HW-7-solution - Solution of HW #7 PDF

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Solution of HW #7...

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Homework #7 1. Singularity in density of states. (a) From the dispersion derived in Chapter 4 for a monatomic linear lattice of N atoms with nearest neighbor interactions, show that the density of modes is 2N 1 , D ( )   2  (m   2 )1 / 2 where  m is the maximum frequency. (b) Suppose that an optical phonon branch has the form (K )  0  AK 2 , near K  0 in three dimension.

Show that

3

 L  2 D ( )     3 / 2 (0   )1 / 2 for    0 and D ( )  0 for    0 .  2  A density of modes is discontinuous.

Here the

Solution: 4C 4C ka ka being the maximum sin   m sin k  0 , with  m  m m 2 2 frequency. With positive k, L dk L L dk . dk  d . Thus D ( )  D ( )d  2  d  d 2 d  m a ka m a ka m a 2 a    m2   2 . cos 1  sin 2 1 2  m 2 dk 2 2 2 2 2 dk 2 .  d  a  m2   2

(a)  

D(  ) 

L dk 2L / a 2N   2 2  d   m    m2   2

(b) For (K )  0  AK 2 ,

  0 d   2 AK   2 A  2 A (0   ) . dk A 3

3

 L  4 k 2 dk  L  4 k 2 dk d    In 3D, D(  ) d      d  2   2  Then 3

3

3

dk  L  1 0    L   L  4 D ( )    4k 2      3/2 0   .  4 d  2  A 2 A(0   )  2  A  2  For    0 , there is no normal modes, so that D ( )  0 .

2. Zero point lattice displacement and strain. (a) In the Debye approximation, show that the mean square displacement of an atom at absolute zero is

R2  3 D2 / 8 2  v3 , where v is the velocity of sound.

u 4( n 1 / 2) / V  2 0

summed

over

the

Start from the result

independent

lattice

modes:

R  (  / 2 V )  . We have included a factor of ½ to go from mean square 1

2

1 amplitude to mean square displacement. (b) Show that   and R 2 diverge for a one-dimensional lattice, but that the mean square strain is finite. Consider ( R / x)2  (1/ 2) K 2u02 as the mean square strain, and show that it is equal to

 D2 L / 4MNv 3 for a line of N atoms each of mass M, counting longitudinal modes

only. The divergence of R 2 is not significant for any physical measurement. Solution: (a) For the motion of frequency  , the average kinetic energy should be half of the  E m 2 1  v  0   total energy: N or v2  . 2 Nm 2 2 2 2 Therefore, we have On the other hand, v2   2R2 and Nm  M  V .

v2

 . 2V   1  D D( ) d . Then R 2   2 V  2 V 0  For 3D, there are three modes (two transverse and one longitudinal) for each  , thus the density of states with the dispersion of   vk is V 4k 2dk V 4 2d  3V 2d D ( )d  3  3    . ( 2 ) 3 ( 2 )3 v3 2 2 v 3 3V D 3D2  D D ( )d  2   d     . Then R 2  V 0 2 V 2 2 v 3 0 8  2v 3  R  2

2



2Ldk 3Ld  . v 2    D D (  )d   3L  D d  3L R2       ln  0 D   .   0 0   2 L v 2 L 2 L v C 2   2 R  For 1D longitudinal wave, N or R  . 2CN 2 4   D R2    D ( )d  . 2 CN 2 CN 0 2Ldk Ld D ( )d  1   . v 2 D L D2  D L 2 D d d ( )        Then R  . 2CN 0 2CN v 0 4CN v

(b) For 1D, D( )d  3 

C  a , where a is the lattice constant, then we have m LD2 a 2  . 4mNv 3

Note the speed of sound v  R2 

 L D2  4CN v

 L D2 v2 m 4 2 N v a

The strain is   R / a , thus

 2  R2 / a 2 

L D2 . 4mNv 3

3. Gruneisen constant. (a) Show that the free energy of a phonon mode of frequency  is kBT ln[ 2 sinh(  / 2kBT )] . It is necessary to retain the zero-point energy   / 2 to obtain this result. (b) If  is the fractional volume change, then the free energy of the crystal may be written as 1 F ( , T )  B 2  k BT  ln[ 2 sinh( K / 2k BT )] , 2 where B is the bulk modulus. Assume that the volume dependence of  K is /     , where  is known as the Grűneisen constant. If  is taken as independent of the mode K, show that F is a minimum with respect to  when B     (  / 2) coth(   / 2k BT ) , and show that this may be written in terms of the thermal energy density as    U (T ) / B . (c) Show that on the Debye model    ln  /  ln V . Note: Many approximations are involved in this theory: the result (a) is valid only if  is independent of temperature;  may be quite different for different modes. Solution: 

(a) Z   e ( n1 / 2 )  / kB T  n0

e   / 2kB T 1    / 2 kB T    / 2 kB T 1 e e  e   / 2 kB T

1     2 sinh   2k BT 

       . F   kB T ln Z  kB T ln 2 sinh   2k BT     1   k   . (b) F (,T )  B 2  k BT  ln  2 sinh k T 2 2 k B    F is minimized when F /   0 .   k   2 cosh  2kB T   d k  B  k BT  0.   k  2k BT d k  2 sinh   2k B T  From the given condition of /     , there is d  / d    .

.

  k   k  Then the F /   0 becomes B    coth  0. k T 2 k B  2    k   k     coth  B k 2  2kB T 

 k

  k   k    coth k T 2 2  B  k

  k    k     exp  exp   k  2 kBT   2 kB T  2   k    k     exp   exp   2 kBT   2 k BT 

   exp  k   1 k k  k  kBT     2 2       k k exp  k   1 k exp  k   1  kB T   kB T  The first term is zero point energy and the 2nd term is thermal energy. If the zero  energy can be ignored, we have   U (T ). B  V D V V V0  ln  D , there is    D / . (c) With           ln V D V D V0 V  ln  D  ln   With  D  k B ,    .  ln V  lnV

4. For d-dimensional crystal, (a) Show that the density of states varies as  d 1 . (b) Deduce from this that the low-temperature specific heat vanishes as Td . Graphite has layered structure with very weak bounding between the layers. What temperature dependence of the specific heat would you expect for graphite at low temperature? (c) Show that if the dispersion were  ~ k  , the low-temperature specific heat would vanish as T d /  . Solution: (a) For d-dimensional system with   vk , Vk d 1 dk V d 1 d D ( )d ~  . Thus D( )   d 1 d d d ( 2 ) ( 2 ) v 

D

(b) The energy of the system is U  

0

Let x 

e  / kB T  1

 D /k BT  xd dx . , U ~ T d 1  x 0 e 1 kB T

D( ) d ~



D

0

 d e  /k BT  1

d

At low temperature ( k BT   D ), the up limit of the integral can be replaced by  .  xd U d 1 dx  T d 1 , and C  T d . Then U ~ T  0 x e 1 T The layered structure of graphite behaves like a 2D system  C ~ T 2 . 1 Vk d 1dk V ( d 1) /  d 1 /  ~   d .  d d d ( 2 ) ( 2 ) v D D    d / U  D ( )d  ~  d k T  /k BT  0 0 e e  / B 1 1 d 1 D / kB T  xd Let x  , U ~T   dx . 0 k BT ex  1 At low temperature ( k BT   D ), the up limit of the integral can be replaced by  . d

(c) For  ~ k  , D ( )d ~

d

Then U ~ T 

1





0

d

1 U xd  T d / . dx  T  , and C  x T e 1...