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 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 following table can be made listing all the possible arrangements.

 Compartment (1)      ↓ Compartment (2)             ↓

 Distribution Number of Particles Name(Type) of particles Number of Particles Name(Type) of Particles Total Number of Arrangements Distribution (1) 0 – 4 {a,b,c,d} 1 Distribution (2) 1 {a} {b}  {c} {d} 3 {b,c,d} {c,d,a} {d,a,b} {a,b,c} 4 Distribution (3) 2 {a,b} {b,c} {c,d} {a,c} {b,d}{a,d} 2 {c,d} {a,d} {a,b}{b,d}{a,c} {b,c} 6 Distribution (4) 3 {b,c,d} {c,d,a} {d,a,b} {a,b,c} 1 {a} {b}  {c} {d} 4 Distribution (5) 4 {a,b,c,d} 0 – 1

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 these four arrangements are shown in table. 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:

Distribution (1) = $d_1$ = (0,4)
Distribution (2) = $d_2$ = (1,3)
Distribution (3) = $d_3$ = (2,2)
Distribution (4) = $d_4$ = (3,1)
Distribution (5) = $d_1$ = (4,0)

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 actually the following distribution of particles:

$d_1= (0,n)$
$d_2= (1,n-1)$
$d_3= (2,n-2)$
$\ldots$
$d_k= (k-1,n-k+1)$
$\ldots$
$d_{n+1} = (n,0)$.

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, 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 Probability

In 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 Macrostate

The 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 $\frac{1}{2^n}$. $\Box$

Remark: The probability of a macrostate (P) is directly proportional to the thermodynamic probability (W).

##### 1 comment
1. Yesica says:

I would like to know what would happen if you have 8 particles or any number of particles in a container that has 3 or more compartments. You mentioned that this works for a two compartment system.

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