04.08.2013 Views

Statistical Mechanics - Physics at Oregon State University

Statistical Mechanics - Physics at Oregon State University

Statistical Mechanics - Physics at Oregon State University

SHOW MORE
SHOW LESS

Create successful ePaper yourself

Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.

114 CHAPTER 5. FERMIONS AND BOSONS<br />

this is impossible! For such a combin<strong>at</strong>ion of density and temper<strong>at</strong>ure we need<br />

the first term in equ<strong>at</strong>ion (5.133). For large values of the volume , the value of<br />

λ will be close to one, and the value of the second term in equ<strong>at</strong>ion (5.133) can<br />

be replaced by (2S + 1)G. In the thermodynamic limit we have<br />

or<br />

lim<br />

V →∞<br />

2S + 1<br />

V<br />

λ = 1 −<br />

λ<br />

1 − λ = n − (2S + 1)nQ(T )G (5.136)<br />

2S + 1<br />

V (n − (2S + 1)nQ(T )G)<br />

(5.137)<br />

Since in most experiments the temper<strong>at</strong>ure is varied and not the density, it<br />

is customary to define a critical temper<strong>at</strong>ure TE, the Einstein temper<strong>at</strong>ure, by<br />

n = nQ(TE)(2S + 1)G, or<br />

2 2π n<br />

TE =<br />

MkB (2S + 1)G<br />

2<br />

3<br />

(5.138)<br />

The Einstein temper<strong>at</strong>ure for Helium is about 3K. The first term in the formula<br />

for n, equ<strong>at</strong>ion (5.133), gives the number of particles in the orbital with the<br />

lowest energy. If the temper<strong>at</strong>ure is higher than the Einstein temper<strong>at</strong>ure,<br />

there are essentially no particles in this orbital. If the temper<strong>at</strong>ure is below the<br />

Einstein temper<strong>at</strong>ure, the number of particles in this orbital becomes very large.<br />

More precisely, the Einstein temper<strong>at</strong>ure is the dividing point between having<br />

or not having a macroscopically large number of particles in the ground st<strong>at</strong>e<br />

orbital. In the thermodynamic limit this is the only orbital with a macroscopic<br />

number of particles, but for finite volumes other orbitals can be important, too.<br />

As we have shown before, for a volume of 1 m 3 we need a temper<strong>at</strong>ure less than<br />

10 −18 K for all particles to be in the ground st<strong>at</strong>e.<br />

The rel<strong>at</strong>ive number of particles in the ground st<strong>at</strong>e orbital, (2S+1)fBE(ɛ111)/N,<br />

for a temper<strong>at</strong>ure below the Einstein temper<strong>at</strong>ure follows from<br />

(2S + 1) fBE(ɛ111)<br />

V<br />

= n − (2S + 1)λ −3<br />

T G (5.139)<br />

and since the thermal wavelength is proportional to T − 1<br />

2 we can write this as<br />

the fraction of particles in the ground st<strong>at</strong>e orbital:<br />

(2S + 1) fBE(ɛ111)<br />

N<br />

3<br />

2 T<br />

= 1 −<br />

TE<br />

(5.140)<br />

This fraction is zero for T > TE. At T = 0, all particles are in the ground st<strong>at</strong>e<br />

orbital.<br />

Note th<strong>at</strong> there is some confusion in defining an orbital. If we define an<br />

orbital by the quantum numbers nx, ny, nz only, the total number of particles<br />

in th<strong>at</strong> orbital is given by (2S + 1)fBE(ɛ(nx, ny, nz)). If we define an orbital by<br />

the quantum numbers n plus a quantum number s for spin, the total number<br />

of particles in th<strong>at</strong> orbital is fBE(ɛ(nx, ny, nz, s)). But now we have 2S + 1

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!