Physical pendulum in phase space

[lakmus]

Hi,
I found out this paper
http://www.pha.jhu.edu/~javalab/pendula/pendula.files/users/olegt/pendulum.pdf
with this animation
http://www.pha.jhu.edu/~javalab/pendula/pendula.files/users/olegt/pendula.html

At first there is written there, that the area of possible states in some range of energies
of pendulum in phase space is conserved due to Liouville's theorem. But at the end there
is all new bigger area uniformly filled up.
So both can't be right, and I thing that the area in the phase should be conserved. But don't
know hot to solve the argument with uniformly filled up the new bigger area which is in agreement
with equillibrium statistical physics of microcanonical ensembles . .
Thanks a lot for any response!

 

[Dale]

At first there is written there, that the area of possible states in some range of energies
of pendulum in phase space is conserved due to Liouville's theorem. But at the end there
is all new bigger area uniformly filled up.
So both can't be right,

It is uniformly but not completely filled up. Think of it like fine stripes across phase space.

 

[lakmus]

Ok, well, what if I choose just some (not all) states with exact value of energy. These would be bounded on movement on this equipotencial (in pendulum case some curve). Will these also spread over whole curve, or not?

 

[Dale]

I would have to work the math, but I believe not. I think that it is the variation in energy that causes the spread. Without that I think they would just cycle around.

 

[lakmus]

I think the same also. But in this case the distribution would not be an uniform - in the definiton of microranonical ensamble, is that the system have constant energy and in that case, in equilibrium all state have the same probability, so the equilibrium distribution function is uniform . . . so this doesn't work for the pendulum (or more clearly for the aproximative math pendulum).
So where is the problem?

 

[Dale]

I think the same also. But in this case the distribution would not be an uniform - in the definiton of microranonical ensamble, is that the system have constant energy and in that case, in equilibrium all state have the same probability, so the equilibrium distribution function is uniform . . . so this doesn't work for the pendulum (or more clearly for the aproximative math pendulum).
So where is the problem?

I am not sure what you are concerned about. If you have a microcanonical ensemble for a given energy then you have all of the states at a given energy. As those evolve over time you will continue to have all of the states for the same energy. So a microcanonical ensemble at one point in time looks the same as the microcanonical ensemble at any point in time. In that case, Louiville's theorem is trivially and clearly satisfied.

 

[lakmus]

And what about and izolated box with a gas of given energy, where at first is not equilibrium - half slow molecules and half fast, will evolving into more "homogenized" state, right?

 

[Dale]

Yes, but an isolated box with a gas of a given energy with half slow and half fast molecules is not a microcanonical ensemble. It is a single microstate within that ensemble. It will evolve to another microstate within that ensemble, and there are many more microstates which are "homogenized" than not.

 

[lakmus]

But microcanonical ensamble define by box of gas of certain enenergy will evolving in phase diagram to uniform distribution function, but the microcanonical ensamble define by pendulum at certain energy will not. This is how i get it know, and my intuition says, that not correct . . .

 

[Dale]

A single box of gas at a certain energy is not a microcanonical ensemble. The microcanonical ensemble is the set of all possible boxes of gas with the same energy. One box is just one microstate out of the ensemble.

 

[lakmus]

Yes, sure your right, wrong terminology. So let's have two izolated systems - pendulum and box with gas. We choose some subset of all posiblle states with constant given energy and watch its development in phase space.
We agreed that the pendulum will just rotate and nothink else happen.
But what will happen with the subset of all posiblle states from the box with gas?

 

[Dale]

In both cases you start with a subset of the microcanonical ensemble, and the systems both evolve to a different subset of the same microcanonical ensemble. There is no difference between the pendulum and the box in that respect.

Now, let's suppose that we partition the phase space into two regions, one much larger than the other, and let's further suppose that the original subset of each microcanonical ensemble comes entirely from the smaller region. Now, if we let each state evolve for a random amount of time and test to see if it winds up in the small or the large region then we will find that it is much more likely to end up in the large region, despite the carefully chosen initial conditions. Again, there is no difference between the pendulum and the box in this respect.

The only difference between the pendulum and the box is that the phase space for the box can be partitioned into "high entropy" and "low entropy" partitions, while the pendulum cannot.

 

[lakmus]

It kind a look like, that treating the pendulum from a statistical point of view does not have any sense . . .
But anyway, thanks a lot. Discussion with you really helped a lot.

 

[Dale]

treating the pendulum from a statistical point of view does not have any sense . . .

I agree. The phase space is too simple for a statistical treatment to be worthwhile.