1 Kinetics 1 Kinetics 1.2 Estimates of ideal gasses and the equipartition theorem

1.1 Big numbers and probability

(September 1, 2023)

1.1.1 Big numbers

Take a number like Avogadro’s number, $N_{A}=6\times 10^{23}$ . The number of rearrangements is of Avogadro’s number $N_{A}!$ is exponentially big, meaning the logarithm is also big. We proved the Stirling approximation¹¹ 1 $\log(x)$ is the same as $\ln(x)$ throughout this course! If we ever need the log base 10 we will write $\log_{10}(x)$ .

\log(N!)\simeq N\log N-N

(1.1)

The Stirling approximation can also be written

\log(N!)\simeq N\log(N/e)\qquad\mbox{or}\qquad\qquad N!\simeq\left(\frac{N}{e}% \right)^{N}

(1.2)

Given $N$ objects, the number off ways I can choose $r_{1}$ objects for group 1, and the remaining $r_{2}$ objects in group 2 (with $r_{2}+r_{2}=N$ ) is given by the ‘‘binomial” coefficients ²² 2 While we wont need it, the reason why its called the binomial coefficient is because the binomial $x+y$ raised to a power is $\displaystyle(x+y)^{N}=$ $\displaystyle\underbrace{(x+y)(x+y)\ldots(x+y)}_{\mbox{$N$ factors}}$ (1.3) $\displaystyle=$ $\displaystyle\sum_{r_{1}=0}^{N}{}^{N}C_{r_{1}r_{2}}\,x^{r_{1}}y^{r_{2}}$ (1.4) In passing to the second line I have to choose $r_{1}$ terms out of the $N$ terms in the first line to take $x$ and the remaining $r_{2}$ terms will take $y$ . Try it out for $N=2$ and $N=3$ . The multinomial coefficients are similar, and expanding $(x+y+z)^{N}$ will lead to a similar expansion involving $x^{r_{1}}y^{r_{2}}z^{r_{3}}$ .

{}^{N}C_{r_{1}r_{2}}=\frac{N!}{r_{1}!r_{2}!}

(1.5)

You should be able to explain this formula. This generalizes – if I have $N$ objects, and I select $r_{1}$ objects into group 1, $r_{2}$ objects into group 2, and the remaining $r_{3}$ objects into group three (with $r_{1}+r_{2}+r_{3}=N$ ), the number of ways to do this is given by the “multinomial” coefficient:

{}^{N}C_{r_{1}r_{2}r_{3}}=\frac{N!}{r_{1}!r_{2}!r_{3}!}

(1.6)

You should be able to explain this formula.

1.1.2 Probability

First consider a set of discrete outcomes $i=1\ldots N$ , each with probability $\mathscr{P}_{i}$ (like a weighted six sided die). The sum of probabilities is unity

\sum_{i}\mathscr{P}_{i}=1

(1.7)

Associated with each outcome is a quantity $x_{i}$ , e.g. $x_{3}$ the money you get for rolling a three. Then the mean of $x$ (the mean money you get by rolling the die)

\left\langle x\right\rangle=\sum_{i}\mathscr{P}_{i}x_{i}

(1.8)

For a given quantity $x$ we define the deviation from the average

\delta x\equiv x-\left\langle x\right\rangle

(1.9)

and the average deviation is zero $\left\langle\delta x\right\rangle=0$ . Then the variance is the mean of the squared deviation

\left\langle\delta x^{2}\right\rangle=\left\langle(x-\left\langle x\right% \rangle)^{2}\right\rangle=\left\langle x^{2}\right\rangle-\left\langle x\right% \rangle^{2}

(1.10)

The standard deviation is

\sigma_{x}=\sqrt{\left\langle x^{2}\right\rangle}

(1.11)

For continuous variable we need the concept of a probability distribution. The probability, ${\rm d}\mathscr{P}$ , to find a particle with position in a range between $x$ and $x+{\rm d}x$ , which we denote $[x,x+{\rm d}x]$ , is denoted

{\rm d}\mathscr{P}=P(x){\rm d}x\,,

(1.12)

where the probability density $P(x)$ is

P(x)=\frac{{\rm d}\mathscr{P}}{{\rm d}x}\,.

(1.13)

A very important probability density is the Gaussian or “normal” distribution which you should try to memorize:

P(x)=\frac{1}{\sqrt{2\pi\sigma^{2}}}e^{-x^{2}/2\sigma^{2}}

(1.14)

It is also called the Bell shaped curve and you should be able to sketch it. In class and in homework we showed:

\displaystyle\int_{-\infty}^{\infty}P(x){\rm d}x=

\displaystyle 1

(1.15)

And worked out a number of integrals

\displaystyle\left\langle x^{n}\right\rangle=\int_{-\infty}^{\infty}P(x)x^{n}{% \rm d}x=\sigma^{n}C_{n}

(1.16)

The numbers are $C_{0}=1$ , $C_{2}=1$ , $C_{4}=3$ with odd moments, such as $\left\langle x\right\rangle$ , being zero.

1.1.3 Independence and the central limit theorem

Consider a two dimensional probability distribution

{\rm d}\mathscr{P}_{x,y}=P(x,y){\rm d}x{\rm d}y

(1.17)

This is the probability of $x$ in $[x,x+{\rm d}x]$ and $y$ in $[y,y+{\rm d}y]$ .

We say that $x$ and $y$ are independent if $P(x,y)=P(x)P(y)$ factorizes so that the probability of finding $x$ and $y$ (in interval ${\rm d}x\,{\rm d}y$ ) is probability of $x$ (in interval ${\rm d}x$ ) times the probability of in $y$ (in interval ${\rm d}y$ )

{\rm d}\mathscr{P}_{x,y}=P(x){\rm d}x\times P(y){\rm d}y

(1.18)

The constants can be arranged so that $P(x)$ and $P(y)$ are separately normalized, e.g.

\int P(x){\rm d}x=1\quad\mbox{and}\quad\int P(y){\rm d}y=1

(1.19)

When the distributions are independent

\left\langle xy\right\rangle=\left\langle x\right\rangle\left\langle y\right\rangle

(1.20)

For definiteness, consider a sequence of random steps in position $x$ . Assume $x_{1}$ , the step in position from step number one, is drawn from the probability distribution $P(x)$ . Also assume second step $x_{2}$ is drawn from the same distribution, and that the choice of $x_{2}$ is no way dependent on $x_{1}$ . Similarly, the third step $x_{3}$ is drawn from $P(x)$ and is no way dependent on $x_{1}$ or $x_{2}$ ; and so on for $x_{4},x_{5},x_{6}\ldots$ . Then we want to know what is the mean, variance, and probability distribution of the sum

Y=x_{1}+x_{2}+\ldots+x_{N}

(1.21)

The answer is for the mean and variance are

	$\displaystyle\left\langle Y\right\rangle=$	$\displaystyle N\left\langle x\right\rangle$		(1.22)
	$\displaystyle\left\langle\delta Y^{2}\right\rangle=$	$\displaystyle N\left\langle\delta x^{2}\right\rangle$		(1.23)

In general the probability of $Y$ depends on $P(x)$ , and nothing much can be said about $P(Y)$ . However, if $N$ is large $N\gg 1$ , then, remarkably, the probability of $Y$ takes on a universal form of a Normal distribution

P(Y)=\frac{1}{\sqrt{2\pi\sigma_{Y}^{2}}}\exp\left[-(Y-\left\langle Y\right% \rangle)^{2}/2\sigma_{Y}^{2}\right]

(1.24)

with $\sigma_{Y}=\sqrt{\left\langle\delta Y^{2}\right\rangle}$ . We did not go over the proof, and it is enough at this level to just accept it as a statement of fact