Trouble understanding Poisson Brackets
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I'm looking at page 94 here - I understand the definition of Poisson brackets at the top of the page (which uses summation convention) but I don't get why the calculations in (4.61) are true. I'm substituting in the generalised coordinate $q_i$ as $f$ and the generalised momentum $p_i$ as $g$, but I must be doing something wrong - do the subscripts in (4.61) not correspond to the subscripts in the definition?
partial-derivative mathematical-physics classical-mechanics
asked Aug 25 '15 at 11:56
jl2jl2
500313
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|
show 8 more comments
$begingroup$
I'm looking at page 94 here - I understand the definition of Poisson brackets at the top of the page (which uses summation convention) but I don't get why the calculations in (4.61) are true. I'm substituting in the generalised coordinate $q_i$ as $f$ and the generalised momentum $p_i$ as $g$, but I must be doing something wrong - do the subscripts in (4.61) not correspond to the subscripts in the definition?
partial-derivative mathematical-physics classical-mechanics
asked Aug 25 '15 at 11:56
jl2jl2
500313
$endgroup$
$begingroup$
What is $q_i$? In the link you provided, authors claim ${q_i, q_j}=0$, where are $q_i, q_j$ defined?
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– 5xum
Aug 25 '15 at 12:02
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$q_i$ for $i=1,...,n$ are the generalised coordinates used in the Lagrangian/Hamiltonian formulations
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– jl2
Aug 25 '15 at 12:07
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That's not much of a definition. Are they solutions to some differential equation or something like that?
$endgroup$
– 5xum
Aug 25 '15 at 12:09
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You need to use the Hamiltonian equation.
$endgroup$
– user40276
Aug 25 '15 at 12:11
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@5xum: It's part of the generic framework which goes: Suppose we have some abstract space $X$ of states parameterized by $q_1,q_2,ldots,q_n$. Then the operands to the Poisson bracket are functions $Xtimes Ytimes mathbb Rtomathbb R$, and the notation $q_i$ is abused to also mean the function that simply returns the $i$'th coordinate of its $X$ input.
$endgroup$
– Henning Makholm
Aug 25 '15 at 12:13
|
show 8 more comments
$begingroup$
I'm looking at page 94 here - I understand the definition of Poisson brackets at the top of the page (which uses summation convention) but I don't get why the calculations in (4.61) are true. I'm substituting in the generalised coordinate $q_i$ as $f$ and the generalised momentum $p_i$ as $g$, but I must be doing something wrong - do the subscripts in (4.61) not correspond to the subscripts in the definition?
partial-derivative mathematical-physics classical-mechanics
asked Aug 25 '15 at 11:56
jl2jl2
500313
$endgroup$
I'm looking at page 94 here - I understand the definition of Poisson brackets at the top of the page (which uses summation convention) but I don't get why the calculations in (4.61) are true. I'm substituting in the generalised coordinate $q_i$ as $f$ and the generalised momentum $p_i$ as $g$, but I must be doing something wrong - do the subscripts in (4.61) not correspond to the subscripts in the definition?
partial-derivative mathematical-physics classical-mechanics
partial-derivative mathematical-physics classical-mechanics
asked Aug 25 '15 at 11:56
jl2jl2
500313
asked Aug 25 '15 at 11:56
jl2jl2
500313
asked Aug 25 '15 at 11:56
jl2jl2
500313
asked Aug 25 '15 at 11:56
jl2jl2
500313
asked Aug 25 '15 at 11:56
jl2jl2
500313
500313
$begingroup$
What is $q_i$? In the link you provided, authors claim ${q_i, q_j}=0$, where are $q_i, q_j$ defined?
$endgroup$
– 5xum
Aug 25 '15 at 12:02
$begingroup$
$q_i$ for $i=1,...,n$ are the generalised coordinates used in the Lagrangian/Hamiltonian formulations
$endgroup$
– jl2
Aug 25 '15 at 12:07
$begingroup$
That's not much of a definition. Are they solutions to some differential equation or something like that?
$endgroup$
– 5xum
Aug 25 '15 at 12:09
$begingroup$
You need to use the Hamiltonian equation.
$endgroup$
– user40276
Aug 25 '15 at 12:11
$begingroup$
@5xum: It's part of the generic framework which goes: Suppose we have some abstract space $X$ of states parameterized by $q_1,q_2,ldots,q_n$. Then the operands to the Poisson bracket are functions $Xtimes Ytimes mathbb Rtomathbb R$, and the notation $q_i$ is abused to also mean the function that simply returns the $i$'th coordinate of its $X$ input.
$endgroup$
– Henning Makholm
Aug 25 '15 at 12:13
|
show 8 more comments
$begingroup$
What is $q_i$? In the link you provided, authors claim ${q_i, q_j}=0$, where are $q_i, q_j$ defined?
$endgroup$
– 5xum
Aug 25 '15 at 12:02
$begingroup$
$q_i$ for $i=1,...,n$ are the generalised coordinates used in the Lagrangian/Hamiltonian formulations
$endgroup$
– jl2
Aug 25 '15 at 12:07
$begingroup$
That's not much of a definition. Are they solutions to some differential equation or something like that?
$endgroup$
– 5xum
Aug 25 '15 at 12:09
$begingroup$
You need to use the Hamiltonian equation.
$endgroup$
– user40276
Aug 25 '15 at 12:11
$begingroup$
@5xum: It's part of the generic framework which goes: Suppose we have some abstract space $X$ of states parameterized by $q_1,q_2,ldots,q_n$. Then the operands to the Poisson bracket are functions $Xtimes Ytimes mathbb Rtomathbb R$, and the notation $q_i$ is abused to also mean the function that simply returns the $i$'th coordinate of its $X$ input.
$endgroup$
– Henning Makholm
Aug 25 '15 at 12:13
$begingroup$
What is $q_i$? In the link you provided, authors claim ${q_i, q_j}=0$, where are $q_i, q_j$ defined?
$endgroup$
– 5xum
Aug 25 '15 at 12:02
$begingroup$
What is $q_i$? In the link you provided, authors claim ${q_i, q_j}=0$, where are $q_i, q_j$ defined?
$endgroup$
– 5xum
Aug 25 '15 at 12:02
$begingroup$
$q_i$ for $i=1,...,n$ are the generalised coordinates used in the Lagrangian/Hamiltonian formulations
$endgroup$
– jl2
Aug 25 '15 at 12:07
$begingroup$
$q_i$ for $i=1,...,n$ are the generalised coordinates used in the Lagrangian/Hamiltonian formulations
$endgroup$
– jl2
Aug 25 '15 at 12:07
$begingroup$
That's not much of a definition. Are they solutions to some differential equation or something like that?
$endgroup$
– 5xum
Aug 25 '15 at 12:09
$begingroup$
That's not much of a definition. Are they solutions to some differential equation or something like that?
$endgroup$
– 5xum
Aug 25 '15 at 12:09
$begingroup$
You need to use the Hamiltonian equation.
$endgroup$
– user40276
Aug 25 '15 at 12:11
$begingroup$
You need to use the Hamiltonian equation.
$endgroup$
– user40276
Aug 25 '15 at 12:11
$begingroup$
@5xum: It's part of the generic framework which goes: Suppose we have some abstract space $X$ of states parameterized by $q_1,q_2,ldots,q_n$. Then the operands to the Poisson bracket are functions $Xtimes Ytimes mathbb Rtomathbb R$, and the notation $q_i$ is abused to also mean the function that simply returns the $i$'th coordinate of its $X$ input.
$endgroup$
– Henning Makholm
Aug 25 '15 at 12:13
$begingroup$
@5xum: It's part of the generic framework which goes: Suppose we have some abstract space $X$ of states parameterized by $q_1,q_2,ldots,q_n$. Then the operands to the Poisson bracket are functions $Xtimes Ytimes mathbb Rtomathbb R$, and the notation $q_i$ is abused to also mean the function that simply returns the $i$'th coordinate of its $X$ input.
$endgroup$
– Henning Makholm
Aug 25 '15 at 12:13
|
show 8 more comments
1 Answer
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${q_i,q_j}=frac{partial q_i}{partial q_k}frac{partial q_j}{partial p_k} - frac{partial q_j}{partial q_k}frac{partial q_i}{partial p_k}=delta_{ik}cdot 0 - delta_{jk}cdot 0=0$
answered Jan 17 at 12:16
tommycauterotommycautero
867
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
${q_i,q_j}=frac{partial q_i}{partial q_k}frac{partial q_j}{partial p_k} - frac{partial q_j}{partial q_k}frac{partial q_i}{partial p_k}=delta_{ik}cdot 0 - delta_{jk}cdot 0=0$
answered Jan 17 at 12:16
tommycauterotommycautero
867
$endgroup$
add a comment |
$begingroup$
${q_i,q_j}=frac{partial q_i}{partial q_k}frac{partial q_j}{partial p_k} - frac{partial q_j}{partial q_k}frac{partial q_i}{partial p_k}=delta_{ik}cdot 0 - delta_{jk}cdot 0=0$
answered Jan 17 at 12:16
tommycauterotommycautero
867
$endgroup$
add a comment |
$begingroup$
${q_i,q_j}=frac{partial q_i}{partial q_k}frac{partial q_j}{partial p_k} - frac{partial q_j}{partial q_k}frac{partial q_i}{partial p_k}=delta_{ik}cdot 0 - delta_{jk}cdot 0=0$
answered Jan 17 at 12:16
tommycauterotommycautero
867
$endgroup$
${q_i,q_j}=frac{partial q_i}{partial q_k}frac{partial q_j}{partial p_k} - frac{partial q_j}{partial q_k}frac{partial q_i}{partial p_k}=delta_{ik}cdot 0 - delta_{jk}cdot 0=0$
answered Jan 17 at 12:16
tommycauterotommycautero
867
answered Jan 17 at 12:16
tommycauterotommycautero
867
answered Jan 17 at 12:16
tommycauterotommycautero
867
answered Jan 17 at 12:16
tommycauterotommycautero
867
867
add a comment |
add a comment |
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$begingroup$
What is $q_i$? In the link you provided, authors claim ${q_i, q_j}=0$, where are $q_i, q_j$ defined?
$endgroup$
– 5xum
Aug 25 '15 at 12:02
$begingroup$
$q_i$ for $i=1,...,n$ are the generalised coordinates used in the Lagrangian/Hamiltonian formulations
$endgroup$
– jl2
Aug 25 '15 at 12:07
$begingroup$
That's not much of a definition. Are they solutions to some differential equation or something like that?
$endgroup$
– 5xum
Aug 25 '15 at 12:09
$begingroup$
You need to use the Hamiltonian equation.
$endgroup$
– user40276
Aug 25 '15 at 12:11
$begingroup$
@5xum: It's part of the generic framework which goes: Suppose we have some abstract space $X$ of states parameterized by $q_1,q_2,ldots,q_n$. Then the operands to the Poisson bracket are functions $Xtimes Ytimes mathbb Rtomathbb R$, and the notation $q_i$ is abused to also mean the function that simply returns the $i$'th coordinate of its $X$ input.
$endgroup$
– Henning Makholm
Aug 25 '15 at 12:13