Recover Hamilton equation from 2-form defined on phase space

  • #1
polology
7
0
TL;DR Summary
I give a summary of hamiltonian mechanics on manifold and my goal is to relate the symplectic 2-form of the phase space to the hamilton's equation by means of the flow equation.
Following Steinacker's book we can say that given the manifold ##N## of a configuration space and its tangent bundle ##TN## we define a differentiable function ##L(\gamma,\dot\gamma): TN\rightarrow \mathbb{R}## and call it the Lagrangian function. We know there's always an isomorphism between a vector space and its dual, so we call it ##I## and we obtain the cotangent bundle ##T^*N##. For future notation simplicity, I'll call it ##M:=T^*N##. This bundle is the phase space and it has ##dim(M)=2n##, so it's locally homeomorphic to ##\mathbb{R}^{2n}##. An element of ##M## is a one-form ##\tilde\omega=\sum\alpha_i(\tilde\omega)dq^i##. We then define ##p_i:M\rightarrow\mathbb{R}## s.t. ##p_i(dq^j)=\delta_i^j##. Taking it as a coefficient to define another one-form, we obtain ##\theta=p_idq^i##. It is easy to show that ##\omega:=d\theta=dp_i\wedge dq^i## and ##d\omega=0##. Since ##\omega## is closed and nondegenerative, it is a symplectic form that makes the phase space ##M## a symplectic manifold. ##p_i## and ##q^i## are then canonically related and Darboux's theorem assures us that there always exists a form defined in such canonical variables for every point of a symplectic manifold. The magic feature of this is that in local coordinates ##(x^a,x^b)## we can write ##\omega=\frac{1}{2}\omega_{ab}dx^a\wedge dx^b## and define the Poisson tensor ##\theta^{ab}## as the inverse of the antisymmetric matrix ##\omega_{ab}##: ##\theta^{ab}\omega_{bc}=\delta^a_c##. This defines a bivector ##\theta^{ab}\partial_a\otimes\partial_b## that led us to the Poisson brackets: ##\{f,g\}=\theta^{ab}\partial_a f\partial_b g## (from this we can obtain the well-known Poisson brackets for canonical variables).

Now, my goal is to relate the last expression to the Euler-Lagrange equation in Hamilton form:

$$\dot f = \{f,H\}$$

We could start by saying that for every function defined in a manifold, we can associate a vector field (projection of vector bundle on the manifold) and a one-parameter group of diffeomorphisms through the flow equation. Let then ##H(p,q): M\rightarrow\mathbb{R}## be the Hamiltonian function in canonical variables, that is the Legendre transform ##C^{\infty}(TN)\rightarrow C^{\infty}(T^*N)## of the Lagrangian defined above. I can construct a vector field ##X_H## associated with it and call it a Hamiltonian vector field. Say, it can satisfy the flow equation:

$$\frac{d}{dt}(g^t_H(x_0))^{\mu}=X^{\mu}_H(g^t_H(x_0))$$

Here, ##g_H^t(x_0)## is the one-parameter group of diffeomorphisms generated by ##H## with respect to the parameter ##t## at the point ##x_0##. Actually, (up to some notations) this is the definition of flow given by Nakahara and I want to stress that it concerns the components because this is going to be a central issue in a moment.

Nevertheless, Steinacker defines the Hamilton vector field as any field s.t.:

$$\{f,g\}=V_f[g]$$

1. I really can't understand this. The Poisson algebra is closed with respect to Poisson brackets (i.e. ##\{\cdot ,\cdot\}:A\times A\rightarrow A##), so I expect that the Poisson bracket of two functions is again a function. How is it possible that we have a vector field? Steinacker then gives proof of this going to local coordinates, but in this way, he demonstrates that ##(V_f)^a=\{f,x^a\}##. Well, now that it is an equation for the components for me it's okay, but the field? I would probably understand if what we were saying was that the function obtained through the Poisson bracket operation generates a vector field that is related to the Lie bracket of the Vector fields of f and g, but it's not what is written in that formula.

On the other hand, given two functions ##f,H\in C^{\infty}(M)## one can take the differential ##dH## and ##df## and do an isomorphism obtaining two vector fields ##IdH## and ##Idf##. Now that we have two vector fields on ##M##, we can use the symplectic form defined previously and so the Poisson bracket. So I think I can construct the 2-form ##\omega^2(IdH,Idf)##.

2. The last step would be to use these results to obtain that ##\{f,H\}=\frac{d}{dt}f(g^t_H(x_0))##. (This would probably mean that we can associate the brackets with a component(!) of the vector flow field. Is this true?) Actually, Arnold starts from this as a definition of the Poisson bracket and then uses it to demonstrate that ##\omega^2(IdH,Idf)=\{f,H\}##. I want to do the opposite, starting from the 2-form naturally existing in the symplectic manifold to obtain that relationship by means of the flow equation. How to do it?
 
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  • #2
Please edit your post to properly display the LaTeX. Note that, unlike regular compiled LaTeX, the math delimeters on the forum are double hashes ## for inline math and double dollars $$ for displayed math. Not single dollars $.

Moderator's note: post has been edited.
 
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  • #3
That’s a lot of text for a forum post, I will try to address a few points.
polology said:
Since ##\omega## is closed and nondegenerative,
Not that important on the whole, but the term is non-degenerate.

polology said:
1. I really can't understand this. The Poisson algebra is closed with respect to Poisson brackets (i.e. ##\{\cdot ,\cdot\}:A\times A\rightarrow A##), so I expect that the Poisson bracket of two functions is again a function.
So is the result of a vector field acting on a function, which is how ##V_H## is defined here:
$$
V_Hf = \{f,H\}
$$
A vector field is a derivative operator ##A \to A## and what is being said is that ##V_H## is obtained by fixing the second argument in the Poisson bracket to ##H##.

polology said:
How is it possible that we have a vector field?
See above. What you need to check - which should be trivial - is that ##V_H## defined in this way is a linear differential operator.

polology said:
Steinacker then gives proof of this going to local coordinates, but in this way, he demonstrates that ##(V_f)^a=\{f,x^a\}##. Well, now that it is an equation for the components for me it's okay, but the field?
The field is ##V_f^a\partial_a## in every point. I am not sure I understood your problem here. Perhaps you can elaborate?
polology said:
On the other hand, given two functions ##f,H\in C^{\infty}(M)## one can take the differential ##dH## and ##df## and do an isomorphism obtaining two vector fields ##IdH## and ##Idf##.
It is unclear to me what your isomorphism is here. The natural isomorphism on a symplectic manifold is defined by the symplectic form ##\omega## (and its inverse) and this is essentially how the Hamiltonian vector field is defined:
$$
\omega(V_H) = dH
$$
or, equivalently,
$$
V_H = \theta(dH)
$$

There are lots of things to address, but lets start there.
 
  • #4
Thank for your answer.

About the first question, I figured it out. I've had just a bit of confusion about the vector field, since I know that its components are functions, but I used to see it as a functional rather than a function, but I forgot the trivial thing that the functional is itself a function.

Regarding the definition of $$I$$, I followed the Arnold's notation:
Screenshot 2024-06-12 215036.png

Ok then, we have a vector field $$V_H$$ tangent to the integral curve given by the flow equation written in the question. Then we take another function $$f$$ that has perhaps another flow equation associated, is this right? For me, this makes sense considering that the Lie brackets measure the non-commutativity of two fluxes, and Poisson brackets are related to Lie brackets. So there are two vector fields associated with points in the manifold, and they can be taken as arguments of a two-form naturally existing in the manifold. Going on with my confusion about this point, is this two-form again a function because it takes an element of a manifold (the vector of a tangent bundle) and gives a real number? But how could Poisson bracket give both vectors and forms?
 

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