Kinetic energy always time independent?! Where is my mistake? [closed]

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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• The Hamiltonian isn't always of the form $T+V$ though. – jacob1729 Jun 9 at 10:02
• @jacob1729 Can you elaborate this in an answer? (e.g. by giving an example) – Stephan Kulla Jun 9 at 10:07
• 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. – 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}
• 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? – Stephan Kulla Jun 10 at 10:57