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

In summary, the conversation is about the Poisson bracket of two observables f and g, both of which start off as zero. The question is whether the Poisson bracket, {f,g}, will always be equal to zero if f and g start off as zero. There is also a mention of a research paper that suggests if both f and g start off as zero, then the Poisson bracket will always be zero, but if only one of them starts off as zero, it may not be zero. The person is seeking clarification on this concept.
  • #1
tut_einstein
31
0
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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  • #2
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?
 
  • #3
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!
 

What is a Poisson Bracket?

A Poisson Bracket is a mathematical concept used in classical mechanics to describe the relationship between two physical quantities, such as position and momentum, in a dynamical system. It is represented by curly brackets and is used to calculate the time evolution of a system.

What are the properties of a Poisson Bracket?

There are several properties that define a Poisson Bracket, including bilinearity, skew-symmetry, and the Jacobi identity. Bilinearity means that the bracket is linear in each of its arguments, while skew-symmetry means that the sign of the bracket changes when the order of the arguments is switched. The Jacobi identity states that the bracket is associative, meaning that the order of operations does not matter when calculating the bracket of three quantities.

How is a Poisson Bracket used in Hamiltonian mechanics?

In Hamiltonian mechanics, the Poisson Bracket is used to calculate the time evolution of a system by expressing the equations of motion in terms of the Hamiltonian, which is defined as the sum of the products of the system's position and momentum. This allows for a more elegant and concise formulation of the equations of motion compared to using Newton's laws of motion.

What is the relationship between a Poisson Bracket and a commutator?

The Poisson Bracket is analogous to the commutator in quantum mechanics, with the main difference being that the Poisson Bracket operates on classical quantities, while the commutator operates on quantum operators. Both concepts represent the non-commutativity of physical quantities and play a crucial role in the dynamics of their respective systems.

How is a Poisson Bracket related to symplectic geometry?

Poisson brackets and symplectic geometry are closely related, as symplectic geometry provides the mathematical framework for understanding the properties of Poisson brackets. In particular, the Poisson Bracket can be seen as a Lie bracket on the space of smooth functions on a symplectic manifold. This connection allows for the application of powerful geometric tools and techniques in analyzing the dynamics of a system described by a Poisson Bracket.

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