Ensembles in phase-space

Ensembles in phase-space#

What you will learn

Counting microstates in quantum systems or lattice models is natural as these systems are already discretized
Defining microstates in classical mechanics requires us to divide phase-space into small volumes of size $\sim h^{N}$
Using classical hailtonians for non-interacting particles we can derive a number of useful relations between free energies chemical potentials.
Equipartition theorem says ever quadratic degree term in Hamiltonian contributes $k T$ of energy
De Broglie wavelength demarcates domain of applicability of classical and quantum mechanics

Quantization of Phase Space in Units of $h$ #

../_images/1d-pspace.png — Fig. 20 Illustration of phase space for a 1D particle in a box.#

In classical mechanics, a system’s state is represented as a point in phase space, with coordinates $(x, p)$ for each degree of freedom. The number of microstates is, in principle, infinite because both position and momentum can vary continuously.
However, in quantum mechanics, the Heisenberg uncertainty principle imposes a fundamental limit on how precisely position and momentum can be specified:
This implies that we cannot resolve phase space with arbitrary precision. Instead, each distinguishable quantum state occupies a minimum phase space volume on the order of:

Δ x Δ p \sim h

For a system with $f$ degrees of freedom (e.g $3 N$ for $N$ particles in three dimensions), the smallest resolvable cell in phase space has volume:

(Δ x Δ p)^{f} \sim h^{f}

To compute the number of accessible microstates $Ω$ in a given region of phase space, we divide the classical phase space volume $Γ$ by the smallest quantum volume $h^{f}$
Since N particles are indistingusihable and classical mechanics has no such concept we have to manually divide numer of microstates by $N!$

Ω (E) = \frac{Γ}{h^{f} N!}

An explicit expression for number of microstates for a given energy $E$ is written by using delta function which counts all microstates in phase-space where $H (p^{N}, x^{N}) = E$

Microcanonical partition function in phase-space

Ω (E) = \frac{1}{N! h^{N}} \int d p^{N} d x^{N} δ (H (p^{N}, x^{N}) - E)

Phase-space of a harmonic oscillator

H (x, p) = \frac{p^{2}}{2 m} + \frac{1}{2} m ω^{2} x^{2}

$x$ = position, $p$ = momentum, $m$ = mass, $ω$ = angular frequency
This Hamiltonian gives the total energy $E$ of a particle undergoing simple harmonic motion.
Setting $H (x, p) = E$ , we get the constant-energy curve:

\frac{x^{2}}{{(\sqrt{\frac{2 E}{m ω^{2}}})}^{2}} + \frac{p^{2}}{{(\sqrt{2 m E})}^{2}} = 1

This is the equation of an ellipse in the $(x, p)$ phase space, with:
- Semi-major axis: $a = \sqrt{\frac{2 E}{m ω^{2}}}$ (in $x$ )
- Semi-minor axis: $b = \sqrt{2 m E}$ (in $p$ )

Compute Area of the Ellipse (Classical Action)

A = π a b = π \sqrt{\frac{2 E}{m ω^{2}}} \cdot \sqrt{2 m E} = \frac{2 π E}{ω}

Plugging in quantum mechanical expression of quantized energies of harmonic oscillator

E = ℏ ω (n + \frac{1}{2})

This area corresponds to the classical action integral $\oint p d x$ , which in semiclassical quantization becomes:

\oint p d x = (n + \frac{1}{2}) h

../_images/87d045616f253e07bc6f2295dfa22389d0d9e494fba0569b718053c214a18ab9.png

Computing Z via classical mechanics#

In quantum mechanics counting microsites is easy, we have discrete energies and could simply sum over microstates to compute partition functions

Z = \sum_{i} e^{- β E_{i}}

Counting microstates in classical mechanics is therefore done by discretizing phase space for N particle system $x^{N}, p^{N}$ into smallest unit boxes of size $h^{N}$ .

\sum_{i} \to \int \frac{d p d x}{h}

Once agan to correct classical mechanics for “double counting” of microstates if we have N indistinguishable “particles” with a factor of $N!$

Classical partition function

Z (β) = \frac{1}{N! h^{N}} \int e^{- β H (x^{N}, p^{N})} d x^{N} d p^{N}

The power and utility of NVT: The non-interacting system#

For the independent particle system, the energy of each particle enter the total energy of the system additively.

E (ϵ_{1}, ϵ_{2}, . . . ϵ_{N}) = ϵ_{1} + ϵ_{2} + . . . ϵ_{N}

The exponential factor in the partition function allows decoupling particle contributions $e^{- β (ϵ_{1} + ϵ_{2})} = e^{- β ϵ_{1}} e^{- β ϵ_{2}}$ . This means that the partition function can be written as the product of the partition functions of individual particles!

Distinguishable states:

Z = z_{1} \cdot z_{2} \cdot z_{3} . . . z_{N}

Indistinguishable states:

Z = \frac{1}{N!} z^{N}

General strategy for using NVT for simple non-interacting systems

Compute the single particle partition function $z = e^{- β ϵ_{1}} + e^{- β ϵ_{2}} + . . .$
To compute full partition function, raise z to the power of N and apply factorial in case of indistinguishable particles $Z = \frac{1}{N!} z$ or take product in case of distinguishable identical particles $z^{N}$
Compute the free energy $F = - k_{B} T l o g Z$
Take derivatives of free energy to get the thermodynamic quantities. E.g one is often interested in computing temperature dependence of $μ (T), U (T), S (T)$

Maxwell-Boltzman distribution#

Since the kinetic and potential energy functions depend separately on momenta and positions, the total probability distribution factorizes into a product of position and momentum distributions:

H (p^{N}, r^{N}) = U (r^{N}) + K (p^{N}) = U (r^{N}) + \sum_{i = 1}^{N} \frac{p_{i}^{2}}{2 m}

This separation leads to:

P (r^{N}, p^{N}) = P (r^{N}) \cdot P (p^{N}) = \int_{V} d r^{N} e^{- β U (r^{N})} \cdot \int_{- \infty}^{+ \infty} d p^{N} e^{- β \sum_{i = 1}^{N} \frac{p_{i}^{2}}{2 m}}

The intergarlwith position-dependent potential for interacting particle system is impossible to evaluate except by simulations or numerical techniques
The momentum-dependent part splits into individaul integrals for every particle and is always a simple quadratic function, making it analytically tractable:

P (p^{N}) = {[\int_{- \infty}^{+ \infty} d p e^{- β \frac{p^{2}}{2 m}}]}^{N}

The momentum (or velocity) distribution follows the Maxwell-Boltzmann distribution in all equilibrium systems, regardless of the complexity of the molecular interactions.

Maxwell-Boltzmann Velocity Distribution: Full Derivation

Separation of variables

In equilibrium, the distribution of velocities in the $x$ , $y$ , and $z$ directions are independent and identical, because there’s no preferred direction in an ideal gas. Therefore, we can write the joint probability distribution as:

f (v_{x}, v_{y}, v_{z}) = f (v_{x}) f (v_{y}) f (v_{z})

Furthermore, the probability of finding a particle with velocity components between $(v_{x}, v_{x} + d v_{x})$ , $(v_{y}, v_{y} + d v_{y})$ , and $(v_{z}, v_{z} + d v_{z})$ is:

f (v_{x}, v_{y}, v_{z}) d v_{x} d v_{y} d v_{z}

Form of the velocity component distribution

The system is in thermal equilibrium, so the probability of a particle having a certain velocity corresponds to the Boltzmann factor for its kinetic energy:

Probability \propto e^{- E_{kin} / k_{B} T}

For a single particle of mass $m$ , the kinetic energy is:

E_{kin} = \frac{1}{2} m (v_{x}^{2} + v_{y}^{2} + v_{z}^{2})

So we postulate:

f (v_{x}, v_{y}, v_{z}) \propto e^{- m (v_{x}^{2} + v_{y}^{2} + v_{z}^{2}) / 2 k_{B} T}

Using the independence of the components:

f (v_{x}) \propto e^{- m v_{x}^{2} / 2 k_{B} T}

We require that:

\int_{- \infty}^{\infty} f (v_{x}) d v_{x} = 1

This is a standard Gaussian integral of form $p (x) = \frac{1}{\sqrt{2 π σ^{2}}} e^{- \frac{x^{2}}{2 σ^{2}}}$ where $σ^{2} = \frac{k_{B} T}{m}$ and therefore we write down:

f (v_{x}) = \sqrt{\frac{m}{2 π k_{B} T}} e^{- \frac{1}{2} m v_{x}^{2} / k_{B} T}

Distribution of speeds

Now, let’s consider the speed $v = \sqrt{v_{x}^{2} + v_{y}^{2} + v_{z}^{2}}$ .
The probability density for the speed (regardless of direction) is obtained by converting to spherical coordinates in velocity space. The volume element in velocity space becomes:

d v_{x} d v_{y} d v_{z} = v^{2} \sin θ d v d θ d ϕ

Integrating over angles gives a factor of $4 π$ :

f (v) d v = \underset{Jacobian}{\underset{⏟}{4 π v^{2}}} \cdot f (v_{x}) f (v_{y}) f (v_{z}) d v

Substitute in the expressions for $f (v_{x})$ , $f (v_{y})$ , and $f (v_{z})$ , and recall that the total exponent is:

f (v_{x}) f (v_{y}) f (v_{z}) = {(\frac{m}{2 π k_{B} T})}^{3 / 2} e^{- \frac{1}{2} m (v_{x}^{2} + v_{y}^{2} + v_{z}^{2}) / k_{B} T} = {(\frac{m}{2 π k_{B} T})}^{3 / 2} e^{- \frac{1}{2} m v^{2} / k_{B} T}

Therefore, the speed distribution is:

f (v) = 4 π v^{2} {(\frac{m}{2 π k_{B} T})}^{3 / 2} e^{- \frac{1}{2} m v^{2} / k_{B} T}

This is the Maxwell-Boltzmann speed distribution. It gives the probability density that a particle in a gas has speed $v$ .

Maxwell-Boltzmann distribution

Velocity component distribution (1D Gaussian):

f (v_{x}) = \sqrt{\frac{m}{2 π k_{B} T}} e^{- \frac{1}{2} m v_{x}^{2} / k_{B} T}

Speed distribution (Maxwell-Boltzmann form):

f (v) = 4 π v^{2} {(\frac{m}{2 π k_{B} T})}^{3 / 2} e^{- \frac{1}{2} m v^{2} / k_{B} T}

$v$ is the speed of a particle,
$m$ is the mass of a particle,
$k_{B}$ is Boltzmann’s constant,
$T$ is the absolute temperature.

../_images/2eca3706c13d6d566c127e5900c58f1af422805a488f2019ede5a86cf32f5bb6.png

Ideal Gas#

A useful approximation is to consider the limit of a very dilute gas, known as an ideal gas, where interactions between particles are neglected $V = 0$ .
Lets evaluate partition function for a single atom of idea gas:

z (β) = {[\frac{1}{h} \int_{0}^{L} d x \int_{- \infty}^{+ \infty} d p e^{- \frac{p^{2}}{2 m k_{B} T}}]}^{3} = V {(\frac{2 π m k_{B} T}{h^{2}})}^{3 / 2}

z = \frac{V}{λ_{T}^{3}} = V n_{Q}

De Broglie Thermal Wavelength and quantum concentration

λ_{T} = \frac{h}{\sqrt{2 π m k_{B} T}}

n_{Q} = \frac{1}{λ_{T}^{3}}

On the Validity of the Classical Approximation

The classical approximation is valid when the average interatomic spacing $\sim n^{- 1 / 3}$ is much larger than $λ_{T}$ , i.e., when:

n λ_{T}^{3} ≪ 1

This ensures that wavefunction overlap is negligible and particles behave classically.

Classical partition function for N Atom Ideal gas

Z = \frac{Z^{N}}{N!} = \frac{(V n_{Q})^{N}}{N!}

F = - k_{B} T l o g Z = - k_{B} T l o g Z = - N k_{B} T l o g (\frac{n_{Q}}{n}) - N k_{B} T

μ = \frac{\partial F}{\partial N} = k_{B} T \cdot l o g (\frac{n}{n_{Q} (T)})

We can also calculate entropy and other thermodynamic quantities and compare with expressions obtained in $N V E$ ensemble

S = - (\frac{\partial F}{\partial T})_{N, V} = N k_{B} [l o g (\frac{n_{Q}}{n}) + \frac{5}{2}]

U = \frac{\partial l o g Z}{\partial (- β)} = \frac{3}{2} N k_{B} T

p = - \frac{\partial F}{\partial V} = \frac{N k_{B} T}{V}

Equipartition Theorem#

Consider a classical system with a Hamiltonian containing a quadratic degree of freedom $x$ , such that:

E (x) = \frac{1}{2} a x^{2}

The canonical partition function for this degree of freedom is:

Z = \int_{- \infty}^{\infty} e^{- β \frac{1}{2} a x^{2}} d x = \sqrt{\frac{2 π}{β a}}

The average energy is:

⟨ E ⟩ = - \frac{\partial \log Z}{\partial β}

\log Z = \frac{1}{2} \log (\frac{2 π}{β a}) = const - \frac{1}{2} \log β

⟨ E ⟩ = - \frac{\partial}{\partial β} (- \frac{1}{2} \log β) = \frac{1}{2 β} = \frac{1}{2} k_{B} T

equipartition function for $f$ independent quadratic degrees of freedom

⟨ E ⟩ = \frac{f}{2} k_{B} T

Examples#

Monatomic ideal gas: 3 translational degrees → $⟨ E ⟩ = \frac{3}{2} k_{B} T$
Diatomic molecule at high $T$ : 3 translational + 2 rotational + $n$ active vibrational modes
→ $⟨ E ⟩ = (\frac{5}{2} + n) k_{B} T$

Note: Equipartition only applies when the quadratic modes are thermally accessible, i.e., when $k_{B} T ≫$ energy level spacing (e.g., $ℏ ω$ for vibrations).

Partition functions for non-interacting molecules#

Total energy and partition function for a molecule with separable degrees of freedom:

Partition Function for a Molecular Gas

E = E_{transl} + E_{vib} + E_{rot} + E_{elec}

Z = Z_{transl} \cdot Z_{vib} \cdot Z_{rot} \cdot Z_{elec}

Translational Degrees of Freedom: Particle in a Box#

For a particle in a 3D box of volume $V = L^{3}$ , the energy levels are:

E_{n_{x}, n_{y}, n_{z}} = \frac{ℏ^{2} π^{2}}{2 m L^{2}} (n_{x}^{2} + n_{y}^{2} + n_{z}^{2})

To approximate the partition function at high temperatures (many levels populated), we replace the discrete sum with an integral:

Z = \sum_{n_{x}, n_{y}, n_{z}} e^{- β E_{n_{x}, n_{y}, n_{z}}} \approx {[\int_{0}^{\infty} d n e^{- β \frac{ℏ^{2} π^{2}}{2 m L^{2}} n^{2}}]}^{3}

This is a Gaussian integral:

\int_{0}^{\infty} e^{- a n^{2}} d n = \frac{1}{2} \sqrt{\frac{π}{a}}

Z = {[\frac{1}{2} \sqrt{\frac{π}{\frac{ℏ^{2} π^{2}}{2 m L^{2}} β}}]}^{3} = {(\frac{L}{h} \sqrt{2 π m k_{B} T})}^{3} = V {(\frac{m k_{B} T}{2 π ℏ^{2}})}^{3 / 2} = V n_{Q}

For $N$ indistinguishable molecules we obtain same result we go for ideal gas!

Z = \frac{(V n_{Q})^{N}}{N!}

Rotational Degrees of Freedom: Rigid Rotor Model#

For a linear molecule (diatomic rotor), the rotational energy levels are:

E_{J} = \frac{ℏ^{2}}{2 I} J (J + 1), with degeneracy (2 J + 1)

The partition function is:

Z = \sum_{J = 0}^{\infty} (2 J + 1) e^{- β \frac{ℏ^{2}}{2 I} J (J + 1)}

At high temperatures ( $k_{B} T ≫ ℏ^{2} / 2 I$ ), this sum can be approximated as an integral:

Z \approx \int_{0}^{\infty} (2 J + 1) e^{- β \frac{ℏ^{2}}{2 I} J (J + 1)} d J

Let $x = J (J + 1) \approx J^{2}$ for large $J$ :

Z \approx \int_{0}^{\infty} 2 J e^{- β \frac{ℏ^{2}}{2 I} J^{2}} d J

This integral is:

\int_{0}^{\infty} J e^{- a J^{2}} d J = \frac{1}{2 a}

Z \approx \frac{1}{β \frac{ℏ^{2}}{2 I}} = \frac{2 I k_{B} T}{ℏ^{2}} = \frac{T}{θ_{rot}}

Where $θ_{rot} = \frac{ℏ^{2}}{2 I k_{B}}$ is the rotational temperature.

Vibrational Degrees of Freedom: Harmonic Oscillator Model#

E_{n} = ℏ ω (n + \frac{1}{2})

Partition function for one oscillator:

z = \sum_{n = 0}^{\infty} e^{- β ℏ ω (n + \frac{1}{2})} = e^{- \frac{1}{2} β ℏ ω} \sum_{n = 0}^{\infty} {(e^{- β ℏ ω})}^{n} = \frac{e^{- \frac{1}{2} β ℏ ω}}{1 - e^{- β ℏ ω}}

For $N$ independent oscillators:

Z = z^{N}

Average energy would then be:

E = - \frac{\partial \log Z}{\partial β} = - N \frac{\partial \log z}{\partial β}

\log z = - \frac{1}{2} β ℏ ω - \log (1 - e^{- β ℏ ω})

\frac{\partial \log z}{\partial β} = - \frac{1}{2} ℏ ω + \frac{ℏ ω e^{- β ℏ ω}}{1 - e^{- β ℏ ω}} = - \frac{1}{2} ℏ ω + \frac{ℏ ω}{e^{β ℏ ω} - 1}

E = N ℏ ω (\frac{1}{2} + \frac{1}{e^{β ℏ ω} - 1})

Low and high temperature behaviours of harmonic oscillators#

Notice the two important limits. At low temperature quantization of energy becomes important while at high temperature result matches with classical mechanics predictions;
As $T \to 0$ , $e^{β ℏ ω} ≫ 1$ , so:

E \to N \cdot \frac{1}{2} ℏ ω

As $T \to \infty$ , $e^{β ℏ ω} \approx 1 + β ℏ ω$ :

\frac{1}{e^{β ℏ ω} - 1} \approx \frac{1}{β ℏ ω} \Rightarrow E \approx N k_{B} T

../_images/996a82d262dbadf80d8a8a0d1ffcdcdc657b404d1963caecc84dd62081323243.png

On heat capacity of harmonic-oscillators#

Einstein Model of solids#

A simple model for solids due to Einstein describes it as $N$ independent harmonic oscillators representing localized atomic vibrations.
The average energy of a quantum harmonic oscillator is:

⟨ E ⟩ = ℏ ω (\frac{1}{2} + \frac{1}{e^{β ℏ ω} - 1})

The heat capacity is the derivative:

C_{V} = \frac{\partial ⟨ E ⟩}{\partial T} = k_{B} (\frac{x^{2} e^{x}}{(e^{x} - 1)^{2}}), x = \frac{ℏ ω}{k_{B} T}

Key predictions of Einstein model of solids#

At low T: $C_{V} \to 0$ — the oscillator is frozen in its ground state.
At high T: $C_{V} \to k_{B}$ — classical equipartition is recovered.

Debye Model of Solids#

The Debye model improves upon Einstein’s and matches experimental data much better than the Einstein model, especially at low temperatures.
Main distinguishing features of Debye model are:
- Treating atomic vibrations as a continuum of phonon modes.
- Including a full spectrum of frequencies up to a Debye cutoff $ω_{D}$ .
- Predicting the correct low-temperature behavior of solids.

U = 9 N k_{B} T {(\frac{T}{T_{D}})}^{3} \int_{0}^{T_{D} / T} \frac{x^{3}}{e^{x} - 1} d x

C_{V} = 9 N k_{B} {(\frac{T}{T_{D}})}^{3} \int_{0}^{T_{D} / T} \frac{x^{4} e^{x}}{(e^{x} - 1)^{2}} d x

Behavior of Debye Model#

Low T: $C_{V} \propto T^{3}$ — from long-wavelength (acoustic) phonons.
High T: $C_{V} \to 3 R$ — classical limit.

Derivation of Debye Model and phonons

Instead of treating each atom as an independent oscillator (like in Einstein’s model), Debye considered collective vibrations of the entire lattice — i.e., phonons, which are quantized sound waves.
In 3D, the number of vibrational modes with frequencies less than $ω$ is proportional to $ω^{3}$ :

g (ω) d ω = \frac{9 N}{ω_{D}^{3}} ω^{2} d ω

$g (ω)$ : phonon density of states
$N$ : number of atoms
$ω_{D}$ : Debye cutoff frequency, chosen so total modes = $3 N$
Each mode contributes energy:

⟨ E (ω) ⟩ = \frac{ℏ ω}{e^{β ℏ ω} - 1}

Total internal energy:

U = \int_{0}^{ω_{D}} g (ω) ⟨ E (ω) ⟩ d ω = \int_{0}^{ω_{D}} \frac{9 N}{ω_{D}^{3}} ω^{2} \cdot \frac{ℏ ω}{e^{β ℏ ω} - 1} d ω

x = \frac{ℏ ω}{k_{B} T} \Rightarrow ω = \frac{k_{B} T}{ℏ} x, d ω = \frac{k_{B} T}{ℏ} d x

x_{D} = \frac{ℏ ω_{D}}{k_{B} T} = \frac{T_{D}}{T}

Then the internal energy becomes:

U = 9 N k_{B} T {(\frac{T}{T_{D}})}^{3} \int_{0}^{x_{D}} \frac{x^{3}}{e^{x} - 1} d x

Now take the derivative:

C_{V} = \frac{d U}{d T} = 9 N k_{B} {(\frac{T}{T_{D}})}^{3} \int_{0}^{x_{D}} \frac{x^{4} e^{x}}{(e^{x} - 1)^{2}} d x

/tmp/ipykernel_2245/174881513.py:15: RuntimeWarning: overflow encountered in scalar power
  integrand = lambda x: (x**4 * np.exp(x)) / (np.exp(x) - 1)**2

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Problems#

Problem-1 1D energy profiles#

Given an energy function $E (x) = A x^{4} - B x^{2} + C$ with constants $(A = 1, B = 4, C = 1)$

Compute free energy difference between minima at temperature $k B T = 1, 10, 100$
Compute free energy around minima of size $δ x = 0.1, 0.5, 1$

Problem-2 Maxwell-Boltzman#

You are tasked with analyzing the behavior of non-interacting nitrogen gas molecules described by Maxwell-Boltzman distirbution:

f (v) = 4 π {(\frac{m}{2 π k_{B} T})}^{3 / 2} v^{2} e^{- \frac{m v^{2}}{2 k_{B} T}}

Where:

$f (v)$ is the probability density for molecular speed $v$
$m$ is the mass of a nitrogen molecule
$k_{B} = 1.38 \times 10^{- 23} J/K$ (Boltzmann constant)
$T$ is the temperature in Kelvin

Step 1: Plot the Maxwell-Boltzmann Speed Distribution

At room temperature (T = 300, 400, 500, 600 K), plot the distribution $f (v)$ for speeds ranging from 0 to 2000 m/s.

Step 2: Identify Most Probable, Average, and RMS Speeds

Compute and print:

The most probable speed $v_{m p} = \sqrt{\frac{2 k_{B} T}{m}}$
The mean speed $\bar{v} = \sqrt{\frac{8 k_{B} T}{π m}}$
The root mean square (RMS) speed $v_{r m s} = \sqrt{\frac{3 k_{B} T}{m}}$

Plot vertical lines on your plot from Step 1 to show these three characteristic speeds.

Step 3: Speed Threshold Analysis

Use numerical integration to calculate the fraction of molecules that have speeds:

(a) Greater than 1000 m/s
(b) Between 500 and 1000 m/s

Hint: Use scipy.integrate.quad to numerically integrate the Maxwell-Boltzmann distribution.

Problem-3 Elastic Collisions#

Complete the lab

In this problem need to modify the code to add elastic collisions to the ideal gas simulations

Problem-4 Elementary derivation of Boltzmann’s distirbution#

Let us do an elementary derivation of Boltzman distribution showing that when a macroscopic system is in equilibrium and coupled to a heat bath at temperatere $T$ we have a universal dependence of probability for finding system at different energies:

P (r^{'}) / P (r) = e^{- β (U (r) - U (r^{'}))}

The essence of the derivation is this. Consider a vertical column of gas somehre in the mountains.

On one hand we have graviational force which acts on a column between $h, h + d h$ with cross section $A$ .
On the other hand we have pressure balance which thankfully keeps the molecules from dropping on the ground.
This means that we have a steady density of molecules at each distance $n (h)$ for a fixed $T$ . Write down this balance of forces (gravitational vs pressure ) and find show how density at $h$ , $n (h)$ is related to density at $h = 0$ , $n (0)$ .

Tip: you may use $P = n k T$ for pressure and $m g h$ for the gravitational force)

Problem-5 2D diploes on a lattice#

Consider a 2D square lattice with $M$ lattice points. On each point we have a mangeetic moment that can point in four possible directions: $+ x, - x, + y, - y$ . Along $y$ axis, the dipole has the energy $ϵ > 0$ and along the x axis $ϵ = 0$ .

Dipoles are not interacting with each other and we also ingnore kinetic energy of diplos since they are fixed at lattice positions.

Write down parition function for this system $Z$
Compute the average energy.
Compute entropy per dipole $s (T)$ . Evaluate the difference $S (T = \infty) - S (T = 0)$ ? Can you see a link with number of arrangements of dipols?
Compute microcanonical partition function $Ω (N ϵ)$
Show that we get the same entropy expression by using $N V E$ and $N V T$ ensembles.