# Basics Statistical Mechanics

#### Contents

## What and why?

**In most realistic situations, we can only predict the motion and future of a particle if**

1) We are given the initial conditions

2) We are given the present forces acting upon it and the particles with which it is interacting

\(However\), when it comes to objects like a box of gas which can be visualised classically as hard spheres going all about , We cannot know the initial conditions since it is too complicated and ofcourse hard to measure

Even , if we observe a gas being made (say by vapourising a solid), we cannot be sure or even get a vague idea of the initial velocity of each and every gas atom/molecule.

In such scenarios, we are facing with an uncertainty because of lack of information and hence we lose our ability to predict.

**But even such situations can be analysed if we assume the following reasonable assumption**

1) For an arbitary system, the system will assume the state that is most likely to happen given enough time

(example, toss a thousand coins and you will more or less end up with half heads and half tails )

So now we have to explain what it means by a state and how to define which state is more likely than others

We have two forms of state-

1) Microstate - microstate is defined as the exact arrangement (distribution of velocities, etc ) of each and every individual particle

**example -1)** if we have a single particle in free space which has an energy 'E' , then it can have two velocities

+V and -V

these two are the microstates of the system

**example -2)** if we have 10 coins in a row, each labelled with a number then

the series of individual states (say - HTTHHHTHHT ) is a microstate

each microstate is equally likely (evidently) assuming that each particle/coin has is identical

2) Macrostate - a macrostate is the set of microstates that correspond to a common quantity

in the particle example above, The Energy 'E' can be said to be the macrostate for the microstates +V amd -V ,

similarly, 6 heads and 4 tails is a macrostate.

However, each macrostate is not necessarily equi probable and infact its probability is proportional to the number of microstates it posseses .

So for a system of gas molecules, the energy it has is a **Macrostate** and the number of ways it can distribute that energy amongst its particles is its **Microstate**

With the definitions clear, we need to quantise our ideas

## Thermal equillibrium

Consider the two systems joined to each other by a fixed conducting wall (Energy can pass, material cannot)

and let the total energy of the system be 'E'

and the energy of each to be \( E_1 \quad and \quad E_2 \) respectively

then let \(\Omega _{ E1 }\) and \(\Omega _{E 2 }\) be the number of microstates corresponding to each system with respective energies

then for each microstate of 1, we can have \(\Omega _{E 2 }\) microstates of 2 and hence the total number of microstates is

\(\Omega _{ E1 }\Omega _{ E2 }\) = Total

as per our assumptions this must be maximised to get the most probable distribution

so differentiating with respect to \(E_1\) and equating to 0

we get

\({ \Omega }_{ E2 }\frac { d{ \Omega }_{ E1 } }{ d{ E }_{ 1 } } +{ \Omega }_{ E1 }\frac { d{ \Omega }_{ E2 } }{ d{ E }_{ 1 } } =0\\ \)

now we invoke conversation of energy to claim

\(\\ d{ E }_{ 1 }+d{ E }_{ 2 }=0\)

and substitute to get

\(\frac { 1 }{ { \Omega }_{ E1 } } \frac { d{ \Omega }_{ E1 } }{ d{ E }_{ 1 } } =\frac { 1 }{ { \Omega }_{ E2 } } \frac { d{ \Omega }_{ E2 } }{ d{ E }_{ 2 } } \)

which is equivalent to

\(\\ \frac { d\quad { ln(\Omega }_{ E1 }) }{ d{ E }_{ 1 } } =\frac { d\quad ln({ \Omega }_{ E2 }) }{ d{ E }_{ 2 } } \)

Now, we have a very powerful relation, in general, for 'n' systems in thermal equillibrium, the quantity that must remain invariant for all of them is

\(\frac { d\quad { ln(\Omega }_{ E }) }{ d{ E } } \)

let us call this quantity

\(\frac { 1 }{ kT } \) (the reasons will soon be apparent)

so for 'n' systems in thermal equillibirum

T must be the same for all, let us call this the temperature

## Getting to the boltzmann distribution

Now, we use the results of the previous section to prove very powerful and general result using which we can deal with most statistical systems

Imagine a small tank connected to a large reservoir as shown in the second figure

Once again let the total energy of the system be E

since the reservoir is so large, most of the overall energy comes from the reservoir

so Let the energy of the small tank be \(\epsilon \) and the energy of the Reservoir be \(E-\epsilon \)

now, **we make a special assumption**

Let the number of microstates for each macrostate of small tank be fixed and equal to 'n'

then

\(\\ P(\epsilon )\quad \propto \quad n{ \Omega }_{ E-\epsilon }\)

(where P is the probability of having that energy) ( it is proportional to number of microstates)

now as 'n' fixed, we can further write

\(\\ P(\epsilon )\quad \propto \quad { \Omega }_{ E-\epsilon }\\ \)

taking 'ln' on both sides

\(\\ ln(P(\varepsilon ))\quad \propto \quad ln({ \Omega }_{ E-\epsilon })\quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \approx \quad ln({ \Omega }_{ E })-\epsilon \frac { dln({ \Omega }_{ E }) }{ d\epsilon } =({ \Omega }_{ E })-\frac { \epsilon }{ kT } \)

so

\(P(\epsilon )\quad =\quad A{ e }^{ -\frac { \epsilon }{ kT } }\)

the importance of this result cannot be overstressed despite the delicate assumption we have made about the fixed number of microstates

for example- Let us imagine a particle in a 1-D universe, for each energy E , it can have velocities +V and -V, so here 'n' value is 2

## How this applies to gas molecules

The small tank need not be a tank, any single molecule in a gas in steady state is interacting with several other molecules through collisions thus gaining and losing energy ,

Here the gas molecule can be called as the small tank and the rest of the vast number of gas molecules as the reservoir, and hence probability of the gas molecule having energy \(\epsilon \) is proportional to \({ e }^{ -\frac { \epsilon }{ kT } }\)

In future edits or in another wiki, we shall derive the maxwell boltzmann distribution using this.

**Cite as:**Basics Statistical Mechanics.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/basics-statistical-mechanics/