Re: Symplectic property
- From: DP <pfennige@xxxxxxxxxxxx>
- Date: Mon, 24 Apr 2006 12:53:11 +0200
The flaw is to assume that a triangle is
transformed into a triangle.
Instead, you should check that any infinitesimal surface
elements conserves its area. This is achieved by
linearizing the map, and checking that the two
eigenvalues l1, l2 of the linear map are reciprocal
(l1 = 1/l2). For higher dimensional maps the
reciprocal eigenvalues should come in pairs.
Jos Groot wrote:
A rather detailed question, I'm afraid....
I tried to check the symplectic property
for the standard map as given in
Edward Ott's 'Chaos in dynamical systems'.
This map is proven to be symplectic in
the book. However, I did not succeed. Maybe
one of you can help me out?
My example: I have a triangle defined
by the three points (0,0), (0,1) and (2,0)
The standard map is (Eq.7.15a and b):
theta(n+1)= (theta(n)+p(n)) mod 2*pi
p(n+1)= p(n)+K*sin(theta(n+1))
Assume K=1. Applying the map equations
once to each of the vertices gives:
(0,0): (0,0).
(0,1): theta= 1, p= 1+sin(1)= 1.84.
(2,0): theta= 2, p= sin(2)= 0.91.
This gives a second triangle with vertices
(0,0), (1, 1.84) and (2, 0.91).
Because this is a simplectic map the map
is area preserving and the area of the
two triangles should be the same, I
think. However, the area of the first
triangle is 1, that of the second one
1.39.
What am I doing wrong?
Jos Groot
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