1549380323-Statistical Mechanics Theory and Molecular Simulation

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Ideal gas 87

S(N,V,E) =kln Ω 1 (N 1 ,V 1 ,E ̄ 1 ) +kln Ω 2 (N 2 ,V 2 ,E ̄ 2 )
=S 1 (N 1 ,V 1 ,E ̄ 1 ) +S 2 (N 2 ,V 2 ,E ̄ 2 ) +O(lnN) + const. (3.4.14)

Finally, in order to compute the temperature of each system, we vary the energyE ̄ 1
by a small amount dE ̄ 1. But sinceE ̄ 1 +E ̄ 2 =E, dE ̄ 1 =−dE ̄ 2. Also, this variation is
made such that the total entropySand energyEremain constant. Thus, we obtain


0 =

∂S 1


∂E ̄ 1


+


∂S 2


∂E ̄ 1


0 =


∂S 1


∂E ̄ 1



∂S 2


∂E ̄ 2


0 =


1


T 1



1


T 2


, (3.4.15)


from which it is clear thatT 1 =T 2. Thus, when two systems in thermal contact reach
equilibrium, their temperatures become equal. Note that this resultholds independent
of the relative sizes of systems 1 and 2, a fact we will need in the nextchapter where
we will consider systems in contact with large heat baths.


3.5 The free particle and the ideal gas


Our first example of the microcanonical ensemble is a single free particle in one spatial
dimension and its extension to an ideal gas ofNparticles in three dimensions. The
microcanonical ensemble for a single particle in one dimension is not especially inter-
esting (it has only one degree of freedom). Nevertheless, it is instructive to go through
the pedagogical exercise in order to show how the partition function is computed so
that the subsequent calculation for the ideal gas becomes more transparent.
The particle is described by a single coordinatex, confined to a region of the real
line betweenx= 0 andx=L, and a single momentump. The free particle Hamiltonian
is just


H=

p^2
2 m

. (3.5.1)


The phase space is two-dimensional, and the microcanonical partition function is


Ω(L,E) =


E 0


h

∫L


0

dx

∫∞


−∞

dp δ

(


p^2
2 m

−E


)


. (3.5.2)


Note that sinceN= 1, Ω only depends onEand the one-dimensional “volume”L.
We see immediately that the integrand is independent ofxso that the integral overx
can be done immediately, yielding


Ω(L,E) =


E 0 L


h

∫∞


−∞

dp δ

(


p^2
2 m

−E


)


. (3.5.3)


In order to perform the momentum integral, we start by introducing a change of
variables,y=p/



2 mso that the integral becomes
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