MIT OpenCourseWare http://ocw.mit.edu

5.62 Physical Chemistry IISpring 2008

For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.

5.62 Spring 2008 Lecture #22 Page 1

Einstein and Debye Solids

Reading: Hill, pp. 98-105, 490-495

The Einstein (quantum) model (all vibrational modes have the same frequency) gave much better agreement with experiment than the Dulong-Petit (classical) model (equipartition). But as T decreases, the Einstein model CV decreases too fast relative to the experimentally observed (approximately T+3 dependence) behavior of CV. Perhaps it would be more realistic to allow the vibrational frequencies to follow a plausible, computationally convenient, but non-constant probability distribution, ρ(ν)

Debye Treatment

Debye derived an improved model for the thermodynamic properties of solids by assuming that the distribution of normal mode frequencies is equivalent to that for sound waves.

CV = ki=1

3N!6

" h#ikT

$%&

'()2

e!h#i kT

1! e!h#i kT( )2

CV

Debye= k

0

#max* d#+(#) h#kT

$%&

'()2

e!h# kT

1! e!h# kT( )2

density of vibrational frequencies

For sound waves traveling in a three-dimensional solid ! "( )# "2 [see Nonlecture

derivation, below]. One way of seeing this is that ! = c

2"k~=c

2"kx2+ ky

2+ kz

2( )1/2

, where

k~

is the wave vector such that the wave is described by ei k~•r~ and

! =2"

|!

k |. The density

of states with a given ν is the number of ways of choosing k~

with the corresponding

magnitude, ⎟k~⎟.

Just as the degeneracy for a given speed state is proportional to c2 in the kinetic k~

theory of gases (as you'll see later in 5.62), the number of ways of picking with magnitude k = | k

~| is proportional to k2.

revised 3/21/08 9:01 AM

5.62 Spring 2008 Lecture #22 Page 2

So ρ(ν) ∝ k2 ∝ ν2

This figure is supposed to show a spherical shell of radius

!k

|k| and thickness dk. The number of states with |k| between |k| and |k| + dk is proportional to the volume of this shell, 4π|k|2.

Problem: What is the distribution of acoustic frequencies in an elastic solid? We are interested in the 3N lowest frequencies.

Solution: Find the harmonic frequencies which satisfy the boundary condition that the displacements are zero at the surface of a crystal of volume V. The wave equation for this problem is very similar to the Schrödinger Equation for a particle in a 3D infinite cubical well.

Consider the initial wave at t = 0, with displacements as function of position, x:

Φ0(x) = Φ(x,t = 0).

At t ≠ 0, the initial wave has moved in the +x direction by vst, where vs is the speed of sound in this medium

Φ(x,t) = Φ0(x - vst).

Making the harmonic approximation:

!0 (x) = Acos2"x#

$%&

'()

!(x,t) = Acos2"(x * v

st)

#+,-

./0

where vs

#= 1

= Acos2"1x

vs

* 2"1t+

,-.

/0

Find the values of ν for which Φ(x,t) = 0, where x is at the surface of the crystal of volume, V.

revised 3/21/08 9:01 AM

5.62 Spring 2008 Lecture #22 Page 3

Schrödinger Equation:

!2" x, y, z( ) =#2m$

!2

" x, y, z( ).

For 3-D infinite cubical box of length a on each edge

2m!nxnynz

!2

="2

a2nx2+ ny

2+ nz

2( )

where satisfy the boundary condition that ψ(x,y,z) = 0 on all six surfaces of the

cube.

Now use the required values of {ε} to solve for allowed ν's:

!nxnynz{ }

d2!(x,t)dx

2=d2

dx2Acos

2"#xvs

$ 2"#t%

&'

(

)*

+

,-

.

/0

= $ 2"#vs

%

&'

(

)*2

!(x,t).

Generalize to 3 dimensions

!2"(x, y, z,t) = # 2$%vs

&

'(

)

*+2

"(x, y, z,t).

Comparing prefactors for Schrödinger equation and wave equation

2!"vs

#

$%

&

'(2

=!2

a2nx2+ ny

2+ nz

2( ).

So now we know how ν depends on the number of standing waves in each of the three crystal directions. We want to know the density of vibrational modes as a function of frequency, ρ(ν), but it is easier to derive the density of modes as a function of n, ρ(n), where

n2! nx

2+ ny

2+ nz

2( ).

This is the equation for a sphere. So the number of modes between n and n + dn is given by the volume of one octant (nx, ny, nz, and n are all positive) of a spherical shell of radius n and thickness dn

!(n)dn = 1

84"n2dn

revised 3/21/08 9:01 AM

5.62 Spring 2008 Lecture #22 Page 4

We want ρ(ν)

!(") = !(n)dn

d".

To find the value of the Jacobian, dnd!

, for the transformation between n and ν as the

independent variable,

n = nx2+ ny

2+ nz

2( )1/2

2!"

vs=!

an =

!

anx2+ ny

2+ nz

2( )1/2

n =2a"

vs

dn

d"=2a

vs

#(") = #(n)dn

d"=1

84!n2( ) 2a

vs=4!a3"2

vs3

a3=V .

There are 3 polarizations (x, y, or z) for each lattice mode, thus

!(") = 34#V

vs

3"2

This is the frequency distribution function that goes into the Debye model.

So we have a physically reasonable model for the density of vibrational states as a function of frequency, ρ(ν).

But Debye had one more trick up his sleeve before inputting ρ(ν) to a statistical mechanical calculation of macroscopic thermodynamic properties.

There cannot be an infinite number of modes; only 3N–6 ≈ 3N. So Debye cut off the mode distribution arbitrarily at νmax to give the correct number of modes. ρ(ν) = Aν2 where A is determined by Debye's cutoff at νmax

3N =0

!max

" #(!)d! = A0

!max

" !2d! =A!

max

3

3$ A =

9N

!max

3$ # !( ) =

9N

!max

3!2

NOTE: We still don't know what νmax is, only that the mode distribution is normalized to this parameter.

revised 3/21/08 9:01 AM

5.62 Spring 2008 Lecture #22 Page 5

Now calculate some bulk properties:

Cv = k 0

!" h# / kT( )2e$h# /kT

1$ e$h# /kT( )2

%(#)d#

CV

Debye= k

0

#max

" 9N

#max

3#2 h#

kT

&'(

)*+

2e$h# /kT

1$ e$h# /kT( )2d#

Debye Temp. , -D , h#max

k x =

h#kT

d# = kThdx

Change of variable from ν to x

CV

Debye= k

0

!D T

" 9N

#max

3x2 kT

h

$%&

'()2

x2 e

*x

1* e*x( )2

kT

h

$%&

'()dx

= 9NkkT

h#max

$

%&

'

()3

0

!D T

" x4e*x

1* e*x( )2dx

= 9NkT

!D

$

%&

'

()3

0

!D T

" x4ex

ex *1( )

2dx

y=!D T+ ,++ 9Nk

y3

0

y

" x4ex

ex *1( )

2dx

not a generallytabulated function

Integrate by parts

u = x4

v =!1

ex!1( )

du = 4x3dx dv =

ex

ex!1( )

2dx

CV

debye=9Nk

y3

!x4

ex !1 0

y

+4x

3dx

ex !1

0

y

"#

$%%

&

'((= 3Nk •

3

y34

0

y

"x3dx

ex !1

!y4

ey !1

#

$%

&

'(

= 3Nk • 4 •3

y3

0

y

"x3

ex !1

dx !3y

ey !1

#

$%

&

'(

CV

Debye= 3Nk • 4D(y)!

3y

ey !1

"

#$%

&'

Debye EinsteinFunction Function

revised 3/21/08 9:01 AM

5.62 Spring 2008 Lecture #22 Page 6

Check high and low temperature limits of C VDebye :

high T limit

Cv

Debye

3Nk=12

y3

0

y

!x3

ex "1

dx "3y

ey "1

x =h#

kTy =

$DT

=h#

max

kT

%12

y3

0

y

!x3

1+ x +…"1( )dx "

3y

1+ y+…"1( )T ! $D & x, y' 0

=12

y3

0

y

! x2dx " 3 =

12

y3•y3

3" 3 = 4 " 3 =1

!CV

Debye"#" 3Nk for T ! $

D

(agrees with classical and Einstein treatments)

low T limit

Cv

Debye

3Nk=12

y3

0

y

! x3

ex "1

dx " 3y

ey "1

T # 0$ x, y#%

& 12y3 0

%! x3

ex "1

dx " 0

=12

y3

'4

15

()*

+,-

!CV

Debye= 3Nk•

4

5"4 T

#D

$

%&

'

()

3

for T * 0 low T limit

(Note the correct T3 behavior that agrees with experiment)

Note: The Debye T3 heat capacity law is in excellent agreement with actual data atall temperatures!

OTHER THERMODYNAMIC FUNCTIONS CALCULATED IN DEBYE MODEL

revised 3/21/08 9:01 AM

5.62 Spring 2008 Lecture #22 Page 7

U ! E0( )vib

= kTi=1

3N!6

" h# / kTeh# /kT !1

= kTi=1

3N!6

" h# / kT( )e!h# /kT

1! e!h# /kT

= kT0

$

% h# / kT( )e!h# /kT

1! e!h# /kT&(#)d#

!(") =9N

"max

3"2

U ! E0( )

vib= kT

0

"max

# 9N

"max

3"2 h" / kT( )e

!h" /kT

1! e!h" /kTd" x = h" / kT ,$

D=h"

max

k

= 3NkT •3

"max

3

0

$D T

#x3 kT

h

%&'

()*3

e!x

1! e!xdx

= 3NkT • 3T

$D

%

&'

(

)*

0

$D T

# x3e!x

1! e!xdx

= 3NkT3

y3

0

y

!x3

ex "1

dx = 3NkT • D(y)

y =!DT

D(y) = 3y"3

0

y

#x3

ex "1

dx

A ! E0( )vib

= kTi=1

3N!6

" ln 1! e!h#i /kT( ) = kT0

#max

$ %(#)ln 1! e!h# /kT( )d#

=9NkT

#max3

0

#max

$ #2 ln 1! e!h# /kT( )d#

= 3NkT •3

#max3

0

&D T

$ x2 kT

h

'()

*+,3

ln 1! e!x( )dx

= 3NkT •3

y3

0

y

$ x2ln 1! e!x( )dx

= 3NkT •3

y3

x3

3ln 1! e!x( )

0

y

!0

y

$ x3

3

dx

ex !1( )

-

./

0

12

revised 3/21/08 9:01 AM

5.62 Spring 2008 Lecture #22 Page 8

Integrate by parts:

u = ln 1! e!x( ) v =x

3

3

du =e!x

1! e!xdx =

dx

ex!1

dv = x2dx

= 3NkT ln 1! e!y( ) !1

3

3

y3

0

y

"x3dx

ex !1

#

$%

&

'(= 3NkT ln 1! e!y( ) !

1

3D(y)

#

$%&

'(

Svib =U ! A( )vib

T=U ! E0( )

vib! A ! E0( )

vib

T

= 3Nk D(y)! ln 1! e!y( ) +1

3D(y)

"

#$%

&'

= 3Nk4

3D(y)! ln(1! e!y )

"

#$%

&'

NOTE: The Debye model does not fit the phonon mode distribution of actual solidsterribly well, but Cv is not too sensitive to these differences. It works well for insulating crystals but fails badly for metals. What is special about metals? Also, fails near melting point of solid because the harmonic approximation fails. Why? Large displacements are necessarily anharmonic.

In the actual ρ(ν), what are the resonances at high ν?

revised 3/21/08 9:01 AM