Statistical Mechanics - Homework Assignment 4: Alejandro G Omez Espinosa March 24, 2013
Statistical Mechanics - Homework Assignment 4: Alejandro G Omez Espinosa March 24, 2013
Statistical Mechanics - Homework Assignment 4: Alejandro G Omez Espinosa March 24, 2013
Pathria 7.14 Consider an n-dimensional Bose gas whose single-particle energy spectrum is given by ps , where s is some positive number. Discuss the onset of Bose-Einstein condensation in this system, especially its dependence on the numbers n and s. Study the thermodynamic behavior of this system and show that, P = sU , nV CV (T ) = n N k, s and CP (T ) = n + 1 Nk s (1)
To follow the procedure of section 7.1, we have to calculate the density of states in an n-dimensional Bose gas. For this, we have to nd the number of microstates available (eq. 2.4.4): (P ) 1 hn ...
pP
dn qdn p
V Pn hn
(2)
where V is the total volume and p the momentum. Since the energy is proportional to p, then: = Cps pn = C
n/s
(3)
Using (3) into equation (7.1.4) of the density of states, we found: a() = d() V n d = n/s n n/s1 d d C h s (4)
Now, let us follow the procedure of the section 7.1 using (4): P kT n n/s1 ln(1 ze )d C n/s hn s 0 V n s n/s s = n/s n ln(1 ze ) n C h s n 0 e V z n/s = d 1 ze C n/s hn 0 V n/s = d C n/s hn 0 z 1 e 1 = V
n/s
0
ze d 1 ze
gomez@physics.rutgers.edu
if x = and dx = d:
V dx 1 x n/s n C h 0 z e 1 xn/s dx V (C )n/s hn 0 z 1 ex 1 n V n g ( z ) +1 +1 s (C )n/s hn s x n/s
= = P kT =
where g (z ) are the Bose-Einstein functions. Then, let us calculate the internal energy of the system: U = kT 2 = kT 2 T V
2
PV kT
z,V
(kT )n/s
n k n/s T n/s1 s n +1 s
sU nV
that is the expression need it. To compute the specic heat of the gas, since T , the pressure and the specic heat of the gas approach their classical values, therefore we can use P V = N kT in the expression above: n n (5) U = P V = N kT s s to calculate the specic heat: Cv Nk = = Cv = Finally, Cv Cp = N k Cp = N k + Cp = n + 1 Nk s n Nk s 1 Nk 1 Nk n Nk s U T n N kT T s
n s
Pathria 7.21 Show that the mean energy per photon in a blackbody radiation cavity is very nearly 2.7 kT . To calculate the mean energy per photon in a blackbody radiation cavity, we need the total energy density in the cavity (eq. 7.3.12): 2 k4 4 U =V T (6) 15h3 c3 and the number of photons in the cavity (eq. 7.3.23): N =V therefore: = = = = U N 2 k 4 T 4 2 h3 c3 15h3 c3 2 (3)(kT )3 4 kT 30( (3)) 4 kT = 2.7kT 30(1.2020) 2 (3)(kT )3 2 h3 c3 (7)
Pathria 7.34 Assuming the excitations to be phonons ( = Ak ), show that their contribution toward the specic heat of an n-dimensional Debye system is proportional to T n . Note that the elements selenium and tellurium form crystals in which atomic chains are arranged in parallel so that in a certain sense they behave as one-dimensional; accordingly, over a certain range of temperatures, the T 1 -law holds. For a similar reason, graphite obeys a T 2 -law over certain range of temperatures. The specic heat in the Debye approximation (eq. 7.4.17) is given by: CV (T ) = 3N kD(x0 ) where x0 =
D kT
(8)
and D(x0 ) is the Debye function, which in the general case is: Dn (x0 ) = n xn 0
x0 0
xn dx ex 1
(9)
nk n T n xn dx = nn ex 1 D
xn dx ex 1
(10)
Since we need to check the proportionality in the temperature and the integral does not depend upon T , is not necessary to calculate the integral. Plugging (10) into (8): CV (T ) = 3N k nk n T n nn D
0
xn dx ex 1
CV T n
(11)