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seen May 1 at 14:52

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.


Mar
16
comment Hysteresis in the Lorenz Equations
Crossposting is not recommended: math.stackexchange.com/q/1191856
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.