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Suppose we have a canonical transform $(q,p)\to(Q,P)$ generated by a function $F=F(q,Q,t)$ of type 1. This gives us the following partial derivative relations:

$$p(Q,P,t)=\frac{\partial F}{\partial q}(q(Q,P,t),Q,t)\qquad P(q,p,t)=-\frac{\partial F}{\partial Q}(q,Q(q,p,t),t)\qquad K(Q,P,t)=H(q,p,t)+\frac{\partial F}{\partial t}(q,Q,t).\tag{1}$$

I am struggling to verify that

$$\dot{p}=\frac{d}{dt}\left(\frac{\partial F}{\partial q}\right)=-\frac{\partial H}{\partial q}\qquad \dot{P}=-\frac{d}{dt}\left(\frac{\partial F}{\partial Q}\right)=-\frac{\partial K}{\partial Q}\tag{2}$$

For the first one, I found that

$$\dot{p}=\frac{d}{dt}\left(\frac{\partial F}{\partial q}\right)=\frac{\partial}{\partial q}\left(\frac{dF}{dt}\right)=\frac{\partial}{\partial q}\left(\frac{\partial F}{\partial q}\dot{q}+\frac{\partial F}{\partial Q}\dot{Q}+\frac{\partial F}{\partial t}\right)=\frac{\partial}{\partial q}\left(\frac{\partial F}{\partial q}\dot{q}+\frac{\partial F}{\partial Q}\dot{Q}+K-H\right).\tag{3}$$

I am not sure how to justify that the first three terms inside the parentheses vanish. My issue is that I am not sure what does it mean to take a partial derivative with respect to $q$ mean. It means to differentiate w.r.t $q$ while treating variables $Q$ and $t$ constant? I am confused what variables the terms $\frac{\partial F}{\partial q}\dot{q}$ and $\frac{\partial F}{\partial Q}\dot{Q}$ depend on.

Qmechanic
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Jimmy Yang
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1 Answers1

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TL;DR: The canonical transformation (CT) does not by itself impose Hamilton's and Kamilton's equations; the CT just ensures that they are equivalent.

  1. In a nutshell, the $2n+1$ type-1 conditions (1) with a generating function $F_1(q,Q,t)$ imply that the Hamiltonian Lagrangian $L_H$ and the Kamiltonian Lagrangian $L_K$ are equal off-shell up to a total time-derivative: $$ \begin{align}\underbrace{p_k\dot{q}^k-H}_{=:L_H}~\equiv~&\underbrace{\left(P_k+\frac{\partial F_1}{\partial Q^k}\right)}_{\equiv 0}\dot{Q}^k+\frac{\partial F_1}{\partial q^k}\dot{q}^k-\left(K-\frac{\partial F_1}{\partial q^k}\right)\cr ~\equiv~&\underbrace{P_k\dot{Q}^k-K}_{=:L_K}+\frac{dF_1}{dt}, \end{align}\tag{A}$$ so that Hamilton's and Kamilton's equations correspond to the same stationary configuration in the phase space, i.e. it is a canonical transformation (CT).$^1$

  2. In more detail: Assuming that $q$ and $Q$ are independent variables, the Euler-Lagrange (EL) equations for $L_H$ becomes$^2$ $$ \begin{align} 0~\approx~&\frac{\partial L_H}{\partial Q^j}-\frac{d}{dt}\frac{\partial L_H}{\partial \dot{Q}^j} \cr ~=~&\frac{\partial p_k}{\partial Q^j}\left\{\dot{q}^k-\left(\frac{\partial H}{\partial p_k}\right)_{\!q}\right\} \cr 0~\approx~&\frac{\partial L_H}{\partial q^j}-\frac{d}{dt}\frac{\partial L_H}{\partial \dot{q}^j} \cr ~=~&-\dot{p}^j +\frac{\partial p_k}{\partial q^j}\dot{q}^k -\left(\frac{\partial H}{\partial q^j}\right)_{\!Q}\cr ~=~& -\dot{p}^j+\frac{\partial p_k}{\partial q^j}\left\{\dot{q}^k-\left(\frac{\partial H}{\partial p_k}\right)_{\!q}\right\}-\left(\frac{\partial H}{\partial q^j}\right)_{\!p}. \end{align} \tag{B}$$ Assuming furthermore that the matrix $$\frac{\partial p_k}{\partial Q^j}~=~\frac{\partial^2 F_1 }{\partial Q^j\partial q^k}\tag{C}$$ is invertible, then eq. (B) is equivalent to Hamilton's equations.

  3. An analogous calculation for $L_K$ leads to Kamilton's equations.

$^1$ Concerning the issue of boundary conditions, see e.g. this Phys.SE post.

$^2$ Notation: The $\approx$ symbol means equality modulo eqs. of motion.

Qmechanic
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