# Can anyone help me make the connection here

## Main Question or Discussion Point

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 - $$\partial$$F/$$\partial$$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?

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cristo
Staff Emeritus
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
Staff Emeritus
Gold Member
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

$$\frac{d \tilde{F}}{dt} = \frac{\partial F}{\partial q} \frac{dq}{dt} + \frac{\partial F}{\partial p} \frac{dp}{dt} + \frac{\partial F}{\partial t}.$$

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

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