280
4 Thermodynamics and Statistical Physics
4.46 For stationary waves, in the x-direction
k x a = n x π
or n x = k x a/π
dn x = (a/π )dk x
Similar expressions are obtained for y and z directions.
dn = dn x dn y dn z
= (a/π )
3 d
3 k
However only the first octant of number space is physically meaningful.
Therefore
dn = (1/8)(a/π )
3 d
3 k
Taking into account the two possible polarizations
dn =
2V
(2π ) 3 d
3 k =
2V
8π 3 .4πk
2 dk
But k =
ω
c
; dk = dω/c
∴ dn =
V ω
2 dω
π 2 c 3
4.47 n! = n(n − 1)(n − 2) . . . (4)(3)(2)
Take the natural logarithm of n!
ln n! = ln 2 + ln 3 + ln 4 + · · · + ln(n − 2) + ln(n − 1) + ln n
= Σ
n
n=1 ln n
=
n
1
ln n dn
= n ln n − n + 1
≈ n ln n − n
where we have neglected 1 for n 1
4.48 p(E) = (2J + 1)e
−J (J +1)
2 /2ikT
The maximum value of p(E) is found by setting d p(E)/dJ = 0
2 −
(2J + 1)
2
2
2I 0 kT
e
−J (J +1)
2 /2I0kT
= 0
Since the exponential factor will be zero only for J = ∞,
2 −
(2J + 1)
2
2
2I 0 kT
= 0
Solving for J , we get
J max =
√
I 0 kT
−
1
2
4 Thermodynamics and Statistical Physics
4.46 For stationary waves, in the x-direction
k x a = n x π
or n x = k x a/π
dn x = (a/π )dk x
Similar expressions are obtained for y and z directions.
dn = dn x dn y dn z
= (a/π )
3 d
3 k
However only the first octant of number space is physically meaningful.
Therefore
dn = (1/8)(a/π )
3 d
3 k
Taking into account the two possible polarizations
dn =
2V
(2π ) 3 d
3 k =
2V
8π 3 .4πk
2 dk
But k =
ω
c
; dk = dω/c
∴ dn =
V ω
2 dω
π 2 c 3
4.47 n! = n(n − 1)(n − 2) . . . (4)(3)(2)
Take the natural logarithm of n!
ln n! = ln 2 + ln 3 + ln 4 + · · · + ln(n − 2) + ln(n − 1) + ln n
= Σ
n
n=1 ln n
=
n
1
ln n dn
= n ln n − n + 1
≈ n ln n − n
where we have neglected 1 for n 1
4.48 p(E) = (2J + 1)e
−J (J +1)
2 /2ikT
The maximum value of p(E) is found by setting d p(E)/dJ = 0
2 −
(2J + 1)
2
2
2I 0 kT
e
−J (J +1)
2 /2I0kT
= 0
Since the exponential factor will be zero only for J = ∞,
2 −
(2J + 1)
2
2
2I 0 kT
= 0
Solving for J , we get
J max =
√
I 0 kT
−
1
2