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Re: [Phys-l] Classical Adiabatic Invariant



The average energy of such a pendulum ~ [omega*position]^2, and the ratio
E/omega

*is* an invariant (I'm not sure as to what you mean by "classical"), based on the constancy of the phase-space-trajectory area traced out by {x(t), (v(t)}.


 
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________________________________
From: John Denker <jsd@av8n.com>
To: Forum for Physics Educators <phys-l@carnot.physics.buffalo.edu>
Sent: Wednesday, June 8, 2011 12:00 PM
Subject: Re: [Phys-l] Classical Adiabatic Invariant

On 06/06/2011 10:53 AM, Bernard Cleyet wrote:
I'm attempting to understand this as applied to a simple pendulum.
The examples I've  found apply to shortening the rod.  Wherein
E/omega is a constant.  Shortening the rod increases the frequency,
so the E must increase.

On 06/06/2011 06:26 PM, we got a clarification:
The changes are very slow.  i.e.  The increase  of v. ~ 60
microgals, (in ~9.8E+8micro. ) which "takes" ~12 hours.  The period
of the pendulum is 2Hz.

In the limit of very slow changes, this is an easy problem.
It involves nothing more than high-school physics.

If you approach it the right way, you should be able to do
it in your head, in less time than it takes to tell about it.

Hint:  The energy is *not* proportional to omega.

Bigger hint:  There are some classical invariants that can
be invoked.  E/omega is just not one of them.

Ginormous hint:
Ercynpr gur ebq ol n fgevat gung cnffrf guebhtu n ubyr va
na vzzbinoyr gbc-cyngr.  Chyyvat ba gur fgevat unf gur fnzr
arg rssrpg nf fubegravat gur ebq ... ohg vg znl or rnfvre
gb nppbhag sbe gur sbepr ba gur fgevat.
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