- #1
UrbanXrisis
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I need to show using Poisson brackets that:
[tex]\left( \frac{\partial}{\partial t} \right) {f,g} = \left( \frac{\partial f}{\partial t} , g} \right)+ \left( {f, \frac{\partial g}{\partial t} \right)[/tex]
I know that:
[tex] (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) [/tex]
To show the above statement, I will expand using Poisson Brackets:
[tex] \left( \frac{\partial f}{\partial f} , g \right) [/tex] and [tex] \left( f, \frac{\partial g}{\partial t} \right) [/tex]
[tex] \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} [/tex]
[tex] \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} [/tex]
[tex] \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} [/tex]
[tex] \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)[/tex]
Am I using Poisson brackets correctly? Not sure how I got double the terms I wanted.
[tex]\left( \frac{\partial}{\partial t} \right) {f,g} = \left( \frac{\partial f}{\partial t} , g} \right)+ \left( {f, \frac{\partial g}{\partial t} \right)[/tex]
I know that:
[tex] (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) [/tex]
To show the above statement, I will expand using Poisson Brackets:
[tex] \left( \frac{\partial f}{\partial f} , g \right) [/tex] and [tex] \left( f, \frac{\partial g}{\partial t} \right) [/tex]
[tex] \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} [/tex]
[tex] \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} [/tex]
[tex] \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} [/tex]
[tex] \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)[/tex]
Am I using Poisson brackets correctly? Not sure how I got double the terms I wanted.
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