Chemical Potential a Summary Definition and interpretations KK chap 5. Thermodynamics definition Concentration Normalization Potential Law of mass action KK chap 9 Saha Equation The density of baryons in the universe Phys 112 S2006 7 Chemical Potential 1
Chemical Potential Definition Microcanonical method Isolated systems 1 and 2 put in contact Thermodynamics functions U 1 2 1 N 1 2 N 2 F=U-τσ Note: Equilibrium of isolated system 12= Probability distribution sharply peaked around configuration of maximum entropy " T <=> = 0 with N 1 N 2 = 0 N 1 du = d" PdV µdn µ = "U "N,V df = "d PdV µdn µ = "F "N,V µ = F N F N 1 at constant ",V " 1 N 1 U,V = " 2 N 2 U,V = µ G=FPV Does not apply to U since entropy changes with the number of particles dg = "d VdP µdn Note that G = Nµ P, µ = "G "N,P Proof: divide system into two subsystems. Subsystems are in equilibrium G 1 = µ P,", N N 1 = G 2 = µ P,", N 1 N 2 "µ G 2 "N = 0 N = µ P," G = Nµ P," f P," N = 0 G = 0 f P," = 0 Phys 112 S2006 7 Chemical Potential 2
Chemical Potential <=> Concentration Canonical method Probability => => U = Ideal gas: increases with temperature increases with concentration Note: p s = 1 " s Z e Z = e " s " s p s = 2 logz s µ = F N ",V = " log Z N ",V Z = 1 N n QV Review examples of attachment to a molecule or trapping by an impurity site KK p.141, Notes chap. 3, p.12 Phys 112 S2006 7 Chemical Potential 3 = " p s log p s s N n Q = exp " 0 log Z " N log N " N µ = " log Z = " log N log n Q V N ",V G = F PV = N log s = log Z D " d" = 1 2m 4, 2 2 0 = " log n Nlog n Q V " " n Q n 1 N = N log " n n Q = n Q N log 3 / 2 " exp ", d" = 2 " F = U " = " log Z exp " "d" = 0 P,, n Q 3 / 2 = Nµ P, x 2 2x 2 e 2 dx 0 " with x = 2" m 2, 2 = 2-2 3/2,, 3 / 2
Chemical Potential = normalization method Grand Canonical method Probability = 1 Z p s, s, N, N In order to determine µ, impose that <N> is the given N N = Np s," s, N, N = N system Examples: Ideal gases Fermi Dirac or ose Einstein. With our convention for the density of states which does not include volume KK 1 0 exp " µ Classical limit non relativistic Proof: s,n µn " s, N exp s, N = N s 0 same result as in the canonical case Phys 112 S2006 7 Chemical Potential 4 ± 1 D Z = d = n exp µ " D " s,n d" = n, µ = log n exp " D " d" = 1 2m 0 4, 2 2 3 / 2 exp µ " exp " "d" = 0 m" 2 2 3 /2 µn " s, N exp n Q m 2, 2 3/2 = n µ = " log = n Q n n Q
Chemical Potential as a Potential Raising the potential energy of a system Let us consider an isolated system at zero potential energy U o µ o = U o N ",V Let us then raise it at uniform potential energy per particle The entropy is not changed by uniform potential number of states not changed Example: arometric pressure equation U o U = U o N" µ = U = N µ o ",V internal external Phys 112 S2006 7 Chemical Potential 5
Difference of Concentration and Potential Let us consider two isolated systems in interaction with each other. If concentration 1 > concentration 2 µ int 1 > µ int 2 The only way to maintain the difference of concentration is to give some potential energy to system 2 with respect to system 1 " µ int 1 1 = µ int 2 2 µ int 1 1 µ int 2 2 µ int 1 µ int 2 = " 2 " 1 µ int = " In practice, if you attempt to prevent an evolution of the concentration, you will have to generate a difference of potential energy and vice versa. At a deeper level the constancy of the total chemical potential in a system in equilibrium reflects the balance between the drift current generated by the external potential and the diffusion current due to the random thermal velocity. cf. Chapter 10 of the notes Phys 112 S2006 7 Chemical Potential 6
alance between drift and diffusion Examples: attery KK p.129 Consider an electrolyte A: ions A - In the middle of the cell, equilibrium between positive and negative ions. ut on electrodes, difference of behavior => selective depletion repulsion - - - - - - - - N A A A A A A A P. E = " o Stops A - neutralization E A N " AN e P e " P If the two electrodes are not connected N P x " V = E.d r V N P p-n Diode F p p hhhhhhhhhh eeeeeeeeeeeeee n F n p ----- F p hhhhhhhhhh eeeeeeeeeeeeee n F n On n side, the donors give their electrons and positive charges remain behind On p side, the acceptors capture the electrons, generating fixed negative charges. The resulting field generate a potential barrier which prevents current to flow in one direction Phys 112 S2006 7 Chemical Potential 7
Calculation Method Consider 2 species of opposite charge q ± : number density n ± x Combine µ int x q n " x x = Constant µ int and. E = " 2 = q n x q " n " x x q " n " x x = Constant 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 8
Several species Equilibrium with several species i If the two systems are in equilibrium, each kind separately has to be in equilibrium => Conserved quantities In a reaction between species, the number of disappearing particles or molecules is related to the number of produced particles or molecules 1 A 1 2 A 2 " 3 A 3 4 A 4 N " i A i 0 with " 3 = 3, " 4 = A1 4 = N A 2 = N A 3 = N A 4 i " 1 " 2 " 3 " 4 The probability distribution at equilibrium will be sharply peaked around the configuration of maximum total entropy : or 1 2 µ i 1 µ i 2 with the constraints µ i 1 = µ i 2 i " = " N A1 N A1 " N A2 N A2 " N A3 N A3 " N A4 N A4 = 0 µ 1 N A1 µ 2 N A2 µ 3 N A3 µ 4 N A4 = 0 N A1 " 1 = N A 2 " 2 = N A 3 " 3 = N A 4 " 4 Phys 112 S2006 7 Chemical Potential 9 " i µ i = 0 i or " i µ i = i µ i initial Final Conservation of chemical potential
Consequences Photons have µ=0 <= they can disappear by interaction with electrons If there is no asymmetry particles and antiparticles have opposite µ s. Phys 112 S2006 7 Chemical Potential 10
Important Note The energies of all the states have to be measured from the same origin. The reaction A A can be exothermic < " ground state of A " ground state of > " ground state of A " ground state of " ground state of A or endothermic " ground state of A This is taken automatically into account by including in the internal partition function the ground state energies i.e. rest mass e.g. Z = exp " s = exp " Ki s exp " ground state i " exp " int j j " Kinetic= Z K Internal= Z int Important to take into account threshold/energy release effects Phys 112 S2006 7 Chemical Potential 11
Classical Ideal Gas Recall: 1 particle Kinematic part N particles Law of mass action Internal e.g. multiplicity g rotation, spin, binding energy -see below Law of Mass Action Consider the reaction => Z = n " i µ i = 0 with µ i = log i n Qi Z int i i Z = exp " s = exp " Ki exp " int j s i " j " Z K = n Q V " n i i = K i 1 N n QV N Z int or Kinetic= Z K Internal =Z int N => F = " log Z µ = F n = " log N ",V n Q Z int i " i A i 0 i n i i = 1 " n Qi Z int i " with K = n Qi Z int i i i Phys 112 S2006 7 Chemical Potential 12
Kinetic View of Mass Action Detailed balance Equilibrium A A " A A 0 dn A dt = C n A n " D n A n A = C n A n D = K n A = K n A n Does not say anything about reaction time e.g. independent of catalyst Phys 112 S2006 7 Chemical Potential 13
Examples ph and the Ionization of Water Define => H 2 O In normal conditions: [ H ] << H 2 O [ H ] OH Definition H OH " H 2 O [ A] concentration of A in mole/liter [ H ] OH ph>7 basic ph<7 acidic [ ] [ ] = K " [ ] = 56 moles/liters [ ] = 10 14 mole/liter ph = log 10 H 2 independently of H [ ] = 14 log 10 [ OH ] Non degenerate semiconductors classical limit n e n Qe exp " " µ c withn Qe = 2 m e 2 2 2 Phys 112 S2006 7 Chemical Potential 14 3 3 [ ] n h n Qh exp " µ " v withn Qh = 2 m h 2 2 2, ne n = n h Qen exp " " c v Qh
Saha Equation: Ionization of Hydrogen n e n p e p H = n QeZ int en QpZ int pe n H n QH Z int H 3/2 m n Qe Z int e = 2 e 2" 2 exp e n Qp Z int p = 2 m p 2" 2 Where we identify the rest energies of the proton, electron. Similarly for hydrogen n QH Z int H = 4 m H 2" 2 3/2 exp H there are 4 spin states Taking into account that the masses of the proton and the hydrogen are very close, the mass action law can be written: 3/2 m n e n p = n e H 2" 2 exp p e H = n m e H 2" 2 exp I where I is the ionization energy of atomic hydrogen if we neglect excited states If the concentrations of protons and electrons are the same: 3/4 m n e = n p = n e H 2" 2 exp I 2 3/2 3/2 Not a oltzmann factor exp p I = p e " H Phys 112 S2006 7 Chemical Potential 15
How to build Nuclei? Potential p Nuclear attraction n Deuterium = 2 H r n-p = 2.22 MeV E pot E kin = constant Potential well γ Requires emission of a photon p n 2 H " Need low enough temperature for n and p to stay bound. Otherwise p n 2 H " More generally equilibrium function of temperature Coulomb repulsion e.g. 4 He potential barrier <- Coulomb repulsion p n 2 H " 2 H 2 H 4 He " Potential Need s enough energy to penetrate barrier => needs high enough temperature r but not too high lest its dissociates DD γ inding energy 28MeV Large amount of He in the universe: hot ig ang Phys 112 S2006 7 Chemical Potential 16
Reaction Rates and Expansion Strong dependence of reaction rate on temperature cf. ordinary cooking Dependence on density The particles have to find each other reaking of bounds tenderness Pressure cooker, refrigerator Establishing new bounds e.g. custard rate " 1 " 2 In an expanding universe The 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 Phys 112 S2006 7 Chemical Potential 17
uilding Nuclei by Fusion Need protons and neutrons ut do it fast enough as neutron will decayhalf life time 10.6 minutes 2 body reactions Density not large enough for 3 body Hydrogen p n Deuterium= 2 H 3 He 2 H 4 He 3 He 7 Li a number of other possible reactions Process stopped by Freeze-Out Temperature and density become too small No element heavier than Lithium Phys 112 S2006 7 Chemical Potential 18
Same formalism as Saha equation but expanding universe Primordial Nucleosynthesis n depleted by low T Freeze out 2 H bottleneck 4 He very much more bound than 2 H => higher equilibrium concentration at low temperature ut 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 carbon Higher A elements will have to be produced in stars and supernovae Dependent on expansion rate and density of pn = Ω b Phys 112 S2006 7 Chemical Potential 19
ig ang Nucleosynthesis 4 He Self consistent Confirmed by Cosmic Microwave ackground D Density of baryons 5 Non aryonic Dark Matter 7 Li Phys 112 S2006 7 Chemical Potential 20