Can anyone help me make the connection here

[rayrey]

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 - \partialF/\partialt

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?

 

[cristo]

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 \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}.

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]

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

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

<br /> \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}.<br />

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