Mixing
Intuition about Poisson bracket
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I've been reading about Hamiltonian mechanics which in its mathematical description uses Poisson manifolds
From my limited understanding, on a Poisson manifold $M$ we can look at the Poisson bracket as a gadget that gives a smooth vector field ${f,- }$ for every smooth function on $M$.
This gives a nice way to write Hamilton's equations of motion.
My questions are: how should I visualize this vector field ${f,- }$? What's its connection to the function $f in C^infty(M)$? What's the connection of the flow of ${f,- }$ to the function $f$?
Am I correct in saying that ${f,g } = 0$ means that $g$ is constant along the flow of ${f,- }$?
If that helps, my background is primarily in algebra, so I'm asking about a physicist's/geometer's way of thinking about this.
abstract-algebra physics symplectic-geometry hamilton-equations poisson-geometry
asked Jan 11 at 14:05
ante.cepericante.ceperic
1,85711432
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add a comment |
$begingroup$
I've been reading about Hamiltonian mechanics which in its mathematical description uses Poisson manifolds
From my limited understanding, on a Poisson manifold $M$ we can look at the Poisson bracket as a gadget that gives a smooth vector field ${f,- }$ for every smooth function on $M$.
This gives a nice way to write Hamilton's equations of motion.
My questions are: how should I visualize this vector field ${f,- }$? What's its connection to the function $f in C^infty(M)$? What's the connection of the flow of ${f,- }$ to the function $f$?
Am I correct in saying that ${f,g } = 0$ means that $g$ is constant along the flow of ${f,- }$?
If that helps, my background is primarily in algebra, so I'm asking about a physicist's/geometer's way of thinking about this.
abstract-algebra physics symplectic-geometry hamilton-equations poisson-geometry
asked Jan 11 at 14:05
ante.cepericante.ceperic
1,85711432
$endgroup$
$begingroup$
I'm in the same boat as you are (re: the last paragraph). The physicists never even mentioned the Poisson bracket (they only ever talk about the Lie bracket), but I ran across it when reading this: fisica.net/mecanicaclassica/… Two friends and I have been struggling to get over the big gap between mathematical texts description and physics text descriptions.
$endgroup$
– rschwieb
Jan 11 at 15:11
add a comment |
$begingroup$
I've been reading about Hamiltonian mechanics which in its mathematical description uses Poisson manifolds
From my limited understanding, on a Poisson manifold $M$ we can look at the Poisson bracket as a gadget that gives a smooth vector field ${f,- }$ for every smooth function on $M$.
This gives a nice way to write Hamilton's equations of motion.
My questions are: how should I visualize this vector field ${f,- }$? What's its connection to the function $f in C^infty(M)$? What's the connection of the flow of ${f,- }$ to the function $f$?
Am I correct in saying that ${f,g } = 0$ means that $g$ is constant along the flow of ${f,- }$?
If that helps, my background is primarily in algebra, so I'm asking about a physicist's/geometer's way of thinking about this.
abstract-algebra physics symplectic-geometry hamilton-equations poisson-geometry
asked Jan 11 at 14:05
ante.cepericante.ceperic
1,85711432
$endgroup$
I've been reading about Hamiltonian mechanics which in its mathematical description uses Poisson manifolds
From my limited understanding, on a Poisson manifold $M$ we can look at the Poisson bracket as a gadget that gives a smooth vector field ${f,- }$ for every smooth function on $M$.
This gives a nice way to write Hamilton's equations of motion.
My questions are: how should I visualize this vector field ${f,- }$? What's its connection to the function $f in C^infty(M)$? What's the connection of the flow of ${f,- }$ to the function $f$?
Am I correct in saying that ${f,g } = 0$ means that $g$ is constant along the flow of ${f,- }$?
If that helps, my background is primarily in algebra, so I'm asking about a physicist's/geometer's way of thinking about this.
abstract-algebra physics symplectic-geometry hamilton-equations poisson-geometry
abstract-algebra physics symplectic-geometry hamilton-equations poisson-geometry
asked Jan 11 at 14:05
ante.cepericante.ceperic
1,85711432
asked Jan 11 at 14:05
ante.cepericante.ceperic
1,85711432
asked Jan 11 at 14:05
ante.cepericante.ceperic
1,85711432
asked Jan 11 at 14:05
ante.cepericante.ceperic
1,85711432
asked Jan 11 at 14:05
ante.cepericante.ceperic
1,85711432
1,85711432
$begingroup$
I'm in the same boat as you are (re: the last paragraph). The physicists never even mentioned the Poisson bracket (they only ever talk about the Lie bracket), but I ran across it when reading this: fisica.net/mecanicaclassica/… Two friends and I have been struggling to get over the big gap between mathematical texts description and physics text descriptions.
$endgroup$
– rschwieb
Jan 11 at 15:11
add a comment |
$begingroup$
I'm in the same boat as you are (re: the last paragraph). The physicists never even mentioned the Poisson bracket (they only ever talk about the Lie bracket), but I ran across it when reading this: fisica.net/mecanicaclassica/… Two friends and I have been struggling to get over the big gap between mathematical texts description and physics text descriptions.
$endgroup$
– rschwieb
Jan 11 at 15:11
$begingroup$
I'm in the same boat as you are (re: the last paragraph). The physicists never even mentioned the Poisson bracket (they only ever talk about the Lie bracket), but I ran across it when reading this: fisica.net/mecanicaclassica/… Two friends and I have been struggling to get over the big gap between mathematical texts description and physics text descriptions.
$endgroup$
– rschwieb
Jan 11 at 15:11
$begingroup$
I'm in the same boat as you are (re: the last paragraph). The physicists never even mentioned the Poisson bracket (they only ever talk about the Lie bracket), but I ran across it when reading this: fisica.net/mecanicaclassica/… Two friends and I have been struggling to get over the big gap between mathematical texts description and physics text descriptions.
$endgroup$
– rschwieb
Jan 11 at 15:11
add a comment |
1 Answer
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Yes, you are correct. Let $X ={f,-}$. Then for all $xin M$ and all test functions $gin C^infty(M)$ you have $X_x(g)={f,g}_x$. Let $Phi$ denote the flow of $X$. Then $$frac{rm d}{{rm d}t} g(Phi_t(p)) = {rm d}g_{Phi_t(p)}(X_{Phi_t(p)}) = X_{Phi_t(p)}(g) = {f,g}_{Phi_t(p)}.$$An arbitrary Poisson Bracket need not have any relation to $f$ whatsoever. For example, one can always consider the zero bracket in any manifold. The picture is different if the Poisson Bracket comes from a symplectic form, that is, ${f,g}=omega(X_f,X_g)$, where $X_f$ and $X_g$ are the Hamiltonian vector fields of $f$ and $g$ (in practice, that's what ${f,-}$ is in your context). I think that a more interesting question would be about the Casimir functions for a given bracket, the functions in the kernel of $fmapsto {f,-}$. For example, the Casimir functions for a bracket coming from a symplectic form are the locally constant maps. Also, there are brackets which do not come from symplectic forms but still have some physical relevance, for example, the so called Nambu brackets (which in $Bbb R^3$ supposedly can be used to model some things about rigid bodies). I'm also not a physicist so I don't really know further details, but you can look up Holm's Geometric Mechanics book.
answered Jan 23 at 22:25
Ivo TerekIvo Terek
46.1k953142
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Yes, you are correct. Let $X ={f,-}$. Then for all $xin M$ and all test functions $gin C^infty(M)$ you have $X_x(g)={f,g}_x$. Let $Phi$ denote the flow of $X$. Then $$frac{rm d}{{rm d}t} g(Phi_t(p)) = {rm d}g_{Phi_t(p)}(X_{Phi_t(p)}) = X_{Phi_t(p)}(g) = {f,g}_{Phi_t(p)}.$$An arbitrary Poisson Bracket need not have any relation to $f$ whatsoever. For example, one can always consider the zero bracket in any manifold. The picture is different if the Poisson Bracket comes from a symplectic form, that is, ${f,g}=omega(X_f,X_g)$, where $X_f$ and $X_g$ are the Hamiltonian vector fields of $f$ and $g$ (in practice, that's what ${f,-}$ is in your context). I think that a more interesting question would be about the Casimir functions for a given bracket, the functions in the kernel of $fmapsto {f,-}$. For example, the Casimir functions for a bracket coming from a symplectic form are the locally constant maps. Also, there are brackets which do not come from symplectic forms but still have some physical relevance, for example, the so called Nambu brackets (which in $Bbb R^3$ supposedly can be used to model some things about rigid bodies). I'm also not a physicist so I don't really know further details, but you can look up Holm's Geometric Mechanics book.
answered Jan 23 at 22:25
Ivo TerekIvo Terek
46.1k953142
$endgroup$
add a comment |
$begingroup$
Yes, you are correct. Let $X ={f,-}$. Then for all $xin M$ and all test functions $gin C^infty(M)$ you have $X_x(g)={f,g}_x$. Let $Phi$ denote the flow of $X$. Then $$frac{rm d}{{rm d}t} g(Phi_t(p)) = {rm d}g_{Phi_t(p)}(X_{Phi_t(p)}) = X_{Phi_t(p)}(g) = {f,g}_{Phi_t(p)}.$$An arbitrary Poisson Bracket need not have any relation to $f$ whatsoever. For example, one can always consider the zero bracket in any manifold. The picture is different if the Poisson Bracket comes from a symplectic form, that is, ${f,g}=omega(X_f,X_g)$, where $X_f$ and $X_g$ are the Hamiltonian vector fields of $f$ and $g$ (in practice, that's what ${f,-}$ is in your context). I think that a more interesting question would be about the Casimir functions for a given bracket, the functions in the kernel of $fmapsto {f,-}$. For example, the Casimir functions for a bracket coming from a symplectic form are the locally constant maps. Also, there are brackets which do not come from symplectic forms but still have some physical relevance, for example, the so called Nambu brackets (which in $Bbb R^3$ supposedly can be used to model some things about rigid bodies). I'm also not a physicist so I don't really know further details, but you can look up Holm's Geometric Mechanics book.
answered Jan 23 at 22:25
Ivo TerekIvo Terek
46.1k953142
$endgroup$
add a comment |
$begingroup$
Yes, you are correct. Let $X ={f,-}$. Then for all $xin M$ and all test functions $gin C^infty(M)$ you have $X_x(g)={f,g}_x$. Let $Phi$ denote the flow of $X$. Then $$frac{rm d}{{rm d}t} g(Phi_t(p)) = {rm d}g_{Phi_t(p)}(X_{Phi_t(p)}) = X_{Phi_t(p)}(g) = {f,g}_{Phi_t(p)}.$$An arbitrary Poisson Bracket need not have any relation to $f$ whatsoever. For example, one can always consider the zero bracket in any manifold. The picture is different if the Poisson Bracket comes from a symplectic form, that is, ${f,g}=omega(X_f,X_g)$, where $X_f$ and $X_g$ are the Hamiltonian vector fields of $f$ and $g$ (in practice, that's what ${f,-}$ is in your context). I think that a more interesting question would be about the Casimir functions for a given bracket, the functions in the kernel of $fmapsto {f,-}$. For example, the Casimir functions for a bracket coming from a symplectic form are the locally constant maps. Also, there are brackets which do not come from symplectic forms but still have some physical relevance, for example, the so called Nambu brackets (which in $Bbb R^3$ supposedly can be used to model some things about rigid bodies). I'm also not a physicist so I don't really know further details, but you can look up Holm's Geometric Mechanics book.
answered Jan 23 at 22:25
Ivo TerekIvo Terek
46.1k953142
$endgroup$
Yes, you are correct. Let $X ={f,-}$. Then for all $xin M$ and all test functions $gin C^infty(M)$ you have $X_x(g)={f,g}_x$. Let $Phi$ denote the flow of $X$. Then $$frac{rm d}{{rm d}t} g(Phi_t(p)) = {rm d}g_{Phi_t(p)}(X_{Phi_t(p)}) = X_{Phi_t(p)}(g) = {f,g}_{Phi_t(p)}.$$An arbitrary Poisson Bracket need not have any relation to $f$ whatsoever. For example, one can always consider the zero bracket in any manifold. The picture is different if the Poisson Bracket comes from a symplectic form, that is, ${f,g}=omega(X_f,X_g)$, where $X_f$ and $X_g$ are the Hamiltonian vector fields of $f$ and $g$ (in practice, that's what ${f,-}$ is in your context). I think that a more interesting question would be about the Casimir functions for a given bracket, the functions in the kernel of $fmapsto {f,-}$. For example, the Casimir functions for a bracket coming from a symplectic form are the locally constant maps. Also, there are brackets which do not come from symplectic forms but still have some physical relevance, for example, the so called Nambu brackets (which in $Bbb R^3$ supposedly can be used to model some things about rigid bodies). I'm also not a physicist so I don't really know further details, but you can look up Holm's Geometric Mechanics book.
answered Jan 23 at 22:25
Ivo TerekIvo Terek
46.1k953142
answered Jan 23 at 22:25
Ivo TerekIvo Terek
46.1k953142
answered Jan 23 at 22:25
Ivo TerekIvo Terek
46.1k953142
answered Jan 23 at 22:25
Ivo TerekIvo Terek
46.1k953142
46.1k953142
add a comment |
add a comment |
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$begingroup$
I'm in the same boat as you are (re: the last paragraph). The physicists never even mentioned the Poisson bracket (they only ever talk about the Lie bracket), but I ran across it when reading this: fisica.net/mecanicaclassica/… Two friends and I have been struggling to get over the big gap between mathematical texts description and physics text descriptions.
$endgroup$
– rschwieb
Jan 11 at 15:11