Timeline for A Simple Stochastic Dynamic Billiard
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 12, 2017 at 9:45 | history | edited | Maurizio Barbato | CC BY-SA 3.0 |
added 504 characters in body
|
| Nov 12, 2017 at 8:15 | history | edited | Maurizio Barbato | CC BY-SA 3.0 |
added 44 characters in body
|
| Nov 12, 2017 at 7:48 | history | edited | Maurizio Barbato | CC BY-SA 3.0 |
edited body
|
| Nov 11, 2017 at 11:02 | comment | added | Maurizio Barbato | @BenoîtKloeckner Dear Benoît, I have emended my model according to your suggestions and I have created a new post A Really Simple Stochastic Dynamic Billiard in order to clarify definitively the matter. Iosif Pinelis's argument below shows that my conjecture holds in this new model. I kindly invite you and Iosif Pinelis to answer my new post, since you deserve the merit of the right formulation of the model. | |
| Nov 11, 2017 at 9:31 | history | edited | Maurizio Barbato | CC BY-SA 3.0 |
deleted 407 characters in body
|
| Nov 11, 2017 at 9:21 | comment | added | Maurizio Barbato | @BenoîtKloeckner You are absolutely right: the problem is the law of reflection. I have no experience in dynamic billiards (this is not my field of research): I would be very grateful to you if you could give me some reference on the stochastic models with Knudsen law you quoted. In any case thank you very very much for your insightful comments on this problem. | |
| Nov 11, 2017 at 9:12 | vote | accept | Maurizio Barbato | ||
| Nov 11, 2017 at 9:11 | history | edited | Maurizio Barbato | CC BY-SA 3.0 |
added 363 characters in body
|
| Nov 10, 2017 at 21:43 | comment | added | Benoît Kloeckner | Last remark: you can reduce to a band $\{0\le x \le a\}$ up to some harmless change of sign in half the angles, by unfolding the rectangle by the symetries with respect to the horizontal sides. This simplifies a bit the model but has no effect on the limit you are interested in. | |
| Nov 10, 2017 at 21:40 | comment | added | Benoît Kloeckner | This choice of reflection law seems to be the problem causing Iosif Pinelis negative answer to your question. By putting more weight toward $-\pi/2$ and $\pi/2$ than a typical gas inside your domain would do, you are creating many very steep trajectories, which take a very long time to hit the $a$-wall. By the way, having had a quick look at your other question, the Liouville measure ($dx dy d\theta$ up to normalization) is certainly not invariant (let alone ergodic) if you use a uniform reflection angle. | |
| Nov 10, 2017 at 21:32 | comment | added | Benoît Kloeckner | Your random reflection law is somewhat odd. It is most usual in this type of problems (especially with a physics motivation) to use the Knudsen law (replace uniform out angle by the law with density proportional to the cosine of the angle with the normal). The reason is that Knudsen law makes the Liouville measure (i.e. uniform measure on pairs (position, direction) stationary (as does deterministic elastic reflection). There are many works on such random billiard. | |
| Nov 10, 2017 at 20:15 | history | edited | Iosif Pinelis |
edited tags
|
|
| Nov 10, 2017 at 20:13 | answer | added | Iosif Pinelis | timeline score: 3 | |
| Nov 10, 2017 at 16:33 | history | asked | Maurizio Barbato | CC BY-SA 3.0 |