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Lie's criterion tells us that $(q,p) \to (Q,P)$ is a canonical transformation, for a system with Hamiltonian $H$ and "Kamiltonian" $K$, if and only if the identity $$\sum_k p_k dq_k -Hdt = \sum_k P_k dQ_k - Kdt + dF \tag{1}$$ is satisfied, where $F$ is a suitable generating function. All textbooks I have consulted so far tell me that if the transformation is time independent, i.e. $$\frac{\partial Q_k}{\partial t} = \frac{\partial P_k}{\partial t} = 0, \quad \forall k, \tag{2}$$ then we may choose $F$ such that $$\frac{\partial F}{\partial t} =0, \quad \text{and thus} \quad K = H. \tag{3}$$ However, I have not yet found a thorough explanation of why exactly this should be the case, and I'm having a hard time determining the necessary steps – which I nevertheless expect to be very trivial. How can this be justified?


EDIT: After pondering on both answers some time, I might have found a proof that $K = H$. Take $F = F_2(q,P,t)$ as generating function (second type): then for $ k = 1,\dots,n$ $$\begin{cases} p_k = \dfrac{\partial F}{\partial q_k}, \\ Q_k = \dfrac{\partial F}{\partial P_k}, \\ K = H + \dfrac{\partial F}{\partial t}. \end{cases} $$ Differentiating the last relation w.r.t. $P_k$ for some $k$ and applying Schwarz's lemma yields $$\frac{\partial^2 F}{\partial P_k \partial t} = \frac{\partial K}{\partial P_k} - \frac{\partial H}{\partial P_k} = \frac{\partial^2 F}{\partial t \partial P_k } = \frac{\partial Q_k}{\partial t} = 0 $$ so we get $$\frac{\partial K}{\partial P_k} = \frac{\partial H}{\partial P_k}. \tag4 $$

By similar arguments, using the generating function $\tilde F = F_3(p,Q,t)$, one obtains the similar identity $$ \frac{\partial K}{\partial Q_k} = \frac{\partial H}{\partial Q_k}. \tag{5} $$ Integrating $(4)$ and applying the FTC we obtain $$\int \frac{\partial K}{\partial P_k} dP_k = \int \frac{\partial H}{\partial P_k} dP_k \quad \implies K = H + c(Q,t), \tag{6} $$ where we have iterated the integration over $P_k$ for all $k = 1,\dots,n$ in order to completely remove the dependency of $c$ from $P$. Differentiating $(6)$ w.r.t. $Q_k$ yields $$\frac{\partial K}{\partial Q_k} = \frac{\partial H}{\partial Q_k} + \frac{\partial c}{\partial Q_k}, $$ which, by comparison with $(5)$, implies $$\frac{\partial c}{\partial Q_k} = 0 \quad \forall k, \quad \implies \quad c = c(t). $$ By the "third" (or $(2n+1)$-th) relation for $F$, $$\frac{\partial F}{\partial t} = c(t), $$ so that $$F(q,P,t) = \Phi(t) + C(q,P),$$ where $\Phi$ is such that $d\Phi/dt = c$. However, by the variational principle, the $F$ had to be chosen such that $\delta F|_{t_1}^{t_2} = 0$, for two consecutive times $t_1 < t_2$, hence we must have $$\delta F|_{t_1}^{t_2} = \delta \Phi|_{t_1}^{t_2} + \delta C|_{t_1}^{t_2} = \delta \Phi|_{t_1}^{t_2} = 0 $$ One such choice of $\Phi$ is the identically vanishing function; in other words, the new function $\hat F = F - \Phi$ generates the same canonical transformation as $F$ (which may be seen by differentiating $\hat F$ w.r.t. $q_k$ and $P_k$). This completes the proof.

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  • $\begingroup$ i've seen you're located in Italy so I guess you can read this document; math.unipd.it/~benettin/links-MA/ma-17_10_25.pdf from a cursory read I guess you should find at least some hints, if not a full answer, around page 23. The chapter on time dependent canonical transf. begins a couple of pages earlier $\endgroup$ – tbt Sep 6 '18 at 21:36
  • $\begingroup$ Note that you do not need $\delta F = 0$ but rather $\delta\int \dot F\operatorname dt=0$. That's why $F$ is determined up to a function $\Phi$ of just the time variable, regardless of whether its variation vanishes at the end points of the integration. $\endgroup$ – Phoenix87 Sep 7 '18 at 22:21
  • $\begingroup$ @Phoenix87 Wouldn’t the variation operator go through both the integral and the time derivative, since it is a virtual differential (acts with time frozen)? $\endgroup$ – giobrach Sep 8 '18 at 13:49
  • $\begingroup$ The functional is $\Sigma[\eta] := \int_{t_1}^{t_2}\dot F(\eta(t),t)\operatorname dt$ and one has to take the variation of this quantity. Since this is a constant for any differentiable function $F$, its variation is exactly 0. The variation of $F$ makes no sense, since $F$ is a function and not a functional. $\endgroup$ – Phoenix87 Sep 8 '18 at 18:18
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The key observation is that the function $F$ is defined up to a differentiable function of $t$ alone. That's because, if $F$ satisfies the Lie condition, then

$$\delta\int[P\dot Q - K]\operatorname dt = \delta\int[p\dot q - H - \dot F - \phi]\operatorname dt=\delta\int[p\dot q - H]\operatorname dt - \delta\int[\dot F + \phi]\operatorname dt$$

which equals zero whenever $\phi$ is a function of $t$ alone (the second term is the variation of a constant).

For concreteness, assume that we have $F=F(q,Q,t)$. Since the transformation is time-independent, we have

$$\partial_Q\partial_tF=-\partial_tP=0$$

and

$$\partial_q\partial_tF=\partial_tp=0$$

Therefore there are functions $f(Q,t)$ and $g(q,t)$ such that $\partial_t F = f = g$. This is only possible if $f$ and $g$ depend only on $t$. We then find that $F$ has the structure

$$F(q,Q,t) = \Phi(t) + F'(q,Q).$$

By the assumptions on $F$, $\Phi$ is the primitive of a function $\phi$, and for the previous argument, we can forget about it, effectively reaching to the conclusion that $F$ can be chosen so that $\partial_t F=0$.

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  • $\begingroup$ Please correct me if I’m mistaken but it seems your proof depends quite weakly on the form of $F$ so presumably would be valid for all types, i.e. $F_1\ldots F_4$. $\endgroup$ – ZeroTheHero Oct 10 '18 at 1:34
  • $\begingroup$ You are right. I have chosen a particular type just for concreteness, but the argument applies to all generating functions. $\endgroup$ – Phoenix87 Oct 10 '18 at 6:47
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    $\begingroup$ This is really an very very nice answer. $\endgroup$ – ZeroTheHero Oct 10 '18 at 16:04
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OP is essentially asking the following question.

For a given canonical transformation (CT) $z=(q,p)\to Z=(Q,P)$ without explicit time-dependence [and given$^1$ a Hamiltonian $H(z,t)$ and a Kamiltonian $K(Z,t)$] that satisfy the Lie criterion (1), how do we know that there locally exists a generating function $F$ without explicit time dependence?

That's a good question. The only (somewhat cumbersome) proof I know uses the fact that we may assume that the generating function $F$ (modulo terms quadratic in $z^I$ and $Z^J$) is locally a generating function $F_a$ for a CT of type $a$, which only depends on half the old variables $z^I$ and half the new variable $Z^J$ and time $t$. Lie's criterion (1) then becomes $2n$ conditions and $K-H=\frac{\partial F_a}{\partial t}$. One may use these $2n$ conditions together with OP's eq. (2) to show that the only allowed $t$-dependence in the generating function $F_a$ must be confined to a term that only depends on time $t$ and nothing else. Such term may be dropped (cf. footnote 1), so that $F_a$ (and $F$) does not depend explicitly on time $t$ and hence the Kamiltonian $K(Z,t)=H(z,t)$ becomes just the old Hamiltonian in new variables. $\Box$

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$^1$ Since Kamilton's equations are invariant if we add a term to $K$ that only depends on time $t$ and nothing else, we will allow such redefinitions of the given $K$. This freedom will be needed in the proof.

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  • $\begingroup$ Does one need to compute the total time derivative of $F$? $\endgroup$ – giobrach Sep 7 '18 at 17:46
  • $\begingroup$ Yes, that is done at some point in the proof. $\endgroup$ – Qmechanic Sep 7 '18 at 18:19
  • $\begingroup$ Why exactly can the time-dependent term be ignored? Is it related to the fact that the generating function $F$ must be such that $F(t_2) - F(t_1) = 0$, where $t_1,t_2$ are the boundaries of integration of the action functional in the variational principle? $\endgroup$ – giobrach Sep 7 '18 at 20:09
  • $\begingroup$ I updated the answer. $\endgroup$ – Qmechanic Sep 7 '18 at 21:02
  • $\begingroup$ And I have updated my question to include a tentative proof. I had to use the stronger hypothesis that the transformation could be generated by a type 2 and a type 3 function, so I'm still in doubt... $\endgroup$ – giobrach Sep 7 '18 at 21:27

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