Descripción: Statistical Mechanics - Pathria Solutions...
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 ε ∝ 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, n n sU , CV (T → ∞) = N k, and CP (T → ∞) = + 1 Nk (1) P = nV s s 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 find the number of microstates available (eq. 2.4.4): Z Z 1 V Pn Σ(P ) ≈ n ... dn qdn p ≈ n (2) h h p≤P
where V is the total volume and p the momentum. Since the energy ε is proportional to p, then: ε n/s ε = Cps ⇒ pn = C where C is the constant of proportionality. Replacing p with ε: V ε n/s Σ(ε) ≈ n h C
(3)
Using (3) into equation (7.1.4) of the density of states, we found: a(ε) =
dΣ(ε) V n dε = n/s n εn/s−1 dε dε C h s
Now, let us follow the procedure of the section 7.1 using (4): Z P V n ∞ n/s−1 = − n/s n ε ln(1 − ze−βε )dε kT C h s 0 ∞ s Z ∞ V n s n/s zβe−βε = − n/s n ε ln(1 − ze−βε ) − εn/s dε n 0 1 − ze−βε C h s n 0 Z ∞ e−βε V zβ n/s = ε dε 1 − ze−βε C n/s hn 0 Z ∞ Vβ εn/s = dε C n/s hn 0 z −1 eβε − 1 ∗
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(4)
if x = βε and dx = βdε: Z
= = P kT
=
∞
n/s x
β Vβ dx −1 x n/s n C h 0 z e −1 β Z ∞ xn/s dx V (Cβ)n/s hn 0 z −1 ex − 1 β n V n g (z)Γ + 1 +1 s (Cβ)n/s hn s
where gµ (z) are the Bose-Einstein functions. Then, let us calculate the internal energy of the system: ∂ PV 2 U = kT ∂T kT z,V n ∂ 2 V = kT 2 n/s n g ns +1 (z)Γ +1 (kT )n/s s ∂T c h n 2 n V + 1 k n/s T n/s−1 = kT 2 n/s n g ns +1 (z)Γ s s c h n 2 n V = (kT )n/s+1 n/s n g ns +1 (z)Γ +1 s s c h n PV n = kT = PV s kT s sU = P nV that is the expression need it. To compute the specific heat of the gas, since T → ∞, the pressure and the specific 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 specific heat: 1 ∂U Cv = Nk N k ∂T 1 ∂ n n = N kT = N k ∂T s s n Cv = Nk s Finally, Cv − Cp = N k n Cp = N k + N k s n Cp = + 1 Nk s
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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): π2 k4 4 U =V T (6) 15h3 c3 and the number of photons in the cavity (eq. 7.3.23): N =V
2ζ(3)(kT )3 π 2 h3 c3
(7)
therefore: ε = = = =
U N π 2 k 4 T 4 π 2 h3 c3 15h3 c3 2ζ(3)(kT )3 π4 kT 30(ζ(3)) π4 kT = 2.7kT 30(1.2020)
Pathria 7.34 Assuming the excitations to be phonons (ω = Ak), show that their contribution toward the specific 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 specific heat in the Debye approximation (eq. 7.4.17) is given by: CV (T ) = 3N kD(x0 ) where x0 =
~ωD kT
and D(x0 ) is the Debye function, which in the general case is: Z x0 n xn Dn (x0 ) = n dx x0 0 e x − 1
Replacing the value of x0 in (9) we found: Z ∞ Z n nk n T n ∞ xn ~ωD xn n Dn = dx = n n dx ~ωD kT ex − 1 ~ ωD 0 ex − 1 0
(8)
(9)
(10)
kT
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): Z nk n T n ∞ xn CV (T ) = 3N k n n dx ⇒ CV ∝ T n (11) ~ ωD 0 ex − 1
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