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Phys 112 (S2006) 7 Chemical Potential 1
Chemical Potential (a Summary)
Definition and interpretations K&K chap 5.• Thermodynamics definition• Concentration• Normalization• Potential
Law of mass action K&K chap 9
Saha EquationThe density of baryons in the universe
Phys 112 (S2006) 7 Chemical Potential 2
Chemical Potential DefinitionMicrocanonical method
Isolated systems 1 and 2 put in contactEquilibrium of isolated system 1+2= Probability distribution sharply
peaked around configuration of maximum entropy
<=>
Thermodynamics functions• U
• F=U-τσ
Note:
Does not apply to U since entropy changes with the number of particles• G=F+PV
Note that
Proof: divide system into two subsystems. Subsystems are in equilibrium
!"T
!N1
= 0 with N1+ N
2= 0
!"1
!N1 U,V
=!"
2
!N2 U,V
= #µ
$
1 2!1 N1 ! 2 N2
dF = !"d# ! PdV + µdN
µ = F N( ) ! F N !1( ) at constant " ,V
! µ ="F
"N # ,V
dG = !"d# +VdP + µdN! µ =
"G
"N # ,PG = Nµ P,!( )
dU = !d" # PdV + µdN ! µ ="U
"N # ,V
!G1
!N1
= µ P," , N1( ) =!G
2
!N2
= µ P," , N2( ) !"µ
"N= 0
!G
!N= µ P,"( )# G = Nµ P,"( )+ f P,"( ) N = 0!G = 0! f P,"( ) = 0
Phys 112 (S2006) 7 Chemical Potential 3
Chemical Potential <=> ConcentrationCanonical method
Probability
=>
=>
Ideal gas:
increases with temperatureincreases with concentrationNote:
Review examples of attachment to a molecule or trapping by an impurity site (K&K p.141, Noteschap. 3, p.12)
ps =1
Ze!"s
#Z = e
!" s
#
s
$
U = !s pss
" = #2 $ logZ
$#! = " ps log ps
s
# =$ % log Z( )
$%F =U !"# = !" log Z
µ =!F
!N " ,V
= #"! log Z
!N ",V
Z =1
N!nQV( )
N
�
nQ
= exp !"#
$
% &
'
( ) D "( )d"0
*+ =1
4, 2
2m
!2
$
%
& &
'
(
) )
3 / 2
exp !"#
$
% &
'
( ) "d"
0*+ =
m#2,! 2$
% &
'
( )
3/2
µ = !"# log Z#N ",V
= " logN ! log nQV( )( ) = " logn
nQ
$
% &
'
( )
log Z ! " N logN " N( ) + Nlog nQV( )
G = F + PV = N! logn
nQ
"
# $
%
& ' ( 1
"
# $
%
& ' + N! = N! log
n
nQ
"
# $
%
& ' = N! log
P
!nQ
"
# $
%
& ' = Nµ P,!( )
�
! "0
#
$ exp %"&
'
( )
*
+ , d" =
&2
'
(
)
)
*
+
,
,
3 / 2
2x2
0
#$ e
x2
2 dx
with x =2"&
! " # $ #
= 2-&2
'
(
)
)
*
+
,
,
3 / 2
Phys 112 (S2006) 7 Chemical Potential 4
Grand Canonical methodProbability
In order to determine µ, impose that <N> is the given N
Examples: Ideal gases• Fermi Dirac or Bose Einstein. With our convention for the density of states
which does not include volume (≠K&K)
• Classical limit (non relativistic)
same result as in the canonical case!Proof:
Chemical Potential = normalization method
p s,! s,N( ),N( ) =1
Zexp
µN " ! s,N( )#
$%&
'()
Z = expµN ! " s,N( )
#$%&
'()
s,N
*
N = Ns,N
! p s," s,N( ), N( ) = Nsystem
! s,N( ) = N!s
1
exp! " µ
#( ) ± 1
D !( )0
$
% d! = n
expµ !"#
$
% & '
( ) D "( )
0
*
+ d" = n, µ = # log n
nQ
$
% &
'
( )
! expµ
"( ) m"
2#!2$ %
& '
3 /2
= n ! µ = " logn
nQ
$
% (
&
' )
�
exp !"#
$
% &
'
( ) D "( )d"0
*+ =1
4, 2
2m
!2
$
%
& &
'
(
) )
3 / 2
exp !"#
$
% &
'
( ) "d"
0*+ =
m#2,! 2$
% &
'
( )
3/2
= nQ
Phys 112 (S2006) 7 Chemical Potential 5
Chemical Potential as a PotentialRaising the potential energy of a system
Let us consider an isolated system at zero potential energy
Let us then raise it at uniform potential energy per particleThe entropy is not changed by uniform potential (number of states not changed)
Example: Barometric pressure equation
Uo µo =!Uo
!N " ,V
Uo !U = Uo + N"#
µ =!U
!N " ,V
= µointernal
!+ #$
external
!
Phys 112 (S2006) 7 Chemical Potential 6
Let us consider two isolated systems in interaction with each other.If concentration 1 > concentration 2
The only way to maintain the difference of concentration is to give somepotential energy to system 2 with respect to system 1
In practice, if you attempt to prevent an evolution of the concentration, youwill have to generate a difference of potential energy (and vice versa).
At a deeper level the constancy of the total chemical potential in asystem in equilibrium reflects the balance between the drift currentgenerated by the external potential and the diffusion current due tothe random thermal velocity. cf. Chapter 10 of the notes
Difference of Concentration and Potential
µint 1 > µint 2
µint 1 +!1 = µint 2 +!2
µint 1
!1
µint 2
!2
!"
µint 1 ! µint 2 ="2 !"1
!µint = "!#
Phys 112 (S2006) 7 Chemical Potential 7
Examples: Battery (K&K p.129)Consider an electrolyte AB: ions A- B+ In the middle of the cell, equilibrium between positive and
negative ions. But on electrodes, difference of behavior => selective depletion + repulsion
If the two electrodes are not connected
p-n Diode
On n side, the donors give their electrons and positive charges remain behindOn p side, the acceptors capture the electrons, generating fixed negative
charges. The resulting field generate a potential barrier which preventscurrent to flow in one direction
Balance between drift and diffusion
A!+ N " AN + e
!B++ P + e
!" BP
N P
B
+
B
+B
+
B
+
B
+
B
+
B
+
B
+
B
+
B
+
A
!
A
!
A
!
A
!
A
!
A
!
A
!
A
!
++++++++
--------
! ! .! E =
"
##o
!
E
N P
V
P
V = !! E .d! r
x
"N
Stops A- neutralization
p n
eeeeeeeeeeeeee
hhhhhhhhhh
!Fn( )
!Fp( )
p
n
eeeeeeeeeeeeee
hhhhhhhhhh!Fn( )!
Fp( ) -----+++
Phys 112 (S2006) 7 Chemical Potential 8
Calculation Method
Consider 2 species of opposite charge q±: number density n±x( )
Combine µint
+x( ) + q+
n+x( )! x( ) = Constant µ
int
"x( ) + q"
n"x( )! x( ) = Constant
and !#.!E = "#
2! =
q+n+x( ) + q"
n"x( )
$%o
!!!!!& 3 equations for 3 functions n+!n
"!!
($ ,often also written %, is the relative dielectric constant of the medium)
Note that in semiconductor books, the constancy of the chemical potential
is expressed in terms of the sum of the drift and diffusion currents being zero.
This is the same physics expressed in different ways!
Phys 112 (S2006) 7 Chemical Potential 9
Several speciesEquilibrium with several species i
If the two systems are in equilibrium, each kind separately has to be inequilibrium
=>
Conserved quantitiesIn a reaction between species, the number of disappearing particles or
molecules is related to the number of produced particles or molecules
The probability distribution at equilibrium will be sharply peaked around theconfiguration of maximum total entropy :
or
with the constraints
µi 1( ) = µi 2( ) !i21
µi 1( ) µi 2( )
!1A1 +!2A2 "# 3A3 +# 4A4
! "i Ai
i
# $ 0 with "3 = %&3 , "4 = %& 4
!" =#"
#NA1
!NA1+
#"
#NA2
!NA2+
#"
#NA3
!NA3+
#"
#NA4
!NA4= 0
µ1!NA1+ µ2!NA
2+ µ3!NA
3+ µ4!NA
4= 0
!NA1
"1
=!N
A2
"2
=!NA
3
"3
=!N
A4
"4
!NA1
"1
=!N
A2
"2
=!NA
3
"3
=!N
A4
"4
! "iµi
i
# = 0
or "iµi
initial
# = $iµi
Final
#
Conservation of chemical potential
Phys 112 (S2006) 7 Chemical Potential 10
ConsequencesPhotons have µ=0 <= they can disappear by interaction with electronsIf there is no asymmetry particles and antiparticles have opposite
µ’s.
Phys 112 (S2006) 7 Chemical Potential 11
Important Note!
The energies of all the states have to be measuredfrom the same origin.
This is taken automatically into account by including in the internalpartition function the ground state energies (i.e. rest mass)
Important to take into account threshold/energy release effects
The reaction A + B! AB can be exothermic
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!" ground state of AB( ) < " ground state of A( ) + " ground state of B( )
or endothermic " ground state of AB( ) > " ground state of A( ) + " ground state of B( )
e.g. Z = exp !"s
#$%&
'()
s
* = exp !"Ki
#$%&
'()
i
*Kinetic=ZK
! "## $##
exp !"ground state
#$
%&'
()exp !
"int j
#$%&
'()j
*Internal=Zint
! "###### $######
Phys 112 (S2006) 7 Chemical Potential 12
Law of mass actionClassical Ideal Gas
Recall: 1 particle
Kinematic part
Internal e.g. multiplicity g (rotation, spin, binding energy -see below) N particles
=>
Law of Mass ActionConsider the reaction
or
=>
Z = exp !"s#
$
% &
'
( )
s
* = exp !"
Ki
#$
% &
'
( )
i
*Kinetic=Z
K
! " # # $ # #
exp !"int j
#$
% &
'
( )
j
*
Internal =Zint
! " # # $ # #
�
ZK
= nQV
�
Z =1
N!nQV( )
N
Zint
N F = !" log Z# µ =$F$N " ,V
= " logn
nQZint
%
& '
(
) *
!iAii
" # 0
!iµii
" = 0 with µi = # logni
nQiZint i
$
% &
'
( )
ni
nQiZint i
!
" #
$
% &
i
'( i
= 1
ni!i
i
" = K #( ) with K #( ) = nQiZint i( )i
"!i
Phys 112 (S2006) 7 Chemical Potential 13
Kinetic View of Mass ActionDetailed balance
Equilibrium
Does not say anything about reaction time (e.g. independent of catalyst)!
dnAB
dt= C !( )nAnB " D !( )nAB
nAB
nAnB
=C !( )
D !( )= K !( )
A + B! AB" AB # A # B! 0 nAB
nAnB
= K !( )
Phys 112 (S2006) 7 Chemical Potential 14
ExamplespH and the Ionization of Water
Define
=>In normal conditions:
Definition
pH>7 basicpH<7 acidic
Non degenerate semiconductors (classical limit)
H++OH
!" H2O
A[ ] ! concentration of A in mole/liter
H+[ ] OH![ ]H2O[ ]
= K "( )
�
H+
[ ] << H2O[ ] = 56 moles/liters
�
H+[ ] OH ![ ] = 10
!14mole/liter( )
2 independently of H
+[ ]
pH = ! log10 H+[ ]( ) = 14 + log10 OH![ ]( )
ne ! nQe exp "#c " µ$
%&'
()*
with!nQe = 2me
*$2+!2
%&'
()*
3
2
!!
nh ! nQh exp "µ " #v$
%&'
()*
with!nQh = 2mh
*$2+!2
%&'
()*
3
2
!!!!!!!!, nenh = nQenQh exp "#c " #v
$%&'
()*
Phys 112 (S2006) 7 Chemical Potential 15
Ionization of HydrogenSaha Equation:
Where we identify the rest energies of the proton, electron. Similarly for hydrogen
there are 4 spin states!
Taking into account that the masses of the proton and the hydrogen are very close, themass action law can be written:
where I is the ionization energy of atomic hydrogen (if we neglectexcited states)
If the concentrations of protons and electrons are the same:
Not a Boltzmann factor!
e + p! H
nQeZ
int e= 2
me!
2"!2#$%
&'(3/2
exp )*e
!#$%
&'(
nQpZint p = 2mp!2"!2
#$%
&'(
3/2
exp )* p!
#$%
&'(
nQHZint H = 4
mH!
2"!2
#$%
&'(
3/2
exp )*H
!#$%
&'(
nenp= n
H
me!
2"!2#$%
&'(3/2
exp )*p+ *
e) *
H
!#$%
&'(= n
H
me!
2"!2#$%
&'(3/2
exp )I
!#$%
&'(
ne= n
p= n
H
me!
2"!2#$%
&'(3/4
exp )I
2!#$%
&'(
�
nenp
nH
=nQeZ int enQpZ int pe
nQHZ
int H
I = !p+ !
e" !
H
Phys 112 (S2006) 7 Chemical Potential 16
How to build Nuclei?
Nuclear attraction
Requires emission of a photon!Need low enough temperature for n and p to stay bound.Otherwise
More generally equilibrium function of temperature
Coulomb repulsion e.g. 4He
Need s enough energy to “penetrate” barrier=> needs high enough temperature(but not too high lest its dissociates)
Large amount of He in the universe: hot Big Bang
Pote
ntia
l
rn-p
B
“Potential well”
p + n!2H +"
p + n!2 H + "
p + n!2H +"
Epot + Ekin = constant
Deuterium =2H
B= 2.22 MeV
p
n
γ
Pote
ntia
l
rDD
potential barrier <- Coulomb repulsion
�
2H
++
2H
+!
4He
++ + "
γBinding energy (28MeV)
Phys 112 (S2006) 7 Chemical Potential 17
Reaction Rates and ExpansionStrong dependence of reaction rate on temperature
cf. ordinary cooking Breaking of bounds (tenderness)Pressure cooker, refrigerator
Establishing new bounds (e.g. custard)
Dependence on densityThe particles have to find each other!
In an expanding universeThe reaction rate has to be greater than the expansion rate
otherwise nuclei are diluted away before having time of reacting!Freeze-out
3 temperature regimes• At high temperature, nucleus cannot exist <- dissociation• At low temperature: nucleus is stable but not enough energy to be
formed The reactions proceed too slowly :Freeze-out• Intermediate: nucleus can be formed and is stable enough to survive
�
rate!"1"2
Phys 112 (S2006) 7 Chemical Potential 18
Need protons and neutronsBut do it fast enough as neutron will decay(half life time 10.6
minutes)
2 body reactionsDensity not large enough for 3 body
+ a number of other possible reactions!
Process stopped by Freeze-OutTemperature and density become too smallNo element heavier than Lithium
Building Nuclei by Fusion
p+n
HydrogenDeuterium=2H
3He
4He
7Li
+
+3He
2H
Phys 112 (S2006) 7 Chemical Potential 19
Primordial Nucleosynthesis
2H bottleneck
Freeze out
n depleted by low T
• 4He very much more bound than 2H => higher equilibrium concentration at low temperatureBut cannot be reached because we have to go through 2-body reactions: deuterium bottleneck!Really starts at 0.150MeV ≈1 minute
• everything is over in 5 minutes: stops below carbonHigher A elements will have to be produced in stars and supernovaeDependent on expansion rate and density of p&n = Ωb
Same formalism asSaha equation butexpanding universe