Appendix/Ramblings/51BiPolytropeStability/BetterInterfacePt2

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Better Interface for 51BiPolytrope Stability Study

This is Part 2 of an extended chapter discussion. For Part 1, go here.

Discretize for Numerical Integration (continued)

General Discretization

Fourth Approximation

Let's assume that we know the four quantities, xJ−1,xJ,(xJ)'≡(dx/dr~)J, and (xJ−1)'≡(dx/dr~)J−1 and want to project forward to determine, xJ+1. We should assume that, locally, the displacement function x is cubic in r~, that is,

x =

a+br~+cr~2+er~3

⇒dxdr~ =

b+2cr~+3er~2,

where we have four unknowns, a,b,c,e. These can be determined by appropriately combining the four relations,

(xJ)' =

b+2cr~J+3er~J2,

(xJ−1)' =

b+2c(r~J−Δr~)+3e(r~J−Δr~)2,

xJ =

a+br~J+cr~J2+er~J3,

xJ−1 =

a+b(r~J−Δr~)+c(r~J−Δr~)2+e(r~J−Δr~)3,

The difference between the first two expressions gives,

(xJ)'−(xJ−1)' =

[2cr~J+3er~J2]−[2c(r~J−Δr~)+3e(r~J−Δr~)2]

  =

2cr~J+3er~J2−[2cr~J−2cΔr~+3e(r~J2−2r~JΔr~+Δr~2)]

  =

2cΔr~+6er~JΔr~−3eΔr~2

⇒2cΔr~ =

[(xJ)'−(xJ−1)']+3eΔr~2−6er~JΔr~

⇒c =

[(xJ)'−(xJ−1)'2Δr~]+3e[Δr~2−r~J].

And the difference between the last two expressions gives,

xJ−xJ−1 =

[br~J+cr~J2+er~J3]−[b(r~J−Δr~)+c(r~J−Δr~)2+e(r~J−Δr~)3]

  =

bΔr~+c(2r~JΔr~−Δr~2)+er~J3−e(r~J−Δr~)(r~J2−2r~JΔr~+Δr~2)

  =

bΔr~+c(2r~JΔr~−Δr~2)+er~J3−e[(r~J)(r~J2−2r~JΔr~+Δr~2)−(Δr~)(r~J2−2r~JΔr~+Δr~2)]

  =

bΔr~+c(2r~JΔr~−Δr~2)−e[−3r~J2Δr~+3r~JΔr~2−Δr~3]

  =

bΔr~+c(2r~JΔr~−Δr~2)+e[3r~J2Δr~−3r~JΔr~2+Δr~3]

⇒bΔr~ =

[xJ−xJ−1]+2cΔr~[Δr~2−r~J]−e[3r~J2Δr~−3r~JΔr~2+Δr~3]

  =

[xJ−xJ−1]+{[(xJ)'−(xJ−1)']+3eΔr~[Δr~−2r~J]}[Δr~2−r~J]−3eΔr~[r~J2−r~JΔr~+Δr~23]

  =

[xJ−xJ−1]+[(xJ)'−(xJ−1)'][Δr~2−r~J]+3eΔr~[Δr~22−r~JΔr~]−3eΔr~[r~JΔr~−2r~J2]−3eΔr~[r~J2−r~JΔr~+Δr~23]

  =

[xJ−xJ−1]+[(xJ)'−(xJ−1)'][Δr~2−r~J]−3eΔr~{[r~JΔr~−Δr~22]+[r~JΔr~−2r~J2]+[r~J2−r~JΔr~+Δr~23]}

  =

[xJ−xJ−1]+[(xJ)'−(xJ−1)'][Δr~2−r~J]+eΔr~[3r~J2−3r~JΔr~+Δr~22].

Summary #1:

In terms of the coefficient, e …

bΔr~ =

[xJ−xJ−1]+[(xJ)'−(xJ−1)'][Δr~2−r~J]+eΔr~[3r~J2−3r~JΔr~+Δr~22],

2cΔr~ =

[(xJ)'−(xJ−1)']+eΔr~[3Δr~−6r~J].

Hence, from the first of the four relations, we find that,

(xJ)'Δr~−3er~J2Δr~ =

(bΔr~)+(2cΔr~)r~J

  =

[xJ−xJ−1]+[(xJ)'−(xJ−1)'][Δr~2−r~J]+eΔr~[3r~J2−3r~JΔr~+Δr~22]+{[(xJ)'−(xJ−1)']+eΔr~[3Δr~−6r~J]}r~J

  =

[xJ−xJ−1]+[(xJ)'−(xJ−1)'][Δr~2]+eΔr~[3r~J2−3r~JΔr~+Δr~22]+eΔr~[3r~JΔr~−6r~J2]

  =

[xJ−xJ−1]+[(xJ)'−(xJ−1)'][Δr~2]+eΔr~[−3r~J2+Δr~22]

⇒(xJ)'Δr~ =

[xJ−xJ−1]+[(xJ)'−(xJ−1)'][Δr~2]+e[Δr~32]

⇒e[Δr~32] =

−[xJ−xJ−1]−[(xJ)'−(xJ−1)'][Δr~2]+(xJ)'Δr~

  =

[xJ−1−xJ]+[(xJ−1)'+(xJ)']Δr~2

⇒eΔr~3 =

2[xJ−1−xJ]+[(xJ−1)'+(xJ)']Δr~.

Finally, from the third of the four relations, we can evaluate the coefficient, a; specifically,

xJ−a−er~J3 =

br~J+cr~J2

  =

r~JΔr~{bΔr~}+r~J22Δr~{2cΔr~}

  =

r~JΔr~{[xJ−xJ−1]+[(xJ)'−(xJ−1)'][Δr~2−r~J]+eΔr~[3r~J2−3r~JΔr~+Δr~22]}+r~J22Δr~{[(xJ)'−(xJ−1)']+eΔr~[3Δr~−6r~J]}

  =

{r~JΔr~[xJ−xJ−1]+r~JΔr~[(xJ)'−(xJ−1)'][Δr~2−r~J]+er~J[3r~J2−3r~JΔr~+Δr~22]}+{r~J22Δr~[(xJ)'−(xJ−1)']+er~J22[3Δr~−6r~J]}

  =

r~JΔr~[xJ−xJ−1]+[(xJ)'−(xJ−1)'][r~J2−r~J22Δr~]+e[3r~J3−3r~J2Δr~+r~JΔr~22]+e[3r~J2Δr~2−3r~J3]

  =

r~JΔr~[xJ−xJ−1]+12Δr~[(xJ)'−(xJ−1)'][r~JΔr~−r~J2]+eΔr~2[r~JΔr~−3r~J2].

That is,

a =

xJ−r~JΔr~[xJ−xJ−1]−12Δr~[(xJ)'−(xJ−1)'][r~JΔr~−r~J2]−e{Δr~2[r~JΔr~−3r~J2]+r~J3}

  =

xJ−r~JΔr~[xJ−xJ−1]−12Δr~[(xJ)'−(xJ−1)'][r~JΔr~−r~J2]−eΔr~3{12Δr~2[r~JΔr~−3r~J2]+(r~JΔr~)3}

Summary #2:

In terms of the coefficient, e …

a =

xJ−r~JΔr~[xJ−xJ−1]−12Δr~[(xJ)'−(xJ−1)'][r~JΔr~−r~J2]−eΔr~3{12Δr~2[r~JΔr~−3r~J2]+(r~JΔr~)3},

bΔr~ =

[xJ−xJ−1]+[(xJ)'−(xJ−1)'][Δr~2−r~J]+eΔr~3[3r~J2Δr~2−3r~JΔr~+12],

cΔr~2 =

[(xJ)'−(xJ−1)']Δr~2+eΔr~3[32−3r~JΔr~],

eΔr~3 =

2[xJ−1−xJ]+[(xJ−1)'+(xJ)']Δr~.

This is test ...

r~J=r~i+Δr~ Δr~ xJ xJ−1 (xJ)' (xJ−1)'
0.01740039 0.001936393 -4.695376 -4.547832 -116.0119 -76.19513
a =

−3.36955−2.76645−eΔr~3(608.9698)=−232.7874,

bΔr~ =

0.5067329+eΔr~3(215.7856)=+80.819698,

cΔr~2 =

−0.0385505+eΔr~3(−25.45794)=−9.51370,

eΔr~3 =

0.3721883.

Hence,

xJ =

a+bΔr~(r~JΔr~)+cΔr~2(r~JΔr~)2+eΔr~3(r~JΔr~)3

  =

−232.7874+726.2442−768.2108+270.0593=−4.68369.

Higher precision value (from Excel) is xJ=−4.695376, which precisely matches the input value. Also from Excel, xJ−1=−4.547832 and xJ+1=−3.803455.

As a result,

xJ+1 =

{a}+(r~J+Δr~){b}+(r~J+Δr~)2{c}+(r~J+Δr~)3{e}

  =

{a}+(r~JΔr~+1){bΔr~}+(r~JΔr~+1)2{cΔr~2}+(r~JΔr~+1)3{eΔr~3}

  =

xJ−r~JΔr~[xJ−xJ−1]−12Δr~[(xJ)'−(xJ−1)'][r~JΔr~−r~J2]−eΔr~3{12Δr~2[r~JΔr~−3r~J2]+(r~JΔr~)3}

   

+(r~JΔr~+1){[xJ−xJ−1]+[(xJ)'−(xJ−1)'][Δr~2−r~J]+eΔr~3[3r~J2Δr~2−3r~JΔr~+12]}

   

+[(r~JΔr~)2+2r~JΔr~+1]{[(xJ)'−(xJ−1)']Δr~2+eΔr~3[32−3r~JΔr~]}

   

+(r~JΔr~+1)[(r~JΔr~)2+2r~JΔr~+1]{eΔr~3}

⇒xJ+1 =

xJ−r~JΔr~[xJ−xJ−1]−12Δr~[(xJ)'−(xJ−1)'][r~JΔr~−r~J2]+[xJ−xJ−1](r~JΔr~+1)+[(xJ)'−(xJ−1)'][Δr~2−r~J](r~JΔr~+1)+[(r~JΔr~)2+2r~JΔr~+1][(xJ)'−(xJ−1)']Δr~2

   

+eΔr~3[3r~J2Δr~2−3r~JΔr~+12](r~JΔr~+1)+eΔr~3[32−3r~JΔr~][(r~JΔr~)2+2r~JΔr~+1]+eΔr~3(r~JΔr~+1)[(r~JΔr~)2+2r~JΔr~+1]−eΔr~3{12Δr~2[r~JΔr~−3r~J2]+(r~JΔr~)3}

  =

2xJ−xJ−1−12[(xJ)'−(xJ−1)'][r~J−r~J2Δr~]+12[(xJ)'−(xJ−1)'][Δr~−2r~J](r~JΔr~+1)+[r~J2Δr~+2r~J+Δr~][(xJ)'−(xJ−1)']12

   

+eΔr~3{[3r~J2Δr~2−3r~JΔr~+12](r~JΔr~+1)+[32−3r~JΔr~][(r~JΔr~)2+2r~JΔr~+1]+(r~JΔr~+1)[(r~JΔr~)2+2r~JΔr~+1]−[12Δr~2(r~JΔr~−3r~J2)+(r~JΔr~)3]}

  =

2xJ−xJ−1+12[(xJ)'−(xJ−1)']{[Δr~−2r~J](r~JΔr~+1)+[r~J2Δr~+2r~J+Δr~]−[r~J−r~J2Δr~]}

   

+eΔr~3{[3(r~JΔr~)3−3(r~JΔr~)2+12(r~JΔr~)]+[3(r~JΔr~)2−3(r~JΔr~)+12]+32[(r~JΔr~)2+2r~JΔr~+1]−3[(r~JΔr~)3+2(r~JΔr~)2+(r~JΔr~)]

   

+[(r~JΔr~)3+2(r~JΔr~)2+(r~JΔr~)]+[(r~JΔr~)2+2r~JΔr~+1]+[−r~J2Δr~+32(r~JΔr~)2−(r~JΔr~)3]}

Continuing …

xJ+1 =

2xJ−xJ−1+[(xJ)'−(xJ−1)']Δr~

   

+eΔr~3{[r~J2Δr~]+[−3(r~JΔr~)+12]+32[(r~JΔr~)2+2r~JΔr~+1]−[6(r~JΔr~)2+3(r~JΔr~)]

   

+3(r~JΔr~)2+[2r~JΔr~+1]+[r~J2Δr~+32(r~JΔr~)2]}

  =

2xJ−xJ−1+[(xJ)'−(xJ−1)']Δr~

   

+eΔr~3{r~JΔr~−6(r~JΔr~)+3+3r~JΔr~−6(r~JΔr~)2+3(r~JΔr~)2+2r~JΔr~+3(r~JΔr~)2}

  =

2xJ−xJ−1+[(xJ)'−(xJ−1)']Δr~+3eΔr~3

Finally we may write,

xJ+1 =

2xJ−xJ−1+[(xJ)'−(xJ−1)']Δr~+3{2[xJ−1−xJ]+[(xJ−1)'+(xJ)']Δr~}

  =

2xJ−xJ−1+6[xJ−1−xJ]+[(xJ)'−(xJ−1)']Δr~+3[(xJ−1)'+(xJ)']Δr~

  =

[5xJ−1−4xJ]+[4(xJ)'+2(xJ−1)']Δr~.

This is test ...

Δr~ xJ xJ−1 (xJ)' (xJ−1)'
0.001936393 -4.695376 -4.547832 -116.0119 -76.19513
xJ+1 =

[5xJ−1−4xJ]+[4(xJ)'+2(xJ−1)']Δr~=−5.15132.

Fifth Approximation

Let's assume that we know the three quantities, xJ−1,xJ,(xJ)'≡(dx/dr~)J, and want to project forward to determine, xJ+1. Here we will assume that, locally, the displacement function x has only an even-power dependence on r~, that is,

x =

a+br~2+cr~4

⇒dxdr~ =

2br~+4cr~3,

where we have three unknowns, a,b,c. These can be determined by appropriately combining the three relations,

(xJ)' =

2br~J+4cr~J3,

xJ =

a+br~J2+cr~J4,

xJ−1 =

a+b(r~J−Δr~)2+c(r~J−Δr~)4,

The difference between the last two expressions gives,

xJ−xJ−1 =

[br~J2+cr~J4]−[b(r~J−Δr~)2+c(r~J−Δr~)4]

  =

br~J2+cr~J4−b(r~J2−2r~JΔr~+Δr~2)−c(r~J2−2r~JΔr~+Δr~2)(r~J2−2r~JΔr~+Δr~2)

  =

cr~J4+2br~JΔr~−bΔr~2−c[r~J2(r~J2−2r~JΔr~+Δr~2)−2r~JΔr~(r~J2−2r~JΔr~+Δr~2)+Δr~2(r~J2−2r~JΔr~+Δr~2)]

  =

cr~J4+2br~JΔr~−bΔr~2−c[r~J4−2r~J3Δr~+r~2Δr~2−2r~J3Δr~+4r~J2Δr~2−2r~JΔr~3+r~J2Δr~2−2r~JΔr~3+Δr~4]

  =

b[2r~JΔr~−Δr~2]+c[4r~J3Δr~−6r~J2Δr~2+4r~JΔr~3−Δr~4]

  =

2br~JΔr~[1−Δr~2r~J]+c[4r~J3Δr~−6r~J2Δr~2+4r~JΔr~3−Δr~4].

From the first expression, we also see that,

2br~JΔr~

=

(xJ)'Δr~−4cr~J3Δr~.

Therefore we have,

xJ−xJ−1 =

[(xJ)'Δr~−4cr~J3Δr~][1−Δr~2r~J]+c[4r~J3Δr~−6r~J2Δr~2+4r~JΔr~3−Δr~4]

  =

[(xJ)'Δr~][1−Δr~2r~J]+[2cr~J2Δr~][Δr~−2r~J]+c[4r~J3Δr~−6r~J2Δr~2+4r~JΔr~3−Δr~4]

  =

[(xJ)'Δr~][1−Δr~2r~J]+c[2r~J2Δr~2−4r~J3Δr~]+c[4r~J3Δr~−6r~J2Δr~2+4r~JΔr~3−Δr~4]

  =

[(xJ)'Δr~][1−Δr~2r~J]+c[−4r~J2Δr~2+4r~JΔr~3−Δr~4]

⇒cΔr~2[4r~J2−4r~JΔr~+Δr~2]

=

[(xJ)'Δr~][1−Δr~2r~J]−xJ+xJ−1

⇒4cr~J2Δr~2[1−Δr~r~J+14(Δr~r~J)2]

=

[(xJ)'Δr~][1−Δr~2r~J]−xJ+xJ−1

⇒4cr~J4[(Δr~r~J)2−(Δr~r~J)3+14(Δr~r~J)4]

=

[(xJ)'Δr~][1−Δr~2r~J]−xJ+xJ−1.

Hence also,

2br~JΔr~(Δr~r~J)

=

(xJ)'Δr~(Δr~r~J)−{4cr~J2Δr~2}

⇒2br~JΔr~(Δr~r~J)[1−Δr~r~J+14(Δr~r~J)2]

=

(xJ)'Δr~(Δr~r~J)[1−Δr~r~J+14(Δr~r~J)2]−{[(xJ)'Δr~][1−Δr~2r~J]−xJ+xJ−1}

 

=

xJ−xJ−1+(xJ)'(Δr~2r~J)[1−Δr~r~J+14(Δr~r~J)2]−[(xJ)'Δr~][1−Δr~2r~J]

 

=

xJ−xJ−1+(xJ)'(Δr~2r~J)[1−Δr~r~J+14(Δr~r~J)2]−(xJ)'Δr~+12(xJ)'(Δr~2r~J)

⇒2br~J2[(Δr~r~J)2−(Δr~r~J)3+14(Δr~r~J)4]

=

xJ−xJ−1+(xJ)'Δr~[−1+32(Δr~r~J)−(Δr~r~J)2+14(Δr~r~J)3]

Finally,

a =

xJ−br~J2−cr~J4

⇒a[(Δr~r~J)2−(Δr~r~J)3+14(Δr~r~J)4] =

xJ[(Δr~r~J)2−(Δr~r~J)3+14(Δr~r~J)4]−br~J2[(Δr~r~J)2−(Δr~r~J)3+14(Δr~r~J)4]−cr~J4[(Δr~r~J)2−(Δr~r~J)3+14(Δr~r~J)4]

  =

xJ[(Δr~r~J)2−(Δr~r~J)3+14(Δr~r~J)4]

   

−12{xJ−xJ−1+(xJ)'Δr~[−1+32(Δr~r~J)−(Δr~r~J)2+14(Δr~r~J)3]}

   

−14{[(xJ)'Δr~][1−Δr~2r~J]−xJ+xJ−1}

  =

xJ[(Δr~r~J)2−(Δr~r~J)3+14(Δr~r~J)4]−12{xJ−xJ−1}−14{−xJ+xJ−1}

   

−12{(xJ)'Δr~[−1+32(Δr~r~J)−(Δr~r~J)2+14(Δr~r~J)3]}−14{(xJ)'Δr~[1−12(Δr~r~J)]}

  =

xJ[(Δr~r~J)2−(Δr~r~J)3+14(Δr~r~J)4]+xJ−1−xJ4

   

+{(xJ)'Δr~[12−14+18(Δr~r~J)−34(Δr~r~J)+12(Δr~r~J)2−18(Δr~r~J)3]}

  =

xJ−1−xJ4+xJ[(Δr~r~J)2−(Δr~r~J)3+14(Δr~r~J)4]+(xJ)'Δr~[14−58(Δr~r~J)+12(Δr~r~J)2−18(Δr~r~J)3].


Summary:
a⋅𝒜 =

xJ−1−xJ4+xJ[(Δr~r~J)2−(Δr~r~J)3+14(Δr~r~J)4]+(xJ)'Δr~[14−58(Δr~r~J)+12(Δr~r~J)2−18(Δr~r~J)3],

2br~J2⋅𝒜

=

xJ−xJ−1+(xJ)'Δr~[−1+32(Δr~r~J)−(Δr~r~J)2+14(Δr~r~J)3],

⇒4cr~J4⋅𝒜

=

(xJ)'Δr~[1−Δr~2r~J]−xJ+xJ−1,

where,

𝒜 ≡

[(Δr~r~J)2−(Δr~r~J)3+14(Δr~r~J)4].

Let's now determine the expression for xJ+1. We begin by writing …

xJ+1 =

a+b(r~J+Δr~)2+c(r~J+Δr~)4

  =

a+b(r~J2+2r~JΔr~+Δr~2)+c[r~J2(r~J2+2r~JΔr~+Δr~2)+2r~JΔr~(r~J2+2r~JΔr~+Δr~2)+Δr~2(r~J2+2r~JΔr~+Δr~2)]

  =

a+br~J2[1+2Δr~r~J+(Δr~r~J)2]+c[(r~J4+2r~J3Δr~+r~J2Δr~2)+(2r~J3Δr~+4r~J2Δr~2+2r~JΔr~3)+(r~J2Δr~2+2r~JΔr~3+Δr~4)]

  =

a+br~J2[1+2Δr~r~J+(Δr~r~J)2]+c[r~J4+4r~J3Δr~+6r~J2Δr~2+4r~JΔr~3+Δr~4]

  =

a+br~J2[1+2(Δr~r~J)+(Δr~r~J)2]+cr~J4[1+4(Δr~r~J)+6(Δr~r~)2+4(Δr~r~J)3+(Δr~r~J)4]

  =

a+2br~J2[12+(Δr~r~J)+12(Δr~r~J)2]+4cr~J4[14+(Δr~r~J)+32(Δr~r~)2+(Δr~r~J)3+14(Δr~r~J)4].

This means that,

𝒜⋅xJ+1 =

a⋅𝒜

   

+2br~J2⋅𝒜[12+(Δr~r~J)+12(Δr~r~J)2]

   

+4cr~J4⋅𝒜[14+(Δr~r~J)+32(Δr~r~)2+(Δr~r~J)3+14(Δr~r~J)4]

  =

xJ−1−xJ4+xJ[(Δr~r~J)2−(Δr~r~J)3+14(Δr~r~J)4]+(xJ)'Δr~[14−58(Δr~r~J)+12(Δr~r~J)2−18(Δr~r~J)3]

   

+{xJ−xJ−1+(xJ)'Δr~[−1+32(Δr~r~J)−(Δr~r~J)2+14(Δr~r~J)3]}[12+(Δr~r~J)+12(Δr~r~J)2]

   

+{(xJ)'Δr~[1−Δr~2r~J]−xJ+xJ−1}[14+(Δr~r~J)+32(Δr~r~)2+(Δr~r~J)3+14(Δr~r~J)4]

  =

14xJ−1−14xJ+xJ[(Δr~r~J)2−(Δr~r~J)3+14(Δr~r~J)4]+(xJ)'Δr~[14−58(Δr~r~J)+12(Δr~r~J)2−18(Δr~r~J)3]

   

+{xJ−xJ−1}[12+(Δr~r~J)+12(Δr~r~J)2]

   

+(xJ)'Δr~[−1+32(Δr~r~J)−(Δr~r~J)2+14(Δr~r~J)3][12+(Δr~r~J)+12(Δr~r~J)2]

   

+{−xJ+xJ−1}[14+(Δr~r~J)+32(Δr~r~)2+(Δr~r~J)3+14(Δr~r~J)4]

   

+(xJ)'Δr~[1−Δr~2r~J][14+(Δr~r~J)+32(Δr~r~)2+(Δr~r~J)3+14(Δr~r~J)4]

Keep going …

𝒜⋅xJ+1 =

xJ[(Δr~r~J)2−(Δr~r~J)3+14(Δr~r~J)4]

   

+{xJ−xJ−1}[−14+12+(Δr~r~J)+12(Δr~r~J)2−14−(Δr~r~J)−32(Δr~r~)2−(Δr~r~J)3−14(Δr~r~J)4]

   

+(xJ)'Δr~[14−58(Δr~r~J)+12(Δr~r~J)2−18(Δr~r~J)3]

   

+(xJ)'Δr~[−1+32(Δr~r~J)−(Δr~r~J)2+14(Δr~r~J)3][12(Δr~r~J)2]

   

+(xJ)'Δr~[−1+32(Δr~r~J)−(Δr~r~J)2+14(Δr~r~J)3][(Δr~r~J)]

   

+(xJ)'Δr~[−1+32(Δr~r~J)−(Δr~r~J)2+14(Δr~r~J)3][12]

   

+(xJ)'Δr~[14+(Δr~r~J)+32(Δr~r~)2+(Δr~r~J)3+14(Δr~r~J)4]

   

+(xJ)'Δr~[−Δr~2r~J][14+(Δr~r~J)+32(Δr~r~)2+(Δr~r~J)3+14(Δr~r~J)4]

midpoint
  =

xJ[(Δr~r~J)2−(Δr~r~J)3+14(Δr~r~J)4]+{xJ−xJ−1}[−(Δr~r~)2−(Δr~r~J)3−14(Δr~r~J)4]

   

+(xJ)'Δr~[14−58(Δr~r~J)+12(Δr~r~J)2−18(Δr~r~J)3]

   

+(xJ)'Δr~[−12(Δr~r~J)2+34(Δr~r~J)3−12(Δr~r~J)4+18(Δr~r~J)5]

   

+(xJ)'Δr~[−(Δr~r~J)+32(Δr~r~J)2−(Δr~r~J)3+14(Δr~r~J)4]

   

+(xJ)'Δr~[−12+34(Δr~r~J)−12(Δr~r~J)2+18(Δr~r~J)3]

   

+(xJ)'Δr~[14+(Δr~r~J)+32(Δr~r~)2+(Δr~r~J)3+14(Δr~r~J)4]

   

+(xJ)'Δr~[−18(Δr~r~J)−12(Δr~r~J)2−34(Δr~r~)3−12(Δr~r~J)4−18(Δr~r~J)5]

---- next in line ----
  =

xJ[−2(Δr~r~J)3]+xJ−1[(Δr~r~)2+(Δr~r~J)3+14(Δr~r~J)4]

   

+(xJ)'Δr~[−18(Δr~r~J)3]

   

+(xJ)'Δr~[34(Δr~r~J)3−12(Δr~r~J)4+18(Δr~r~J)5]

   

+(xJ)'Δr~[32(Δr~r~J)2−(Δr~r~J)3+14(Δr~r~J)4]

   

+(xJ)'Δr~[−12(Δr~r~J)2+18(Δr~r~J)3]

   

+(xJ)'Δr~[32(Δr~r~)2+(Δr~r~J)3+14(Δr~r~J)4]

   

+(xJ)'Δr~[−12(Δr~r~J)2−34(Δr~r~)3−12(Δr~r~J)4−18(Δr~r~J)5]

Interface

CORE:   When J=(i−1) (where i means interface), we can obtain the fractional displacements at the interface, xi and pi, via the expressions,

xi =

xi−2−2Δr~{1r~[3x+pγg]}i−1,

      and       pi =

pi−2+2Δr~{ρ~P~⋅M~rr~2[(4x+p)+σc2(2π3⋅ρ~cr~3M~r)x]}i−1.

Then, setting J=i, the pair of radial derivatives at the interface and as viewed from the perspective of the core is given by the expressions,

(dxdr~)i|core =

−1r~i[3xi+pi6/5],

      and       (dpdr~)i|core =

(ρ~i)coreP~i⋅M~corer~i2[(4xi+pi)+σc2(2π3⋅ρ~cr~i3M~core)xi].

It is important to recognize that, throughout the core, (dx/dr~) has been evaluated by setting γg=6/5. If we continue to use this value of γg at the interface, we are determining the slope as viewed from the perspective of the core.


ENVELOPE:   On the other hand, as viewed from the perspective of the envelope, all parameters used to determine (dx/dr~)i at the interface (and throughout the entire envelope) are the same except γg, which equals 2 instead of 6/5. Specifically at the interface, we have,

(dxdr~)i|env =

−1r~i[3xi+pi2],

      and       (dpdr~)i|env =

(ρ~i)envP~i⋅M~corer~i2[(4xi+pi)+σc2(2π3⋅ρ~cr~i3M~core)xi].

(See, for example, our related discussion.) Hence, we appreciate that there is a discontinuous change in the value of this slope at the interface. We note as well — for the first time (8/17/2023)! — that there must also be a discontinuous jump in the slope of the "pressure perturbation." All of the variables used to evaluate (dp/dr~)i are the same irrespective of your core/envelope point of view except the leading density term. As viewed from the perspective of the core, (ρ~i)|core=msurf5(μe/μc)−10θi5 whereas, from the perspective of the envelope, (ρ~i)|env=msurf5(μe/μc)−9θi5ϕi. Appreciating that ϕi=1, this means that the slope of the "pressure perturbation" is a factor of μe/μc smaller as viewed from the perspective of the envelope.

Then the value of the fractional radial displacement and the value of the pressure perturbation at the first zone outside of the interface are obtained by setting J=i. That is,

xi+1 =

xi−1−2Δr~{1r~[3x+p2]}i,

      and       pi+1 =

pi−1+2Δr~{(ρ~i)|envP~⋅M~rr~2[(4x+p)+σc2(2π3⋅ρ~cr~3M~r)x]}i.

But, as written, these two expressions are unacceptable because the values just inside the interface, xi−1 and pi−1, are not known as viewed from the perspective of the envelope. However, we can fix this by drawing from the "average" expressions as replacements, namely,

xi =

12(xi−1+xi+1)⇒xi−1=(2xi−xi+1),

      and       pi =

12(pi−1+pi+1)⇒pi−1=(2pi−pi+1),

in which case we have,

2xi+1 =

2xi−2Δr~{1r~[3x+p2]}i,

      and       2pi+1 =

2pi+2Δr~{(ρ~i)|envP~⋅M~rr~2[(4x+p)+σc2(2π3⋅ρ~cr~3M~r)x]}i.


See Also

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