Talk:Microstate (statistical mechanics)

Why finite number of microstates?
Why is there a finite number of microstates?? Quantum physics wasn't around when the idea was invented....help please.

There isn't classically, (since even infinitesimally close states in phase space are possible) - that's why you show that it's reasonable to assume that they're uniformly distributed through a given volume of phase space.

Entropy
I always think of entropy of the i-th macrostate as $$S_i=\ln(W_i)\,$$ where $$W_i\,$$ is the number of microstates in the i-th macrostate. Can you remind me how this relates to the use of the $$p_i$$ which from your definition is $$p_i=W_i/\sum_i W_i\,$$? Thanks PAR 05:57, 4 November 2005 (UTC)
 * In fact I'm not sure exactly what you mean by "i-th macrostate." But I think I understand the question though. The expression I gqve in the article works for just any system, including far away from thermodynamic equilibrium (have yet to mention that, but it's in order).


 * Then an isolated system keeps the same energy, so that it fluctuates between those of his microstates that have a given energy. Such microstates number $$\Omega(E)~$$. In this framework the general expression of S given in the article is maximal when all states are equally likely:
 * $$p_1 = \ldots = p_N = \frac{1}{\Omega (E)}$$
 * And in this framework the expression of the entropy from the article reduces to:
 * $$S = -k_B log (\Omega (E))~$$, which is Boltzman's definition of entropy. In fact it is only valid at thermodynamic equilibrium, for an isolated system. For a non-isolated system, even at thermodynamic equilibrium, it's not Boltzman's entropy that works.ThorinMuglindir 13:06, 4 November 2005 (UTC)


 * oh my god... I understand why your question was somehow unclear to me: I did a mistake yesterday in defining Ei and pi. In reality p_i is the proba associated with microstate i (and not macro as I first wrote), and Ei is the energy level of this microstate, which is also an energy level of the system. Thanks for helping me realize this. will correct at once.ThorinMuglindir 13:17, 4 November 2005 (UTC)

Internal energy
The internal energy is the mean of the system's energy This definition is the traduction of the first law of thermodynamics

Traduction gives me problems, my dictionary gives the root meaning as traduce - to misrepresent, speak ill of. Can someone clarify/correct this please?--Damorbel 10:06, 12 August 2007 (UTC)

According to wiktionary, "traduction" is not an english word. That should definitely be fixed, but I'm not qualified to do that myself. MassimoH (talk) 23:38, 26 February 2009 (UTC)

I'm doing research on microbial thermodynamics and have no idea what I'm supposed to be looking for. Any help? —Preceding unsigned comment added by 69.123.215.4 (talk) 23:41, 22 July 2008 (UTC)

-Article needs variable identification-

The variables in the equations need to be identified. For example what is the U in the Internal Energy equation? Without such identification this article is useless to anyone not already familiar with the subject, and is of no use to someone who is already familiar with the subject because the article is very basic. In other words this article as is seems useless. —Preceding unsigned comment added by 63.240.143.147 (talk) 20:58, 22 July 2010 (UTC)

Question
Do different macrostates have different internal energy, volume and no. of particles? please clarify with example.

Microstate and Macrostate
In the article, it looks like, the term macrostate has been identified with a thermodynamic state specified by thermodynamic variables like temperature, pressure etc.

However, in statistical mechanics, the term macrostate is most often used to refer to a state with specific  Energy, Number of particles and Volume. This is not really a thermodynamic state. Thermodynamic variables are averages over several microstates with appropriate Boltzmann weight determined by the particular macrostate they belong to. In this sense the article needs a revamp in order to make it in agreement with most treatments of statistical mechanics. Prathambhu (talk) 09:48, 21 March 2013 (UTC)

does it make sense to include thermodynamics here?
Currently the article is occupied by a large section on "Microscopic definitions of thermodynamic concepts", however almost none of the concepts therein can be defined for a microstate. Things like entropy cannot be defined for a microstate --- they are only defined for ensembles of microstates. As far as I know, the only macroscopic quantities associated with a microstate are particle number, energy, volume, magnetization, and so on. The main point about a microstate however is that it is not only defined by macroscopic quantities, but rather defined by the position and momentum of every single particle therein. --Nanite (talk) 10:27, 19 August 2013 (UTC)

Configuration
"a specific microscopic configuration of a thermodynamic system" — what does this configuration include? Thank you. - 89.110.6.217 (talk) 17:31, 19 November 2014 (UTC)
 * I actually think that it'd be right to write what it includes (maybe depending on some special cases) in the article, otherwise the definition looks mystical, as if it defines some thing in itself. - 89.110.6.217 (talk) 18:38, 19 November 2014 (UTC)

Quantum/classical heat vs work
For a closed system (no transfer of matter), heat in statistical mechanics is the energy transfer associated with a disordered, microscopic action on the system, associated with jumps in occupation numbers of the energy levels of the system, without change in the values of the energy levels themselves.

Isn't this true just for quantum systems? I don't know what energy levels are for classical systems.

Hmmm, that's not quite true. I can also understand energy levels for "abstract" discrete systems, such as noninteracting spins and noninteracting harmonic oscillators, defining and analyzed purely as an abstract system of energy levels without reference to quantum mechanics.

But I know no mechanism for changing the energy levels in a classical system or the kind of "abstract" system described above. I only know about the adiabatic theorem of quantum systems, which gives some justification for identifying the energy levels of one quantum system with the energy levels of a gradually perturbed version of it. Yet this theorem requires a Hilbert space operator setup (quantum mechanics) for its proof.

ADDED: I see that the quantum/classical distinction is addressed a few lines later in the article. I put in a qualification at the start of the section so the reader is not misled. I also removed the reference to sums versus integrals and just emphasized classical limits. I hope what I wrote is accurate.

178.38.117.139 (talk) 21:27, 5 February 2015 (UTC)

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micromacrostate of a black hole
The informational holographic horizon of a black hole has (or correctly is) a microstate which is also a macrostate.

Some Dutch physicists lectured that the informational holographic horizon of a black hole is pixelated but that's wrong. Information isn't lost but it's converted into momentum, rotational speed, degrees of change of the original spin etc.

Black holes come only in quantized sizes, but that's not astrophysically detectable because astronomically isn't observable, also black holes carry a surrounding halo around their informational horizon. Inside the informational horizon we have the actual black hole, inside the event horizon (different notion especially in a non rotating black hole - non rotating black holes don't exist in nature, because of the way space twirls) one simply is doomed to merge with the black hole.