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Statistical Mechanics - Physics at Oregon State University

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74 CHAPTER 4. STATISTICS OF INDEPENDENT PARTICLES.<br />

4.2 Boltzmann gas again.<br />

An ideal gas is defined as a system of (1) free, (2) non-interacting particles in<br />

the (3) classical regime. Restriction (1) means th<strong>at</strong> there are no external forces,<br />

beyond those needed to confine the particles in a box of volume V. Restriction (2)<br />

means th<strong>at</strong> there are no internal forces. Restriction (3) means th<strong>at</strong> we can ignore<br />

the quantum effects rel<strong>at</strong>ed to restrictions in occup<strong>at</strong>ion numbers. This means<br />

th<strong>at</strong> we use the Maxwell-Boltzmann distribution function. Restrictions (1) and<br />

(3) are easy to relax. Most of solid st<strong>at</strong>e physics is based on such calcul<strong>at</strong>ions,<br />

where electrons can only singly occupy quantum st<strong>at</strong>es in a periodic potential.<br />

Restriction (2) is much harder to relax, and many discussions in physics focus<br />

on these correl<strong>at</strong>ion effects.<br />

Ideal gas again.<br />

In this section we derive the results for the ideal gas again, but now from<br />

the independent particle point of view. This implicitly resolves the deb<strong>at</strong>e of<br />

the factor N! The energy levels of the particles in the box are the same as we<br />

have used before. The ideal gas follows the Boltzmann distribution function<br />

and hence we find<br />

N = <br />

fMB(ɛo) = <br />

orb<br />

orb<br />

e µ−ɛo<br />

k B T = e µ<br />

k B T Z1<br />

We again introduce the quantum concentr<strong>at</strong>ion nQ(T ) and solve for µ :<br />

<br />

MkBT<br />

nQ(T ) =<br />

2π2 3<br />

2<br />

<br />

n<br />

µ = kBT log<br />

nQ(T )<br />

(4.25)<br />

(4.26)<br />

(4.27)<br />

The requirement µ ≪ −kBT is indeed equivalent to n ≪ nQ(T ). The internal<br />

energy is also easy to obtain:<br />

U = <br />

orb<br />

ɛoe µ−ɛo<br />

µ<br />

kB T k = e B T 2 ∂<br />

kBT<br />

∂T<br />

2 NkBT<br />

U =<br />

Z1<br />

<br />

∂Z1<br />

∂T N,V<br />

<br />

orb<br />

ɛo − k e B T (4.28)<br />

= 3<br />

2 NkBT (4.29)<br />

as we had before. To get to the last step we used th<strong>at</strong> Z1 is proportional to T 3<br />

2 .<br />

Always the same str<strong>at</strong>egy.<br />

In many cases one uses the formula for N in the same way we did here.<br />

From it we solved for µ(T, N, V ) and we also obtained U(N,T,V) accordingly.

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