140 reputation
6
bio website
location
age
visits member for 3 years, 7 months
seen Nov 29 at 13:54

As somebody used to say:

Does research. Smokes. Battles administration. Smokes. Wishes he could stop battling administration so that he could have more time to do research. Smokes some more.

The same. Except I do not smoke.


This paragraph is for my personal use but freely available:

Welcome to Math.SE! Please, consider updating your question to include what you have tried and where you are getting stuck. That way, people on this site will know exactly what help you need.


Nov
29
awarded  Commentator
Sep
24
awarded  Autobiographer
Jun
9
comment Brownian motion and physical meaning
The question seems to be "What is BM in physics?" You could try physics.SE.
Apr
7
answered Random walks on resistive network
Feb
8
comment Probability of finding n particles in a volume v
Sorry but to believe that, for 10 indistinguishable particles and 2 equal volumes, the split 0-10 has as much chances than the split 5-5 to happen, is to exhibit a deep misconception of what is going on. (And I do not plan to comment further on this question.)
Feb
7
comment Probability of finding n particles in a volume v
Do you really believe that? Wow.
Feb
7
comment Probability of finding n particles in a volume v
Honestly I did not try to check the computations in your revised version but I see no reason why $p_n$ should not be eactly equal to $p_n={N\choose n}x^n(1-x)^{N-n}$ with $x=v/V$, for every $n$.
Feb
7
comment Probability of finding n particles in a volume v
Three comments, and I still do not see where you address the point I (after others) made. (Maybe a problem of language?) Anyway, first, $V$ is irrelevant, only the ratio $V/v$ matters. Second, as already said, for $V/v=2$ fixed (no limit here), the distribution of $n$ is not uniform (for no $N$ except $N=2$). On the contrary it is concentrating pretty quickly arounf $N/2$ when $N$ grows. Already for $N=16$ (hardly of Avogadro size), the range $(25\% N,75\% N)$ receives more than $95\%$ of the total mass, not the $50\%$ your solution predict.
Feb
6
comment Probability of finding n particles in a volume v
OP: No idea what made you accept the posted answer but be aware that it is not correct, for the reason explained by Yvan above.
Feb
6
awarded  Critic
Feb
6
comment Probability of finding n particles in a volume v
Hmmm... So, what you are saying is that, with N particles and two cells, the first cell has twice as many chances to receive between 0% and 20% of the N particles than between 45% and 55% of them because there are N/5 states in the first case and N/10 in the second case? This seems as wrong as can be since, when N grows large, the probability of the 45%-55% range goes to 1.
Jul
31
awarded  Supporter
May
7
revised Collision time of Brownian particles
added 52 characters in body; added 27 characters in body
May
7
revised Collision time of Brownian particles
added 22 characters in body
May
7
comment Zero magnetization of spin model without external magnetic field
@Marek: Good job. A minor quibble: if Gibbs measures are phases then $\mu_0^\beta=\frac12(\mu_-^\beta+\mu_+^\beta)$ is also a phase. As you know $\mu_0^\beta$ is the thermodynamic limit for a free boundary just like $\mu_\pm^\beta$ are thermodynamic limits for $\pm$ boundaries. Hence $\mu_0^\beta$ is not only a mathematical artefact (contrarily to what a too fast reading of your post could imply).
May
7
awarded  Teacher
May
7
awarded  Editor
May
7
revised Collision time of Brownian particles
added 632 characters in body
May
7
comment Zero magnetization of spin model without external magnetic field
At $\beta=\beta_c$, there is only one phase.
May
7
answered Collision time of Brownian particles