- #Lecture 17
- #Lecture 20: Classical Statistical Mechanics
-
#Lecture 21
- [[#What is \(N!\) (for large \(N\)):-]]
- #Poisson Distribution
- #Lecture 22
- #Lecture 22
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(Q) Does the Liapunov exponent depend on the initial condition or
not?
\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=1.1]
\draw[thick] plot [smooth cycle, tension=1] coordinates {(0,2) (1.8,2.4) (3.2,1.6) (3,-0.2) (1,-0.8) (-0.8,0.4)};
\draw[thick] plot [smooth cycle, tension=1] coordinates {(1.5,1.7) (2.2,1.6) (2.1,0.5) (1.4,0.7)};
\draw[thick] plot [smooth cycle, tension=1] coordinates {(0.3,1.1) (0.7,1.3) (0.6,0.4) (0.1,0.5)};
\node at (0.6,1.9) {$x_0$};
\draw[->,thick] (0.6,1.75) -- (0.55,1.15);
\node[right] at (2.6,1.9) {attractor$_1$};
\node[left] at (-0.3,0.7) {attractor$_2$};
\end{tikzpicture}
If the initial condition is such that the trajectory comes to an
arbitrary point in phase space, since the Liapunov exponent is
calculated by taking time steps to infinity, and the point has
explored the whole phase space, therefore in that sense the
Liapunov exponent should not depend on the initial
condition.
Though there could be multiple attractors, which suggests it
depends on \(x_0\), but generally, otherwise, wherever we start,
\(x_0\) is going to fill up the whole phase space. So the Liapunov
exponent now becomes the property of the trajectory
rather than of a point.
(measure (area, volume) is preserved; still the system is
chaotic), conservative system.
It is like the Bernoulli map where we have to cut the graph above
\(x_{n+1} = 1\) and put it back. We could have also folded the
piece back rather than cutting, to gain non-linearity. (analogy)
\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=2.6]
\draw[->] (0,0) -- (1.2,0) node[right] {$x_n$};
\draw[->] (0,0) -- (0,1.2) node[above] {$x_{n+1}$};
\draw[thick] (0,0) -- (0.5,1) -- (1,0);
\node[below] at (0,0) {$0$};
\end{tikzpicture}
Tent map (non-linear, continuous)
\(\lambda(x_0) = \ln 2\), all rational numbers are periodic unstable
orbits.
So the idea is to stretch the interval, then put the second half
part on top of the first-half map.
\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=1]
\draw[thick] (0,2) -- (4,2);
\node at (0,2.3) {$0$};
\node at (4,2.3) {$1$};
\draw[->] (2,1.6) -- (2,1.1);
\draw[thick] (0,0.5) -- (8,0.5);
\node at (0,0.8) {$0$};
\node at (4,0.8) {$1$};
\node at (8,0.8) {$2$};
\draw[->] (4,0.1) -- (2,-0.8);
\draw[thick] (0,-1.3) -- (4,-1.3);
\node at (0,-1) {$0$};
\node at (4,-1) {$1$};
\end{tikzpicture}
Baker's map takes one more dimension (increasing dimension by
one) in such a way that measure is preserved.
Let's draw \(x_{n+1}\) vs \(y_{n+1}\), and \(x_n\) vs \(y_n\).
\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=1.6]
\draw[thick] (0,0) rectangle (1,1);
\node[left] at (0,0.5) {$y$};
\node[below] at (0.5,0) {$x$};
\node[below] at (0,0) {$0$};
\node[below] at (0.5,0) {$1/2$};
\node[below] at (1,0) {$1$};
\draw[dashed] (0.5,0) -- (0.5,1);
\fill (0.2,0.2) circle (0.02);
\fill (0.7,0.3) circle (0.02);
\node at (0.9,1.1) {$A = 1{\times}1 = 1$};
\draw[->] (1.2,0.5) -- (1.9,0.5) node[midway,above] {expanding $x$ by $2$} node[midway,below] {contracting $y$ by $1/2$};
\draw[thick] (2.1,0.25) rectangle (4.1,0.75);
\draw[dashed] (3.1,0.25) -- (3.1,0.75);
\node[below] at (2.1,0.25) {$0$};
\node[below] at (3.1,0.25) {$1$};
\node[below] at (4.1,0.25) {$2$};
\fill (2.4,0.35) circle (0.02);
\fill (3.4,0.4) circle (0.02);
\end{tikzpicture}
\((A = 1{\times}\tfrac12) = \tfrac12\), folded piece placed on top \(\Rightarrow\) total area
\((A_f) = 1\)
- Map is invertible, still chaotic because (\(\lambda_1 = \ln 2\))
- Area is preserved (\(A_i = 1\), \(A_f = 1\))
- \(\Rightarrow\) Jacobian of transformation is \(1\)
- Liapunov exponents are \(\ln 2,\ \ln\frac12\) \[ \text{total Liapunov exponent} = \ln 2 + \ln \tfrac12 = 0 \]
As long as measure is preserved, the sum of Liapunov exponents is
zero.
All Liapunov exponents are zero for integrable Hamiltonian
systems since it is integrable; none of \(\lambda(x_0)\) should be
\(+\)ve, \(-\)ve (expanding/contracting volume space), sum of these
also will be zero, (otherwise...).
For non-integrable Hamilton systems, Liapunov exponents occur in
pairs such that the total Liapunov exponent is zero. (\(-\)ve for
each \(+\)ve).
Note: unlike differential dynamics, in discrete time
dynamics we can write an explicit solution to a map, but it will
still be chaotic.
\(\hookrightarrow\) still chaotic.
We can see how we do not see loss of information in Baker's map,
if we represent \(x_n, y_n, x_{n+1}, y_{n+1}\) in binary:
Chaotic but (measure preserving and invertible).
(wow!)
Note: The general rule to produce chaos in a 1-D map is
that it has to be non-invertible; for 2 and higher dimensions,
invertible maps can also produce chaos.
determinant of transformation matrix \(= 1\) (orientation
preserving)
\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=1.8]
\draw[thick] (0,0) rectangle (1,1);
\draw[thick,->] (0,0) -- (1,1);
\node[below] at (0,0) {$0$};
\node[below] at (1,0) {$1$};
\node[left] at (0,1) {$1$};
\node[left] at (0,0.5) {$y$};
\node[below] at (0.5,0) {$x$};
\node at (0.5,0.55) {\textit{[cat face sketch]}};
\end{tikzpicture}
The reason it is named Arnold's cat map is because he put the
face of a cat in this graph, and after a few iterations we can't
recognize the face of the cat.
\(\to\) every point is having a stretching direction and a
contracting direction.
\(\hookrightarrow\) notice the `\(+\)' sign is written down.
If the continued fraction terminates, \(x_0\) = rational; else
\(x_0\) is an irrational number.
So to get \(x_1\): take the reciprocal of \(x_0\), throw away the
integer part \(a_0\). So,
\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=3.2]
\draw[->] (0,0) -- (1.15,0) node[right] {$x_n$};
\draw[->] (0,0) -- (0,1.15) node[above] {$x_{n+1}$};
\draw[dashed] (0,0) -- (1,1);
\foreach \a in {1,2,3,4}{
\draw[thick,domain={1/(\a+1)}:{1/\a},samples=30] plot (\x,{1/\x - \a});
}
\node[below] at (0,0) {$0$};
\node[below] at (0.25,0) {$1/4$};
\node[below] at (0.333,0) {$1/3$};
\node[below] at (0.5,0) {$1/2$};
\node[below] at (0.75,0) {$3/4$};
\node[below] at (1,0) {$1$};
\end{tikzpicture}
for \(\left(\dfrac12 < x_n < 1\right)\), \(\left(1 < \dfrac{1}{x_n} < 2\right)\)
for any given \(x_{n+1}\) there lie infinite pre-images.
slope at each point \(|{\cdot}| > 1\)
\(\to\) unstable at each point.
8.1) Let's take any arbitrary number from \(0\) to \(1\). Take its
reciprocal, throw away the integer, and repeat the process for a
long time. What is the probability that the final number is less
than \(\alpha\), \((0,\alpha)\)?
Before we calculate, let's find the fixed point of the Gauss map.
f.p\(_1\) \(\Rightarrow\) \(x = \dfrac{1}{x} - 1\)
So f.p\(_{1,2} = \dfrac{\sqrt5 - 1}{2}\) as \(0 < x < 1\)
\(\to x + 1 = \dfrac{1}{x} \;\Rightarrow\; x = \dfrac{1}{1+x}\)
Continued fraction can be written as
For f.point 2, \(x = \dfrac{1}{x} - 2\)
continued fraction \(\Rightarrow\) \(\dfrac{1}{x} = x + 2\)
Q.) What sort of number is this?
Ans. It would be the fixed point of one iteration of the map, or
a period-2 cycle.
- Statistical mechanics helps in understanding the behaviour
of systems which involve a large number of degrees of freedom
(which are interacting non-linearly with one another).
- It is the science of averages. Also it is the study at the
largest/longest length scales and time scales.
-
Statistical mechanics is partitioned into two parts:
- Equilibrium statistical mechanics.
- Non-equilibrium statistical mechanics.
- Equilibrium statistical mechanics arose from the question
that: are there any steady states (equilibrium states) which the
system will finally settle down to? If there is, how does it
behave, and so on?
- Thermodynamics is the study of systems at their longest
length scales and time-scales.
- So all fluctuations which happen at very small length
scales and time scales are neglected, and we look at things at
very long time scales and very long length scales.
Let us start with the Hamiltonian of a system. The system
contains many particles and is isolated from the rest of the
universe.
(exchange of matter & energy are prohibited).
\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=1.1]
\draw[thick] plot [smooth cycle, tension=1] coordinates {(0,1.6) (1.6,2) (2.6,1) (2.2,-0.6) (0.6,-0.8) (-0.8,0.4)};
\draw[thick,domain=0:540,samples=100,variable=\t]
plot ({0.9*cos(\t)/(1+0.02*\t)+0.7},{0.9*sin(\t)/(1+0.02*\t)+0.6});
\end{tikzpicture}
\[
H(q,p)
\]
If the rules are time-translation invariant, then we know that
the Hamiltonian is a constant of motion. \(\left(\dfrac{dH}{dt}=0\right)\)
{with uncertainty \(\delta E\)}
Phase space becomes \(6N\)-dimensional for \(N\) particles. So we
need to know \(3N\) constants of motion which are in involution
with each other.
\(\{C_i, C_j\} = 0\) if \(i \neq j\) [?]
Constants of motion will be:
- \(H\) \(1\)
- total linear momentum (\(P\)) \(3\)
- Total angular momentum (\(L\)) \(3\)
- C.O.M position (\(\vec R\)) \(3\)
Total \(10\) c.o.m [constants of motion].
To specify the state of the system we need a point in phase space
\((q_1,q_2,\ldots,q_N,\,p_1,p_2,\ldots,p_N)\).
And that point will move on some \(6N\)-dimensional hypersurface.
- Due to lack of a sufficient number of constants of motion
which are in involution with each other, we know that the
differential equation of motion is not integrable, and the system
of such kind is going to show chaotic behaviour with some
positive Liapunov exponents. It is likely to wander around
completely in a strange fashion, such that the system is ergodic
on the energy surface.
- It does not have to be strictly ergodic; there can be a
great deal of mixing (exponential separation of the volume
element).
Any way, exponential separation \(\Rightarrow\) mixing \(\Downarrow\)
ergodicity (implies)
- The volume element of phase space will come closer to
every point of phase space in a given sufficient time. {definition
of ergodicity}.
- The whole phase space is not accessible to the system because we specified its total energy.
- we become ignorant in knowing what happens to system at time \(t \neq t_0\), and assume that every point/volume element of phase space is as equally likely as other points of phase space. We can term this principle as principle of maximum ignorance.
- \(\rangle\) _"In a state of thermal eq\(^m\), all accessible micro-states of the system are equally probable."_
_"This single statement is enough to derive all equations of stat. mech, whereas thermal eq\(^m\) turns out to be a special case of it."_
Two systems are said to be in thermal equilibrium if there is no heat flow between them when they are connected to each other.
A system is said to be in thermodynamic eq\(^m\) if it is in mechanical, physical and thermal equilibrium.
- \(\rangle\) state of thermal eq\(^m\) also means that long time average of macroscopic variables reach a steady state and do not change with time. Such macroscopic quantities are Energy (internal), Total pressure, etc.
_"All physical macroscopic quantities are time independent in a system of thermal eq\(^m\) state."_
Note:-
- \(\rangle\) Microscopic variables like velocity of a particle in certain d.o.f is not time independent, whereas the average velocity of all/each particles is time independent, which is zero, i.e. system is not moving anywhere.
- \(\rangle\) The average values can only be independent of time if the probability distribution (over which the average is defined) is itself independent of time.
micro state:-
Specifying the state of every constituent of the system. The set \((q_1 \cdots q_n,\, p_1 \cdots p_n)\) is the microstate of the system, or we can say that position in phase space is microstate, whereas the total momentum (avg) of all particles is macrostate.
By accessible microstate we restrict the velocity a particle can have, i.e. \(H(q,p) = E_{\text{total}}\).
So a particle of such an isolated system can not have \(K.E > E\). So some microstates are not accessible anymore.
"All those micro-states which are compatible with all restrictions applied on the system are termed as accessible micro-state."
"microstate tells what each and every particle is doing and macrostate is giving some gross/overall information."
If microstate energy is the only parameter of the microstate \(\to\) (in quantum stat. mech.)
_"If an isolated system has 10 J of total energy then we can have a single particle which has energy close to 10 J and rest of all are close to zero Energy. (microstate) This is equally probable as other states; followed from postulate of eq\(^m\) stat. mech."_
In phase-space, probability density obeys
if probability distribution is independent of time
remember \(f_{eq}\) is not a physical observable.
Since such an isolated system uses microstates, such an ensemble is termed as the microcanonical ensemble. A stat. ensemble with \(f_{eq} = \delta(H(q,p)-E)\), which corresponds to an isolated system in thermal eq\(^m\), is called the microcanonical ensemble (in phase space).
- \(\rangle\) If we assume we can not have a point in phase space but a volume shell of finite resolution, then only can we define the probability of occupying a certain state (volume element in phase space/shell). \[ \Delta q\, \Delta p \geq \frac{\hbar}{2} \]
for 1 particle, volume element
for \(N\) particles the volume element would be \(\sim h^{3N}\).
If volume of accessible phase space is \(M\),
let probability of any microstate \(= P\), \(\sum_i P_i = 1\)
Set of \(N\) coins, where each coin (distinguishable/identifiable) has \(\{H,T\}\) two states.
(still a state of the system as a whole, but describes what each object is doing.)
Let, Total no. of heads \(= H\)
Total no. of tails \(= T\)
Let, \(H - T = M\).
\((N,M)\) or \((N,H)\) or \((H,T)\) are macrostate of the system.
For given \(N\),
ex for \(N=3\), let \((N,H)\) represent macrostate; we have \((3,0),(3,1),(3,2),(3,3)\) — 4 macrostates.
no. of microstates are much larger than no. of macrostates.
- \(\rangle\) Probability of any possible/specific microstate would be \(\left(\dfrac{1}{2^N}\right)\).
- \(\rangle\) Probability of attaining a given macrostate:- we have \(0 \leq H \leq N\) \[ P(H) = {}^N C_H \cdot \left(\frac{1}{2}\right)^H \left(\frac{1}{2}\right)^{(N-H)} \]
Suppose the coin is biased, then
then,
\(P(H) = \) coefficient of \(x^H\) in the "expanded" generating function.
Property of G.F.:-
to find \(\langle H \rangle = \sum_{H=0}^{N} P(H) H\) we differentiate the G.F.
so,
find
equating (1) and (2)
Also, now we can find variance,
Therefore \(\Delta H = \) std. deviation \(=\) uncertainty \(= \sqrt{Npq}\)
Therefore relative fluctuation will be:-
- \(\rangle\) It says relative fluctuation becomes smaller as no. of coin tosses increases / no. of coins increases.
In thermodynamics we deal with \(\sim 10^{23}\) d.o.f, so relative fluctuations become very small. This is the main reason "why thermodynamics works".
_"In statistical mechanics the generating function is called the partition function (\(\sim\) laplace transformation / green fn.)"_
Note:-
out of 100 coin tosses, \(P(50H)\) or \(P(50T)\) is most probable, because \({}^{100}C_{50}\) is biggest for all \(H\) in \(\left({}^{100}C_H\right)\).
each microstate is equally probable, but \(P(1H \,\&\, 99T)\) has contribution — but each macrostate is made up of many microstates. So attaining \(1H\) and \(99T\) has a lesser no. of microstates than that of \(50H, 50T\); so one example is that one molecule can have nearly all the total energy and rest of all are nearly at rest; indeed such microstates are equally probable, but the macrostate has contribution from very few microstates, so such a macrostate is less likely to exist than other possible macrostates.
\(N\) coins, \(H\) heads, \(T\) tails
What does this distribution look like?
\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=0.9]
\draw[->] (0,0) -- (9,0) node[right] {$H$};
\draw[->] (0,0) -- (0,3) node[above] {$P(H)$};
\foreach \x/\y in {0.5/0.2,1.5/0.5,2.5/1.0,3.5/1.7,4.5/2.4,5.5/1.7,6.5/1.0,7.5/0.5,8.5/0.2}
\draw (\x,0) -- (\x,\y);
\draw[dashed] (4.5,0) -- (4.5,2.6);
\node at (4.5,-0.4) {$N/2$};
\node at (8.5,-0.4) {$N$};
\end{tikzpicture}
\[
P(H) = {}^N C_H\, p^H q^{N-H}
\]
we would like to find out what does \({}^N C_H\) do, at large no. of \(N\).
Sterling formula:-
def:-
\(n!\) is defined as \(\displaystyle\int_0^{\infty} dx\, e^{-x} x^n = \Gamma(n+1) \;\; \textbf{[?]} \quad (n = 1,2,3,\dots)\)
\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=1]
\draw[->] (-0.2,0) -- (5,0) node[right] {$x$};
\draw[->] (0,-0.2) -- (0,3) node[above] {};
\draw[blue, domain=0:5, samples=50] plot (\x, {2.5*exp(-\x)});
\node[blue] at (3.6,0.5) {$e^{-x}$};
\draw[blue, domain=0:2.3, samples=50] plot (\x, {0.35*\x*\x});
\node[blue] at (2.1,2.0) {$x^n$};
\draw[thick, domain=0.3:4.5, samples=80] plot (\x, {2.2*exp(-((\x-2)*(\x-2))/1.2)});
\draw[dashed] (2,0) -- (2,2.2);
\node at (2,-0.4) {$n$};
\end{tikzpicture}
\[
g(x) = e^{-x} x^n = e^{-x} e^{\ln x^n} = e^{-x + n \log_e x} = e^{-(x - n \ln x)}
\]
\[
g'(x) = e^{-(x-n\ln x)} \cdot (-1)\left(1 - \frac{n}{x}\right) = 0 \implies \boxed{g'(n) = 0}
\]
let, \(f(x) = x - n\ln x\)
for very large \(n\)
Gaussian integral
also
So for large \(N\):
\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=0.9]
\draw[->] (0,0) -- (7,0) node[right] {$H$};
\draw[->] (0,0) -- (0,3) node[above] {$P(H)$};
\draw[thick, domain=0.2:6.5, samples=100] plot (\x, {2.6*exp(-((\x-1.8)*(\x-1.8))/0.6)*(1/(1+exp(-3*(\x-0.6))))});
\end{tikzpicture}
Take an ideal gas contained in volume \(V\). \(N = \) total no. of particles.
\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}
\draw (0,0) ellipse (2.3 and 1.6);
\node at (-1.1,0.9) {$N,V$};
\draw (0.2,-0.3) rectangle (0.9,0.3);
\node at (0.55,0.7) {$n,v$};
\end{tikzpicture}
What is the probability that \(v\) contains \(n\) particles?
\(P(n) = \) Probability that \(v\) contains \(n\) particles.
Probability that a given particle is in volume \(v\) \(= v/V\).
Assume that all the particles are moving independent of each other; therefore the probability that there are \(n\) of them inside \(v\) is
Here \(n\) is a random variable whose value can vary from \((0 \leq n \leq N)\).
here \(p = v/V\), \(q = 1 - v/V\).
this is the binomial distribution.
let's put \(\rho = N/V\)
so,
What happens to \(P(n)\), when \(N \to \infty\) and \(V \to \infty\) but keeping \(\rho\) fixed.
_(we can't use \(n! = e^{-n}n^n\sqrt{2\pi n}\) as \(n\) can take \(0\) too)_
and \(\bar n = \rho v\), \(=\) avg. no. of particles in sub-volume \(v\).
We have assumed that particles behave like Newtonian mech. particles and are distinguishable from one another.
In \(\left(P(n) = \dfrac{e^{-\rho v}(\rho v)^n}{n!}\right)\) formula, \((N,V)\) don't appear anymore, as \((N,V \to \infty)\) such that their ratio \(\rho = N/V\) is finite (this is called the thermodynamic limit of the system, when \(N \to \infty, V \to \infty\) such that \(\rho = N/V\) is finite). Statistical mech. reduces to thermodynamics in the thermodynamic limit.
for random variable \(n = 0,1,2,\dots \infty\)
Generating function
In poisson distribution, variance = mean.
for poisson's distribution, \(\langle n \rangle = \lambda\), \((\Delta n)^2 = \lambda\), & all higher moments \(=\lambda\), so it's a single-parameter distribution;
In gaussian distribution, except for \(\langle n \rangle\) & \((\Delta n)^2\), all higher moments vanish.
\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=0.9]
\draw[->] (0,0) -- (8,0) node[right] {$n$};
\draw[->] (0,0) -- (0,3) node[above] {$P(n)$};
\draw[dashed, domain=0.2:7.5, samples=100] plot (\x, {2.5*exp(-((\x-3.8)*(\x-3.8))/2.2)});
\foreach \x/\y in {0.5/0.15,1.3/0.4,2.1/0.9,2.9/1.6,3.7/2.3,4.5/2.4,5.3/1.9,6.1/1.2,6.9/0.6,7.5/0.25}
\draw (\x,0) -- (\x,\y);
\draw[dashed] (3.8,0) -- (3.8,2.5);
\node at (3.8,-0.4) {$\bar n$};
\end{tikzpicture}
\(\bar n = \) avg. no. of particles in subvolume \(v\).
If \(\bar n\) itself becomes very large, then the deviation from mean \((n - \bar n)\) takes a gaussian curve, and we call that a continuous variable if \(\bar n\) is very large.
from \(\displaystyle P(n) = \frac{e^{-\bar n}(\bar n)^n}{n!}\)
using Sterling's formula again, and replacing the variable to \(x = n - \bar n\)
_"The whole point of the above discussion is that if we start with a binomial distribution of Bernoulli trials, and then from that we take the no. of particles (trials) to be very large such that \(p(H) \to 0\) (probability of success in a trial is vanishingly small), then the Binomial distribution goes over into the Poisson distribution. Then if the mean value of the Poisson distribution is very large compared to unity, the deviation from the mean \((n-\bar n)\) is approximately a continuous variable that has a Gaussian shape."_
So a probability distribution can shift over to another probability distribution.
Generating function:-
let's go back to our problem of an isolated system in thermal equilibrium.
\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}
\draw (0,0) circle (2);
\draw[dashed] (-0.3,1.98) to[out=-85,in=95] (0.3,-1.98);
\node at (-1,0.6) {$\tilde A$};
\node at (1,0.6) {$\tilde B$};
\node at (-1,-0.7) {$N,V,E$};
\node at (1,-0.7) {$N',V',E'$};
\node at (0,2.3) {$N_{\text{tot}},\, V_{\text{tot}},\, E_{\text{tot}},\, \Omega_{\text{tot}}$};
\node[align=center] at (0,-2.6) {no.\ of accessible microstate.};
\end{tikzpicture}
Imagine the system is made of two sub-systems, in eq. thermal eq\(^m\) with each other, such that
(not necessarily: \(E+E' \neq E_{\text{tot}}\) always if pot. energy plays a role.)
\(E+E' \cong E_{\text{tot}}\) to a very good approximation.
_"as no. of d.o.f which are interacting near the partition is \(\left(\dfrac{1}{10^8}\right)\)th of the total d.o.f"_ (assuming short range forces).
Then at any instant of time, what is the probability that Energy of \(A\) is \(E\); \(P(E)\)
\(P(E) = \) no. of microstates of this entire system such that \(A\) has energy \(E\). \((\Omega(E))\) (also \(B\) has energy \(E'\).)
\(\Omega(E)\) & \(\Omega'(E')\) can have totally different functions [?].
As \(A\) can be a jar of a large container of oil and \(B\) can be our atmosphere. So \(A,B\) have different d.o.f.
Note that Energy of \(A\) and \(B\) are not fixed. Because there are fluctuations.
\(E, V, N, N', V', E'\) are variable quantities.
What we were trying to do is to find, is
_"Given the postulate of equal a priori probabilities of all the accessible microstates of the total system, what can we say about the probability distribution of (for instance say) Energy in system \(A\)."_
If \(A\) & \(B\) are assumed as independent systems (they actually are not) then \(\Omega_{\text{tot}}(E) = \Omega(E)\cdot\Omega'(E')\)
In the above discussion we have assumed that the interaction Energy is fairly negligible.
\(\Omega(E)\) is actually a function of \(V,N\) also.
Remember always:
Let's see how big these numbers are:
Suppose we take a single particle and put it into volume \(V\), and the particle is moving around with some energy \(\leq E\). Let's calculate how many microstates it has.
\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}
\draw (0,0) rectangle (2,1.4);
\draw (1,0.7) node {$\bullet$};
\node at (1,-0.4) {$V$};
\node at (0.4,1.1) {$\varepsilon$};
\end{tikzpicture}
No. of microstates is the no. of shells in phase-space (true in any dimensions)
Let no. of such microstates be
where \(d^3p = 4\pi p^2\,dp\) is the volume element, and \(\dfrac{\text{phase space volume}}{\text{min. volume }(h^3)}\) gives the number of microstates.
Relation between \(E\) and \(p\) for a free particle is
\(m = \) rest mass always.
Mass does not increase with speed, it's just \(p = \gamma m v\) instead of \(mv\). \(\left(\gamma = \dfrac{1}{\sqrt{1-v^2/c^2}}\right)\)
If \(p \ll mc\) \(\{mc = \text{Compton momentum}\}\)
In \(n\) dimensions, \((n-1)\) angles vary for \((0-\pi \text{ or } 0-2\pi)\).
In \(n\)-spatial dimensions,
No. of microstates with energy in \((\varepsilon,\varepsilon+d\varepsilon) \sim \varepsilon^{1/2}d\varepsilon = d\phi(\varepsilon)\)
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\begin{tikzpicture}[scale=1]
\draw[dashed] (0,0) circle (1.5);
\draw (0,0) circle (1.3);
\draw[->] (0,0) -- (1.1,0.9) node[midway,above] {$p=\sqrt{2m\varepsilon}$};
\node at (2.3,1) {$d\varepsilon$};
\node at (0,-2) {(given energy shell)};
\end{tikzpicture}
\[
d\Omega = 4\pi p^2\,dp \qquad d\nu \simeq 8\pi\, m^{3/2}\varepsilon^{1/2}\,d\varepsilon
\]
\[
\omega(\varepsilon) \sim \varepsilon^{1/2}d\varepsilon = f(\varepsilon)\,d\varepsilon
\]
Density of states \(f(\varepsilon) = \varepsilon^{1/2} = g_i\) (degeneracy at energy \(\varepsilon\)) {no. of microstates per unit energy interval at energy \(\varepsilon\)}
For \(n\)-D: \(\ \Omega \propto \displaystyle\int p^{n-1}dp\ \ (\Omega\propto \varepsilon^{n/2})\) for \((n\geq 2)\).
Case for 2-D container
density of states becomes constant \(\propto\) 2-D.
Case for 1-D
Density of states \(\uparrow\) as \(\varepsilon\downarrow\)
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\begin{tikzpicture}[scale=0.9]
\draw[->] (0,0) -- (5,0) node[right] {$\varepsilon$};
\draw[->] (0,0) -- (0,3) node[above] {$\phi(\varepsilon)$ (No. of microstates)};
\draw[thick] (0.4,0.1) .. controls (2,0.3) and (3,1.5) .. (4.5,2.7) node[right] {classical};
\draw[thick] (0.6,0.1) -- (1,0.1) -- (1,0.3) -- (1.6,0.3) -- (1.6,0.6) -- (2.3,0.6) -- (2.3,1.1) -- (3,1.1) -- (3,1.8) -- (3.8,1.8);
\node at (3.6,1.2) {quantum mech. states};
\node at (4,0.9) {$\varepsilon^{3/2}$};
\node at (0.6,-0.3) {$\varepsilon_{gm}$};
\node at (1,-0.4) {1st excited state};
\end{tikzpicture}
As \(\varepsilon\uparrow\) no. of states increases [3-D case] (microscopic)
Now, we want to find \(\Omega(E)\), so we will like to assume again that particles are free
and
Sum of energy of individual particles (with index \(i\)) = total energy.
For any two particles having \(z\) states, together they occupy \(z^2 = q\) states.
If one particle can take \(\phi(\varepsilon)\) states with energy \(\varepsilon\) to \(\varepsilon+d\varepsilon\): \(= k\varepsilon^{3/2}\);
independent \(N\) particles will take \(\ -\ k\varepsilon^{(3/2)N}\).
If all \(\varepsilon_i\) are of same order of magnitude \(\varepsilon\) {approximation}
So,
For very large \(N\), \(\Omega(E)\) is astronomically large number.
\(\alpha = \tfrac{3}{2}\, n\, \tfrac{1}{2}\, n\, \tfrac{5}{2}\cdots\)
lecture 22 (time 30:15)
Note: For Most probable "macro" state i.e. \(P(E)\) is maximum \(\equiv \{\text{eqm state}\}\)
So the system which has largest no. of microstates, that macrostate is the most probable state.
For Most probable macrostate it implies:
As \(P(E)\) is very large, let's take \(\log P(E)\) and differentiate.
for eqm \(\left(\dfrac{\partial P(E)}{\partial E}=\dfrac{\partial P}{\partial V}=\dfrac{\partial P}{\partial N}=0\right)\)
therefore
cond. for thermal eqm./most probable macrostate.
\(A\,|\,B\) possible, among two different kinds of matter, fluid, (gas \(\to\) liquid)...
or,
defn of temperature:
Let's define temp so for this property, a function \(\beta(E)\), of \(A/B\).
By defn, in equilibrium
"Any two parts have same temp in large isolated system under thermal equilibrium." (large d.o.f)
"Of course each part must be large enough to talk about temp, as temp. is not defined for a single particle."
If \(\Omega(E)\) is not strictly increasing function, and if so
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\begin{tikzpicture}[scale=0.9]
\draw[->] (0,0) -- (4,0) node[right] {$E$};
\draw[->] (0,0) -- (0,2.2);
\draw[thick] (0.3,0.2) .. controls (1.2,2) and (2,2) .. (2.6,1.2) .. controls (3.1,0.5) and (3.4,0.3) .. (3.7,0.1);
\node at (1,-0.3) {$(T>0)$};
\node at (3.2,-0.3) {$(T<0)$};
\end{tikzpicture}
(+ve slope implies positive temp; -ve slope implies negative temp.) So possibility of \((-\text{ve})\) temp is built in our defn. of temp.
Since \(\Omega = \Omega(N,V,E)\), from \(\left(\dfrac{\partial \ln\Omega}{\partial V}=0,\ \dfrac{\partial \ln\Omega}{\partial N}=0\right)\)
(in thermodynamic terms)
(Entropy representation)
(Energy representation)