NVE#

What you need to know

  • Understand that the essence of equilirbium statistical mechanics is connecting thermodynamics operating at macroscopic scales with the microscopic states of atoms and molecules.

  • The macro-to-micro connection is established using probability arguments with Entropy playing the central role (Boltzmann equation) From the thermodynamics point of view, an isolated system with N particles, fixed volume, and Energy is a single macrostate. Such a system, however, would consist of a vast number of microstates. The number can be quantified for some simple systems analytically but is generally intractable even using computers!

  • In StatMech we replace the dynamics of N particle system with an ensemble of all possible microstates that the system can be consistent with a macrostate (NVE).

  • Ensemble essentially gives us a probability distribution over microstates, which allows the calculation of Entropy.

  • Through entropy, all kinds of substance-specific thermodynamic quantities can be computed from the first principles.

  • E.g., heat capacities, magnetic susceptibilities, free energy differences. These quantities serve as an input for thermodynamics because thermodynamics does not refer to the microscopic world.

  • Thermodynamics while agnostic to atomic details is an exact theory. Predictions of statistical mechanics must obey the rules of thermodynamics.

Boltzmann’s equation: “A bridge” between microscopic dynamics and macroscopic thermodynamics#

  • Consider an isolated fluid system with \(N=const\), \(V=const\) and \(E=const\) which we will refer to as NVE system. What should be the probabilities of various states sampled by this system?

  • It is clear that in the absence of any other information, no states should have higher probabilities than others. This is known as “principle of equal a priory probability”

  • Mathematically, this can be expressed by maximizing entropy \(S=-\sum p_i log p_i\) with no constraints other than normalization. We end up with \(p_i = \frac{1}{\Omega}\) for all \(i\) microstates since we have no objective reason to favor any one microstate.

\[\boxed{S = k_B log \Omega}\]

  • \(\Omega\) is the number of micro-states consistent with a macrostate of our system.

  • \(S(N, V, E)\) is the entropy of an isolated system (we are in the NVE).

  • \(k_B =1.380649\cdot 10^{-23} J/K\), Boltzmann’s constant

Computing number of microstates \(\Omega\) using quantum mechanics#

A collection of up/down spins#

  • Consider a single spin atom under an external magnetic field \(B\). The system can be in two states with energy \(\epsilon = −m_z B\) where \(m_z=\pm \mu\) is the spin magnetic number and \(\mu\) magnetic moment of the spin. Generalizing to N spins is straightforward

  • \(N= N_{\uparrow} + N_{\downarrow}\) total number of spins

  • \(M = N_{\uparrow} - N_{\downarrow}\) overall magnetization

  • \(m = M/n\) magnetization per spin ranging from -1 to +1.

  • The energy is sum of independent spin energies \(E = \sum^N_i \epsilon_i = -N_{\uparrow}\mu B + N_{\downarrow}\mu B = -\mu B (N_{\uparrow} - N_{\downarrow}) = -\mu M B\).

  • To find number of microstates for a given \(E\) we need to find a number of ways to partition \(N_{\uparrow}\) and \(N_{\downarrow}\) spins.

\[\Omega(N, N_{\uparrow}) = \frac{N!}{N_{\uparrow}! N_{\downarrow}!} = \frac{N!}{[1/2(N+M)!] [1/2(N-M)!]}\]
\[\boxed{\Omega (E,N)= \frac{N!}{\Big[\frac{1}{2}(N-E/B\mu)\Big]! \Big[\frac{1}{2}(N+E/B\mu)\Big]!}}\]
\[S = Nk_B \Big[log 2 - \frac{1}{2}log(1-m^2)-\frac{m}{2} log \frac{1+m}{1-m} \Big]\]

  • Find temperature dependence by using \(\frac{1}{T} = \frac{\partial S}{\partial E}|_N\)

\[U = -\mu N B tanh \Big(\frac{\mu B}{k_B T} \Big)\]

  • magnetization per spin \(m=-U/\mu NB\) is given by:

\[m = tanh \Big(\frac{\mu B}{k_B T} \Big)\]

A collection of harmonic oscilaltors#

  • Consider a single 1D harmonic oscillator \(E_n = \hbar \omega (n+1/2)\) with \(n = 0, 1, 2, ...\) This system has infinitely many but still countable number of states.

  • Consider three harmonic oscillators \(E(n_1, n_2, n_3) = \hbar \omega (n_1+ n_2+n_3) + \frac{3}{2}\hbar \omega\). System with energy \(5/2 \hbar\omega\) can be in (1, 0, 0) or (0,1,0) or (0,0,1) microstates.

  • Now consider N harmonic oscillators. This can describe N atoms in a crystal, each vibrating around its mechanical equilibrium state independet of each other (“Einstein model”).

  • The question boils down to finding how many states correspond to \(n_1+n_2+...n_N = M\). In other words what is the degeneracy of the state correspodning to the total quantum number M?

\[E(n_1,n_2,...n_N) = \sum^N_{i=1} \hbar \omega \Big(n_i+\frac{1}{2}\Big) = \sum^N_{i=1} n_i +\frac{N}{2}\hbar \omega \]

The question is answered by partitioning M quanta of energy among N oscillators. This is same as finding number of ways of putting M ideantical balls into \(N-1\) boxes.

\[\boxed{\Omega(E,N) = \frac{(M+N-1)!}{M!(N-1)!} = \frac{(E/\hbar\omega+N/2-1)!}{(E/\hbar\omega-N/2)!(N-1)!}}\]

Monoatomic ideal gas#

Quantum density of states of an ideal gas

\[E({\bf n}) = \frac{h^2}{8mL^2} {\bf n^2}\]
\[\Omega(N,V, E) = \int_{E(n)=E} dn = \frac{C_{3N-1}}{2^{3N}} \int \delta \Big(\frac{(8mE)^{1/2}L}{h}-{\bf n} \Big) d {\bf n}\]

  • A sum over all total quantum number of N 3D particles in a box ends up being a problem of finding the volume of an N-dimensional sphere

Classical density of states of an ideal gas

\[H(p,q) =\frac{p^2}{2m} = E\]
\[{\Omega(E) = \frac{1}{N! h^{3N}}\int_{p^N,q^N} dp^N dq^N \delta(H(p,q) -E) =\frac{V^N}{N! h^{3N}} \int dp^N \delta(p^2/2m -E)}\]
\[ \Omega(E) = \frac{V^N}{N! h^{3N}} \delta V(R)\]

Where \(\delta V(R)\) is a volume of a spherical shell with radius \(R = (2mE)^{1/2}\) and thickness \(\delta R = 1/2 (2m/E)^{1/2} \delta E\)

Volume of a sphere in N dimesnional space

\[V(R) = \frac{\pi^{D/2}}{(D/2)!} R^D\]

For \(D\rightarrow \infty\) we discover that most of the volume of the sphere is concentrated at its surface!

\[\delta V(R) = V(R) -V(R-\delta R) = C [R^D - (R-\delta R)^D] = CR^D [1-(1- (\delta R/R)^D)] = CR^D = V(R)\]

Classical density of states of an ideal gas#

\[\boxed{\Omega(E) = \frac{V^N}{h^{3N} N!} \cdot \frac{(2m\pi E)^{3N/2}}{(3N/2)!}}\]
\[\boxed{ S = log \Omega(E) = k_B N \cdot \Big [ log \Big(\frac{V}{N \lambda^3}\Big) + \frac{5}{2}\Big]}\]

  • Note linear dependence on N. Entropy is an extensive quantity!

  • exponent 3/2 reflexts that each particle has 3 degrees of freedom

  • \(\lambda = \Big(\frac{3h^2 N}{4\pi m E}\Big)^{1/2}\) thermal de Broglie wavelength.

  • This result know as “Sackur Tetrode equation” was known long before statistical mechanics.

Thermal, Mechanical and chemical equilibrium#

  • The number of microstate for a system consisting of two parts can be written as a sum of all posisble energy partitionings among the two states.

\[\Omega(E) = \sum_{E_1} \Omega_1(E_1) \Omega_2 (E-E_1) \approx \Omega_1(U_1) \Omega_2 (E-U_1)\]

  • Since \(\Omega(E)\) is a sharply increasing function of energy we expect the maximum term for some energy \(E_1=U_1\) to dominate the sum. Let us find what state this maximum value should correspond to?

\[\frac{\partial log \Omega (E_1)}{\partial E_1} \Big |_{E_1=U_1}= \frac{\partial log\Omega_1(E_1)}{\partial E_1} - \frac{\partial log\Omega(E_2)}{\partial E_2}=0\]

Example: thermal contact between two ideal gases#

Recall that for ideal gas we have \(\Omega(E) = C V^N E^{3/2 N}\). Therefore for an ideal gas we will have:

\[\frac{\partial log \Omega (E_1)}{\partial E_1} \Big |_{E_1=U_1}= \frac{3}{2}\frac{N_1}{U_1} - \frac{3}{2}\frac{N_2}{U_2} = 0\]

Using definition of entropy we get the most probable value of energy \((U_1)\) corresponding to the equilibirum state

\[\frac{1}{k_B T} = \frac{3}{2}\frac{N_1}{U_1} = \frac{3}{2}\frac{N_2}{U_2}\]

Equations of state for ideal gas#

\[\frac{1}{T} = \frac{\partial S}{\partial E} = \frac{3}{2}k_B \frac{N}{E}\]

\[\frac{p}{T} = \frac{\partial S}{\partial V} = k_B N \frac{1}{V}\]

\[\frac{\mu}{T} = -\frac{\partial S}{\partial N} = k_B T \cdot log \frac{N}{V} \lambda^3\]

Quasistatic process and Irreversibility#

  • From the NVE ensemble reasoning, we can now state that a quasistatic process corresponds to a process that does not change in the number of microstates. That is removal of a constraint keeps the “volume” of microstates intact.

\[\Delta S = k_B log \frac{\Omega_f}{\Omega_i} = 0\]

  • On the other hand, once the number of microstates grows upon removal of a constraint, then re-instating the constraint will not shrink the “volume” of microstate space.

\[\Delta S = k_B log \frac{\Omega_f}{\Omega_i} > 0\]

Problems#

Shottky defects#

Schotky defects are vacancies in a lattice of atoms. Creating a single vacancy costs an energy \(\epsilon\). Consider a lattice with \(N\) atoms and \(n\) vacacnies. In this model the total energy is solely a function of defects: \(E=n\epsilon\)

  • Write down number of states and compute the entropy via Boltzmann formula. Plot number of states as a function of energy. You can use log of number of states for plotting.

  • Compute how the temperature would affect the fraction of vacancies on the lattice. Plot fraction of vacancies as a function of temperature.

  • How would the total energy depend on temperature \(T\). Derive expression for the high temeprature limit (\(\frac{\epsilon}{k_b T} \gg 1\)).

  • Plot total energy as a function of temperature E(T)

Lattice gas#

Consider a lattice gas of N particles distributed among V cells (with \(N\geq V\)). Suppose that each cell may be either empty or occupied by a single particle. The number of microscopic states of this syste will be given by:

\[\Omega (N, V) = \frac{V!}{N! (V-N)!}\]

  • Obtain an expression for the entropy per particle \(s(v)=\frac{1}{N} \cdot S(N,V)\) where \(v=\frac{V}{N}\).

  • From this simple fundamental equation, obtain an expression of equation of state \(p/T\).

    • Write an expansion of \(p/T\) in terms of density \(1/v\). Show that the first term gives Boyle law of ideal gases.

  • Sketch a graph of \(\mu/T\), where \(\mu(\rho)\) is a chemical potential as a function of density. Comment on \(\rho\rightarrow 0\) and \(\rho\rightarrow 1\) limits.

Polymer Elasticity#

Solve the problem 2.7 from the book.