1

I am having a hard time time understanding the meaning behind Microstates and Macrostates, I am in search of a simple, intuitive yet accurate definition of these terms .

And why does the formula $$N!/(n_1!*n_2!*n_3!.....).$$ give you the number of Microstates. I am familiar with Permutations and Combinations, so what is actually happening here though?

And why is the entropy given by $$S=k_b\ln \Omega,$$ where $\Omega$ is the number of Microstates and $k_b$ the Boltzmann constant.

Why should we get the value of Entropy by multiplying the Boltzmann Constant with the number of Microstates ?

I am introductory Physics Student who is interested in learning Physics. I am familiar with Calc 1, 2 and AP physics courses. I hope the explanations are at that level.

And is there any book in Statistical Menchanics that is akin to Theoretical Minimum by Leonard Susskind.

devCharaf
  • 699
Harry Case
  • 377

2 Answers2

1

Considering a system of $N$ particles, we will have at lest $3N$ degrees of freedom to deal with. These are the 3 possible translation direction ($x,y,z$) for each particle. We could have more if the particles have som internal degrees of freedom, like spin in quantum mechanics. If one wants to specify these exact quantities, then one should solve the Shrödinger equation for this system. The resulting state is what is referred to as the microstate.

It's obvious to say that solving the microstate of a system might be complicated and irrelevant.

The macrostate on the other hand is what could be observed on the macroscopic level, like temperature, pressure ... . In general, a macrostate is enough to describe a certain system.

Now for the Boltzmann formula for entropy, there are many things we can deduce from it. First of all, it captures the additivity of entropy. Say that you put to systems with $\Omega_i$ number of possible microstates each. then the total number of microstates is $\Omega=\Omega_1\times\Omega_2$. Given the additivity property of the $\log$, you can see that $$S = S_1 + S_2$.$ Another thing is that when the number of combinations is $\Omega=1$ then the entropy is $0$. There is a good simple proof here for the formula.

As of the Boltzmann constant, once we prove the linearity $S\propto\ln\Omega$, then this constant is found through experiments.

devCharaf
  • 699
  • Why does two systems of Microstates $Ω_1$ & $Ω_2$ when brought together ,have their total Microstates as $Ω_1$×$Ω_2$ ? – Harry Case Mar 24 '22 at 07:57
  • It's just basic combinatorics. For each microstate $i\in\Omega_1$ in which the first system is in, then system 2 can be in $\Omega_2$ microstates. Therefore the number of total microstates of the whole system is the product. – devCharaf Mar 24 '22 at 17:03
  • ,I get the combinatorial part ,but I don't understand how the Microstates fit into this ,(ie) if by Microstates you mean arrangement of particles ,are you trying to say that for each arrangement of particles in system 1 ,since we are combining the system ,there is a corresponding $Ω_2$ arrangement of particles in system 2 for every arrangement of particles in system 1 ,the combination of the system 1 with system 2 ,creates a new arrangement where for each arrangement of 1 we have $Ω_2$ arrangement of 2 . – Harry Case Mar 24 '22 at 17:23
1

There are many questions here so I will stick to answering the one about the multinomial coefficient $$\binom{N}{n_1,n_2,\cdots,n_m}\equiv\frac{N!}{n_1!n_2!\cdots n_m!},\quad N=n_1+n_2+\cdots+n_m.$$ This quantity may be interpretted as the number of unique ways of arranging $N$ items into $m$ categories with $n_i$ items in the $i$th category. For example, we may have $N=6$ balls with $m=3$ different colours and want to know the number of ways of arranging $n_1=1$ orange balls, $n_2=3$ blue balls, and $n_3=2$ purple balls; the answer is $6!/(1!3!2!)=60$ unique arrangements.

This can be derived directly from the knowledge that the binomial coefficient tells us how many ways there are of arranging $N$ into one category with $n_1$ items and another category with $n_2=N-n_1$ items, $\binom{N}{n_1}=\frac{N!}{n_1!(N-n_1)!}=\binom{N}{n_1,n_2}$. Then, one may realize that there is actually another category with which we want to subdivide our $n_2$ items, say into $n_2^*$ items with one label and $n_3\equiv n_2-n_2^*$ with another, and regular binomial coefficients tell us that there are $\binom{n_2}{n_2^*}=n_2!/(n_2^*! n_3!)$ ways of doing that, so the total number of possibilities is now $$\binom{N}{n_1}\times\binom{n_2}{n_2^*}=\frac{N!}{n_1! n_2!}\times\frac{n_2!}{n_2^*!n_3!}=\frac{N!}{n_1!n_2^*!n_3!}=\binom{N}{n_1,n_2^*,n_3}.$$ Continuing and subdividing $n_3$ into two categories, by induction, we find that the regular binomial coefficients lead us to this definition and interpretation of multinomial coefficients as above.

So in the case of finding the number of microstates, we want to know the number of unique methods in which the particles can be arranged. Consider that we have $n_1$ particles of type $1$, $n_2$ particles of type $2$, all the up until $n_m$ particles of type $m$. Then the number of unique ways in which they can be arranged is $\binom{N=n_1+n_2+\cdots +n_m}{n_1,n_2,\cdots,n_M}$, which tells you the number of microstates.

Quantum Mechanic
  • 4,921
  • Ok ,I get the idea of *Multinomial Coefficients* ,but here if $n_1$ an arrangement of molecules and $n_1!$ is the number of ways in which that arrangement can be attained , Similarly $n_2$ is another arrangement and $n_2!$ is the number of ways in which the arrangement $n_2$ can be attained and N! no of ways in which the molecules can be arranged ,how does it help in finding the number of Microstates ? – Harry Case Mar 22 '22 at 02:28
  • @HarryCase I've updated the answer: the number of unique arrangements of your molecules is exactly what the number of microstates is. – Quantum Mechanic Mar 22 '22 at 13:33
  • ,but is $N!/n_1!n_2!n_3!.....$ the no of Microstates (or) something else ? ,Because the no of Microstates for Maxwell Boltzmann Distribution is something different . – Harry Case Mar 23 '22 at 03:40
  • @HarryCase well where did you find this expression?? The number of microstates depends on the physical scenario at hand. In Maxwell-Boltzmann we care about the distribution of different velocities, with multinomial coefficients we are just looking at permuting the arrangements of a bunch of particles of different types. – Quantum Mechanic Mar 23 '22 at 13:07
  • this expression popped up when I searched for formula for Microstates this popped up ,my bad ,I don't understand how the the formula arrived ,I understand till the one with multinomial coefficient ,btw what does the Multinomial coefficient represent then, $N!$ in the numerator is arrangement of particles considering the particles as distinguishable ,but this isn't true ,so if $n_1$ is a Macrostate ,then $n_1!$ is the no of Microstates that lead to the n_1 Macrostate ,so what does $N!/n_1!n_2!n_3!....$ represent . – Harry Case Mar 23 '22 at 16:06
  • The multinomial coefficient tells you the number of unique ways of arranging $N$ items into $m$ categories with $n_i$ items in the $i$th category... I am repeating myself. So if there are $m$ types of particles and you know your macrostate has $n_i$ particles of type $i$ then this tells you the number of microstates, nothing more. You'll probably benefit from asking a larger number of specific questions on this site, as opposed to one question with lots of different sub-questions – Quantum Mechanic Mar 23 '22 at 17:53
  • ,I have attached the link to a specific question ,It would be great if you could answer it .https://physics.stackexchange.com/questions/700388/question-regarding-the-intuition-for-calculating-the-number-of-microstates-in-ma – Harry Case Mar 24 '22 at 02:48