I have some problems understanding the Lagrangian and the Hamiltonian formalism. Those can be condensed in the following "derivation" of $\frac{\partial T}{\partial t} = 0$ from the equation $\frac{\partial H}{\partial t} = - \frac{\partial L}{\partial t}$. Since the kinetic energy might be time dependent (for example when our frame of reference accelerates), it seems that I missed something really important.

Question: . Where is my mistake or where did I missunderstood the Lagrangian / Hamiltonian formalism?

Derivation: One of the Hamiltonian equations is $\frac{\partial H}{\partial t} = - \frac{\partial L}{\partial t}$ (see section "Deriving Hamilton's equations" of the Wikipedia article "Hamiltonian mechanics"). With $H=T+V$ and $L=T-V$ we get ($T$ stands for kinetic energy and $V$ for potential energy):

$$\begin{array}{rrl} & \frac{\partial H}{\partial t} & = - \frac{\partial L}{\partial t} \\ \iff & \frac{\partial (T+V)}{\partial t} & = - \frac{\partial (T-V)}{\partial t} \\ \iff & \frac{\partial T}{\partial t} + \frac{\partial V}{\partial t} & = - \frac{\partial T}{\partial t} + \frac{\partial V}{\partial t} \\ \iff & 2\frac{\partial T}{\partial t} & = 0 \\ \iff & \frac{\partial T}{\partial t} & = 0 \end{array}$$

I want to elaborate the given answer by Qmechanic, since it took me some time to understand it.

Extended Answer: We have to keep in mind, that $H = H(p,q,t)$ and $L(\dot q, q, t)$ have different signatures. While the Hamiltonian depends on the generalized impuls $p$, the Lagrangian depends on the velocity $\dot q$.

Therefore $\frac{\partial H}{\partial t}$ and $\frac{\partial L}{\partial t}$ are different partial derivations. In $\frac{\partial H}{\partial t}$ the variables $p$ and $q$ are held constant while in $\frac{\partial L}{\partial t}$ the variables $\dot q$ and $q$ are held constant. When we notate the first derivation with $\partial_t^{p,q}$ and the second with $\partial_t^{\dot q,q}$ we see were I made a mistake in the above derivation:

$$\begin{array}{rrl} & \partial_t^{p,q} H & = - \partial_t^{\dot q,q} L \\ \iff & \partial_t^{p,q} (T+V) & = - \partial_t^{\dot q,q} (T-V) \\ \iff & \partial_t^{p,q} T + \partial_t^{p,q} V & = - \partial_t^{\dot q,q} T + \partial_t^{\dot q,q} V \\ \iff & \partial_t^{p,q} T + \partial_t^{\dot q,q} T & = \partial_t^{\dot q,q} V - \partial_t^{p,q} V\\ \end{array}$$

Since $V$ doesn't depend on $p$ nor $\dot q$ we have $\partial_t^{\dot q,q} V - \partial_t^{p,q} V = 0$:

$$\partial_t^{p,q} T + \partial_t^{\dot q,q} T = 0$$

However, $\partial_t^{p,q} T$ do not have to be the same as $\partial_t^{\dot q,q} T$. This is not the case, when the impuls-velocity-connection is time dependent, i.e. $p = p(\dot q, t)$. An example is a launching rocket whose mass decreases with time. Here we have $T=\frac 12 m(t)\dot q^2$ and thus $p=m(t)\dot q$ (see example in the answer by Qmechanic).

Both partial derivations are only the same, when the following property is fulfilled:

$$p \text{ is constant over time} \iff \dot q \text{ is constant over time}$$


closed as off-topic by David Z Jun 10 at 10:17

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – David Z
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    $\begingroup$ The Hamiltonian isn't always of the form $T+V$ though. $\endgroup$ – jacob1729 Jun 9 at 10:02
  • $\begingroup$ @jacob1729 Can you elaborate this in an answer? (e.g. by giving an example) $\endgroup$ – Stephan Kulla Jun 9 at 10:07
  • 2
    $\begingroup$ I'm not convinced that solves your problem fully, but if $T$ isn't a quadratic function of velocities then $H$ won't be in the form $T+V$. However you could have a time dependant quadratic $T$ (eg $1/2 m(t)\dot{q}^2$) and your issue would remain. $\endgroup$ – jacob1729 Jun 9 at 10:14
  1. In a nutshell, even if we assume the non-generic relations $$L(q,v,t)~=~T(v,t)~-~V(q,t)\quad\text{and}\quad H(q,p,t)~=~T(p,t)~+~V(q,t),$$ then OP's mistake is to be cavalier about functional dependence of $T$, and in particular, its explicit time dependence.

  2. Perhaps a simple example is in order, cf. above comment by jacob1729: $$\begin{align} L(q,v,t)~=~\frac{m(t)}{2}v^2 \quad &\Rightarrow\quad \frac{\partial L(q,v,t)}{\partial t} ~=~ \color{red}{+}\frac{m^{\prime}(t)}{m(t)}L(q,v,t)\cr\cr \updownarrow\text{identify}\qquad & \qquad\qquad\text{sum up to zero }\updownarrow\cr\cr H(q,p,t)~=~\frac{p^2}{2m(t)}\quad &\Rightarrow\quad \frac{\partial H(q,p,t)}{\partial t} ~=~ \color{red}{-}\frac{m^{\prime}(t)}{m(t)}H(q,p,t). \end{align}$$

  • $\begingroup$ So my mistake was that $\partial_t H$ and $\partial_t L$ are different partial derivations? In $\partial_t H$ the variables $p$ and $q$ are held constant while in $\partial_t L$ the variables $\dot q$ and $q$ are constant. Both partial derivations might be different, since $p(\dot q, t)$ might be time dependent. Is this a valid description of my mistake? $\endgroup$ – Stephan Kulla Jun 10 at 10:57
  • $\begingroup$ Yes. Exactly right. $\endgroup$ – Qmechanic Jun 10 at 11:23
  • $\begingroup$ Thanks a lot for your help! $\endgroup$ – Stephan Kulla Jun 10 at 11:26

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