Does the Poisson Bracket Always Equal Zero When Both Observables Start at Zero?

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The discussion centers on the behavior of the Poisson bracket {f, g} when both observables f and g start at zero. It is established that if both f and g are identically zero at the initial time, the Poisson bracket {f, g} is indeed zero. However, if only one observable, say f, is zero while g is not, the Poisson bracket {f, g} does not necessarily equal zero. This conclusion is supported by a referenced research paper that clarifies the conditions under which the Poisson bracket evaluates to zero.

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tut_einstein
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Hello,

If you have two observables f and g both of which start off as: f =0 and g =0 and you evaluate their possion bracket:

{f,g}, will it necessarily be equal to zero?

Also, if just f=0 and g wasn't zero, would {f,g} =0?

Thanks!
 
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What do you think?

Are f ang g identically zero? or zero at the "start" (= initial time?) only? or zero only when some constraint is applied?
 
f and g start out being zero. so because the poisson bracket consists of derivatives of f and g, it need not necessarily be zero even if f and g started off being zero? this is what i would think.

but I'm reading a research paper that seems to suggest that if f anf g start off being zero, then the poisson bracket {f,g} is always zero.
but if only one of them started out being zero, {f,g} needn't be zero. i don't understand why this would be true.

thanks!
 

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