In this article we will define what are Macrostates and Microstates in Statistical Physics with examples and illustrations. Consider some (4, say) distinguishable particles. If we wish to distribute them into two exactly similar compartments in an open box, then the priori probability for a particle of going into any one of the compartments will be exactly 1/2 as both compartments are identical. If the four particles are named as a , b, c and d and the compartments as compartment (1) and compartment (2), then the following table can be made listing all the possible arrangements.
[Download this table in PDF for better view.] Now in the second distribution, only one particle is chosen to be inside compartment (1) and all others in Compartment (2). There are exactly $\binom{4}{1}$ = 4 ways to do so. All four arrangements are shown in the table above. Similarly, in the distribution (3) exactly two particles enter in each of the compartment (1) and (2) , i.e., if $a$ and $b$ are placed in the compartment (1), $c$ and $d$ must take place in Compartment (2). There are exactly $\binom{4}{2}$=6 ways to do so. Following the same procedure to complete table by increasing the number of particles one by one in Compartment (1) and decreasing them one by one in Compartment (2). We have following 5 different distributions for the system of four particles:
These compartment-wise distributions of a system of particles are known as macrostates of the system. There are five macrostates observed corresponding to a system of four particles. In general, for a system of n particles, there are exactly $n+1$ macrostates (for the system of n particles to be distributed in two identical compartments). These macrostates are the following distribution of particles:
Now, the different possible arrangements for each macrostate are in their own ‘a distribution’, called microstates. For example {{a}, {b,c,d}} is a microstate of macrostate $d_2$ (distribution 2) in the table of system of four particles. From the table, it can be observed that a macrostate may have many corresponding microstates. The distribution $d_1=(0,4)$ has one microstate {{ }{a,b,c,d}}, while the distribution $d_2= (1,3)$ has four, $d_3$ has six , $d_4$ has four and $d_5$ has one. The total number of microstates for the system of four particles is, therefore, $1+4+6+4+1=16=2^4$. In general, the total number of microstates for the system of n-particles in ‘two compartment’ composition is $2^n$. And also, the number of microstates corresponding to a single macrostate with $(r, n-r)$ distribution among $n$ particles is $\binom{n}{r}$. Thermodynamic ProbabilityIn statistical physics, the number of microstates plays very important role at quantum level. The number of microstates corresponding to any macrostate is called the thermodynamic probability of the macrostate, represented by $W$ or $\Omega$. Since, for the macrostate $(r,n-r)$ the number of microstates is $$\binom{n}{r} = \frac{n!}{r! \cdot (n-r)!}$$ . Thus, the thermodynamic probability of this macrostate: $$W_{(r,n-r)} = \binom{n}{r} = \frac{n!}{r! \cdot (n-r)!}$$. In mathematics, we know that $$\binom{n}{r} = \binom{n}{n-r} =\frac{n!}{r! \cdot (n-r)!}$$. Therefore, $$ W_{(r,n-r)} =W_{(n-r,r)}$$. Or, $$ W_{(r,n-r)} =\binom{n}{r} = \frac{n!}{r! \cdot (n-r)!}= W_{(n-r,r)}$$ . That is, for a system of four distinguishable particles, $n=4$, $r=0,1,2,3,4$, provided $$W_{(0,4)} =\binom{4}{0} = \frac{4!}{0! \cdot (4)!}= W_{(4,0)} =1$$ $$W_{(1,3)} =\binom{4}{1} = \frac{4!}{1! \cdot (3)!}= W_{(3,1)}=4$$ $$W_{(2,2)} =\binom{4}{2} = \frac{4!}{2! \cdot (2)!}= W_{(2,2)}=6$$ All agree with the table. Thermodynamic probability is not identical to ‘probability’ in mathematics and statistics, but it gives rise to that. Probability of a MacrostateThe probability of macrostate (i.e., availability of a macrostate) in a system is given by $$P_m = \frac{\textrm{No. of such microstates in the macrostate}}{\textrm{Total no. of microstates in the system}}$$ Thus, $$P_m = \frac{W}{2^n}$$ Or, $$P_{(r,n-r)} =\frac{W_{(r,n-r)}}{2^n} = \binom{n}{r} \times \frac{1}{2^n} = P_{(n-r,r)} $$ Practically, the probability of finding the macrostate (0,4) in a system of four particles is $$P_{(0,4)} =\frac{W_{(0,4)}}{2^4} = \binom{4}{0} \times \frac{1}{2^4} = \frac{1}{16}$$ Similarly, $P_{(1,3)}= \frac{1}{4}$. Macrostate with highest probability The macrostate (r, n-r) has the highest probability only if $$r=\frac{n}{2} \, \textrm{ when n is even}$$ and $$r= \frac{n+1}{2} \, \textrm{or} \frac{n-1}{2} \, \textrm{when n is odd}$$. Then the highest probability is given by: $$P_{max} = \frac{n!}{\frac{n}{2}! \cdot \frac{n}{2}!} \times \frac{1}{2^n}$$ (n is even) $$P_{max} = \frac{n!}{ \frac{n-1}{2}! \frac{n+1}{2}!} \times \frac{1}{2^n}$$ (n is odd) Macrostate with least probability The macrostate (r, n-r) is least probable when $r$ is either zero or n. $$P_{min} = \frac{n!}{0! \cdot n!} \times \frac{1}{2^n}$$. Thus, least probability for a macrostate is $\frac{1}{2^n}$. $\Box$
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First of all, if there are $p^N$ possible states, this means that the particles and the compartments are distinguishable. For indistinguishable particles (and distinguisable compartments, which is probably reasonable physically speaking) the number of possible states, or configurations, has to be divided. For instance the configuration $(n_1,n_2,\dots,n_p)$ of indistinguishable particles corresponds to $$\frac{N!}{n_1!n_2!\dots n_p!}$$ configurations of distinguishable particles. Note that we have of course $$\sum_{{\substack{n_1,n_2,\dots,n_p\\n_1+n_2+\cdots+n_p=N}}} \frac{N!}{n_1!n_2!\dots n_p!}=p^N.$$ Let us start with the answer to the question "how many configurations have no compartment empty ?" With indistinguishable particles, we place $N$ particles in $p$ compartments, with at least one particle in each. To count this number, we use the bullet representation of the integer $N$ like this $$\bullet\bullet\bullet\bullet\cdots\bullet$$ there are $N$ bullets. We now place the compartment separations, denoted as $|$. One possible configuration with $N=10$ into $p=5$ compartments is $$\bullet|\bullet\bullet\bullet|\bullet\bullet\bullet|\bullet|\bullet\bullet$$ which is a reprensentation of $(1,3,3,1,2)$. We have to place $p-1$ divisors at any $N-1$ positions. Therefore the number of possible divisions is $\binom{N-1}{p-1}$ (basically, we have counted the number of partition of the integer $N$ into sums of non-zero integers when order matters). There are therefore $$C_0(N,p)=\binom{N-1}{p-1}\quad\text{(indistinguishable particles)}$$ configurations with no empty compartments. In the case of distinguishable particles, this actually more difficult, as @ndrv pointed out. We have $p^N$ possibilities, but we have to substract configurations with one empty compartment. We can pick up any of the $p$ compartment as empty and substract $p\times (p-1)^N$, but then we have substracted the configurations with two empty compartments two times. So must add $\binom{p}{2}(p-2)^N$. But by doing this we have also added twice the configurations with three empty compartments, so must add them back. This goes on until we have added or substracted the configuration with $p-1$ empty compartments, since there are no further configurations. As a conclusion, we have $$C_0(N,p)=\sum_{j=0}^{p-1}(-1)^j\binom pj(p-j)^N \quad\text{(distinguishable particles)}$$ This number is related to the Stirling number of second kind by $C_0(N,p)=p!\genfrac\{\}{0pt}{}Np$. We can answer to the question "How many configurations have exactly one empty compartment ?" The answer is, in both cases $$pC_0(N,p-1)$$ because there remains only the choice of the empty compartment among the $p$ possible compartments. We can even answer to the question "How many configurations have exactly $k$ empty compartments ?" From the same reasoning, we find $$\binom pkC_0(N,p-k)$$ If we now want to allow one or more empty compartments, we can sum over $k$: $$C(N,p)=\sum_{k=1}^{p-1}\binom pkC_0(N,p-k).$$ but we can also remark that $$C(N,p)=p^N-C_0(N,p).$$ |
The total number of microstates for the random distribution of 10 particles in two boxes are
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