Proving Poisson Brackets Homework Statement

[Nusc]

Homework Statement

$$f(p(t),q(t)) = f_o + \frac{t^1}{1!}\{H,f_o\}+\frac{t^2}{2!}\{H,\{H,f_o\}}+...$$
Prove the above equality. p & q are just coords and momenta

How do we do this if we don't know what H is?
Where do we start?

Homework Equations

The Attempt at a Solution

 

[jdwood983]

Homework Statement

$$f(p(t),q(t)) = f_o + \frac{t^1}{1!}\{H,f_o\}+\frac{t^2}{2!}\{H,\{H,f_o\}}+...$$
Prove the above equality. p & q are just coords and momenta

How do we do this if we don't know what H is?
Where do we start?

Homework Equations

The Attempt at a Solution

Looks like you may not need to know what the Hamiltonian is, just how the Hamiltonian works in the Poisson bracket (assume $f$ is not a constant of motion):

$$\frac{df(q,p)}{dt}&=\frac{\partial f}{\partial t}+\frac{\partial f}{\partial q}\frac{dq}{dt}+\frac{\partial f}{\partial p}\frac{dp}{dt}=\frac{\partial f}{\partial t}+\frac{\partial f}{\partial q}\frac{\partial H}{\partial p}-\frac{\partial f}{\partial p}\frac{\partial H}{\partial q}=\frac{\partial f}{\partial t}+\{f,H\}$$

So, using this, it looks like you would expand $f(q,p)$ as a Taylor series. Note that if $f(q,p)$ is a constant of motion, $f_{,t}=0$ so that you have

$$\frac{df(q,p)}{dt}=\{f,H\}$$

 

[Nusc]

What's the theorem that says mixed partials are commutative? Not Claurait's...

 

[Nusc]

Also

I know

$$\frac{\partial }{\partial t}\{H,f\} = \{\frac{\partial H}{\partial t},f\} + \{H,\frac{\partial f}{\partial t}\}$$

What about for ordinary derivatives?
$$\frac{d}{d t}\{H,f\} =?$$

 

[gabbagabbahey]

Note that if $f(q,p)$ is a constant of motion, $f_{,t}=0$ so that you have

$$\frac{df(q,p)}{dt}=\{f,H\}$$

No, if $f(q,p)$ is a constant of motion, then [tex]\frac{df}{dt}=0[/itex]. The fact that $f$ has no explicit time dependence (it is given as a function of $q$ and $p$ only), tells you that [tex]\frac{\partial f}{\partial t}=0[/itex]

 

[gabbagabbahey]

What about for ordinary derivatives?
$$\frac{d}{d t}\{H,f\} =?$$

You tell us...expand the Poisson bracket and calculate the derivatives...what do you get?

 

[Nusc]

$$\frac{d}{dt}\frac{\partial f}{\partial q}\frac{\partial H}{\partial p}-\frac{d}{dt}\frac{\partial f}{\partial p}\frac{\partial H}{\partial q}$$

 

[gabbagabbahey]

Okay, now use the product rule...

[tex]\frac{d}{dt}\left(\frac{\partial f}{\partial q}\frac{\partial H}{\partial p}\right)=[/itex]

?