**'equiprobability principle'**which states that

... in a state of thermal equilibrium, all the accessible microstates of the system are equally probable

It just looks like a very innocent and trivial sentence but no one states that this is the very foundation of all equilibrium statistical mechanics and I must assure you that it is far from trivial! I had a chance to tackle with this notion this term when I was working on a computational project for my graduate Statistical Mechanics class which involves the

*'Monte Carlo Simulation of Hard Spheres'*focusing on ergodicity of Boltzman Statistical Mechanics and Classical Mechanics. I will be trying to share my*'enlightment'*with a couple of posts starting with this one.
First of all, we need to state that there is no complete axiomatic derivation of equilibrium statistical mechanics from the foundations of classical dynamics. We need additional inputs such as the one above in order to get it. Thus the above statement, the equiprobability principle, plays a significant role in statistical mechanics.

Consider an isolated, closed Hamiltonian system with no energy and particle exchange allowed with the surrounding, for example a gas in a closed container. We can characterize the system with the Hamiltonian function $H(q,p)$ if the involved dynamics has a time translational symmetry and $q$'s and $p$'s are the corresponding degrees of freedom (DOF) of the system. We can measure the value of $H$ with a precision, say, $\delta E$: \[H(q,p) = E + \delta E \tag{1}\]This specification of the energy of the systems can be thought as a constraint, thus once we specify $E$, we are constrained to move on an energy

Consider an isolated, closed Hamiltonian system with no energy and particle exchange allowed with the surrounding, for example a gas in a closed container. We can characterize the system with the Hamiltonian function $H(q,p)$ if the involved dynamics has a time translational symmetry and $q$'s and $p$'s are the corresponding degrees of freedom (DOF) of the system. We can measure the value of $H$ with a precision, say, $\delta E$: \[H(q,p) = E + \delta E \tag{1}\]This specification of the energy of the systems can be thought as a constraint, thus once we specify $E$, we are constrained to move on an energy

*hyper surface*(a shell in phase space) which is defined by $(1)$. In the absence of further information, there is little thing one can do except our experience with general classical systems with many DOF tells that this systems is likely to be chaotic; it is likely to have positive Lyapunov exponents and it wonders around such that energy surface is filled up (i.e*ergodic*- more on that in the next post).*The simple representation of a phase space of the system. The dimension of the whole space in this case is 24, thus collection of configuration and momentum axis are depicted with only one axis for representational purposes (Source).*

Due to this energy constraint, it is clear that the whole space is not accessible to the system. Our assumption is, any one portion of this energy shell is as likely to be occupied as the other part of the shell. Actually, we can regard this statement as a maximal ignorance; since we know nothing about it, we say therefore that all outcomes are equally likely. Now we can specify our

**fundamental postulate of equilibrium statistical mechanics**again as:In a state ofthermal equilibrium, all theaccessible microstatesof the system areequally probable.

This postulate is actually suffices to derive all equilibrium statistical mechanics, furthermore thermodynamics comes as a special case. Now let's go over the highlighted words in the definition one by one, starting with the

**thermal equilibrium**.**Thermal equilibrium**is defined as the case where time averages of macroscopic quantities are independent of time. In our gas in a container example, if we look for the velocity of a particular gas molecule in an instant of time (which we call a

**microscopic quantity**) we don't expect it to be time-independent, since the gas molecule encounters collisions with other particles and the wall. But the average velocity of all the particles (which we call a

**macroscopic quantity**) is zero for all time since the velocities to the right are compensated with the ones to the left and similarly for the other directions. Thus we say that the average velocity in thermal equilibrium is time-independent. Similarly we can say that average energy and pressure is also time-independent macroscopic quantities.

Averages are taken with respect to some probability distribution. The only way some macroscopic quantity becomes time-independent is that the probability distribution itself is also independent of time; because that way you are sure that all the moments of all the variables you take are all independent of time.

The next word to be defined is an

**accessible microstate**. The microstate is defined by specifying the state of each constituents of the systems, i.e all the coordinates $(q,p)$ for each particle. The accessible microstates are the ones which are allowed by the constraint that the total energy is $E$. If you increase E, more states will be accessible, if you decrease $E \to 0$, almost all particles will get to rest, thus there will only be a limited number of accessible states. (This configuration under these constraints is called the**microcanonical ensemble**.)
Now comes the crucial part:

**equally probable**. Before we define it, let's us the question*'Can we derive the equiprobability from classical dynamics?'*. Actually we can derive the converse of it. By that I mean, if for any instant of time, all the microstates of a system are equally probable, then we can show that it remains there for all time, i.e system is in**equilibrium**. But we are in search for other way around: If we have a thermal equilibrium, what is the probability distribution?
In classical dynamics, the phase space density, $\rho$, of the system obeys the

**Liouville Equation**: \[\frac {\partial \rho}{\partial t} = \{H,\rho \} \tag{2}\] This expression is equal to zero in the equilibrium, i.e it is independent of time. \[\frac {\partial \rho}{\partial t} = \{H,\rho \} = 0 \tag{3}\] Thus we can say that $\rho$ is a*Furthermore, the eqn.$(3)$ suggests that $\rho$ must be a function of $H$ and possibly other constants of the motion.***constant of the motion.**
Under equal probability condition, so far we only know that for our closed isolated system is on the energy shell defined by $(1)$. From only this information in hand, our equilibrium phase space density function($\rho_{eq}$) can only be the Dirac delta function of $H(q,p)$ peaked at the value of $E$: \[\rho_{eq}=\rho_{eq}(H) = \delta(H(q,p) - E) \tag{4} \] If we take the energy shell on which our system lies and divide it into little patches, eqn.($4$) does not distinguish one patch or another. This fact definitely shows that equal volume of patches in the phase space have equal probability. Thus our assumption of equal probability seems to be satisfied... But only with one caveat! That is, we need to specify a point in phase space in order to describe a state, but a point in a continuous space is a set of measure zero (the probability in continuum of finding any specific value is zero

**)**!
We need to do a further assumption that we have a finite resolution in order to give a finite equal probability to the patches in our phase space. We can not specify a point in a phase space, we don't have an infinite precision available to us. So we divide our phase space into unit cells; as long as our unit cell is finite, we are in a good shape. The problem of having uncountable many points in our phase space is tamed by discretizing, by giving a resolution. So we define a microstate as something in elementary volume element in phase space. But where does this finite resolution come from?

Nature comes to our rescue at this time and quantum mechanics tells us that due to the

**uncertainty principle**, we only have a resolution up to a constant times $h$, the Planck's constant. This comes from the foundational fact that you can not specify the coordinate $q$ and its conjugate momentum $p$ at the same time, thus there is an inherent resolution in our $(q,p)$ phase space. We could construct whole theory of classical statistical mechanics pretending that there is a finite resolution in our phase space.
Finally, we can compute the number of accessible microstates $\Omega (E)$: \[\Omega (E) = \frac{\mathcal{V}(E)}{h^{3N}}\tag{5}\]where $ \mathcal{V}(E)$ is the volume of the accessible phase space and it is divided by unit cells of $h^{3N}$($h$ for each $(q,p)$ pair). The fact that, our probability density saying that $H(q,p) = E$ and nothing more, implies that each of these microstates has as much weight as another else, hence they are equally probable with the probability of \[\mathcal{P} = \frac{1}{\Omega (E)} \tag{6}\] Thus we have showed that the equiprobability of accessible microstates is the most crucial assumption in equilibrium statistical mechanics and it holds for a discrete phase space governed by the quantum mechanical restriction of finite resolution.

*These notes have been composed with the help of the wonderful Classical Mechanics lectures of Prof.V.Balakrishnan on Youtube.*
Bu yazına daha önce göz gezdirmiştim, az önce Balakrishnan'ın Statistical Mechanics derslerini izliyordum, Arif yazmıştı bununla ilgili deyip tekrar bakayım ve gelmişken yorum bırakayım dedim, şu an tepeden bir ışık iniyor gibi hissediyorum açıkçası.

ReplyDeleteBizim İTÜ'de dersteki faz uzayı çalışmamız, harmonic oscillator için Hamiltonian yazıp, o meşhur elipsleri çizmekten ibaretti. Sonra direkt number of microstates diye yapıştırdık Omega'yı. Ama tabii ki faz uzayında bir volume element alalım,ve onun üzerindeki ayrıştırma gücümüz kuantum mekaniğinin temel ilkelerinden dolayı Planck sabiti kadar olacak ve Omega'daki h^{3N} terimi buradan gelecek demedi kimse.

--SPOILER--

Daha sonra yazı-tura modeli üzerinden binomial distribution'ı ve generating function'u gösteriyor.(Gelecek yazının bunlarla ilgili olacağını tahmin ve ümit ediyorum) Ve çook fazla duymuş olduğumuz partition function'un "aslında" bu binomial dağılım için generating function olduğunu gösteriyor.

En çok şaşırdığım şey ise tüm bunları 1 saat 6 dakika 11 saniyede yapıyor olması. İnanılır gibi değil gerçekten. Yani ciddi bir sıkıntı var, tanımlar ve varsayımlardan sonra pek çok şey "crystal clear" hale geliyor, niye bataklıktan gitmek tercih ediliyor anlayabilmiş değilim.

Diğer yazıları da sabırsızlıkla bekliyorum! Eline sağlık.

Kemal

Haklısın Kemal.. Benim bu notlarım dersin 'transkripti' gibi birşey aslında; o kadar açık ve net anlatıyor ki eklenip çıkarılacak bir şey hiç olmuyor her seferinde... Adam her dersinde dikkat edersen 'faz uzayı' üzerinden yaklaşıyor tüm problemlere; bunda dinamik sistemci olmasının büyük payı var belki ama faz uzayı kavramının fiziğin kalbinde yatıyor olduğu gerçeğini bana kalırsa teyit ediyor bir taraftan da..

DeleteBahsettiğin diğer lecture'ları da izledim ben ve partition'ın bir generating function olması benim için müthiş bir aydınlanmaydı. Olasılık ve istatistikte hatta birçok diferansiyel denklem çözümleriyle ilişkili polinomların üretilmesinde sürekli karşına çıkar üretici fonksiyonlar ve istatistiksel mekanikte sürekli ortalama almak için kullandığımız bu bölüşüm fonksiyonu en doğal haliyle bir üretici fonksiyon olmalı elbette diyor insan ardından..

Konuyu matematiksel olarak en geniş çerçeveden alıp tuğlaları adım adım yerleştirerek ilerliyor oluşu müthiş etkili bir şey.. Matematiksel Metodlar dersinde bunu müthiş hissediyorsun mesela, onlara da göz atmanı kesinlikle tavsiye ederim.