In Goldstein, it says that the Hamiltonian is dependent, in functional form and magnitude, on the chosen set of generalized coordinates. In one set it might be conserved, but in another it might not. But it is also true that $$\frac{\partial L}{\partial t}=-\frac{\partial H}{\partial t}.\tag{1}$$ The Lagrangian is always $L=T-U$. Changing the chosen generalized coordinates should give the same magnitude for the Lagrangian in any set of generalized coordinates at all times. Thus, $\frac{\partial L}{\partial t}$ should be the same for all coordinates, and hence so does $\frac{\partial H}{\partial t}.$ If the Hamiltonian is conserved in a given set of coordinates, it should be conserved in all others. Is there a problem with this argument?
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1$\begingroup$ Even if $L$ does not depend on the used coordinates, $\partial L/\partial t$ does… $\endgroup$– Valter MorettiCommented Jul 2, 2023 at 13:43
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2 Answers
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TL;DR: The explicit derivative $\frac{\partial L}{\partial t}$ could change under a time-dependent coordinate transformation $$q^j \quad\longrightarrow\quad Q^k~=~Q^k(q,t).$$
Example: Consider a 1D point particle $$L~=~\frac{m}{2}\dot{y}^2-mgy$$ in a vertical coordinate system $y$ relative to Earth. The Lagrangian energy function is $$h(y,\dot{y},t)~=~\left(\dot{y}\frac{\partial}{\partial \dot{y}}-1\right)L~=~\frac{m}{2}\dot{y}^2+mgy.$$
Next consider a coordinate system $$Y~=~y-vt$$ relative to an elevator moving up with a constant velocity $v$. Then the Lagrangian $$L~=~\frac{m}{2}(\dot{Y}+v)^2-mg(Y+vt)$$ and the Lagrangian energy function $$\tilde{h}(Y,\dot{Y},t)~=~\left(\dot{Y}\frac{\partial}{\partial \dot{Y}}-1\right)L~=~\frac{m}{2}(\dot{Y}^2-v^2) +mg(Y+vt)$$ now have explicit time-dependence.
Interestingly, while the Lagrangian $L$ is invariant under reparametrizations, the Lagrangian energy function $h$ is not.
Concerning eq. (1), which holds for the Hamiltonian $H(q,p,t)$ [but not necessarily for the Lagrangian energy function $h(q,\dot{q},t)$ despite they take the same value], see e.g. this related Phys.SE post.
answered Jul 2, 2023 at 13:34
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$\begingroup$ @Qmechanik if I take $~Y=y-v\,t^2~$ i obtain that $~\partial_t L\ne -\partial_t H~$ my mistake ? $\endgroup$– EliCommented Jul 3, 2023 at 7:32
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1$\begingroup$ Hi @Eli. I updated the answer. $\endgroup$– Qmechanic ♦Commented Jul 3, 2023 at 10:31
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$\begingroup$ @Qmechanic how does eq. (1) not hold for the energy function if it's identical with the Hamiltonian in form? $\endgroup$– EM_1Commented Jul 3, 2023 at 16:26
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1$\begingroup$ @EM_1: I could be wrong, but I think the distinction is that $$\left( \frac{\partial H}{\partial t} \right)_{q, p} \neq \left( \frac{\partial h}{\partial t} \right)_{q, \dot{q}}$$because you're not holding the same set of variables constant in each case. It's the same sort of distinction that arises in thermodynamics, where the heat capacity of a system depends on whether you hold $P$ or $V$ constant. $\endgroup$ Commented Jul 3, 2023 at 17:33
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The equation $\partial_t L = -\partial_t H$ is perfectly valid. Note that it only says something about the explicit time-dependence of the Lagrangian and Hamiltonian. However, it does not uniquely define the Hamiltonian and also does not imply that the Hamiltonian is conserved in all sets of generalized coordinates. This can be seen from the definition of the Hamiltonian $H = \sum_i p_i q_i - L$: It is not uniquely determined by the Lagrangian alone, but, as you already mentioned, depends on the choice of $\{q_i\}$ and $\{p_i\}$.
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1$\begingroup$ Right but that's not exactly my question. My question is that "if" the Hamiltonian is conserved in one set of generalized coordinates, is it conserved in all others? $\endgroup$– EM_1Commented Jul 2, 2023 at 13:16
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1$\begingroup$ No, it isn't. The reason is because $H$ depends on the choice of $p_i,q_i$, as I wrote in my answer. One can be very clever with the choice of $q_i,p_i$ and find those under which $H$ is invariant, but most often this is not the case. Consider the 1D harmonic oscillator. The Hamiltonian is conserved in phase space $(x,p)$. But in polar coordinates $(r,\theta)=(r\cos(\theta),r\sin(\theta))$ the time-derivative $\frac{dH}{dt}\neq 0$ since it contains terms $\frac{d r}{dt}$ and $\frac{d \theta}{dt}$, which do not cancel each other. $\endgroup$ Commented Jul 2, 2023 at 13:29
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$\begingroup$ Two points: 1st of all, $\frac{dH}{dt} = \frac{\partial H}{\partial t}$. (A remark to be made here is that, surprisingly, the implicit time dependence of the Hamiltonian coming from $p$ and $q$ cancels out such that the equation is true, regardless of coordinate choice, so the argument about a clever choice of coordinate doesn't work) 2nd of all, how are there $r$ and $\theta$ coordinates in a $1D$ oscillator? Shouldn't it be a single coordinate? $\endgroup$– EM_1Commented Jul 2, 2023 at 13:36
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1$\begingroup$ Yes, but what I actually meant is to say that finding $q_i,p_i$ for which $H$ is integrable, is non-trivial. In the HO it is obvious, and thats why almost all textbooks use $(x,p)$ as the generalized coordinates. The HO in polar coordinates can be obtained by using $H = T+U = \frac{mv^2}{2} + kx^2$ with $x=(r\cos(\theta),r\sin(\theta))$ and $v= \dot{x} =( \dot{r} \cos(\theta)-r\sin(\theta)\dot{\theta}, \dot{r} \sin(\theta)+r\cos(\theta)\dot{\theta})$. $\endgroup$ Commented Jul 2, 2023 at 13:47