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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:

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

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    $\begingroup$ 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
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    $\begingroup$ 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
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  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: $$\begin{align} L(q,v,t)~=~\frac{m(t)}{2}v^2 \quad &\Rightarrow\quad \frac{\partial L(q,v,t)}{\partial t} ~=~ \color{red}{+}\frac{m^{\prime}(t)}{m(t)}L(q,v,t)\cr\cr \updownarrow\text{identify}\qquad & \qquad\qquad\text{sum up to zero }\updownarrow\cr\cr H(q,p,t)~=~\frac{p^2}{2m(t)}\quad &\Rightarrow\quad \frac{\partial H(q,p,t)}{\partial t} ~=~ \color{red}{-}\frac{m^{\prime}(t)}{m(t)}H(q,p,t). \end{align}$$

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  • $\begingroup$ 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

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