Eh9

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mind yer p's and q's

Liouville's theorem (Hamiltonian). Earlier versions of this article had this proved (good enough for a simple physicist like me anyway). And it's on the web in some graduate physics/chemistry courses. I recollect you consider the stretch of a little dp.dq ' product and apply Hamilton's relations. Bob aka Linuxlad (talk) 19:50, 21 March 2008 (UTC)

1) Can you spell out how you're thinking through the coupled pendulums please - the phrase 'transfer of phase space' isn't clear to me.

2) The statement is meant to apply to a small aligned hypercube for short times, when I think the effects you're worried about are are second order. Clearly at longer times the volume bloweth where it listeth. Bob aka Linuxlad (talk) 13:00, 23 March 2008 (UTC)

On the double pendulum - can we check we share the same understanding of the basic mechanics? If we talk first about the single pendulum, for which the phase trajectory is just an ellipse I recollect, then I make up a mini ensemble of 4 identical pendulums, slightly displaced in peak amplitude and in position so that I have four marker points in the p, q plane. Then I assert that these 4 points will move, keeping close together such that dp.dq is constant? OK?

• • • NOTE (dp.dq) is just an ordinary numerical product - I don't do wedges!

Turning to the double case, and looking at the Hamiltonian at [1] then that clearly has some egregious behaviour when the singularities at the denominator are hit. So if I choose my marker points too widely spaced they will indeed move away rapidly. (I presume that's what you mean by 'transfer of phase space volume?) But in evaluating a phase space density that doesn't really worry me- if I choose close enough points _almost everywhere_ the differentials are well behaved and the behaviour I suggested follows, for small times... Bob aka Linuxlad (talk) 14:27, 24 March 2008 (UTC)

You can take four marker points if you like, but what is preserved is the area of the quadrilateral they form.

yes, that's what I'd just said, I thought :-) (Is my simplistic notation confusing you?*)

Nothing about the distances is preserved exactly.

I didn't claim that - quite the reverse

Even if you choose them initially as a parallelogram, the configuration won't (in general) stay a parallelogram forever.

I'm not interested in forever, just want to show the change in the product is 2nd order in time.

Thus talking about $\Delta q$ and $\Delta p$ even in one dimension isn't well-defined.

In fact, an initial quadrilateral doesn't even stay a quadrilateral, because the sides deform into curves (again, in general) under a Hamiltonian flow.

So? - I'm talking short times, small dp & dq and almost everywhere,and looking at their product.

It's natural to talk about $dq$ and $dp$, and we're talking differential forms and exterior algebras, and this is the minimum amount of apparatus to make those objects well-defined, but computations are easier. The elementary mathematical object that is well-defined is an area of a region as it moves under the flow.

Second time, already used this idea I thought. *

This is what is typically illustrated with what's called a "cat map"; look it up, you'll see. If you want to do a computation of areas, you'll need to integrate, and you'll end up computing that differential form anyway.

Using the notation from the wolfram.com site you referenced, the coordinate variables are $\theta _{1},\theta _{2}$ and the conjugate variables are $p_{\theta _{1}},p_{\theta _{2}}$. Those denominators are positive-definite, so there's no need to deal with undefined points in this coordinate chart.

OK, conceded

One way of stating Liouville's theorem says that the symplectic form for the whole space, $d\theta _{1}\wedge d\theta _{2}\wedge dp_{\theta _{1}}\wedge dp_{\theta _{2}}$, has norm 1 constant in time. The claim I am challenging is the same statement made about a pair such as $(\theta _{1},p_{\theta _{1}})$. The differential form in question is $d\theta _{1}\wedge dp_{\theta _{1}}$ (in this case, that norm is co-area). If this claim is true, that norm will be constant with value 1. If you work that out for the double pendulum, you'll see that it's not constant.

Let me think about that - I only need it to be 2nd order in dt I think

A consequence of Liouville's theorem for a 4-dimensional phase space is that the corresponding norms for the pairs $(\theta _{1},p_{\theta _{1}})$ and $(\theta _{2},p_{\theta _{2}})$ are multiplicative reciprocals.

Is it?! Well here is a key point for me to try to fathom in your argument - it's my understanding that it's (dp1.dq1) that locally balances at short times, not (dp1.dp2) or similar. And my limited maths seems to show it's true.And, physically, it's the result I'd expect.

When that for $\theta _{1}$ is greater than 1 and thus that for $\theta _{2}$ is less than 1, phase volume is being (instantaneously) transferred from leaf 2 (shrinking) to leaf 1 (gaining). This is what I mean by transfer of phase volume.--Eh9 (talk) 03:32, 25 March 2008 (UTC)

see preliminary comments in line above

• Is my simplistic notation confusing you? (and vice versa) I don't do differential forms, sorry!

bob aka Linuxlad (talk) 08:35, 25 March 2008 (UTC)

• Liouville's Theorem is an exact theorem, not an approximation. It's exactly true about regions, even a region that starts off as a quadrilateral and immediately (in general) becomes not-a-quadrilateral. (I misspoke slightly in a way that might be construed differently than I meant.) If you want to make an argument about second-order corrections, you are simply duplicating---for a single case---the proofs that use differential forms.
• On differential forms: When you compute an area in a non-Cartesian coordinate system (such as polar coordinates), you are using differential forms, even if you don't know how their rigorous definition. Computation of a "volume element" or "area" element is almost the entirety of computing the corresponding differential form. The symbols $dp_{i}$ are differential 1-forms (1-dimensional). The exterior product $\wedge$ is anti-commutative, which is why using a dot for multiplication isn't used. It's anti-commutative because it swaps orientation and thus the sign of an integral. And you probably do know something about differential forms; they're a kind of tensor.
• As for short times: If you only think about approximations, you must operate only locally both temporally (short times) and spatially (small distances). This is all wrapped up within the concept of a differential form. When you look at forms, all those local considerations are taken care of and you can start thinking about global geometry. Liouville's theorem is true for all time, not just for short (local) times.
• And a clarification: A Hamiltonian flow, often denoted $F_{t}$, acts on volume forms. The canonical symplectic form of Darboux's theorem is $\omega =dp\wedge dq$. The statement of Liouville's theorem in this notation is $\forall t>0:F_{t}\omega =\omega$. Let $\omega _{i}=dp_{i}\wedge dq^{i}$. Modulo sign, $\omega =\prod \omega _{i}$. But in general $F_{t}\omega _{i}\neq \omega _{i}$. This is a precise version of what I was saying about forms staying constant.--Eh9 (talk) 17:27, 25 March 2008 (UTC)
• A simpler system that exhibits the same qualitative behavior in question (transfer of phase volume) is a pair of simple harmonic oscillators that are coupled. I don't remember if bilinear coupling is adequate or whether you need nonlinear coupling.--Eh9 (talk) 17:37, 25 March 2008 (UTC)

Unfortunately you edited just as I was stripping down my thoughts. As a result we're in danger of deepening the mutual confusion. Let's keep clear of the differential forms as far as possible and try to share a common maths language please. I don't need to think all space or long time to justify the statement we're querying (which is NOT Liouville in its entirety) - it's a statement of short times I'm interested in so terms of order dt squared don't worry me (see the statement calculating the Jacobian for a bit of Phase Space on the Wolfram site.)

With this in mind, how do _you_ reckon $d\theta _{1}\wedge dp_{\theta _{1}}$ (for your double-pendulum) varies??? (OK it looks like it's O-level or stoking up my old MathCad 7, but you've obviously got it to hand)

At some stage we need to get the final result down to a reasonable level. Few physicists of my generation are really happy with differential forms. (Yes, I do have a copy of Penrose, and I note his mentor was Denis Sciama, who taught his GR in ordinary suffix notation :-)) Bob aka Linuxlad (talk) 17:58, 25 March 2008 (UTC)

If I go through the effort of working out a counterexample in its entirety, I will only do so in order to publicly challenge the accuracy of the article. You have quite sufficient information at this point to work out a counterexample to your own satisfaction, even using an approximation method. Outside people who get involved in the issue will have no problem understand my argument.--Eh9 (talk) 11:24, 26 March 2008 (UTC)