Classical Hamiltonian: Energy Conservation?

[KFC]

If the classical Hamiltonian is define as

$$H = f(q, p)$$

p, q is generalized coordinates and they are time-dependent. But H does not explicitly depend on time. Can I conclude that the energy is conserved (even q, p are time-dependent implicitly)? Namely, if no matter if p, q are time-dependent or not, if H does not contains $$t$$ explicitly, I find that the Poisson bracket

$$\left\{H, H\right\} \equiv 0$$

so the energy is conserved, right?

But what about if H explicitly depend on time? According to the definition of Poisson bracket, $$\left\{H, H\right\} \neq 0$$ ?

 

[jambaugh]

Short answer is YES!

Long answer. Once the hamiltonian is defined the the time evolution of any quantity is given by:

$$\frac{d}{dt}F = [[H,F]] + \frac{\partial}{\partial t}F$$
where [[ ]] is the Poisson bracket and the partial derivative applies to explicit time dependence. Since:
$$[[H,H]] = 0$$
and there is no explicit time dependence you get:
$$\frac{d H}{dt} = 0$$

Expand this in terms of p and q time dependence and you'll get a relationship which becomes the Jacobi identity when you introduce Hamilton's equations.

 

[KFC]

Short answer is YES!

Long answer. Once the hamiltonian is defined the the time evolution of any quantity is given by:

$$\frac{d}{dt}F = [[H,F]] + \frac{\partial}{\partial t}F$$
where [[ ]] is the Poisson bracket and the partial derivative applies to explicit time dependence. Since:
$$[[H,H]] = 0$$
and there is no explicit time dependence you get:
$$\frac{d H}{dt} = 0$$

Expand this in terms of p and q time dependence and you'll get a relationship which becomes the Jacobi identity when you introduce Hamilton's equations.

Thank you so much. I look up the evolution relation in some textbook, but it reads

$$\frac{d}{dt}F = [[F,H]] + \frac{\partial}{\partial t}F$$

I wonder if I should exchange F, H in poisson bracket?

 

[jambaugh]

Thank you so much. I look up the evolution relation in some textbook, but it reads

$$\frac{d}{dt}F = [[F,H]] + \frac{\partial}{\partial t}F$$

I wonder if $$[[F, H]] = [[H, F]]$$ ?

No actually $$[[F,H]] = - [[H,F]]$$

But there is a convention choice which is not quite uniform in the sign of the Poisson bracket. Its a matter of whether you define:

$$[[A,B]] = \frac{\partial A}{\partial p}\frac{\partial B}{\partial q}-\frac{\partial B}{\partial p}\frac{\partial A}{\partial q}$$
vs.
$$[[A,B]] = \frac{\partial A}{\partial q}\frac{\partial B}{\partial p}-\frac{\partial B}{\partial q}\frac{\partial A}{\partial p}$$
(note p and q are reversed.)

To conform with your reference reverse my Poisson brackets. (I recommend you stick to convention and do this.)

I prefer to (buck convention and) reverse the sign/order so that for canonical conjugate variables U and V:

$$\frac{d}{dV} F = [[U,F]]$$

This fits better with the conventions for generators in Lie algebras.

 

[KFC]

No actually $$[[F,H]] = - [[H,F]]$$

But there is a convention choice which is not quite uniform in the sign of the Poisson bracket. Its a matter of whether you define:

$$[[A,B]] = \frac{\partial A}{\partial p}\frac{\partial B}{\partial q}-\frac{\partial B}{\partial p}\frac{\partial A}{\partial q}$$
vs.
$$[[A,B]] = \frac{\partial A}{\partial q}\frac{\partial B}{\partial p}-\frac{\partial B}{\partial q}\frac{\partial A}{\partial p}$$
(note p and q are reversed.)

To conform with your reference reverse my Poisson brackets. (I recommend you stick to convention and do this.)

I prefer to (buck convention and) reverse the sign/order so that for canonical conjugate variables U and V:

$$\frac{d}{dV} F = [[U,F]]$$

This fits better with the conventions for generators in Lie algebras.

Got u. Thanks a lot. X'mas