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newton_1372
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I have a system described by the hamiltonian H in the coordinates [itex]i, p_i [/itex]. A transformation of the type
$Q_i = Q_i(q_i, p_i)$ is called "canonical" if exists a new Hamiltonian, say K(Q,P), and a function F(Q_i,P_i,q_i,p_i) such that is verified
[itex]p\dot q_i-H=P_i\dot Q_i-K+\frac{dF}{dt}[/itex]
How can i prove that a trasformation is canonical (in this sense) if and only if
[itex]\{Q_i,P_j\}=\delta_{ij}[/itex]
when {} are the Poisson's braket?
I'd wish to understand another thing regardind that. I found a derivation that shows that is sufficient that is been conserved the hamiltonian form of equations of motion to conserve poisson brakets...but we know that the transformations satisfing 1). are not the unique to conserve the hamiltonian form of eq. of motion...so, what is the link between these 2 sets of trasformations?
SETS 1. Trasformations such that exists K, and F t.c. the eq. 1). is satisfied.
SETS 2. Trasformations such that exists K satisfing the equation of motion of Hamilton:
[itex] \dot Q = \frac{\partial K}{\partial P}\\ \dot P = -\frac{\partial K}{\partial Q}[/itex]
$Q_i = Q_i(q_i, p_i)$ is called "canonical" if exists a new Hamiltonian, say K(Q,P), and a function F(Q_i,P_i,q_i,p_i) such that is verified
[itex]p\dot q_i-H=P_i\dot Q_i-K+\frac{dF}{dt}[/itex]
How can i prove that a trasformation is canonical (in this sense) if and only if
[itex]\{Q_i,P_j\}=\delta_{ij}[/itex]
when {} are the Poisson's braket?
I'd wish to understand another thing regardind that. I found a derivation that shows that is sufficient that is been conserved the hamiltonian form of equations of motion to conserve poisson brakets...but we know that the transformations satisfing 1). are not the unique to conserve the hamiltonian form of eq. of motion...so, what is the link between these 2 sets of trasformations?
SETS 1. Trasformations such that exists K, and F t.c. the eq. 1). is satisfied.
SETS 2. Trasformations such that exists K satisfing the equation of motion of Hamilton:
[itex] \dot Q = \frac{\partial K}{\partial P}\\ \dot P = -\frac{\partial K}{\partial Q}[/itex]
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