Proving Poisson Brackets Homework Statement

AI Thread Summary
The discussion revolves around proving the equality involving Poisson brackets and a function f expressed as a Taylor series in terms of time t. Participants highlight that knowledge of the Hamiltonian H may not be necessary, as understanding how H operates within the Poisson bracket is key. The approach involves expanding f(q,p) as a Taylor series and considering the implications of f being a constant of motion. There are inquiries about the commutativity of mixed partial derivatives and the differentiation of Poisson brackets over time. The conversation emphasizes the importance of calculating derivatives and applying the product rule to further explore the relationship between the variables.
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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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