Proving Poisson Brackets Homework Statement

Nusc
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Homework Statement



<br /> <br /> f(p(t),q(t)) = f_o + \frac{t^1}{1!}\{H,f_o\}+\frac{t^2}{2!}\{H,\{H,f_o\}}+...<br /> <br />
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

 
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Nusc said:

Homework Statement



<br /> <br /> f(p(t),q(t)) = f_o + \frac{t^1}{1!}\{H,f_o\}+\frac{t^2}{2!}\{H,\{H,f_o\}}+...<br /> <br />
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):

<br /> \frac{df(q,p)}{dt}&amp;=\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\}<br />

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

<br /> \frac{df(q,p)}{dt}=\{f,H\}<br />
 
What's the theorem that says mixed partials are commutative? Not Claurait's...
 
Also

I know

<br /> \frac{\partial }{\partial t}\{H,f\} = \{\frac{\partial H}{\partial t},f\} + \{H,\frac{\partial f}{\partial t}\} <br />

What about for ordinary derivatives?
<br /> \frac{d}{d t}\{H,f\} =?<br />
 
jdwood983 said:
Note that if f(q,p) is a constant of motion, f_{,t}=0 so that you have

<br /> \frac{df(q,p)}{dt}=\{f,H\}<br />

No, if f(q,p) is a constant of motion, then \frac{df}{dt}=0[/itex]. The fact that f has no <i>explicit</i> time dependence (it is given as a function of q and p only), tells you that \frac{\partial f}{\partial t}=0[/itex]
 
Nusc said:
What about for ordinary derivatives?
<br /> \frac{d}{d t}\{H,f\} =?<br />

You tell us...expand the Poisson bracket and calculate the derivatives...what do you get?
 
<br /> \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}<br />
 
Okay, now use the product rule...

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