Physics 556Stellar AstrophysicsProf. James Buckley

Lecture 11Chemical Potential and Ionization State

Tuesday, April 2, 2013

Chemical Potential

“If to any homogeneous mass in a state of hydrostatic stress we suppose an infinitesimal quantity of any substance to be added, the mass remaining homogeneous and its entropy and volume remaining unchanged, the increase of the energy of the mass divided by the quantity of the substance added is the potential for that substance in the mass considered.

”

Josiah Willard Gibbs (1839-1903)

While we may now take the fundamental definition of chemical potential to be the Lagrange multiplier that appears from adding a constraint on the total number of particles to the entropy maximization (or adding an effective term to the entropy), a more physical interpretation, is as follows:

The chemical potential of a thermodynamic system is the amount by which the energy of the system would change if an additional particle were introduced, with entropy and volume held constant. If a system contains more than one species of particles, each species has its own chemical potential.

Tuesday, April 2, 2013

Chemical Potential: Stat Mech

Or substituting d3p = 4πp2dp we have

ρ(p)dp =8πp2dp

h3

ρ(p)dp = 2d3p

h3

For relativistic or non-relativistic particles, the density of states is given by:

Fermi−Dirac occupation probability : f(E) =g

e(E−µ)/kT + 1

And in the classical limit, we get the Maxwell-Boltzmann equation:

f(E) ≈ e−(E−µ)/kT

The total distribution function (differential number of particles in the momen-tum interval (p, p + dp) is given by

n(p)dp =8π

h3eµe−p2/2mkT p2 dp

f(E) =1

e(E−µ)/kT (1 + e−(E−µ)/kT )

Assuming that E is small compared to |µ| and that the chemical potential isnegative (µ < 0) then e−(E−µ)/kT ≈ eµ/kT = e−|µ|/kT � 1

f(E) ≈ e−(E−µ)/kT�1− e−(E−µ)/kT

�(weak degeneracy approximation)

Tuesday, April 2, 2013

Chemical Potential: Stat Mech

One can then find the chemical potential by integrating this expression over all

momenta, equating to the total number density, and solving for µ

n(p)dp =8π

h3eµ/kT e−p2/2mkT p2 dp

n =� ∞

0n(p)dp =

8π

h3eµ/kT

� ∞

0e−p2/2mkT p2dp

� ∞

0x2 e−x2

dx =√

π

4

n =� ∞

0n(p)dp =

8π

h3eµ/kT (2mkT )3/2

� ∞

0x2 e−x2

dx

n =2eµ/kT

h3(2πmkT )3/2

µ = −kT ln(T 3/2/n)− 32

ln�

2πmk

h2

�− kT ln 2

eµ/kT =nh3

2(2πmkT )−3/2

Tuesday, April 2, 2013

Chemical Potential: Thermo

If one also realizes that the energy changes as one adds or removes particles ofeach of n species with number densities n1, n2, ..., nn then we can write the totalinternal energy as:

U = U(S, V, n1, ..., nn)

and then the chemical potential is

µi ≡�

∂U

∂Ni

�

S,V,Nj �=i

Including the extra contribution to the total internal energy from adding orsubtracting particles of different species we have:

dU = TdS − PdV +�

i

µidNi

Note that all of the natural variables of U(yi), yi = {S, V, and {Ni}} areextensive quantities, that is:

U({αyi}) = αU({yi})

are homogeneous functions and therefore simply integrable giving an integralform for the equation of state:

U = TS − PV +�

i

µiNi

Tuesday, April 2, 2013

Thermodynamic Potentials

• In general, a thermodynamic potential is a scalar potential function used to represent the thermodynamic state of a system. For example, the internal energy, U, is the energy of configuration of a given system of conservative forces. Expressions for all other thermodynamic energy potentials are derivable via Legendre transforms from an expression for U. Common thermodynamic energy potentials are:

Potential Formula Natural Variables

Internal Energy U TS − PV +�

i µiNi S,V,{Ni}Helmholtz free energy F U − TS T, V, {Ni}Enthalpy H U + PV S, P, {Ni}Gibbs Free Energy G U + PV − TS T, P, {Ni

Tuesday, April 2, 2013

Gibbs Free Energy

µi =�

∂G

∂Ni

�

T,P,nj �=i

dG = V dP − SdT +�

i

µidNi

Now define a new state variable, the Gibbs free energy G:

G = U + PV − TS

dG = dU + PdV + V dP − TdS − SdT

substituting for dU

Physically, the Gibbs free energy is defined as the maximum amount of workobtainable from an isothermal, isobaric thermodynamic system.

Substituting for U in the expression for the Gibbs free energy, i.e.:

U = TS − PV +�

i

µiNi

one obtains : G = TS − PV +�

i

µiNi + PV − TS or

G =�

i

µiNi

dU = TdS − PdV +�

i

µidNi

Tuesday, April 2, 2013

Classical LimitFor a single species with number density n = N/V , at low density (so we cantreat the particles as classical, distinguishable particles) we have

G = U − TS + PV

and G = µN = µnV

µ =52kT − kT ln(T 3/2/n)− Ts0

Substituting the expression for the internal energy per unit volume u = 32nkT ,

pressure P = nkT and for the entropy per unit volume s = nk ln(T 3/2/n)+ns0:

G =�

32nkT − nkT ln(T 3/2/n)− nTs0

�V + nkTV

µ = −kT ln(T 3/2/n)− 32

ln�

2πmk

h2

�− kT ln 2

The classical expression for the chemical potential should agree with the low-density limit of the QM expression derived previously:

Which is possible, if we redefine the arbitrary constant in the entropy s0 tomatch the QM calculation (which takes into account the minimum volume ofphase space ∆p3∆x3 = h3 from the uncertainty principle)

so =52k +

32k ln

�2πmk

h2

�+ k ln 2

Tuesday, April 2, 2013

NR Gas, Weak Degeneracy

f =1

e(E−µ)/kT (1 + e−(E−µ)/kT )

Weak degeneracy |µ/kT |� 1 and µ < 0 e−(E−µ)/kT is small, and

f ≈ e−(E−µ)/kT (1− e−(E−µ)/kT )

ne =8π

h3

�f p2dp

E =p2

2mfor nonrelativistic energies E < m c2

substitute x ≡ p/√

2mkT E/kT = x2

� ∞

0e−βx2dx =

∂

∂β

� ∞

0e−βx2

dx =∂

∂β

�14

� ∞

0e−βr22πrdr

=∂

∂β

�π

4β

� ∞

0e−udu =

�π

4

�−1

2

�β−3/2 =

√π

4β3/2

ne =8π

h3(2mkT )3/2

�eµ/kT

� ∞

0e−x2

x2dx− e2µ/kT

� ∞

0e−2x2

x2dx

�

ne =2(2πmkT )3/2

h3eµ/kT

�1− eµ/kT

23/2

�

Tuesday, April 2, 2013

NR Gas, Weak Degeneracy

ne =2(2πmkT )3/2

h3eµ/kT

�1− eµ/kT

23/2

�

For a weakly degenerate gas, µ/kT is large and negative, so the second termin parentheses is small compared with 1, and we can approximate by settingµ = µ0 (the chemical potential for a nondegenerate gas). This will allow us toinvert the expression and solve for µ

ne ≈2(2πmkT )3/2

h3eµ/kT

�1− eµ0/kT

23/2

�

where µ0 = −kT ln(T 3/2/n)− 32

ln�

2πmk

h2

�− kT ln 2

Substituting, and solving for µ we obtain (after a fair bit of algebra):

µweak degen = −kT ln�

2πmk

h2

�3/2

− kT ln(2T 3/2/ne)−ne

21/2

�h2

2πmkT

�3/2

Tuesday, April 2, 2013

NR Gas, Weak DegeneracyArmed with the chemical potential:

µweak degen = −kT ln�

2πmk

h2

�3/2

− kT ln(2T 3/2/ne)−ne

21/2

�h2

2πmkT

�3/2

and the occupation probability

f ≈ e−(E−µ)/kT (1− e−(E−µ)/kT )we can solve for thermodynamic quantities like pressure and internal energy byperforming the appropriate moment integrals of the probability density func-tions.for internal energy we can calculate

where E is the kinetic energy, for a NR gas E = p2/2m and for a relativistic gasE = E −m0c2

U =8π

h3

� ∞

0Ef(p)p2dp

Substituting for the chemical potential, and doing lots of math, we obtain

Ue ≈32nekT

�1 +

ne

27/2

�h2

2πmkT

�3/2�

And the heat capacity per unit volume is cv =�

∂U

∂T

�

V

=32nek

�1− ne

29/2

�h2

2πmkT

�3/2�

Tuesday, April 2, 2013

Relativistic Gas, Weak Degen.For an ultrarelativistic gas, with weak degeneracy, we follow the same procedure,but set E = pc and introduce the variable x = pc/kT in performing the integralsover the probability distribution function f .

ne =8π

(hc)3eµr,nd/kT

� ∞

0e−pc/kT c3p2dp

Evaluating the intergral and solving for µ gives

µr,nd = kT ln(ne/16π)− 3kT ln(kT/hc)

We start by calculating the chemical potential for an ultrarelativistic nondegen-erate gas µr,nd

For an ultrarelativistic, but weakly degenerate gas, the total number density isgiven by:

ne ≈ 16π

�kT

hc

�3

eµ/kT

�1− e

µr,nd/kT

23

�

Solving for µ (which we will call µr,d) we obtain

µr,d = −kT ln

�16π

ne

�kT

hc

�3�

+ne

27π

�hc

kT

�3

Ue ≈ 3nekT

�1 +

ne

256π

�hc

kT

�3�

cV ≈ 3nek

�1− ne

128π

�hc

kT

�3�

Tuesday, April 2, 2013

Atoms

If there are additional internal degrees of freedom in our particles (e.g., atomicenergy levels, vibrational states in molecules, etc.) we start with a modifiedformula for the internal energy per particle:

up = 3/2kT +�

n gnEne−En/kT

�n gne−En/kT

this is more compactly written in terms of the partition function, which we candefine as

Z ≡�

n

gne−En/kT

up =32kT + kT 2 ∂ lnZ

∂T

Following the usual steps, one can derive the chemical potential, specific heat

and entropy. Of greatest importance is the chemical potential:

µA = −kT ln(T 3/2/nA)− 32kT

�2πmAk

h2

�− kT ln

�

n

gne−En/kT

Tuesday, April 2, 2013

Reaction Equilibrium

The Gibbs free energy for a system with n different species, of number densitiesni is given by

dG = −SdT + V dP +n�

i=1

µidni

if changes occur under constant P and T then

(dG)P,T =n�

i=1

µidni

And integrating gives the result (derived previously by substituting for U in theexpression G = U + PV − TS)

G =n�

i=1

µini

Say that we have reversible reactions that convert particles of type νi particlesof type χi into other species, e.g.:

ν1χ1 + ν2χ2 ↔ ν3χ3 + ν4χ4

then changes in the number density of the constituents are constrained suchthat

dn1

−ν1=

dn2

−ν2=

dn3

ν3=

dn4

ν4

Tuesday, April 2, 2013

Reaction Equilibrium

(dG)P,T =4�

i=1

µidni = dn1

�µ1 +

ν2

ν1µ2 −

ν3

ν1µ3 −

ν4

ν1µ4

�

or (dG)P,T = dn2

�ν1

ν2µ1 + µ2 −

ν3

ν2µ3 −

ν4

ν2µ4

�, etc.

In equilibrium dG = 0 and, regardless of how we write dG we get the samecondition for equilibrium:

ν1µ1 + ν2µ2 = ν3µ3 + ν4µ4

For other reactions, e.g.,

ν1χ1 ↔ ν2χ2 + ν3χ3

we get similar equations

ν1µ1 = ν2µ2 + ν3µ3

and the prescription for going from any arbitrary reactions (in equilibrium withthe reverse reaction) is straightforward.

Tuesday, April 2, 2013

Ionization Equilibrium

H ↔ e− + p

µe = −kT ln(T 3/2/ne)−32kT ln

�2πmek

h2

�− kT ln 2

µp = −kT ln(T 3/2/np)−32kT ln

�2πmpk

h2

�− kT ln 2

µH = −kT ln(T 3/2/nH)− 32kT

�2πmHk

h2

�− kT ln

�

n

gne−En/kT

µH = µe + µp

assuming mp ≈ mH

nH

nenp

=�

gne−En/kT

4�

2πmkT

h2

�3/2

Giving the Saha equation (after a bit of math!)

Tuesday, April 2, 2013