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First of all, this is a follow-up of my first question. The idea is the same, every state would consist of some particles in a line (with an energy associated) and the particles can not be nearer than $D$ units between each other.

Now I have to add the possibility that the number of particles changes (it is of course not greater than $\frac{M}{D}$) and I thought of using the grand canonical ensemble, but I have yet again some doubts.

First, why is it not more likely for such an ensemble to have 0 particles than more than 0? I mean, I see in the formula that $P \propto e^{\frac{\mu N - E(s)}{kT}}$ but I don't understand why, if it costs energy to get a new particle, the probability is larger. What am I failing to see?

Second, how can $\mu$ be derived, how would you derive it in a normal experiment? This is not totally necessary (I can simply adjust it until my simulations match the real data) but I would like to know as an illiterate.

Finally, for the Boltzmann distribution case, I used a Metropolis-Hastings algorithm to approximate the distribution function. Is it advisable here as well or is there any other better method?

Thank you very much for your help!

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    $\begingroup$ Chemical potential works like potential energy. So, (out of thermodynamical equilibrium) the spontaneous behaviour is the decreasing of chemical potential, and it corresponds to an increasing of entropy. On the other way, increasing internal energy (without work) corresponds to an increasing of entropy. So, the 2 behaviours are different, and it explains the minus sign. $\endgroup$
    – Trimok
    Oct 8 '13 at 9:20
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First, why is it not more likely for such an ensemble to have 0 particles than more than 0? I mean, I see in the formula ... but I don't understand why, if it costs energy to get a new particle, the probability is larger. What am I failing to see?

This is basically the same as the classic "energy versus entropy" thing, except with particles instead of energy. It does cost energy to have a particle in the system, but the states with one or more particles outnumber the states with no particles, and this outweighs the energy cost.

Second, how can be derived, how would you derive it in a normal experiment? This is not totally necessary (I can simply adjust it until my simulations match the real data) but I would like to know as an illiterate.

I'm sure chemists have all sorts of clever ways to measure chemical potentials, but it all comes down to the fact that, similarly to temperature, chemical potentials become equal at equilibrium. (This isn't true if there are chemical reactions in the system, due to the way chemical potential is defined, but in that case they go into exact ratios that can be calculated.) So if you have some kind of standard system that changes in some consistent way depending on the chemical potential of some species, then you can measure chemical potentials by putting this system in contact with other systems, in much the same way that you use a thermometer to measure temperature. pH indicator paper is an example of this. (pH being essentially the chemical potential of H+ ions, but measured in weird units.)

Finally, for the Boltzmann distribution case, I used a Metropolis-Hastings algorithm to approximate the distribution function. Is it advisable here as well or is there any other better method?

Yes, Metropolis-Hastings should be fine in this case as well. Of course, you want to be calculating $e^{(\mu\Delta N + \Delta E)/kT}$ for transitions that change the number of particles, rather than just $e^{\Delta E/kT}$. Apart from that the algorithm is the same.

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  • $\begingroup$ I'm sorry for the short-ish answers - I'm happy to refine them if you have any questions. $\endgroup$
    – Nathaniel
    Oct 8 '13 at 9:25
  • $\begingroup$ The "outnumbering" part always confuses me, but I think I understand! And also, the increase of energy cost of adding a new particle is already counted in $\Delta E$, so it is starting to make sense! Thanks very much for the answers, I wonder why statistical physics is not as famous as other fields, for me it has a lot of intelligence and knowledge behind. $\endgroup$ Oct 8 '13 at 9:33
  • $\begingroup$ Duronman, I think it's because often statistical mechanics is taught in a painful and inelegant approach of the 19th century tradition. I know I would have been much more pleased and edified if my instructors had just taught me straight Gibbs. $\endgroup$
    – Nanite
    Nov 12 '13 at 0:37
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One way to understand why the number of particles is not zero is to see $\mu$ as accounting for the presence of a reservoir of particles in contact with your system in the same way that temperature accounts for the presence of a energy reservoir in contact with your system.

The reason why you get a factor $e^{-\beta E}$ in the canonical ensemble (where $\beta = \partial S/\partial E$ in the reservoir) is because the additional energy you put in your system is taken from the energy reservoir and this decreases the entropy of the reservoir so such a move should be unfavoured and indeed it is exponentially unfavoured.

In the case of a particle reservoir, taking $\delta N$ particles from the reservoir to add them in the system increases the entropy of the reservoir by an amount $\delta N\mu/T$ (because $TdS = dU-\mu dN$). It is thus exponentially likely as $e^{\beta \mu \delta N}$.

Basically, when it comes to Gibbs ensemble with reservoirs what matter is how "happy" will the reservoir be: in the canonical ensemble, the reservoir is happy if you don't take too much energy from it while in the grand canonical case it is happy if you take particles from it because you relieve it of the cost $\mu$ of having these additional particles.

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