\begingroup$ @yuggib:One finds $\frac{\partial L}{\partial \dot{q}} = \dot{q}+f(q)$ and hence $\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}=\ddot{q} + \dot{q}f^{\prime}(q)$, right? Also, $\frac{\partial L}{\partial q} = \dot{q}f^{\prime}(q)-V^{\prime}(q)$. Thus, I think the eom I stated should be correct. $\endgroup$
    – psm
    Jul 26 '14 at 8:42
  • $\begingroup$ yes, you are right...it was too late yesterday night! $\endgroup$
    – yuggib
    Jul 26 '14 at 8:47
  • $\begingroup$ @Qmechanic: Therein they only discuss examples, where e.g. $E \neq H$ but $H$ conserved. I know these examples. But, as I stated in my question, I'm not certain, whether from $T$ being not homogeneous quadratic one can already infer that $\dot{E} \neq 0$ or whether one can still construct an example in which $E$ is conserved. (That in this case $E \neq H$ is clear to me, but I'm not sure whether this tells me already that the total energy is not conserved.) $\endgroup$
    – psm
    Jul 26 '14 at 8:48

It is actually clear that $T$ being NOT homogeneous quadratic in the velocities of the generalized coordinates does not imply that energy is not conserved. E.g. one can consider a boost transformation for a free particle in 1D and then $T=\frac{1}{2}m\dot{x}^2$ becomes $T=\frac{1}{2}m(\dot{x}-c)^2$. Clearly, $E$ is conserved.

Unfortunately, I wasn't thinking of boosts or of adding total time deivatives to the Lagrangian. Embarrassing.



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