Chapter 5: Stellar Structure Modeling - UMassThe Equations of Stellar Structure We consider the modeling of stars in hydrostatic and thermal equilibrium. We hope to know how such global

Chapter 5: Stellar Structure Modeling

The Equations of Stellar Structure

Polytropic Equations of State and Polytropes

Numerical Calculation of the Lane-Emden Equation

The Eddington Standard Model

The Structure of the EnvelopeRadiative envelope structureConvective envelopeConvective stars

Outline






The Equations of Stellar StructureWe consider the modeling of stars in hydrostatic and thermalequilibrium. We hope to know how such global properties as theluminosity and radius of a star depend on its mass and initialchemical composition.

I The set of four differential equations to be solved:I The mass conservation

drdMr

=1

4πr2ρ,

I The hydrostatic equation

dPdMr

= −GMr

4πr4 ,

I The energy equation

dLr

dMr= ε,

I The heat transfer method

∇ ≡ d lnTd lnP

.

The Equations of Stellar StructureWe consider the modeling of stars in hydrostatic and thermalequilibrium. We hope to know how such global properties as theluminosity and radius of a star depend on its mass and initialchemical composition.

I The set of four differential equations to be solved:I The mass conservation

drdMr

=1

4πr2ρ,

I The hydrostatic equation

dPdMr

= −GMr

4πr4 ,

I The energy equation

dLr

dMr= ε,

I The heat transfer method

∇ ≡ d lnTd lnP

.

I To implement the specific heat transfer, we may first assume thatthe transfer is all due to the radiation and compute

∇rad =3

16πacPκT 4

Lr

GMr.

We then set∇ = ∇rad if ∇rad ≤ ∇ad

for pure diffusive radiative transfer, or

∇ = ∇ad if ∇rad > ∇ad

when adiabatic convection is present. Iteration of the above isneeded to reach a converge of the solution, if there is one and ifit is unique.

I Four boundary conditions are required to close the system. Forsimplicity, we choose “zero” conditions, which are r = Lr = 0 atthe center (Mr = 0), and ρ = T = 0 at the surface (Mr = M).

I Of course, we assume that we already know the microscopicconstituent physics: i.e., the quantities P, κ, and ε as functions ofρ,T , and X , where X is shorthand for composition.


∇rad =3

16πacPκT 4

Lr

GMr.








∇rad =3

16πacPκT 4

Lr

GMr.







Outline






Polytropic Equations of State and Polytrope

We define a polytropic stellar model (or polytrope) to be one in whichthe pressure is given by

P(r) = Kρ1+1/n(r),

where both the polytropic index n and K are constant.This model allows us to avoid dealing with both the heat transfer andthermal balance.

Examples that we have encountered:I A zero temperature, completely degenerate electron gas: e.g.,

Pe = 1.0× 1013( ρµe)5/3 dyn cm−2 if non-relativistic.

I For a region with efficient convection, i.e.,

∇ = ∇ad =

(∂lnT∂lnP

)ad

= 1− 1/Γ2.

If Γ2 is assumed constant, then P(r) ∝ T Γ2/(Γ2−1)(r).If in addition, the gas is ideal, then P(r) ∝ ρΓ2 (r).

Polytropic Equations of State and Polytrope

We define a polytropic stellar model (or polytrope) to be one in whichthe pressure is given by

P(r) = Kρ1+1/n(r),

where both the polytropic index n and K are constant.This model allows us to avoid dealing with both the heat transfer andthermal balance.

Examples that we have encountered:I A zero temperature, completely degenerate electron gas: e.g.,

Pe = 1.0× 1013( ρµe)5/3 dyn cm−2 if non-relativistic.

I For a region with efficient convection, i.e.,

∇ = ∇ad =

(∂lnT∂lnP

)ad

= 1− 1/Γ2.

If Γ2 is assumed constant, then P(r) ∝ T Γ2/(Γ2−1)(r).If in addition, the gas is ideal, then P(r) ∝ ρΓ2 (r).

Lane-Emden equationFor a polytrope, we can derive from the hydrostatic and massconservation equations (in Euclidean coordinates) the following

1r2

ddr

(r2

ρ

dPdr

)= −G

r2dMr

dr= −4πGρ.

Now perform the transformations to make the the equationdimensionless: ρ(r) = ρcθ

n(r) and r = rnξ, we then have theLane-Emden equation:

1ξ2

ddξ

(ξ2 dθn

dξ

)= −θn

n

where Pc is defined (from the EoS) as Pc = Kρ1+1/nc and

r2n =

(n + 1)Pc

4πGρ2c.

The solutions, θn(ξ), are called “Lane-Emden solutions”.Note that if an ideal gas with constant µ is assumed, θn measurestemperature T (r) = Tcθn(r), where Tc = Kρ1/n

c (NAk/µ)−1.


1r2

ddr

(r2

ρ

dPdr

)= −G

r2dMr

dr= −4πGρ.



1ξ2

ddξ

(ξ2 dθn

dξ

)= −θn

n


r2n =

(n + 1)Pc

4πGρ2c.


c (NAk/µ)−1.


1r2

ddr

(r2

ρ

dPdr

)= −G

r2dMr

dr= −4πGρ.



1ξ2

ddξ

(ξ2 dθn

dξ

)= −θn

n


r2n =

(n + 1)Pc

4πGρ2c.


c (NAk/µ)−1.

Since the Lane-Emden equation is a second-order differentialequation, we need two real boundary conditions:

I For ρc to really be the central density, we require thatθn(ξ = 0) = 1;

I The spherical symmetry at the center (dP/dr vanishing at r = 0)requires that θ′n(ξ = 0) = 0.

I The surface of a model star is where the first zero of θn occurs,θn(ξ1) = 0, where ξ1 is the location of the surface. Thisinterpretation of the solution is not a boundary condition.

Analytical E-solutions for θn are obtainable forn = 0,1, and 5: e.g., for n = 0, ρ(r) = ρc , the

solution is θ0(ξ) = 1− ξ2

6. Clearly, ξ1 = 61/2.

Numerical methods must be used for general n(e.g., n = 1.5 or 3 for a completely degenerate,non or fully relativistic electron gas (P ∝ ρ5/3 or∝ ρ4/3). Solutions of Lane-Emden

equation for n = 0, 1, 2, 3, 4, 5.






solution is

θ0(ξ) = 1− ξ2

6. Clearly, ξ1 = 61/2.


equation for n = 0, 1, 2, 3, 4, 5.







6. Clearly, ξ1 = 61/2.


equation for n = 0, 1, 2, 3, 4, 5.







6. Clearly, ξ1 = 61/2.


equation for n = 0, 1, 2, 3, 4, 5.

Given n and K , we can find the dependence of P and ρ on ξ.However, to get the absolute physical numbers, we need R = rnξ1,which depends on ρc and the stellar mass, which are related via

M =

∫ R

04πr2ρ(r)dr = 4πr3

n ρc

∫ ξ1

0ξ2θn

ndξ = 4πr3n ρc(−ξ2θ′n)ξ1 .

For a given n, (−ξ2θ′n)ξ1 is known. With the above equation, togetherwith R = rnξ1, we can then solve ρc and rn in terms of M and R. Sincern is a function of ρc and Pc , the latter can be expressed in terms ofρc . Therefore, from Pc = Kρ1+1/n

c , we can get K as

K =

[4π

ξn+1(−θ′n)n−1

]1/n

ξ1

Gn + 1

M1−1/nR−1+3/n, (1)

which will be referred to later when the convective star is discussed.

A useful quantity that depends only on n is the ratio

ρc

< ρ >=

13

(ξ

−θ′n

)ξ1

,

where < ρ > is the volume-averaged mean density of a star.

Given n and K , we can find the dependence of P and ρ on ξ.However, to get the absolute physical numbers, we need R = rnξ1,which depends on ρc and the stellar mass, which are related via

M =

∫ R

04πr2ρ(r)dr = 4πr3

n ρc

∫ ξ1

0ξ2θn

ndξ = 4πr3n ρc(−ξ2θ′n)ξ1 .

For a given n, (−ξ2θ′n)ξ1 is known. With the above equation, togetherwith R = rnξ1, we can then solve ρc and rn in terms of M and R. Sincern is a function of ρc and Pc , the latter can be expressed in terms ofρc . Therefore, from Pc = Kρ1+1/n

c , we can get K as

K =

[4π

ξn+1(−θ′n)n−1

]1/n

ξ1

Gn + 1

M1−1/nR−1+3/n, (1)

which will be referred to later when the convective star is discussed.A useful quantity that depends only on n is the ratio

ρc

< ρ >=

13

(ξ

−θ′n

)ξ1

,

where < ρ > is the volume-averaged mean density of a star.

Outline






Numerical Calculation of the Lane-Emden EquationA convenient way to numerically solve a high-order differentialequation is to cast it into a set of first-order equations. We thus castthe second-order Lane-Emden equation,

1ξ2

ddξ

(ξ2 dθn

dξ

)= −θn

n , (2)

in the form of two first-order equations by introducing the newvariables x = ξ, y = θn, and z = (dθn/dξ) = (dy/dx):

y ′ =dydx

= z; z ′ =dzdx

= −yn − 2x

z

With the two boundary conditions at the center (y = 1, z = 0), onemay use a simple “shooting method”, whereby one “shoots” from astarting point and hopes that the shot will end up at the right place;e.g., using a “Runge-Kutta” integrator.Suppose we know the values of y and z at some point xi and callthese values yi and zi . We can use the above equations to find yi+1and zi+1 at xi+1 = xi + h, where h is called the “step size”.


1ξ2

ddξ

(ξ2 dθn

dξ

)= −θn

n , (2)


y ′ =dydx

= z; z ′ =dzdx

= −yn − 2x

z

With the two boundary conditions at the center (y = 1, z = 0), onemay use a simple “shooting method”, whereby one “shoots” from astarting point and hopes that the shot will end up at the right place;e.g., using a “Runge-Kutta” integrator.

Suppose we know the values of y and z at some point xi and callthese values yi and zi . We can use the above equations to find yi+1and zi+1 at xi+1 = xi + h, where h is called the “step size”.


1ξ2

ddξ

(ξ2 dθn

dξ

)= −θn

n , (2)


y ′ =dydx

= z; z ′ =dzdx

= −yn − 2x

z

With the two boundary conditions at the center (y = 1, z = 0), onemay use a simple “shooting method”, whereby one “shoots” from astarting point and hopes that the shot will end up at the right place;e.g., using a “Runge-Kutta” integrator.Suppose we know the values of y and z at some point xi and callthese values yi and zi . We can use the above equations to find yi+1and zi+1 at xi+1 = xi + h, where h is called the “step size”.

Care needs to be taken at the origin, where z ′ is indeterminatebecause both x and z are equal to zero.

The resolution to this problem is to expand θn(ξ) in a series about theorigin, θn(ξ) = a0 + a1ξ + a2ξ

2..., and to obtain a solution of Eq. 2which keeps at least the first non-zero coefficient of ξ.A quick consideration of the boundary conditions can immediately tellus that a0 = 1 and all the odd terms should be zero (e.g., a1 = 0).Now inserting θn(ξ) = 1 + a2ξ

2 + a4ξ4 into Eq. 2 and comparing the

coefficients of individual ξ terms to establish the constants in theexpansion, we get

y = θn(ξ) = 1− 16ξ2 +

n120

ξ4...

For ξ → 0, we find that y ′ → −1/3x and z ′ → −1/3, which may beused to start the integration.This way, the calculation may march from the origin to the surface,when y = θn cross the zero.Clearly, this shooting method may not be optimal, because the erroraccumulates with the steps. Thus alternative methods may be usedto obtain a more accurate solution.

Care needs to be taken at the origin, where z ′ is indeterminatebecause both x and z are equal to zero.The resolution to this problem is to expand θn(ξ) in a series about theorigin, θn(ξ) = a0 + a1ξ + a2ξ

2..., and to obtain a solution of Eq. 2which keeps at least the first non-zero coefficient of ξ.

A quick consideration of the boundary conditions can immediately tellus that a0 = 1 and all the odd terms should be zero (e.g., a1 = 0).Now inserting θn(ξ) = 1 + a2ξ



y = θn(ξ) = 1− 16ξ2 +

n120

ξ4...






y = θn(ξ) = 1− 16ξ2 +

n120

ξ4...






y = θn(ξ) = 1− 16ξ2 +

n120

ξ4...






y = θn(ξ) = 1− 16ξ2 +

n120

ξ4...


Newton-Raphson and Henyey MethodThis is a more powerful technique to solve the stellar structureequations, using the “integration” over the model (via iterations),instead of shooting from one point to another.

However, the location of the outer boundary is not knownbefore-hand. In the polytropic stellar structure we want to get, theradius ξ1 needs to improve the solution:

I We do a simple conversion, x → x = ξ/λ, where λ = ξ1, or aneigenvalue, is to be determined. This can be done since wehave one more boundary condition, y = 0 at the new x = 1, inaddition to the two at the center (y = 1, z = 0).

I Now, x is within the closed interval [0,1] and can be divided intoa grid. Then the differential equations can be written in a “finitedifference form” and the boundary conditions can be applied.

I The converted equations are generally also depend on λ; e.g.,the Lane-Emden equation becomes

y ′ =dydx

= λz; z ′ =dzdx

= −λyn − 2x

z.

Newton-Raphson and Henyey MethodThis is a more powerful technique to solve the stellar structureequations, using the “integration” over the model (via iterations),instead of shooting from one point to another.However, the location of the outer boundary is not knownbefore-hand. In the polytropic stellar structure we want to get, theradius ξ1 needs to improve the solution:

I We do a simple conversion, x → x = ξ/λ, where λ = ξ1, or aneigenvalue, is to be determined. This can be done since wehave one more boundary condition, y = 0 at the new x = 1, inaddition to the two at the center (y = 1, z = 0).

I Now, x is within the closed interval [0,1] and can be divided intoa grid. Then the differential equations can be written in a “finitedifference form” and the boundary conditions can be applied.

I The converted equations are generally also depend on λ; e.g.,the Lane-Emden equation becomes

y ′ =dydx

= λz; z ′ =dzdx

= −λyn − 2x

z.

The above example has a general form:

dydx

= f (x , y , z, λ);dzdx

= g(x , y , z, λ)

with boundary conditions on y and z specified at the endpoints of theinterval x1 ≤ x ≤ xN . For our example, we have two boundaryconditions at x1 and one at xN :

b1(x1, y1, z1) = 0; b2(x1, y1, z1) = 0; b3(xN , yN , zN) = 0.

Assuming that f ,g,b1, b2, and b3 are well behaved, the differentialequations can be cast in a “finite difference form” over a “mesh” in x ;i.e., x1, x2, ..., xN at which y and z are to be evaluated.

I For simplicity, consider that the mesh interval is constant; i.e.,xi+1 − xi = ∆x for i = 1, ...,N − 1.

I The equations can hence be expressed as

yi+1 − yi

∆x=

12

(fi+1 + fi );zi+1 − zi

∆x=

12

(gi+1 + gi )

I The above expressions then represent 2N − 2 equations, whichtogether with the three boundary conditions can in principle beused to solve 2N + 1 variables yi , zi , and λ.

The difficulty is that the variables are all mixed up among theequations, generally in a nonlinear fashion.One way to get out of this difficulty is to use the Newton-Raphsonmethod: find the solution by linearizing the equations and theboundary conditions.

Suppose that we have a “guessed” solution (e.g., from the shootingmethod) that gives yi and zi for all i , which generally do not satisfy theequations. we may make corrections

yi → yi + ∆yi ; zi → zi + ∆zi

so that the new yi and zi might satisfy both the equations and theboundary conditions.

We now estimate the values of ∆yi and ∆zi for all i by letting the newyi and zi satisfy the linearized equations and boundary conditions.











yi+1 − yi

∆x=

12

(fi+1 + fi ), for example, can be written as

yi+1 + ∆yi+1 − yi −∆yi =

∆x2

[fi+1 +

( ∂f∂y

)i+1

∆yi+1 +( ∂f∂z

)i+1

∆zi+1 +( ∂f∂λ

)i+1

∆λ]

+∆x2

[fi +

( ∂f∂y

)i∆yi +

( ∂f∂z

)i∆zi +

( ∂f∂λ

)i∆λ]

Some manipulation leads to

yi+1 − yi −∆x2

(fi+1 + fi ) =[∆x2

( ∂f∂y

)i

+ 1]∆yi +

[∆x2

( ∂f∂y

)i+1− 1]∆yi+1

+[∆x

2

( ∂f∂z

)i

]∆zi +

[∆x2

( ∂f∂z

)i+1

]∆zi+1+

∆x2

[( ∂f∂λ

)i+( ∂f∂λ

)i+1

]∆λ

Note that the left-hand side of these equations are zero when thedifference equations are satisfied; that is when ∆yi and ∆zi go tozero.

yi+1 − yi

∆x=

12


yi+1 + ∆yi+1 − yi −∆yi =

∆x2

[fi+1 +

( ∂f∂y

)i+1

∆yi+1 +( ∂f∂z

)i+1

∆zi+1 +( ∂f∂λ

)i+1

∆λ]

+∆x2

[fi +

( ∂f∂y

)i∆yi +

( ∂f∂z

)i∆zi +

( ∂f∂λ

)i∆λ]


yi+1 − yi −∆x2

(fi+1 + fi ) =[∆x2

( ∂f∂y

)i

+ 1]∆yi +

[∆x2

( ∂f∂y

)i+1− 1]∆yi+1

+[∆x

2

( ∂f∂z

)i

]∆zi +

[∆x2

( ∂f∂z

)i+1

]∆zi+1+

∆x2

[( ∂f∂λ

)i+( ∂f∂λ

)i+1

]∆λ


yi+1 − yi

∆x=

12


yi+1 + ∆yi+1 − yi −∆yi =

∆x2

[fi+1 +

( ∂f∂y

)i+1

∆yi+1 +( ∂f∂z

)i+1

∆zi+1 +( ∂f∂λ

)i+1

∆λ]

+∆x2

[fi +

( ∂f∂y

)i∆yi +

( ∂f∂z

)i∆zi +

( ∂f∂λ

)i∆λ]


yi+1 − yi −∆x2

(fi+1 + fi ) =[∆x2

( ∂f∂y

)i

+ 1]∆yi +

[∆x2

( ∂f∂y

)i+1− 1]∆yi+1

+[∆x

2

( ∂f∂z

)i

]∆zi +

[∆x2

( ∂f∂z

)i+1

]∆zi+1+

∆x2

[( ∂f∂λ

)i+( ∂f∂λ

)i+1

]∆λ


Similarly, the j th boundary conditions can be linearized into

−bj =(∂bj

∂y

)(1 or N)

∆y(1 or N) +(∂bj

∂z

)(1 or N)

∆z(1 or N).

We can arrange all these equations in a matrix form

M · U = R.

in which,

U ≡ (∆y1,∆z1,∆y2,∆z2, · · · ,∆yN ,∆zN ,∆λ)T

where the superscript “T” indicates transpose;

R = (−b1,−b2,Y3/2,Z3/2, · · · ,YN−1/2,ZN−1/2,−b3)T

where

Yi+1/2 ≡ yi+1 − yi −∆x2

(fi+1 + fi )

Zi+1/2 ≡ zi+1 − zi −∆x2

(gi+1 + gi );

and finally

M =∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

(∂b1

∂y

)1

(∂b1

∂z

)1

0 0 0 0 · · · · · · · · · 0(∂b2

∂y

)1

(∂b2

∂z

)1

0 0 0 0 · · · · · · · · · 0

A1 + 1 B1 A2 − 1 B2 0 0 · · · · · · · · · E1 + E2C1 D1 + 1 C2 D2 − 1 0 0 · · · · · · · · · F1 + F20 0 A2 + 1 B2 A3 − 1 B3 · · · · · · · · · E2 + E3· · · · · · · · · · · · · · · · · · · · · · · · · · · · · ·

0 0 0 0 0 0 · · ·(∂b3

∂y

)N

(∂b3

∂z

)N

0

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣where

Ai ≡∆x2

( ∂f∂y

)i; Bi ≡

∆x2

( ∂f∂z

)i; Ei ≡

∆x2

( ∂f∂λ

)i

Ci ≡∆x2

(∂g∂y

)i; Di ≡

∆x2

(∂g∂z

)i; Fi ≡

∆x2

(∂g∂λ

)i

Once the solution set U is found, then new values of yi and zi areobtained by adding ∆yi and ∆zi to the corresponding old guesses.If all goes well, then the corrections decrease as the square of theirabsolute values (as the original set of the differential equations is notlinear). We iterate the above procedure until ∆yi and ∆zi becomesufficiently small.

and finally

M =∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

(∂b1

∂y

)1

(∂b1

∂z

)1

0 0 0 0 · · · · · · · · · 0(∂b2

∂y

)1

(∂b2

∂z

)1

0 0 0 0 · · · · · · · · · 0

A1 + 1 B1 A2 − 1 B2 0 0 · · · · · · · · · E1 + E2C1 D1 + 1 C2 D2 − 1 0 0 · · · · · · · · · F1 + F20 0 A2 + 1 B2 A3 − 1 B3 · · · · · · · · · E2 + E3· · · · · · · · · · · · · · · · · · · · · · · · · · · · · ·

0 0 0 0 0 0 · · ·(∂b3

∂y

)N

(∂b3

∂z

)N

0

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣where

Ai ≡∆x2

( ∂f∂y

)i; Bi ≡

∆x2

( ∂f∂z

)i; Ei ≡

∆x2

( ∂f∂λ

)i

Ci ≡∆x2

(∂g∂y

)i; Di ≡

∆x2

(∂g∂z

)i; Fi ≡

∆x2

(∂g∂λ

)i

Once the solution set U is found, then new values of yi and zi areobtained by adding ∆yi and ∆zi to the corresponding old guesses.If all goes well, then the corrections decrease as the square of theirabsolute values (as the original set of the differential equations is notlinear). We iterate the above procedure until ∆yi and ∆zi becomesufficiently small.

Outline






The Eddington Standard ModelThis gives a simple example of the use of polytropes in making astellar pseudo-model, which approximately incorporates the energyand radiative transfer equations.Recall that in case of no convection the radiative transfer equationcan be expressed as

∇ =3

16πacPκT 4

Lr

GMr.

On the other hand,

∇ ≡ d lnTd lnP

=14

PPrad

dPrad

dP.

where we have use the radiation pressure Prad = aT 4/3.

The heat transfer equation can then expresses as

dPrad

dP=

κ(r)Lr

4πcGMr=

L4πcGM

κ(r)η(r),

where we have defined

η(r) ≡ < ε(r) >

< ε(R) >=

Lr/Mr

L/M; < ε(r) >≡ Lr

Mr=

∫ r0 εdMr∫ r0 dMr

.

The Eddington Standard ModelThis gives a simple example of the use of polytropes in making astellar pseudo-model, which approximately incorporates the energyand radiative transfer equations.Recall that in case of no convection the radiative transfer equationcan be expressed as

∇ =3

16πacPκT 4

Lr

GMr.

On the other hand,

∇ ≡ d lnTd lnP

=14

PPrad

dPrad

dP.

where we have use the radiation pressure Prad = aT 4/3.The heat transfer equation can then expresses as

dPrad

dP=

κ(r)Lr

4πcGMr=

L4πcGM

κ(r)η(r),

where we have defined

η(r) ≡ < ε(r) >

< ε(R) >=

Lr/Mr

L/M; < ε(r) >≡ Lr

Mr=

∫ r0 εdMr∫ r0 dMr

.

Assuming that the surface pressure is equal to zero, the integration ofthe above equation gives

Prad (r) =L

4πcGM< κ(r)η(r) > P(r),

where

< κ(r)η(r) >=1

P(r)

∫ P(r)

0κ(r)η(r)dP.

If we can assume that < κ(r)η(r) > varies weakly with position in astar, or close to a constant, as Eddington did, then the ratio of1− β ≡ Prad/P is a constant, where β is the gas to total pressureratio and depends on both the opacity and the energy generate rate.This constancy may be translated into a T vs. ρ relation as follows:

If the pressure is contributed by ideal gas plus radiation only, then

Prad = P − Pgas =1− ββ

Pgas =1− ββ

NAkµ

ρT = aT 4/3

T (r) =

(1− ββ

3a

NAkµ

)1/3

ρ1/3(r).

P =Pgas

β=

NAkµ

ρTβ

= Kρ4/3(r),

where

K =

[1− ββ4

3a

(NAkµ

)4]1/3

.

So we have a polytrope with n = 3.

The sold line is the relation through afull-blown model of a ZAMS sun with totalradius of 6.168 × 1010 cm or 0.886R�. Thedashed line shows the standard modelresult with the same total radius.

Outline






The Structure of the Envelope

To construct a more realistic stellar model, we need to deal with theouter layer more carefully. We have considered the structure of thestellar atmosphere in the heat transfer chapter. This considerationassumes that convection plays no role in heat transport between thetrue and photosphere surface, which is consistent with our notion of aradiating, static, and visible surface (although this is not true for thesun).

We now consider the stellar “envelope”, which consists of the portionof a star that starts at the photosphere, and continues inward, butcontains negligible mass, has no thermonuclear or gravitationalenergy sources, and is in hydrostatic equilibrium. We want to seehow deep the radiative envelope may be until the convectiontakes over for heat transfer.

The Structure of the Envelope

To construct a more realistic stellar model, we need to deal with theouter layer more carefully. We have considered the structure of thestellar atmosphere in the heat transfer chapter. This considerationassumes that convection plays no role in heat transport between thetrue and photosphere surface, which is consistent with our notion of aradiating, static, and visible surface (although this is not true for thesun).

We now consider the stellar “envelope”, which consists of the portionof a star that starts at the photosphere, and continues inward, butcontains negligible mass, has no thermonuclear or gravitationalenergy sources, and is in hydrostatic equilibrium. We want to seehow deep the radiative envelope may be until the convectiontakes over for heat transfer.

In such a radiative heat transfer region, ∇ = ∇rad with

∇ =d lnTd lnP

=3κL

16πacGMPT 4 ,

which can be solved to give ∇ as a function of T (as a proxy for thedepth of the envelope).

At the photosphere (the outer boundary of the solution), we caneasily show ∇p = 1/8, using Pp = 2gs/3κp (assuming L << Ledd ),gs = GM/R2, and L = 4πR2σT 4

eff , as well as σ = ac/4.Using the interpolation form of the opacity for ideal gas,κ = κgPnT−n−s, we have

∇ =1

(K ′)n+1(1 + neff )Pn+1T−(n+s+4), (3)

in which

K ′ =

(1

1 + neff

16πacGM3κgL

)1/(n+1)

,

where neff = (s + 3)/(n + 1) is the “effective polytropic index”. At thephotosphere, we have

∇p =1

(K ′)n+1(1 + neff )Pn+1

p T−(n+s+4)p . (4)

In such a radiative heat transfer region, ∇ = ∇rad with

∇ =d lnTd lnP

=3κL

16πacGMPT 4 ,

which can be solved to give ∇ as a function of T (as a proxy for thedepth of the envelope).At the photosphere (the outer boundary of the solution), we caneasily show ∇p = 1/8, using Pp = 2gs/3κp (assuming L << Ledd ),gs = GM/R2, and L = 4πR2σT 4

eff , as well as σ = ac/4.Using the interpolation form of the opacity for ideal gas,κ = κgPnT−n−s, we have

∇ =1

(K ′)n+1(1 + neff )Pn+1T−(n+s+4), (3)

in which

K ′ =

(1

1 + neff

16πacGM3κgL

)1/(n+1)

,

where neff = (s + 3)/(n + 1) is the “effective polytropic index”. At thephotosphere, we have

∇p =1

(K ′)n+1(1 + neff )Pn+1

p T−(n+s+4)p . (4)

Eq. 3 can be re-written as

PndP = (1 + neff )(K ′)n+1T n+s+3dT .

If n + s + 4 is non-zero, this can be integrated to give

Pn+1 − Pn+1p = (K ′)n+1(T n+s+4 − T n+s+4

p ). (5)

Replacing P and Pp with ∇ and ∇p from Eqs. 3 and 4, respectively,the above equation can be expressed as

∇ =1

1 + neff+

(Tp

T

)n+s+4(∇p −

11 + neff

), (6)

which determines at which temperature, convection might occur.

Radiative envelope structureWe first consider a case in which the convection does not occur rightat the photosphere. Assuming that n + 1 and n + s + 4 are bothpositive [e.g., for Kramer’s opacity (n = 1 and s = 3.5) and electronscattering (n = s = 0)] and as we go to deep depths in the envelope,then we have

P → K ′T 1+neff ; ∇ → 11 + neff

.

Thus as far as the interior structure is concerned, we could just aswell have used zero boundary conditions for the density andtemperature.

For monatomic ideal gas with constant composition,

P = Kρ1+1/neff , where K =

(NAkµ

)1+1/neff

(K ′)−1/neff .

We can then have a polytropic-like solution in the envelope. Withappropriate conditions of continuity, one can then connect theenvelope solution to the interior one.


P → K ′T 1+neff ; ∇ → 11 + neff

.




(NAkµ

)1+1/neff

(K ′)−1/neff .



P → K ′T 1+neff ; ∇ → 11 + neff

.




(NAkµ

)1+1/neff

(K ′)−1/neff .


Specifically, if Kramers’ opacity holds in the envelope, thenneff = 3.25 and ∇ = 0.2353 < ∇ad = 1− 1/Γ2 = 0.4, which impliesno convection as we have assumed. The same is true for the electronscattering opacity.

We may determine the temperature distribution as a function of theradius. Rewrite the equation of hydrostatic equilibrium in the form

dPdr

=P∇

1T

dTdr

= −GMr2 ρ.

Using P = ρNAkT/µ to replace the pressure, we have

(1 + neff )dTdr

= − GMµ

NAkr2 .

Assuming that the envelope is radiative with a constant neff , we canintegrate the above equation to get

Specifically, if Kramers’ opacity holds in the envelope, thenneff = 3.25 and ∇ = 0.2353 < ∇ad = 1− 1/Γ2 = 0.4, which impliesno convection as we have assumed. The same is true for the electronscattering opacity.

We may determine the temperature distribution as a function of theradius. Rewrite the equation of hydrostatic equilibrium in the form

dPdr

=P∇

1T

dTdr

= −GMr2 ρ.

Using P = ρNAkT/µ to replace the pressure, we have

(1 + neff )dTdr

= − GMµ

NAkr2 .

Assuming that the envelope is radiative with a constant neff , we canintegrate the above equation to get

T (r)− TP =1

1 + neff

GMµ

NAk

(1r− 1

R

)=

2.3× 107 K1 + neff

µ

(M

M�

)(R

R�

)−1(1x− 1)

where x = r/R.For a reasonable value of neff (e.g., = 3.25 for Kramers’ opacity) andµ (≈ 0.6 for a Pop I star), the term in front of

( 1x − 1

)in the above

equation is very large. Thus the last term has to be very small; i.e.,the temperature drops to a photosphere temperature over a radiusδr . 1% of R.

We can also get a density distribution from the above temperaturedistribution, using P = K ′T 1+neff for ideal gas.It can be shown that for a solar mass and luminosity, for example,traversing 15% of the total radius inward from the surface uses uponly a little less than 1% of the mass, confirming our assumption thatMr ≈ M through the envelope.

T (r)− TP =1

1 + neff

GMµ

NAk

(1r− 1

R

)=

2.3× 107 K1 + neff

µ

(M

M�

)(R

R�

)−1(1x− 1)

where x = r/R.For a reasonable value of neff (e.g., = 3.25 for Kramers’ opacity) andµ (≈ 0.6 for a Pop I star), the term in front of

( 1x − 1

)in the above

equation is very large. Thus the last term has to be very small; i.e.,the temperature drops to a photosphere temperature over a radiusδr . 1% of R.

We can also get a density distribution from the above temperaturedistribution, using P = K ′T 1+neff for ideal gas.It can be shown that for a solar mass and luminosity, for example,traversing 15% of the total radius inward from the surface uses uponly a little less than 1% of the mass, confirming our assumption thatMr ≈ M through the envelope.

Convective envelope

In a cool star, where H− opacity (n = 1/2, and s = −9) is important,the envelope can become convective.We first check where the convection may take place in a suchenvelope.

For H− opacity [hence neff = (s + 3)/(n + 1) = −4], Eq. 6

∇ =1

1 + neff+

(Tp

T

)n+s+4(∇p −

11 + neff

)(7)

(with ∇p = 1/8) can now be expressed as

∇(r) = −13

+1124

(T

Teff

)9/2

.

Since temperature increases steeply with depth, so does ∇.Eventually, when ∇ > ∇ad , the stellar material becomes convective.Where does the convective envelope start for ideal gas?

Convective envelope

In a cool star, where H− opacity (n = 1/2, and s = −9) is important,the envelope can become convective.We first check where the convection may take place in a suchenvelope.

For H− opacity [hence neff = (s + 3)/(n + 1) = −4], Eq. 6

∇ =1

1 + neff+

(Tp

T

)n+s+4(∇p −

11 + neff

)(7)

(with ∇p = 1/8) can now be expressed as

∇(r) = −13

+1124

(T

Teff

)9/2

.

Since temperature increases steeply with depth, so does ∇.Eventually, when ∇ > ∇ad , the stellar material becomes convective.Where does the convective envelope start for ideal gas?

Where does the convective envelope start?

At depths deeper than the criticaldepth, ∇ = ∇ad = 0.4 for ideal gas(Γ2 = 5/3).The transition to convection occursalmost immediately (very close tothe photosphere) at

Tf = (8/5)2/9Teff = 1.11Teff

.

The corresponding pressurePf = 22/3Pp can be found from

The curves are calculated for a modelZAMS sun.(

PPp

)n+1

= 1 +1

1 + neff

1∇p

[(TTp

)n+s+4

− 1

],

which can be easily obtained from Eqs. 4 and 5 (while Tp = Teff ).

Convective starsWe consider the dependence of Teff on such global stellar propertiesas the composition (which affects the mean molecular weight) andopacity (assumed to be uniform), as well as the mass and luminosityof a fully convective star (except for the outermost thin layer justbelow the photosphere).Assuming the convection is effective (adiabatic), the implyingpolytrope of index is 3/2 and

P = K ′n=3/2T 5/2,

where K ′n=3/2 = (NAk/µ)5/2K−3/2.Recall that K is related to the the mass and radius of the star asdefined by the E-solution polytrope (Eq. 1):

K =

[4π

ξn+1(−θ′n)n−1

]1/n

ξ1

Gn + 1

M1−1/nR−1+3/n,

we haveK ′n=3/2 ∝

1µ5/2M1/2R3/2 .

Using Pf = K ′n=3/2T 5/2f , Pp = 2gs/3κp, and gs = GM/R2, as well as

L = 4πσR2T 4eff to replace R in the above relations, we get

Teff ∝ µ13/51κ−4/510 (M/M�)7/51(L/L�)1/102

Therefore, such a completely convective star has a nearly constanteffective temperature, independent of the luminosity, correspondinglyto a nearly vertical evolutionary track of a star in the H-R diagram.

In reality, ionization processes and convection, albeit almostnegligible, occur in the outer layers of nearly all stars and a completeand accurate integration including all effects is necessary in modelingreal stars.

Using Pf = K ′n=3/2T 5/2f , Pp = 2gs/3κp, and gs = GM/R2, as well as

L = 4πσR2T 4eff to replace R in the above relations, we get

Teff ∝ µ13/51κ−4/510 (M/M�)7/51(L/L�)1/102

Therefore, such a completely convective star has a nearly constanteffective temperature, independent of the luminosity, correspondinglyto a nearly vertical evolutionary track of a star in the H-R diagram.

In reality, ionization processes and convection, albeit almostnegligible, occur in the outer layers of nearly all stars and a completeand accurate integration including all effects is necessary in modelingreal stars.

Review

Key concepts: Polytrope, Lane-Emden equation, the EddingtonStandard Model, Newton-Raphson and Henyey Methods

1. What are the basic equations and the boundary conditions thatare needed to construct a normal stellar interior model? Whatmicroscopic physics should be implemented in such a modeling?

2. Why is the polytropic stellar model only a second-orderdifferential equation?

3. Please give two examples of the situations in which a polytropicequation of state may be used?

4. What is the key assumption made in the Eddington StandardModel? Why may this assumption be reasonable?

5. What is the basic approach of the Newton-Raphson or HenyeyMethod in solving the stellar equations? How is it different from asimply “shooting method”?

6. How do different opacity laws affect the structure of the envelopeof a star?

Chapter 5: Stellar Structure Modeling - UMassThe Equations of Stellar Structure We consider the modeling of stars in hydrostatic and thermal equilibrium. We hope to know how such global

Documents