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Statistical Mechanics Study Guide: Solutions to Problems in Physics 353 - Prof. Raghuveer , Study notes of Physics

A study guide for the solutions to suggested problems in physics 353 - statistical mechanics, taught by prof. Raghuveer parthasarathy at the university of oregon during spring 2008. It covers the concepts of barometric pressure and the chemical potential for fermions.

Typology: Study notes

Pre 2010

Uploaded on 07/29/2009

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Prof. Raghuveer Parthasarathy
University of Oregon; Spring 2008
Physics 353 – Statistical Mechanics
Study Guide: Solutions to Suggested Problems
1. Barometric pressure.
I’ll call the height h, to avoid too many “z”s.
Consider “layers” of atmosphere, each with some concentration ()nh . Diffusive equilibrium means
that the chemical potential of all layers is the same; ()h
μ
= constant. The chemical potential has an
“internal” and “external” part, where the external chemical potential is any relevant potential energy:
int ext
μ
μμ
=+. Here, ()
ln
int
Qint
nh
nZ
μτ
=
⎜⎟
⎜⎟
⎝⎠
(given) and ext mgh
μ
=
(gravitational potential). Calling
the concentration at height zero 0
n, wee see that () (0)h
μ
μ
=
implies:
0
()
ln ln 0
Qint Qint
n
nh mgh
nnZZ
ττ
⎛⎞ ⎛⎞
+= +
⎜⎟ ⎜⎟
⎜⎟ ⎜⎟
⎝⎠ ⎝⎠
. Therefore
() ()
0
ln ( ) ln
mgh
nh n
τ
=− + , and
()
0
() exp /nh n mgh
τ
=− .
2. Fermion chemical potential.
pf2

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Prof. Raghuveer Parthasarathy

University of Oregon; Spring 2008

Physics 353 – Statistical Mechanics

Study Guide: Solutions to Suggested Problems

1. Barometric pressure.

I’ll call the height h , to avoid too many “z”s.

Consider “layers” of atmosphere, each with some concentration n h ( ). Diffusive equilibrium means

that the chemical potential of all layers is the same; μ ( ) h = constant. The chemical potential has an

“internal” and “external” part, where the external chemical potential is any relevant potential energy:

μ = μ int + μ ext. Here,

int ln Q int

n h

n Z

(given) and μ ext = mgh (gravitational potential). Calling

the concentration at height zero n 0 , wee see that μ ( ) h = μ(0)implies:

ln ln 0 Q int Q int

n h n mgh n Z nZ

⎜ ⎟ +^ =^ ⎜ ⎟+

. Therefore ln ( ( ) ) ln( 0 )

mgh n h n

= − + , and

n h ( ) = n 0 exp ( − mgh / τ).

2. Fermion chemical potential.