Poisson Brackets Explained: Understanding the Relationship between {x,p} = 1"

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Homework Help Overview

The discussion revolves around the concept of Poisson brackets in classical mechanics, specifically examining the relationship between position (x) and momentum (p) and why their Poisson bracket is equal to 1.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the calculation of Poisson brackets, questioning the assumption that the result should be zero and discussing the independence of x and p.

Discussion Status

Some participants provide calculations to clarify the relationship, while others reflect on misunderstandings regarding derivatives and their implications. There appears to be a productive exchange of ideas, with some guidance offered on the correct interpretation of the derivatives involved.

Contextual Notes

One participant references lecture notes that may have contributed to confusion regarding the derivatives, indicating a potential gap in understanding the definitions or relationships between the variables involved.

gijoe
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can anyone tell me why the poisson brackets for {x,p} = 1 ..from (dx/dx)(dp/dp) - (dx/dp)(dp/dx)... shouldn this equal 0??
 
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I don't know why you think it should be 0.

Calculate each derivative and plug the values into the formula. You should get 1. Remember that x and p are independent.
 
dx/dx = 1. dp/dp = 1. dx/dp = 0. dp/dx = 0. 1 - 0 = 1. Which of these is not clear to you?
 
oh right, my lecture notes said that (dx/dp) = 1... i then assumed that (dp/dx) = 1.. oh yes and p is velocity dependent, now i get it... thanks!
 
{x,p} = 1 therefore
(dx/dx)(dp/dp) - (dx/dp)(dp/dx)=i
 

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