Post by: napdmitry on 07/05/2023 14:30:50
Post by: alancalverd on 07/05/2023 14:47:05
Hence, not all the equal energy states are populated, which makes the system nonergodic.non sequitur
Post by: Eternal Student on 07/05/2023 15:38:31
So, first, I would like to share the information above...It may help if you can provide some references or links. Sadly the forum software won't allow you to provide a working hyperlink until you've made a few posts. That's to reduce the amount of advertising and spam etc. This means your first post just may not be as productive as you hoped - sorry.
A request of mutual benefit:
You could start to increase the number of posts you've made by replying to this forum thread:
https://www.thenakedscientists.com/forum/index.php?topic=86280.msg704124#msg704124
The basic question in the OP (Original Post) was about how or why this site might attract and retain new members. The OP (Original Poster) @Zer0 hasn't had many replies and even just a short sentence from a new member would be appreciated.
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And, second, to ask for any other examples of systems with known uneven laws of energy partitioning.I can at least offer one example of such a system. The behaviour can sometimes depend on the substance rather than on the container in which it is put.
A common example (of ergodicity breaking) is that of spontaneous magnetisation in ferromagnetic systems, whereby below the Curie temperature the system preferentially adopts a non-zero magnetisation even though the ergodic hypothesis would imply that no net magnetisation should exist by virtue of the system exploring all states whose time-averaged magnetisation should be zero.
https://en.wikipedia.org/wiki/Ergodic_hypothesis#Phenomenology
Best Wishes.
Post by: napdmitry on 07/05/2023 16:13:59
Post by: napdmitry on 07/05/2023 20:18:53
fig1.jpg (56.66 kB . 778x341 - viewed 660 times)
Post by: alancalverd on 07/05/2023 23:33:33
Post by: Eternal Student on 08/05/2023 04:52:32
I've re-read your posts but without glancing at the articles or paper your talking about, it's not really all that clear (to me) what is being done.
Consider the usual ideal gas,...You could be using a purely Newtonian model for that. However this later statement...
Hence, not all the equal energy states are populated,...suggests that you could be using a Quantum Mechanical model. The later post with the diagram seems to emphasize angular momentum as a Newtonian concept instead of as an Angular momentum operator acting on a QM system. Your model may be a mixture or hybrid (which is also common).
This statement may be incorrect:
Hence, not all the equal energy states are populated, which makes the system nonergodic.I'm not sure if you mean "energy states" as the term is used in QM, or some reference to paths through position and momentum space.
I think we need to establish what ergodicity is and/or how you are considering it.
Ergodicity can be considered as a property of an observable where an average over time is equal to an average over multiple systems (an ensemble). You might be considering ergodicity as the property of a system suggesting that it will visit (and re-visit) all parts of position-momentum space over large times. The two definitions or approaches are compatible but it's easier to consider the first definition and then the second will follow for Newtonian systems. For Quantum Mechanics the first definition or approach is nearly always easier to use (the second version or definition is just complicated for QM and some discussion appears later).
You ( @napdmitry ) might already know this but if anyone else is reading then a diagram like this one will help:
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Diagram snipped from a video - original creator was the London Mathematical Laboratory, I think.
The diagram shows values for an observable X observed in 5 independent but identical experiments (well, identical as much as possible, but the variable X is inherently random. For a deterministic system, the initial value of X was random). The random variable X(t) is ergodic if the average you obtain over the 5 different systems at one instance of time (the red shaded average of values) is equal to the average you would obtain by just following any one system over time (the green shaded average of values). Formally, you consider the limit as the number of identical systems and the time over which an average is taken → ∞.
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1. Now, if you're using a Quantum mechanical model, then your observables are given by operators (... usual QM..). So, you might be tempted to consider the position and momentum of a gas particle as an observable BUT they are certainly not clearly defined straight line paths like that shown on one of your later diagrams. Particles can be found in all sorts of places with some probability and uncertainty relationships will limit simultaneous position and momentum determinations. Overall it would be a difficult task to demonstrate that these sorts of observables won't be ergodic.
2. If you're using Newtonian mechanics, then particles should have the straight line paths as illustrated in your diagram and you can know the position and momentum simultaneously etc. However, the entire model was idealised. Any perturbation will put the particle onto a different path. For example, the usual assumption in an ideal gas is that collisions between particles are negligible but some assumption about randomness of trajectories is also made. Exactly how you reconcile all of the assumptions is open to interpretation and it doesn't matter, we just want those assumptions and see what consequences follow. However one reasonable and practical interpretation is that there could be collisions or close approaches where some inter-atomic forces deflect the particles. They are rare enough to be ignored in most calculations but sufficient to perturb the trajectories and ensure that over long durations of time, the particles do travel on all paths and visit all parts of the position - momentum space.
But what is the law of partitioning, instead of the even one? That is the problem.You might be talking about partitioning between different ways of accounting for energy macroscopically. For example, an ideal gas will tend to have an equal amount of motion (kinetic energy) in the x-, y- and z- directions.
On a more microscopic level, I have to agree with @alancalverd , the Boltzmann distribution describes the situation quite well in most situations.
Provided a system can be considered to be in contact with a heat bath that maintains temperature T and a given microstate of your system has energy Ei then the probability of the system being in that microstate is given by:
Pi =
with Kb = Boltzmann constant; Z = partition function =
Various developments are possible from there.
---This post is already too long. I'm ending. It's the best I can do for a discussion at the moment, sorry.
Best Wishes.
Post by: alancalverd on 08/05/2023 09:52:01
Here is an illustration explaining the difference between rectangular and round vessels. In the rectangular vessel, the trajectory fills the surface of constant energy, while in the round one only zero measure subset of it.I can't open the attachment but at a glance the illustration is of a finite number of finite objects making elastic collisions with the walls of a vessel but not with each other. An ideal gas consists of a countable infinity of infinitesimal objects that make elastic and indistinguishable random collisions with each other and the walls, so the shape of the vessel is immaterial.
Note that this only applies in principle (and definitely in practice) to a large vessel where the surface to volume ratio is negligible. Getting the last molecule out of a tight corner is actually very difficult - one of the reasons that ultrahigh vacuum equipment and glass food jars are mostly of circular section - inefficient packing but sterilisable with 100% accessible content..
Post by: napdmitry on 08/05/2023 14:17:45
To add, the arguments above are easily checkable with simple numerical modeling. I've been doing it, some other guys I know, and anyone else could reproduce it for no cost. I'm just sharing some information that I'm sure is correct and may be new and interesting to the adepts of physics.
Post by: Bored chemist on 08/05/2023 16:56:38
That's going to make it difficult to talk about thermodynamic properties.
Post by: alancalverd on 08/05/2023 17:22:51
There will be a pressure gradient perpendicular to the axis of rotation if there is no turbulent disequilibrium but that does not degrade the ergodicity of the system: it merely reflects a modification of the Boltzmann distribution.
Post by: Eternal Student on 09/05/2023 16:44:57
I've started looking through the paper you attached. It's fairly hard going (for me). The material itself is complicated and I'm also going to assume it was originally produced in some other language and translated. The two things together make it quite heavy work to read through. I don't speak any other language well, so an original version of the paper won't help me at all, sorry.
I have a passing familiarity with measure theory and I think I've started to get the gist of page 2.
There don't seem to be any references cited for anything after the introduction on page 1, so you've made it quite difficult for anyone other than an expert in precisely that area to follow or fill in any blanks in their own knowledge.
Let's take this sentence as an example:
The filling density of an invariant surface of a Hamilton system is known to be uniform with respect to the special measure, which is called ergodic.
It took me about 15 minutes to make some sense out of that. (The main problem being that a measure shouldn't be ergodic, there should be some Transformation that has that property for that measure).
Overall, this is what I think you were saying:
There is a measure preserving system (X, B, μ, T) where X = the 4N dimensional position-momentum space ( a subset of ℜ4N) ; B = a σ-Algebra of the measurable subsets; μ is a measure defined on B; and T is a transformation T: X → X.
Since you're considering the orbit of a point describing your system it will be the set {x, Tx, T2x, T3x, ....} that you'll be interested in. Where T is some transformation that maps the state at time t to the state at time t+δt. Your reference to the "Hamilton system" was presumably to assert that T is uniquely determined by the Hamiltonian. I don't think the measure can be "called ergodic". T can be μ-ergodic or equivalently you could say that the system (X, B, μ, T) is an ergodic system. So you are asserting that the measure is one which will make T μ-ergodic. Your reference to the measure being "special" could be that it's a probability measure.
That might all be common enough knowledge for an expert in precisley that area. I'm not one of those. I'm just able to recognise some mathematical terms. For someone like me, the original sentence is just technical terms put together in what might as well be random order. They do seem to be the right terms, it's just making sense of it takes time.
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There is also this sentence, in your article, which seeems useful to address the question in the title of your post.
For an ideal gas in a round vessel, this surface is the intersection of surfaces of constant
energy and constant angular momentum. It makes the system non-ergodic in a sense that the surface of constant
energy is not filled densely
It is in only in one very specific sense, defining ergodicity as some requirement that the system should visit all places representing constant total energy.
Two samples each of N gas particles in circular containers which both have the same total angular momentum L and total energy E, should have the same surfaces of constant energy and of constant angular momentum. So that their behaviours should be the same. Any one system evolving over time should visit all the same parts of the momentum-position space as you would observe in an analysis over an ensemble of systems over the same instance of time. That is a common and acceptable understanding of ergodicity (as in post #6 with the diagram). It just so happens that the permitted regions of the momentum-position space didn't include all possible energies - but, that's ok.
To paraphrase that, the angular momentum condition doesn't prevent observables of the system being ergodic, it just reduces the region of the position-momentum space that could be visitied.
Summary:
Some effort has been made to study the paper (see the above example, 15 minutes on just one sentence) but by the time I was at the end of page 2, I was defeated. That's quite probably a lack of expertise on my part. I don't really have enough time to go over all the other pages, sorry.
I'm just sharing some information that I'm sure is correct and may be new and interesting to the adepts of physics.Thank you. I'm not a really an adept of physics, or to paraphrase that - I know little and my opinion isn't one you need to worry about.
Best Wishes.
[Minor note, the Spell Check facilitiy seems to have been removed from the reply interface. I'll anticipate that there are spelling and typing errors, sorry].
Post by: napdmitry on 09/05/2023 17:55:44
Quote
It just so happens that the permitted regions of the momentum-position space didn't include all possible energies - but, that's ok.
That's ok, just all the particle distributions are different from that in other vessels. The equipartition theorem does not work and particles have different mean energies. The time average does not coincide with the average over the microcanonical ensemble. And what ensemble will people use for an isolated ideal gas?
Post by: Zer0 on 18/05/2023 21:08:06
& Welcome to the TNS Forum.
Hope the issue with posting photos/images was Resolved?
Post by: napdmitry on 19/05/2023 11:08:56