A Hamiltonian formulation of classical mechanics as symplectic manifold

AI Thread Summary
In the Hamiltonian formulation of classical mechanics, the phase space is identified as a symplectic manifold characterized by a closed non-degenerate 2-form, denoted as ##\omega##. The discussion clarifies that the non-degeneracy of ##\omega## implies it can be expressed locally as a linear combination of wedge products of covector bases, leading to local canonical coordinates. The non-degenerate condition is equivalent to the determinant of the associated matrix ##a_{ij}## being non-zero. Additionally, it is confirmed that this matrix is skew-symmetric due to the properties of the symplectic structure. Understanding these properties is crucial for grasping the implications of Darboux's theorem in the context of symplectic geometry.
cianfa72
Messages
2,805
Reaction score
296
TL;DR Summary
About the definition of symplectic manifold structure employed in the hamiltonian formulation of classical mechanics
Hi, in the Hamiltonian formulation of classical mechanics, the phase space is a symplectic manifold. Namely there is a closed non-degenerate 2-form ##\omega## that assign a symplectic structure to the ##2m## even dimensional manifold (the phase space).

As explained here Darboux's theorem since ##\omega## is by definition closed from Poincare lemma there exist locally a 1-form ##\theta## such that locally ##\omega = d\theta##.

However I've not a clear understanding why such ##d\theta## fulfills the Darboux's theorem hypothesis hence there are local canonical coordinates such that ##\omega## can be written as
$$\omega = dq_i \wedge dp_i$$
If ##\omega## was a rank ##m## form then by definition ##(d\theta)^m \neq 0## and of course ##\theta \wedge (d\theta)^m = 0## since it would be a ##2m+1## form defined on a 2m-dimensional manifold.

So the question is: why ##\omega## is assumed to be a 2-form with constant rank ##m## ? Thanks.
 
Last edited:
Physics news on Phys.org
cianfa72 said:
So the question is: why ##\omega## is assumed to be a 2-form with constant rank ##m## ? Thanks.
It is not assumed, it follows from the fact that it is non-degenerate. Write it locally as ##\omega=\sum_{i,j} a_{ij}\theta^i\wedge\theta^j##, then ##\wedge^m \omega = det(a_{ij})\theta^1\wedge\cdots\wedge\theta^{2m}## is non-zero exactly when ##det(a_{ij})## is non-zero exactly when ##\omega## is non-degenerate.
 
martinbn said:
Write it locally as ##\omega=\sum_{i,j} a_{ij}\theta^i\wedge\theta^j##
Ah ok, so every 2-form can be always written as linear combination of wedge products of covector (1-form) basis of the dual space at each point on the manifold.

Then the definition of non-degenerate is equivalent (iff condition) to ##det(a_{ij}) \neq 0##.
 
cianfa72 said:
Ah ok, so every 2-form can be always written as linear combination of wedge products of covector (1-form) basis of the dual space at each point on the manifold.

Then the definition of non-degenerate is equivalent (iff condition) to ##det(a_{ij}) \neq 0##.
Yes, if the ##\theta^i## form a basis of 1-forms, then the ##\theta^i\wedge\theta^j## form a basis of 2-forms.
 
martinbn said:
Write it locally as ##\omega=\sum_{i,j} a_{ij}\theta^i\wedge\theta^j##, then ##\wedge^m \omega = det(a_{ij})\theta^1\wedge\cdots\wedge\theta^{2m}## is non-zero exactly when ##det(a_{ij})## is non-zero exactly when ##\omega## is non-degenerate.
Sorry to resume this old thread, in the definition of the 2-form ##\omega##, is the matrix ##a_{ij}## (with even dimension) assumed to be skew-symmetric (with even dimension)? Thanks.
 
Last edited:
Yes.
 
Thread 'Gauss' law seems to imply instantaneous electric field'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Hello! Let's say I have a cavity resonant at 10 GHz with a Q factor of 1000. Given the Lorentzian shape of the cavity, I can also drive the cavity at, say 100 MHz. Of course the response will be very very weak, but non-zero given that the Loretzian shape never really reaches zero. I am trying to understand how are the magnetic and electric field distributions of the field at 100 MHz relative to the ones at 10 GHz? In particular, if inside the cavity I have some structure, such as 2 plates...
Back
Top