HW-6-solution - Solution of HW #6 PDF

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Homework #6 1. Continuum wave equation. Show that for long wavelengths the equation of motion, d 2u M 2 s  C (us 1  us 1  2us ) , reduces to the continuum elastic wave equation dt 2  2u 2  u  v  x2 t 2 where v is the velocity of sound. Solution: For   a , us 1  us is small.

Replacing us with u(x) , we have us  u(sa) ,

d us  u (x ,t )  . Thus t 2 dt 2 1  2u 2 u a us 1  u ((s  1)a )  u (a )  a x 2  x2  2u 2 (( 1 ) ) (( 1 ) ) 2 ( ) a      u s a u s a u sa or x2 d 2u Then M 2 s  C (us 1  us 1  2us ) becomes dt 2 2 2 u 2  u( x, t )  2 u ( x, t ) 2  u a   M C v or t 2 x2 t 2 x2 2

2

us 1  u ((s  1)a) , and

with v 

Ca 2 . M

Note the mass density is   M / a3 and the Yong’s modulus is Y  C / a . Then v

Ca 2 Y  . M 

2. Diatomic chain. Consider the normal modes of a linear chain in which the force constants between nearest-neighbor atoms are alternately C and 10C. Let the masses be equal, and let the nearest-neighbor separation be a/2. Find  (K ) at K  0 and K   / a . Sketch in the dispersion relation by eye. This problem simulates a crystal of diatomic molecules such as H2. Solution: C

10C

C

10C

C

The atoms can be separated into two groups: The first group has spring C at the left and spring 10C at right, and its motion can be described by u  A exp( ikx  it) .

The second group has spring 10C at the left and spring C at right, and its motion can be described by v  B exp(ikx  i t ) . Then the equations of motion for these two groups are:  d 2u s m 2  10C( vs1  us )  C( vs1  us )  dt2 m d vs  C( us 1  vs )  10 C( us 1  vs )  dt 2 Substitute u  A exp( ikx  it ) and v  B exp(ikx  i t ) into these two equations,

  m 2 A  10C ( Be ika / 2  A)  C ( Beika / 2  A) ;  ika / 2 ika / 2 2  ( ) 10 ( )      m B C Ae B C Ae B   (11C  m  2 )A  C (10e ika / 2  e ika / 2 )B  0 or  . ika / 2  eika / 2 ) A  (11C  m  2 ) B  0  C (10e To have non-zero solution, we have  (11C  m 2 ) A  C (10e  ika / 2  e ika / 2) B  0    0 . det 2 ika / 2 ika / 2 C e e A C m B  ( 10 ) ( 11 ) 0        (11C  m 2 ) 2  C 2 (100  1  20 cos ka)  0 , (11C  m 2 )  C 101  20 cos ka C (11  101  20 cos ka ) m 22C , A  B ; and  22  0 , A  B . At k=0,  12  m 20C 2C 2 At k=/a,  1  , A  iB ; and  22  , A  iB . m m

2  

  

k a



a

3. Atomic vibrations in a metal. Consider point ions of mass M and charge e immersed in a uniform sea of conduction electrons. The ions are imagined to be in stable equilibrium when at regular lattice points. If one ion is displaced a small distance r from its equilibrium position, the restoring force is largely due to the electric charge within the sphere of radius r centered at the equilibrium position. Take the number density of ions (or of conduction electrons ) as 3 /( 4R 3 ) , which defines R. (a) Show 2 3 1/ 2 that the frequency of a single ion set into oscillation is   (e / MR ) . (b) Estimate the value of this frequency for sodium, roughly. (c) From (a), (b), and some common sense, estimate the order of magnitude of the velocity of sound in metal. Solution: (a) For a uniformly charged sphere of charge density  , the electrical field at r e 3e 1 4 3 4 r  r . Now    , so within the sphere is E( r)  2   3 3 4R / 3 3 r 4R 3 er E ( r)   3 . R e2 e2 r The force acting on the ion is F  eE   3   Kr with K  3 . R R K e2 .  M MR 3 (b) For Sodium, M  23  1.67  1024 g  3.8  1023 g . have e2 (4.8  10 10 ) 2 14    ~ 10 (1 / s) . MR 3 3.8  10 23  10 24 (c) v ~ a ~ 1014  108  106 cm / s .

Then  

Taking R=1Å, we

4. Soft phonon modes. Consider a line of ions of equal mass but alternating in charge, p with e p  e(1) as the charge on the pth ion. The interatomic potential is the sum of two contributions: (1) a short-range interaction of force constant C1R   that acts between nearest-neighbors only, and (2) a coulomb interaction between all ions. (a) Show that the contribution of the coulomb interaction to the atomic force constants is CpC  2( 1) p e2 / p3a 3 , where a is the equilibrium nearest-neighbor distance. (b) From  2  (2 / M ) C p (1  cos pKa ) , here C includes both nearest neighbor and p 0

other neighbors, show that the dispersion relation may be written as  1   2 / 02  sin2  Ka     ( 1) p (1  cos pKa ) p  3 , 2  p 1 2 3 2 where  0  4 / M and   e / a . (c) Show that  2 is negative (unstable mode) at the zone boundary Ka   if   0.475 or 4 / 7 (3) , where  is a Riemann zeta function. Show further that the speed of sound at small Ka is imaginary if

  ( 2ln 2) 1  0.721 . Thus  2 goes to zero and the lattice is unstable for some value of Ka in the interval (0,  ) if 0.475    0.721 . Notice that the phonon spectrum is not that of a diatomic lattice because the interaction of any ion with its neighbors is the same as that of any other ion. Solution: (a) The force between two ions is  e2 / r 2 . us

us+p pa pa+us+p-us

For two ions at sth and (s+p)th sites, their separation distance is pa without ion displacement, and is pa  us  p  u s after the sth and (s+p)th ions are displaced by us and us p , respectively. Then the force between the two ions will change by an

( 1)p e 2 ( 1) p e2 ( 1) p 2 e2   (us p  us ) . ( pa  us p  us )2 ( pa ) 2 ( pa) 3 accounts for the sign change of the two charges.

amount of

p Here (1)

(b) The equation of motion for the sth ion is  d 2u (1) p 2 e2 m 2 s   (u s1  u s )   (u s1  u s )   (us p  us  us p  us ) . dt ( pa ) 3 p 1 Let us  A exp( ikxs  i t) , and substitute it into the above equation. 

 m 2   (e ika  1)   (e ika  1)   p1



 2 (cos ka  1)   p1

2 

( 1) 4e (cos pka  1) ( pa )3 p

2 4e 2 (1  cos ka )  ma 3 m

2



 p1

(1) p (1  cos pka ) p3

p 4 (1)  ka  4e sin 2    (1  cos pka)  3 m p3  2  ma p1 2



( 1) p 2e 2 ikpa ikpa  1) (e  1  e 3 (pa )

Define  0 



4 e2 and   3 , then m a

p  2 (1) 2  ka  sin (1  cos pka )       p3 02  2 p 1

 ka  (c) At ka   , cos( pka)  (1) p and sin   1 . Then 2  p 2   (1) (1  ( 1) p )  1   2 3 p 0 p 1 1 1  2 2 2     1     3  3  3  ...   1 2  1 3  3  ...  3 5  1 3 5     1  2.106 If   1 / 2.106  0.475 ,  2  0 . For ka  1 , 2 2 p   2  ka  (1)  kpa      2    p3  2  02  2  p 1 2 p   ( 1)   ka  1  2      (1 2 ln 2) p  2  p 1  If   1 / 2 ln 2  0.721 ,  2  0 . Strong Coulomb interaction makes the lattice unstable.

 ka     2 

2

5. For a 1D lattice, if k1  k2  2 n / a , (a) Show that k1 and k 2 describe the same elastic wave. (b) For a special case of k1   / 3a and k 2  7 / 3a , make a plot of cos(k1 x ) and cos(k2 x ) versus x / a . Confirm the conclusion of (a) from the plot. Solution: (a) For u1  A exp( ik1 x  it ) , u2  A exp( ik 2 x i t) , u2 / u1  exp[(ik 2  k1) x] . For k1  k2  2n / a and x  pa , u2 / u1  exp(i2np]  1. (b) 1.0

0.5

0.0

-0.5

-1.0

0

1

2

3

4

5

x/a

Note the two curves cross at x/a=integer.

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9

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