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Natural Science Forum / Physics / Research / April 2008



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Local vs Global Constants of Motion

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dan@frogfly.org - 08 Apr 2008 19:22 GMT
I'm trying to understand the distinction between local and global
constants of motion. Thirring's Classical Mathematical Physics (http://
books.google.com/books?id=NFdStDHPOfcC pg 44) gives the 1-D harmonic
oscillator as an example.

The phase-space trajectories are (x(t), p(t)) = (A sin(w t + phi), B w
cos(w t + phi)), and the constants of motion are energy = x^2 + (p/
w)^2 and phase = arctan(w q / p) - w t.

The energy is clearly global, but he says that the phase angle is only
defined locally. I see that the arctan is discontinuous across the p
axis
so that this particular expression doesn't work. But I keep thinking
that
there ought to be a way to fix this by adding some other "correcting"
discontinuous function to the expression.

Clearly you could do this for any *particular* trajectory. Say phi =
0. Then we could have something like

arctan(w q / p) - w t + ...
  {     0 for     0 < w t <= pi/2
      -pi for  pi/2 < w t <= 3pi/2
    -2 pi for 3pi/2 < w y <= 5pi/2
    etc
  }

But I think that you would need different "correcting" functions for
different initial phase angles. So you couldn't have a function of
*only*
(q,p,t) that is constant for *all* trajectories. And by a *global*
constant
of motion we mean a constant of motion that applies not only at all
times
for given trajectory, but that applies to *all* trajectories.

Do I have this right?

Thanks,

Dan
--
Dan Becker
Igor Khavkine - 11 Apr 2008 01:06 GMT
On Apr 8, 2:22 pm, d...@frogfly.org wrote:
> I'm trying to understand the distinction between local and global
> constants of motion. Thirring's Classical Mathematical Physics
[quoted text clipped - 27 lines]
> applies not only at all times > for given trajectory, but that applies
> to *all* trajectories.

Physically, the phase doesn't change when you add 2pi to it. Therefore,
we must identify phi with phi+2pi. Topologically, the phase variable
lives on a circle. For example, its the same circle as the result of
identifying the endpoints of the [0,2pi] interval. The identify function
on this interval, basically the phase variable phi, is not continuous.
It cannot be continuous for the same reason that the circle needs an
atlas of at least two charts to fully parametrize it.

Note, however that cos(phi) and sin(phi) are perfectly continuous,
smooth functions of the physical phase. They are as well globally
defined as the energy.

Hope this helps.

Igor
dan@frogfly.org - 12 Apr 2008 14:30 GMT
> On Apr 8, 2:22 pm, d...@frogfly.org wrote:
>
> Physically, the phase doesn't change when you add 2pi to it. Therefore,
> we must identify phi with phi+2pi. Topologically, the phase variable
> lives on a circle.

Yes, that's the root of the problem! What's interesting to me about
this example is that you *can* cover all of phase space with the (q,p)
chart, and yet there is still a local constant of motion that can't be
extended globally.

Conversely, Thirring gives an example of a 2-D manifold that *can't*
be covered with a single chart, and yet, for at least some vector
fields,*does* have a global constant of motion: manifold is a torus,
coords (phi1, phi2) with vector field (w1, w2). If w1 and w2 are
rationally related, then sin(w2 phi1 - w1 phi2) is globally defined.

I know it's a basic fact about vector fields on manifolds that they
can be "straightened" locally, but not always globally. This somehow
freaks me out, but I'm not sure why; I'm very comfortable with the
fact that there are tons of coordinates systems on, for example, R^2,
that aren't defined globally. It should probably freak me out more
that vector fields can be straightened at all than that the
straightening isn't always global!

Thanks for the response,

Dan
Chris H. Fleming - 12 Apr 2008 14:30 GMT
On Apr 8, 2:22 pm, d...@frogfly.org wrote:
> I'm trying to understand the distinction between local and global
> constants of motion. Thirring's Classical Mathematical Physics (http://
[quoted text clipped - 39 lines]
> --
> Dan Becker

Tangent has no global inverse function as it isn't bijective. Such a
thing is impossible to construct. However you can find some local
domain where Tangent is bijective and define a local inverse function.
That is what is on your calculator.

http://en.wikipedia.org/wiki/Inverse_function#Partial_inverses
 
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