Homework #4

1. In general, the internal energy U depends on both temperature and volume, U=U(T,V). The volume dependence comes from the potential energy due to the interactions among the particles. For free particles, there are no interactions, thus, the  ∂U  internal energy U should be independent of the volume, i.e.,   = 0 . Verify this ∂  V T result for ideal gas pV = RT .

 ∂U   ∂S  From TdS = dU + pdV , or dU = TdS − pdV , there is   = T   − p . ∂ ∂  V T  V T

 ∂S   ∂p   ∂U   ∂P  Using Maxwell relation   =   , we have   = T   − p . ∂ ∂ ∂ ∂  V T  T V  V T  T V

For ideal gas pV = RT or p = RT /V .

 ∂U   ∂P  R Thus   = T   − p = T − p = p − p = 0 . ∂ ∂  V T  T V V

 ∂C   ∂ 2 P   ∂C   ∂ 2V  2. Prove  V  = T   and  P  = −T   .  ∂  ∂ 2  ∂  ∂ 2 V T  T V P T  T  P

For ideal gas, pV = RT . Show that CV is independent of the volume and CP is independent of the pressure.

 ∂   ∂   ∂  ∂    ∂  ∂   = S CV =  S  =  S  Since CV T   , there is   T     T     .  ∂   ∂  ∂  ∂  ∂  ∂  T V V T  V T V T  T V T V

 ∂S   ∂p  Using Maxwell relation   =   , ∂ ∂  V T  T V  ∂C   ∂  ∂S    ∂  ∂p    ∂ 2 p  we have  V  = T     = T     = T   .  ∂  ∂  ∂  ∂  ∂  ∂ 2 V T  T V T V  T T V V  T V

 ∂S  Similarly, from C = T   , there is P ∂  T  p

 ∂C   ∂  ∂    ∂  ∂    p   S   S    = T     = T     .  ∂p  ∂p  ∂T  ∂T  ∂p  T  p T  T  p

 ∂S   ∂V  Using Maxwell relation   = −  , ∂  ∂   p T T p

 ∂C   ∂  ∂S    ∂  ∂V    ∂ 2V   p  =     = −     = −   we have   T     T   T  2  .  ∂p  ∂T  ∂p  ∂T  ∂T   ∂T  T  T  p  p  p p

RT RT For ideal gas, p = and V = . V p

 ∂p  R  ∂2 p   ∂V  R  ∂ 2V  Then   = ,  = 0 and   = ,  = 0 .  ∂  ∂ 2  ∂  ∂ 2 T V V  T V T p p  T  p

 2  ∂  2   ∂C  ∂ p  C p  ∂ V Thus  V  = T   = 0 and   = −T   = 0.  ∂  ∂ 2 ∂ ∂ 2 V T  T V  p T  T  p

3. For free particles in two dimensions, what is density of states (DOS) in low speed limit (ε=p2/2m), and in high speed limit (ε=pc)?

A2πpdp For a 2D system, D(ε )dε = , where A is the area of the system. h2 p 2 pdp (a) For ε = , dε = or pdp = mdε . Thus 2m m

A2πpdp A2πmdε A2πm D(ε )dε = = or D(ε ) = . h 2 h 2 h 2

dε (b) For ε = pc , dε = cdp or dp = . Thus c

A2πpdp A2πεdε A2πε D(ε )dε = = or D(ε ) = . h2 h2c 2 h2c2

4. If a kind of particles obeys the following dispersion ε=Dpα, what is the density of states (three dimensions)?

V 4πp 2dp D(ε )dε = . Here V is the volume of the system. h3

1 α − α 1/ 1 α  ε  ε dε For ε = Dp , or p =   and dp = . Thus  D  αD1/α

2 α− α − V 4πp 2dp V 4π  ε α ε 1/ 1dε V 4π ε 3/ 1dε D(ε )dε = =   = ⋅ . h3 h3  D  αD1/α h3 αD3/α

4πVε 3/α −1 D(ε ) = . h3αD3/α

5. A system consists of N simple harmonic oscillators (one dimension). In the classical p 2 mω 2 x 2 limit, the energy of a simple harmonic oscillator is ε = + . Get the 2m 2 partition function, entropy, and the internal energy of this system.

The single particle partition function is

 β 2 β ω 2 2  = dxdp −βε = dxdp − p − m x  Z1 ∫ e ∫ exp  h h  2m 2  +∞ +∞ 1  βp 2   βmω 2 x 2  = ∫ exp− dp ∫ exp− dx h −∞  2m  −∞  2  +∞ +∞ 4  βp 2   βmω 2 x 2  = ∫ exp− dp ∫ exp− dx h 0  2m  0  2 

β βmω 2 Let y = p and z = x . 2m 2

+∞  β 2  +∞  β ω 2 2  = 4 − p   − m x  Z1 ∫ exp dp ∫ exp dx h 0  2m  0  2  +∞ +∞ 4 2m 2 = ⋅ exp()− y 2 dy exp ()− z 2 dz β β ω 2 ∫ ∫ h m 0 0 π π π = 4 2m ⋅ 2 ⋅ ⋅ = 2 = 1 = kBT h β βmω 2 2 2 hβω βω ω

The partition function is

 k T  N Z = Z N =  B  . 1  ω 

 k T  The free energy is F = −k T log Z = −Nk T log B  . B B  ω  ∂ log Z ∂ N The energy is U = − = − [−N log()βω ] = = Nk T . ∂β ∂β β B

U − F 1   k T    k T  The entropy is S = = Nk T + Nk T log B  = Nk 1 − log B  . T T  B B  ω  B   ω 