Can anyone help me make the connection here

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if F is any function and H is the hamiltonian, then the Poisson Bracket of F and H, is given by:

[F,H] = dF/dt - [tex]\partial[/tex]F/[tex]\partial[/tex]t

Can someone show me how the right side of this equation comes about?
Also how can the normal derivative of F w.r.t t be different from the partial of F w.r.t t?
 

Answers and Replies

cristo
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In this case, where F is any function of coordinates and momenta (i.e. F=F(t,q,p), where the position and momentum depend upon time) the total derivative of F with respect to time will be [tex]\frac{dF}{dt}=\frac{\partial f}{\partial t}+\frac{\partial F}{\partial q}\frac{d q}{dt}+\frac{\partial F}{\partial p}\frac{dp}{dt}[/tex].

Now, use the Hamilton-Jacobi equations, to get the equation in terms of H, and use the definition of the Poisson Bracket of H and F to get the equation in the required form.
 
George Jones
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Also how can the normal derivative of F w.r.t t be different from the partial of F w.r.t t?
This is a standard abuse of notation.

Suppose, for example, [itex]F = F \left( q, p, t \right)[/itex] ([itex]F:\mathbb{R}^3 \leftarrow \mathbb{R}[/itex]), and [itex]q = q \left( t \right)[/itex] and [itex]p = p \left( t \right)[/itex] (i.e., each is a map from [itex]\mathbb{R}[/itex] to [itex]\mathbb{R}[/itex]).

Define a new function [tex]\tilde{F} \left( t \right) = F\left( q\left(t) , p\left(t)\right), t \right)[/tex] (another map from [itex]\mathbb{R}[/itex] to [itex]\mathbb{R}[/itex]) by composition of functions. The multivariable chain rule then gives

[tex]
\frac{d \tilde{F}}{dt} = \frac{\partial F}{\partial q} \frac{dq}{dt} + \frac{\partial F}{\partial p} \frac{dp}{dt} + \frac{\partial F}{\partial t}.
[/tex]

Even though [itex]F[/itex] and [tex]\tilde{F}[/tex] are different functions, as they have different domains, it is standard to omit the twiddle.
 
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