LECTURE 10

Simple Applications of Statistical Mechanics

We have seen that if we can calculate the partition function

∑ Z = e-βEr r

(1)

then we can derive just about anything we want to know from the partition function such as the mean internal energy, the entropy, the pressure, the Helmholtz free energy, the specific heat, etc. Let’s give some examples of how this works.

Paramagnetism

You may remember from E&M that the magnetic behavior of materials usually arises from the angular momentum of the electrons. The electrons have 2 forms of angular momentum: spin and orbital. Classically their orbital angular momentum is associated with their orbital motion around the nucleus. Quantum mechanically the orbital angular momentum is associated with their spatial wavefunction. Even if an electron has zero orbital angular momentum, it still has acts like a tiny magnet and has a magnetic moment ≈

_B. Associated with this magnetic moment is the spin angular momentum ℏ ⃗
S

of the electron. Protons and neutrons also have angular momentum ℏ ⃗
J

and magnetic moments

. This gives rise to nuclear magnetic moments which are involved in NMR and MRI. The magnetic moment of an atom, ion or single elementary particle such as an electron or proton in free space is proportional to its angular momentum:

⃗μ = gμB ⃗J = - γ¯h ⃗J

(2)

where the proportionality factor g is the “spectroscopic splitting” factor or just the g factor, μ_B the Bohr magneton, γ the gyromagnetic ratio, and ℏ the total angular momentum. Rather than considering the case of arbitrary (see Reif 7.8, pages 257-262), let us begin by considering the simple case of spin-1/2 where J = 1∕2 and g = 2. Then the allowed quantum energy states are

⃗ E = - ⃗μ ⋅ H = - gμBJ⃗⋅ ⃗H ⃗ = - gμBJzH (since H ∥ˆz ) = - 2μ (± 1-)H B 2 = ± μBH (3)

where H is the external magnetic field. Let us also assume that we have N such independent ions or atoms all in contact with a heat reservoir at temperature T. A physical system that approximates this model is a metal with magnetic impurities like Mn²⁺.

If we consider only one atom or ion, the partition function ξ of that one atom or ion becomes

∑ ξ = e-βEr = eβμBH + e-βμBH allstatesr

(4)

and

( )N Z = ξN = eβμBH + e-βμBH

(5)

The mean energy can be calculated from

-- ∂ lnZ ∂ lnξ N ( )N E = - ------ = - N ------= - --μBH eβμBH + e-βμBH ∂β ∂ β ξ

(6)

-- eβμBH----e-βμBH- E = - N μBH eβμBH + e-βμBH

(7)

( ) -- μBH-- E = - N μBH tanh kBT

(8)

The total magnetic moment or magnetization can be calculated from

--- M = N μ- μ ↑ - μ↓ = N -------- μ ↑ + μ ↓ eβμBH---e--βμBH- = N eβμBH + e- βμBH (9)

( ) --- μBH-- M = N μB tanh kBT

(10)

Alternatively we can calculate the generalized force associated with the magnetic field. Recall that

--- dW = Xdx

(11)

and

--- 1-∂ lnZ- X = β ∂x

(12)

¿From Reif 11.1 we find that magnetic work is

dW = -M⃗ ⋅ d⃗H

(13)

where is the magnetization. Suppose is pointing along the ẑ axis. We can identify H with the external parameter x, and M, the z–component of , with the generalized force. Then the generalized force is

1-∂-ln-Z- M = β ∂H N ∂ lnξ = --------- β ∂H eβμBH - e-βμBH = N μB -βμBH------βμBH- e ( + e) = N μB tanh -μB-- -- kBT E = - --- (14) H

We see that

-- --- E = - M H

(15)

We can examine M in the limit of high and low temperature.

( ) ( ) If T → 0 or μBH-- → ∞ tanh -μB-- → 1 and M-- → N μ kBT kBT B

(16)

Physically at low temperatures all the magnetic moments are aligned with the magnetic field.

tanh.eps

magnetization.eps

At high temperatures, μ_BH∕k_BT → 0 and T →∞. Since tanh(x) → x as x → 0, we obtain

---~ (μBH--) N-μ2BH-- 1- M = N μB k T = k T B B

(17)

We expect the magnetization to be proportional to the magnetic field H. This is what characterizes a paramagnet. How easy or hard it is to magnetize the system is reflected in the magnetic susceptibility χ:

M = χH

(18)

¿From eq. (17) we can read off the susceptibility:

--- M N μ2B 1 χ = H-- = -k--- T- T → ∞ (high temperature limit) B

(19)

The fact that χ varies as 1∕T is know as the Curie law. χ is called the Curie susceptibility because it goes as 1∕T.

Curie.eps

We can also calculate the Helmholtz free energy using eqns. (5) and (4):

N F = - kBT ln Z = - kBT ln(ξ ) = - N kBT lnξ

(20)

where

( ) βμBH -βμBH μBH-- ξ = e + e = 2cosh (β μBH ) = 2 cosh kBT

(21)

( ( )) μBH-- F = - N kBT ln 2 cosh kBT

(22)

We can also calculate the heat capacity. Recall from lecture 9 that

∂2F- CV = - T ∂T 2

(23)

Taking two derivatives of (22) yields

( μBH )2 2 (μBH ) CV = N kB k--T- sech -k-T- B B

(24)

The specific heat per spin is

CV ( μBH )2 2 (μBH ) cV = --- = kB ----- sech ----- N kBT kBT

(25)

This is called the Schottky specific heat. It is the specific heat characteristic of two state systems. A two state system has only 2 states available to it. For example a spin–1/2 object has spin–up and spin–down states. Another example is an object that can only access the lowest states in a double well potential. In considering such discrete states, we are thinking quantum mechanically.

schottky.eps

Classical Harmonic Oscillator

Consider a classical harmonic oscillator with a spring constant κ. The typical example is a mass on a spring. The energy is given by

p2 1 E = ----+ -κx2 2m 2

(26)

Assume the harmonic oscillator is in contact with a thermal reservoir at temperature T. The partition function is

Z = & -1 ∫ e- βEdxdp ho ∫ ( ) ∫ ( ) 1-- ∞ βp2- ∞ β-κx2 = ho -∞ dp exp - 2m - ∞ dx exp - 2 (27)

Recall from Reif Appendix A4 that

∫ ( )1∕2 ∞ e- αx2dx = π- -∞ α

(28)

Using this (27) becomes

( ) ( ) -2-π- 1∕2 2-πm- 1∕2 1-2π-( m-)1∕2 Z = h β κ βh = β h κ o o o

(29)

( ) 2π (m )1∕2 lnZ = - ln β + ln h-- -κ o

(30)

and

-- ∂-ln-Z- ∂-ln-β- 1- E = - ∂ β = ∂ β = β = kBT

(31)

This makes sense; we expect the mean energy to be of order k_BT. It turns out that E - k_BT has equal contributions from the kinetic energy and from the potential energy. Each contributes k_BT∕2. One can show this explicitly. The mean kinetic energy is

---- KE =

Classical Equipartition Theorem

Let the energy of a system with f degrees of freedom be E = E(q₁...q_f,p₁...p_f) and assume

the total energy splits additively into the form
$E = ϵi(pi) + E ′(q1,...,pf)$

where ϵ_i involves only p_i and E^′ is independent of p_i.
$2 ϵi(pi) = bpi$

where b is a constant.

If we replaced p_i by q_i, the theorem is still true. We just want ϵ_i to be a quadratic function of one component of q or p.

Assume that the system is in equilibrium at temperature T. Then

∫ - ∞-∞ ϵie-βEdq1...dpf ϵi = -∫∞----βE---------- ∫- ∞ e dq1...dpf ∞-∞ ϵie-β(ϵi+E′)dq1...dpf = -∫∞--e-β(ϵi+E′)dq-...dp-- ∫- ∞ ∫ 1 f′ -∞-∞-ϵie-βϵidpi--∞∞-e-βE-dq1...dpf- = ∫∞ e-βϵidp ∫∞ e-βE ′dq ...dp - ∞ i -∞ 1 f

where the last integrals in the numerator and denominator do not involve dp_i. These integrals cancel leaving

∫ - --∞∞--ϵie-βϵidpi ϵi = ∫∞ e- βϵidpi -∞∂ ∫∞ -βϵ --∂β---∞-e---idpi = ∫∞ e-βϵidp - ∞ (∫ ∞ i ) = - -∂-ln e- βϵidp ∂β -∞ i

Now use the second assumption:

∫ ∞ - βϵ ∫ ∞ - βbp2 - 1∕2 ∫ ∞ -by2 e idpi = e idpi = β e dy -∞ -∞ -∞

where y ≡ β^1∕2p_i. Thus

( ∫ ∞ -βϵi ) 1- ∫ ∞ -by2 ln -∞ e dpi = - 2 ln β + ln - infty e dy

Notice that the integral does not involve β. So when we take the derivative in (36), only the first term is involved.

∂ ( 1 ) 1 ϵi = ---- - --lnβ = --- ∂ β 2 2β

ϵi = 1kBT 2

This is the classical equipartition theorem. It says that each additive quadratic term in the energy (i.e., each degree of freedom) contributes k_BT∕2 to the mean energy of the system.

Note that this theorem is true only in classical statistical mechanics as opposed to quantum mechanics. In quantum mechanics the energy of a system is discretized into energy levels. At low energies the levels are rather far apart. As the energy increases, the energy levels become more closely spaced. The equipartition theorem holds when mean energy is such that the levels near it have an energy level spacing ΔE ≪ k_BT.

Mean kinetic energy of a gas molecule

Let’s consider the simple example of a molecule in a gas (not necessarily an ideal gas) at temperature T. Its kinetic energy is given by

1 ( ) K = ---- p2x + p2y + p2z 2m

The kinetic energy of the other molecules do not involve the momentum of this particular molecule. The potential energy of interaction between the gas molecules also is independent of . So the equipartition theorem tells us that

--- 3- K = 2kBT

For an ideal monatomic gas the energy is solely kinetic energy, so

-- ( ) E = N 3-kBT = 3-νRT 2 2

where N is the number of gas particles, R = N_ak_B is the gas constant, and ν is the number of moles. The molar specific heat at constant volume is

( --) CV ∂ E 3 cV = --- = ---- = -R ν ∂T V 2

Brownian Motion

Brownian motion was discovered by Brown, a botanist, in the 1800’s and was explained by Einstein in 1905. If you put a small macroscopic particle in a liquid and watch it in a microscope, it jiggles around because the molecules in the liquid keep bumping into it. This is called Brownian motion. The molecules in the liquid are moving around because of thermal fluctuations. To see this, let the small macroscopic particle have a mass m and be immersed in a liquid at temperature T. Consider the x-component of the velocity v_x.

v- = 0 x

Even though the mean value of the velocity vanishes, this does not mean that v_x = 0. There are velocity fluctuations so that v_x²≠0. ¿From the equipartition theorem we have

1----- 1 -mv2x = -kBT 2 2

--- v2 = kBT-- x m

The factor of 1∕m means that the fluctuations are negligible for large m like a golf ball. But when m is small (e.g., when the particle is micron sized), then the velocity fluctuations become appreciable and can be observed under a microscope. Notice that the size of the fluctuations are proportional to temperature. The higher the temperature, the larger the fluctuations. This is what we expect of thermal fluctuations.