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5.62 Physical Chemistry II
Spring 2008
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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.
2
3N !6
$ h# '
e!h#i kT
CV = k " & i )
% kT ( (1 ! e!h#i kT )2
i=1
Debye
V
C
=k*
#max
0
$ h# '
e!h# kT
d#+(#) & )
% kT ( (1 ! e!h# kT )2
2
density of vibrational frequencies
For sound waves traveling in a three-dimensional solid ! ( ") # "2 [see Nonlecture
1/2
c
c 2
derivation, below]. One way of seeing this is that ! =
k~ =
kx + ky2 + kz2 ) , where
(
2"
2"
i k~ •r~
2"
and ! =
. The density
k~ is the wave vector such that the wave is described by e
|k|
!
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
theory of gases (as you'll see later in 5.62), the number of ways of picking k~ with
magnitude k = | k~ | is proportional to k2.
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5.62 Spring 2008
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Lecture 22
Page 2
So ρ(ν) ∝ k2 ∝ ν2
This figure is supposed to show a spherical shell of radius |k| and thickness dk. The
!
number of k 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:
$ 2"x '
)
! 0 (x) = A cos &
% # (
+ 2"(x * vs t) .
!(x,t) = A cos 0/
,
#
+ 2"1x
.
= A cos * 2"1t 0
, vs
/
where
vs
=1
#
Find the values of ν for which Φ(x,t) = 0, where x is at the surface of the crystal of
volume, V.
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5.62 Spring 2008
Lecture 22
Page 3
Schrödinger Equation: ! 2 " ( x, y, z ) =
#2m$
" ( x, y, z ) .
!2
For 3-D infinite cubical box of length a on each edge
2m!nx ny nz
!2
=
"2 2
n + ny2 + nz2 )
2 ( x
a
where {!nx ny nz } 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:
% 2"#x
(.
d 2!(x,t) d 2 +
=
A
cos
$
2"#t
'
*0
dx 2
dx 2 ,
& vs
)/
% 2"# (
= $'
* !(x,t).
& vs )
2
Generalize to 3 dimensions
& 2$% )
! 2"(x, y, z,t) = # (
+ "(x, y, z,t).
' vs *
2
Comparing prefactors for Schrödinger equation and wave equation
# 2!" & ! 2 2
2
2
%
( = 2 ( n x + n y + nz ) .
$ vs ' a
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
2
n 2 ! ( nx2 + ny2 + nz2 ) .
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 = 18 4"n 2 dn
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5.62 Spring 2008
Lecture 22
Page 4
We want ρ(ν)
!(") = !(n)
To find the value of the Jacobian,
independent variable,
dn
.
d"
dn
, for the transformation between n and ν as the
d!
n = ( nx2 + ny2 + nz2 )
1/2
1/2
2!" !
!
= n = ( nx2 + ny2 + nz2 )
vs
a
a
2a"
n=
vs
dn 2a
=
d" vs
#(") = #(n)
dn 1 (
2a 4!a 3 "2
= 4!n 2 )
=
d" 8
vs
v3s
a 3 = V.
There are 3 polarizations (x, y, or z) for each lattice mode, thus
4#V 2
"
v3s
This is the frequency distribution function that goes into the Debye model.
!(") = 3
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 =
!max
"
0
#(!)d! = A
!max
"
0
!2 d! =
3
A!max
9N
9N
$ A = 3 $ # ( !) = 3 ! 2
3
!max
!max
NOTE: We still don't know what νmax is, only that the mode distribution is normalized to
this parameter.
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5.62 Spring 2008
Lecture 22
Page 5
Now calculate some bulk properties:
2 $h#/ kT
! (h# / kT ) e
Cv = k "
%(#)d#
2
0
(1 $ e$h#/ kT )
#max
9N 2 & h# )
e$h#/ kT
(
+
C
=k "
#
d#
3
' kT * (1 $ e$h#/ kT )2
#max
0
h#
h#
kT
Debye Temp. , -D , max x =
d# =
dx
k
kT
h
2
Debye
V
Change of variable from ν to x
Debye
V
C
=k
!D T
"
0
9N 2 $ kT ' 2 e* x $ kT '
& )dx
x & ) x
2
3
% h (
#max
(1 * e* x ) % h (
2
$ kT '
= 9Nk &
)
% h#max (
$T '
= 9Nk & )
% !D (
3 !D T
"
(1 * e* x )
0
3 !D T
x 4 e* x
x 4 ex
2
dx
9Nk
dx + ++, 3
2
x
y
( e *1)
"
0
y=! D T
y
"
0
x 4 ex
( e x *1)
2
dx
not a generally
tabulated function
Integrate by parts
u = x4
v=
!1
( e !1)
x
du = 4x 3dx
debye
V
C
y
#
9Nk % !x 4
= 3
+
y %$ e x !1 0
#
3
= 3Nk • %4 • 3
$ y
y
"
0
dv =
ex
( e x !1)
2
dx
y
&
4x 3dx (
3#
x 3dx
y4 &
" ex !1 (' = 3Nk • y 3 %$4 " ex !1 ! ey !1('
0
0
3
x
3y &
dx
!
(
e x !1
e y !1'
y
"
3y %
CVDebye = 3Nk • $4D(y) ! y '
#
e !1&
Debye
Einstein
Function Function
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5.62 Spring 2008
Lecture 22
Page 6
Check high and low temperature limits of C VDebye :
high T limit
y
CvDebye 12
= 3
3Nk
y
!
x3
3y
dx " y
x
e "1
e "1
12
% 3
y
y
!
x3
3y
dx "
(1+ x +… "1)
(1+ y +… "1)
12
y3
y
=
0
0
!
x 2 dx " 3 =
0
x=
h#
kT
y=
$ D h#max
=
T
kT
T ! $ D & x, y ' 0
12 y 3
• " 3= 4 " 3=1
y3 3
!CVDebye ""
# 3Nk for T ! $D
(agrees with classical and Einstein treatments)
low T limit
CvDebye 12
= 3
3Nk
y
y
x3
3y
dx " y
x
e "1
e "1
!
0
T # 0 $ x, y # %
x3
dx " 0
0
e x "1
12 ( ' 4 +
= 3* y ) 15 ,
&
12
y3
!
%
4 $T '
= 3Nk• " 4 & ) for T * 0
5 % #D (
3
!C
Debye
V
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 at
all temperatures!
OTHER THERMODYNAMIC FUNCTIONS CALCULATED IN DEBYE MODEL
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5.62 Spring 2008
Lecture 22
3N !6
h# / kT
eh#/ kT !1
(U ! E0 )vib = kT "
i=1
= kT
(h# / kT ) e!h#/ kT
1 ! e!h#/ kT
3N !6
"
i=1
= kT
(h# / kT ) e!h#/ kT
&(#)d#
1 ! e!h#/ kT
$
%
0
!(") =
Page 7
9N 2
"
3
"max
(U ! E0 )vib = kT
"max
#
0
9N 2 (h" / kT ) e!h"/ kT
"
d"
3
"max
1 ! e!h"/ kT
x = h" / kT ,$ D =
h"max
k
% kT (
x ' * e! x
& h )
dx
1 ! e! x
3
= 3NkT •
$D T
3
3
"max
#
0
3
% T ( $ D T x 3 e! x
= 3NkT • 3' * #
dx
!x
& $D ) 0 1 ! e
= 3NkT
3
y3
y
!
0
y=
3
x
dx = 3NkT • D(y)
e "1
x
!D
T
y
D(y) = 3y"3 #
0
3N !6
( A ! E0 )vib = kT "
ln (1 ! e!h# / kT ) = kT
i
#max
$
x3
dx
e x "1
%(#)ln (1 ! e!h#/ kT ) d#
0
i=1
=
9NkT
3
#max
#max
= 3NkT •
= 3NkT •
$
#2 ln (1 ! e!h#/ kT ) d#
0
&D T
3
3
#max
3
y3
y
$
$
0
' kT *
x ) , ln (1 ! e! x ) dx
( h +
3
2
x 2 ln (1 ! e! x ) dx
0
y
3 - x3
= 3NkT • 3 / ln (1 ! e! x ) 0 !
y .3
y
$
0
x 3 dx 0
2
3 ( e x !1) 1
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5.62 Spring 2008
Lecture 22
Page 8
Integrate by parts:
x3
v=
3
u = ln (1 ! e! x )
du =
e! x
dx
dx = x
!x
1! e
e !1
dv = x 2 dx
#
1 3
= 3NkT %ln (1 ! e! y ) !
3 y3
$
y
"
0
#
&
x 3dx &
1
!y
( = 3NkT %ln (1 ! e ) ! D(y)(
x
$
'
e !1'
3
(U ! A)vib (U ! E0 )vib ! ( A ! E0 )vib
=
T
T
"
%
1
= 3Nk $ D(y) ! ln (1 ! e! y ) + D(y)'
#
&
3
Svib =
"4
%
= 3Nk $ D(y) ! ln(1 ! e! y )'
#3
&
NOTE: The Debye model does not fit the phonon mode distribution of actual solids
terribly 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