# Poisson brackets

1. Oct 2, 2006

### UrbanXrisis

I need to show using Poisson brackets that:

$$\left( \frac{\partial}{\partial t} \right) {f,g} = \left( \frac{\partial f}{\partial t} , g} \right)+ \left( {f, \frac{\partial g}{\partial t} \right)$$

I know that:

$$(f,g) = \left( \frac{\partial f}{\partial q} \frac{\partial g}{\partial p}} \right)- \left( {\frac{\partial f}{\partial p} \frac{\partial g}{\partial q} \right)$$

To show the above statement, I will expand using Poisson Brackets:

$$\left( \frac{\partial f}{\partial f} , g \right)$$ and $$\left( f, \frac{\partial g}{\partial t} \right)$$

$$\left( {\frac{\partial f}{\partial f} , g} \right)= \frac{\partial }{\partial q} \frac{\partial f}{\partial t} \frac{\partial g}{\partial p} - \frac{\partial }{\partial p} \frac{\partial f}{\partial t} \frac{\partial g}{\partial q}$$

$$\left( {f, \frac{\partial g}{\partial t} } \right)= \frac{\partial f}{\partial q} \frac{\partial }{\partial p} \frac{\partial g}{\partial t} - \frac{\partial f}{\partial p} \frac{\partial }{\partial q} \frac{\partial g}{\partial t}$$

$$\frac{\partial}{\partial t} \left( {f,g} \right)= \frac{\partial }{\partial q} \frac{\partial f}{\partial t} \frac{\partial g}{\partial p} - \frac{\partial }{\partial p} \frac{\partial f}{\partial t} \frac{\partial g}{\partial q} + \frac{\partial f}{\partial q} \frac{\partial }{\partial p} \frac{\partial g}{\partial t} - \frac{\partial f}{\partial p} \frac{\partial }{\partial q} \frac{\partial g}{\partial t}$$

$$\frac{\partial}{\partial t} \left( f,g \right)= \frac{\partial}{\partial t} \left( 2 \frac{\partial f}{\partial q} \frac{\partial g}{\partial p}} - 2 \frac{\partial f}{\partial p} \frac{\partial g}{\partial q} \right)$$

Am I using Poisson brackets correctly? Not sure how I got double the terms I wanted.

Last edited: Oct 2, 2006
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