Examples 145

We now extend this derivation to the case ofNparticles in three dimensions, i.e.,
an ideal gas ofNparticles in a cubic box of sideL(volumeV=L^3 ), for which the
Hamiltonian is

H=

∑N

i=1

p^2 i

2 m

. (4.5.5)

Since each momentum vectorpihas three components, we may also write the Hamil-
tonian as

H=

∑N

i=1

∑^3

α=1

p^2 αi

2 m

, (4.5.6)

whereα= (x,y,z) indexes the Cartesian components ofpi. The sum in eqn. (4.5.6)
contains 3Nterms. Thus, the partition function is given by

Q(N,V,T) =

1

N!h^3 N

∫

D(V)

dNr

∫

dNpexp

[

−β

∑N

i=1

p^2 i

2 m

]

. (4.5.7)

Since the Hamiltonian is separable in the each of theN coordinates and momenta,
the partition function can be simplified according to

Q(N,V,T) =

1

N!

[

1

h^3

∫

D(V)

dr 1

∫

dp 1 e−βp

(^21) / 2 m

][

1

h^3

∫

D(V)

dr 2

∫

dp 2 e−βp

(^22) / 2 m

]

···

[

1

h^3

∫

D(V)

drN

∫

dpNe−βp

(^2) N/ 2 m

]

. (4.5.8)

Since each integral in brackets is the same, we can write eqn. (4.5.8)as

Q(N,V,T) =

1

N!

[

1

h^3

∫

D(V)

dr

∫

dpe−βp

(^2) / 2 m

]N

. (4.5.9)

The six-dimensional integral in brackets is just

1

h^3

∫

D(V)

dr

∫

dpe−βp

(^2) / 2 m

1

h^3

∫L

0

dx

∫L

0

dy

∫L

0

dz

×

∫∞

−∞

dpxe−βp

(^2) x/ 2 m

∫∞

−∞

dpye−βp

(^2) y/ 2 m

∫∞

−∞

dpze−βp

(^2) z/ 2 m

. (4.5.10)

Eqn. (4.5.10) can also be written as

1

h^3

∫

D(V)

dr

∫

dpe−βp

(^2) / 2 m

[

1

h

∫L

0

dx

∫∞

−∞

dpe−βp

(^2) / 2 m

] 3

, (4.5.11)