# Canonical Transformations

1. Jul 7, 2013

### MisterX

Background

For which of the invertible transformations $(\mathbf{q}, \mathbf{p}) \leftrightarrow(\mathbf{Q}, \mathbf{P})$
$\mathbf{Q}(\mathbf {q}, \mathbf {p}, t)$
$\mathbf{P}(\mathbf{q}, \mathbf {p}, t)$
is it so that for every Hamiltonian $\mathcal{H}(\mathbf {q}, \mathbf {p}, t)$ there is a $\mathcal{K}$ such that
$$\dot{Q}_i = \frac{\partial\mathcal{K}}{\partial P_i} \;\;\;\;\;\;\;\; \dot{P}_i = -\frac{\partial\mathcal{K}}{\partial Q_i}\; ?$$
Stationary action should correspond, and that condition is met if

$\sum p_i\dot{q}_i - \mathcal{H} = \sum P_i\dot{Q}_i - \mathcal{K} + \frac{dF}{dt}$,

since integrating $\frac{dF}{dt}$ results in something only dependent of the endpoints.

Question
Consider this part of Goldstein's Classical Mechanics.

rearranging 9.13 to make this clear:

$$\mathcal{K} = \mathcal{H} + \frac{\partial F_1}{\partial t} + \sum_i \dot{Q}_i\left(P_i - \frac{\partial F_1}{\partial Q_i} \right) + \sum_i \dot{q}_i\left(\frac{\partial F_1}{\partial q_i} - p_i\right)$$
I guess I might like this explained a more. Why aren't we able to to have the coefficients of $\dot{q}_i$ or $\dot{Q}_i$ be non-zero, and have the difference absorbed into $\mathcal{K}$ ?

2. Jul 7, 2013

### facenian

it's because K should not depend on the derivates of Q and q

3. Jul 8, 2013

### vanhees71

I would argue as follows. The Hamiltonian version of the action is
$$S[x,p]=\int_{t_1}^{t_2} \mathrm{d} t [\dot{q}^k p_k - H(t,q,p)].$$
The trajectory in phase space is determined as the stationary point of this functional with the boundary values $q(t_1)$ and $q(t_2)$ fixed.

Now if you want to determine new phase-space coordinates $(Q,P)$ that describe the same system in the new coordinates by the variational principle, i.e., such that the phase-space trajectories are described by the Hamilton canonical equations, you must have
$$\mathrm{d} q^k p_k - \mathrm{d} Q^k P_k - \mathrm{d} t (H-K)=\mathrm{d} f.$$
From this it is clear that the "natural" independent variables for $f$ are $q$, $Q$, and $t$. Then comparing the differential on each side leads to
$$p_k=\frac{\partial f}{\partial q^k}, \quad P_k=-\frac{\partial f}{\partial Q^k}, \quad K=H+\frac{\partial f}{\partial t}.$$
Then you can go over to other pairs of old and new independent phase-space variables in the "generator" using appropriate Legendre transformations.

4. Sep 23, 2014

### jaydeepsingh

Jul 7, 2013 #1
MisterX

Messages:
585
Background

For which of the invertible transformations (q,p)↔(Q,P)
Q(q,p,t)
P(q,p,t)
is it so that for every Hamiltonian H(q,p,t) there is a K such that
Q˙i=∂K∂PiP˙i=−∂K∂Qi?

Is It show that K is dependent upon (P, Q) not in Derivative of (P, Q)? plz.

also show that
Q˙i=∂K∂PiP˙i=−∂K∂Qi?