Appendix/Ramblings/51BiPolytropeStability/RethinkEnvelope

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Rethink Handling of n = 1 Envelope

Solution Steps

Drawing from an accompanying discussion

  • Step 1: Choose nc and ne.
  • Step 2: Adopt boundary conditions at the center of the core (θ=1 and dθ/dξ=0 at ξ=0), then solve the Lane-Emden equation to obtain the solution, θ(ξ), and its first derivative, dθ/dξ throughout the core; the radial location, ξ=ξs, at which θ(ξ) first goes to zero identifies the natural surface of an isolated polytrope that has a polytropic index nc.
  • Step 3 Choose the desired location, 0<ξi<ξs, of the outer edge of the core.
  • Step 4: Specify Kc and ρ0; the structural profile of, for example, ρ(r), P(r), and Mr(r) is then obtained throughout the core — over the radial range, 0ξξi and 0rri — via the relations shown in the 2nd column of Table 1.
  • Step 5: Specify the ratio μe/μc and adopt the boundary condition, ϕi=1; then use the interface conditions as rearranged and presented in Table 3 to determine, respectively:
    • The gas density at the base of the envelope, ρe;
    • The polytropic constant of the envelope, Ke, relative to the polytropic constant of the core, Kc;
    • The ratio of the two dimensionless radial parameters at the interface, ηi/ξi;
    • The radial derivative of the envelope solution at the interface, (dϕ/dη)i.
  • Step 6: The last sub-step of solution step 5 provides the boundary condition that is needed — in addition to our earlier specification that ϕi=1 — to derive the desired particular solution, ϕ(η), of the Lane-Emden equation that is relevant throughout the envelope; knowing ϕ(η) also provides the relevant structural first derivative, dϕ/dη, throughout the envelope.
  • Step 7: The surface of the bipolytrope will be located at the radial location, η=ηs and r=R, at which ϕ(η) first drops to zero.
  • Step 8: The structural profile of, for example, ρ(r), P(r), and Mr(r) is then obtained throughout the envelope — over the radial range, ηiηηs and rirR — via the relations provided in the 3rd column of Table 1.

Setup

Drawing from the accompanying Table 1, we have …

Core

Envelope

nc=5

ne=1

1ξ2ddξ(ξ2dθdξ)=θ5

sol'n: θ(ξ)

1η2ddη(η2dϕdη)=ϕ

sol'n: ϕ(η)

Specify: Kc and ρ0

ρ

=

ρ0θ5

P

=

Kcρ06/5θ6

r

=

[3Kc2πG]1/2ρ02/5ξ

Mr

=

4π[3Kc2πG]3/2ρ01/5(ξ2dθdξ)

Knowing: Ke and ρe

ρ

=

ρeϕ

P

=

Keρe2ϕ2

r

=

[Ke2πG]1/2η

Mr

=

4π[Ke2πG]3/2ρe(η2dϕdη)


From an accompanying discussion of (nc,ne)=(5,1) bipolytropes, we know that the solution to the pair of Lane-Emden equations is …

θ(ξ)=[1+13ξ2]1/2θi=[1+13ξi2]1/2,

dθdξ=ξ3[1+13ξ2]3/2(dθdξ)i=ξi3[1+13ξi2]3/2;

and,

ϕ=A[sin(ηB)η],

dϕdη=Aη2[ηcos(ηB)sin(ηB)].

Adopting the same normalizations as before, we have,

Core

Envelope

ρ*ρρ0

=

θ5

P*PKcρ06/5

=

θ6

r*r[G1/2ρ02/5Kc1/2]

=

[32π]1/2ξ

Mr*Mr[G3/2ρ01/5Kc3/2]

=

4π[32π]3/2(ξ2dθdξ)

ρ*

=

(ρeρ0)ϕ

P*

=

[Keρe2Kcρ06/5]ϕ2

r*

=

ρ02/5[Ke2πKc]1/2η

Mr*

=

4π[ρeρ01/5][Ke2πKc]3/2(η2dϕdη)

Interface Conditions

Now, at the core-envelope interface …

  • ρ*|c=θi5
  • ρ*|e=(ρe/ρ0)ϕi
  • By choice, ϕi=1
  • ρ*|eμc=ρ*|cμe

Hence,

ρeρ0 = ρ*|e=ρ*|c(μeμc)=(μeμc)θi5.

Also, setting the value of P* equal across the boundary gives us,

θi6 =

[Keρe2Kcρ06/5]ϕi2=ρ04/5(KeKc)(μeμc)2θi10

ρ04/5(KeKc)

=

(μeμc)2θi4

As a result, throughout the envelope,

P* =

(ρeρ0)2(μeμc)2θi4ϕ2=θi6ϕ2;

r* =

(2π)1/2[ρ04/5KeKc]1/2η=(2π)1/2[(μeμc)1θi2]η;

Mr*

=

2(2π)1/2[ρeρ0](ρ04/5KeKc)3/2(η2dϕdη)=2(2π)1/2[(μeμc)θi5][(μeμc)2θi4]3/2(η2dϕdη)

 

=

(2π)1/2(μeμc)2θi1(η2dϕdη).

In summary, then,

Core

Envelope

ρ*

=

θ5

P*

=

θ6

r*

=

[32π]1/2ξ

Mr*

=

4π[32π]3/2(ξ2dθdξ)

 

=

[2433π223π3]1/2{ξ33[1+13ξ2]3/2}

 

=

(6π)1/2ξ3[1+13ξ2]3/2

ρ*

=

(μeμc)θi5ϕ

P*

=

θi6ϕ2

r*

=

(2π)1/2[(μeμc)1θi2]η

Mr*

=

(2π)1/2(μeμc)2θi1(η2dϕdη)

 

=

(2π)1/2(μeμc)2θi1{A[sin(ηB)ηcos(ηB)]}

This matches our earlier derivation.

Profiles of Physical Variables

Let's begin by choosing the value of ξi at which the core-envelope interface will occur. For example, setting ξi=31/2 means that r*=(2π)1/2 and that,

ηi

=

(μeμc)θi2=(μeμc)[1+ξi23]1=910(μeμc).

This means that we will be outside the core — and, hopefully, inside the envelope — for all values of η>ηi, which means for all values of r*>(2π)1/2.

See Also

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