SSC/Stability/n1PolytropeLAWE/Pt3: Difference between revisions

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Latest revision as of 14:58, 11 January 2026

Radial Oscillations of n = 1 Polytropic Spheres (Pt 3)[edit]


Part I:   Search for Analytic Solutions
 

Part II:  New Ideas
 

Part III:  What About Bipolytropes?
 

Part IV:  Most General Structural Solution
 


What About Bipolytropes?[edit]

Here we will try to find an analytic expression for the radial displacement function, x, for a bipolytropic envelope whose polytropic index is, ne=1. As in the above succinct derivation, the relevant LAWE is,

LAWE

=

d2xdξ2+[4(n+1)Q]1ξdxdξ+(n+1)[(σc26γg)ξ2θαQ]xξ2

 

=

d2xdξ2+[42Q]1ξdxdξ+[(σc26)ξ3sinξ2Q]xξ2

 

=

d2xdξ2+[2+2ξcosξsinξ]1ξdxdξ+[2+2ξcosξsinξ]xξ2+[(σc26)ξsinξ]x

First Attempt[edit]

Let's try,

x

=

A+B(ξF)2[1(ξD)cot(ξC)].

First, note that,

ddξ[cot(ξC)]

=

ddξ[cos(ξC)sin(ξC)]

 

=

[1+cot2(ξC)].

Hence,

dxdξ

=

2B(ξF)3[1(ξD)cot(ξC)]B(ξF)2{cot(ξC)(ξD)[1+cot2(ξC)]}

 

=

B(ξF)3{[22(ξD)cot(ξC)](ξF)[cot(ξC)(ξD)[1+cot2(ξC)]]}

 

=

B(ξF)3{2cot(ξC)[2(ξD)+(ξF)]+(ξF)(ξD)[1+cot2(ξC)]}

 

=

B(ξF)3{2[3ξ(2D+F)]cot(ξC)+[ξ2(D+F)ξ+FD]+[ξ2(D+F)ξ+FD]cot2(ξC)}

 

=

B(ξF)3{[ξ2(D+F)ξ+FD+2][3ξ(2D+F)]cot(ξC)+[ξ2(D+F)ξ+FD]cot2(ξC)}.

And,

d2xdξ2

=

3B(ξF)4{[ξ2(D+F)ξ+FD+2][3ξ(2D+F)]cot(ξC)+[ξ2(D+F)ξ+FD]cot2(ξC)}

 

 

B(ξF)3{[2ξ(D+F)]3cot(ξC)[3ξ(2D+F)]dcot(ξC)dξ+[2ξ(D+F)]cot2(ξC)+[ξ2(D+F)ξ+FD]dcot2(ξC)dξ}

[(ξF)4B]d2xdξ2

=

3[ξ2(D+F)ξ+FD+2]3[3ξ(2D+F)]cot(ξC)+3[ξ2(D+F)ξ+FD]cot2(ξC)

 

 

(ξF)[2ξ(D+F)]+3(ξF)cot(ξC)(ξF)[2ξ(D+F)]cot2(ξC)

 

 

+(ξF)[3ξ(2D+F)]dcot(ξC)dξ(ξF)[ξ2(D+F)ξ+FD]dcot2(ξC)dξ

 

=

3[ξ2(D+F)ξ+FD+2](ξF)[2ξ(D+F)]+{3(ξF)3[3ξ(2D+F)]}cot(ξC)

 

 

+{3[ξ2(D+F)ξ+FD](ξF)[2ξ(D+F)]}cot2(ξC)

 

 

(ξF)[3ξ(2D+F)][1+cot2(ξC)]+(ξF)[ξ2(D+F)ξ+FD]2cot(ξC)[1+cot2(ξC)]

 

=

3[ξ2(D+F)ξ+FD+2](ξF)[2ξ(D+F)](ξF)[3ξ(2D+F)]+{3(ξF)3[3ξ(2D+F)]+2(ξF)[ξ2(D+F)ξ+FD]}cot(ξC)

 

 

+{3[ξ2(D+F)ξ+FD](ξF)[2ξ(D+F)](ξF)[3ξ(2D+F)]}cot2(ξC)+2(ξF)[ξ2(D+F)ξ+FD]cot3(ξC).

Let's set C=D=F and see if these expressions match the ones above.

dxdξ|C=D=F

=

Bξ3{2+ξ23ξcotξ+ξ2cot2ξ}.

ξ4Bd2xdξ2|C=D=F

=

3[ξ2+2](ξ)[2ξ]ξ[3ξ]+{3(ξ)3[3ξ]+2ξ[ξ2]}cot(ξ)

 

 

+{3[ξ2]ξ[2ξ]ξ[3ξ]}cot2ξ+2ξ[ξ2]cot3ξ

 

=

3ξ2+62ξ23ξ2+[3ξ9ξ+2ξ3]cot(ξ)+[3ξ22ξ23ξ2]cot2ξ+2ξ3cot3ξ

 

=

62ξ2+[6ξ+2ξ3]cot(ξ)2ξ2cot2ξ+2ξ3cot3ξ

Second Attempt[edit]

Up to this point we have been rather cavalier about the use of ξ (and ξi) to represent the envelope's dimensionless radius (and interface location). Let's switch to η,

r*

=

(μeμc)1θi2(2π)1/2η

0

=

d2xdη2+{4(ρ*P*)Mr*(r*)}1ηdxdη+12πθi4(μeμc)2(ρ*P*){2πσc23γgαgMr*(r*)3}x.

and, throughout the envelope we have,

ρ*P*

=

(μeμc)θi1ϕ(η)1;

Mr*r*

=

(μeμc)2θi1(2π)1/2(η2dϕdη)[(μeμc)1θi2(2π)1/2η]1=2(μeμc)1θiη(dϕdη).

Hence, the LAWE relevant to the envelope is,

0

=

d2xdη2+{4[ρ*P*][Mr*(r*)]}1ηdxdη+12πθi4(μeμc)2[ρ*P*]{2πσc23γgαe(r*)2[Mr*r*]}x

 

=

d2xdη2+{4[(μeμc)θi1ϕ(η)1][2(μeμc)1θiη(dϕdη)]}1ηdxdη+12πθi4(μeμc)2[(μeμc)θi1ϕ(η)1]{2πσc23γgαe[(μeμc)1θi2(2π)1/2η]2[2(μeμc)1θiη(dϕdη)]}x

 

=

d2xdη2+{4[2ηϕ(dϕdη)]}1ηdxdη+12πθi5ϕ(μeμc)1{2πσc23γgαe[(μeμc)θi5(4π)η1](dϕdη)}x

 

=

d2xdη2+{4[2ηϕ(dϕdη)]}1ηdxdη+12πθi5ϕ(μeμc)1{2πσc23γg}xαe[2ηϕ(dϕdη)]xη2.

If we assume that, αe=(34/2)=1 and σc2=0, then the relevant envelope LAWE is,

0

=

d2xdη2+{42Q}1ηdxdη[2Q]xη2,

where,

Qdlnϕdlnη.

Now consider the,

Precise Solution to the Polytropic LAWE

xP

=

b(n1)2n[1+(n3n1)(1ηϕn)dϕdη]

 

=

b[(1ηϕ)dϕdη]

 

=

bη2[dlnϕdlnη]

 

=

bQη2.

From our accompanying discussion, we recall that the most general solution to the n=1 Lane-Emden equation can be written in the form,

ϕ=A[sin(ηB)η],

where A and B are constants whose values can be obtained from our accompanying parameter table. The first derivative of this function is,

dϕdη=Aη2[ηcos(ηB)sin(ηB)].

Hence,

Q=dlnϕdlnη

=

ηϕAη2[ηcos(ηB)sin(ηB)]

 

=

[1ηcot(ηB)]

xP

=

bη2[1ηcot(ηB)].

What is this in terms of the dimensionless radius, r*/R*? Well,

r*R*

=

(μeμc)1θi2(2π)1/2η[2πθi2ηs(μeμc)]

 

=

ηηs=η(π+B)

η

=

r*R*(π+B).

Also,

ηB

=

r*R*(π+B)B=π(r*R*)B[1(r*R*)]

 

=

π+π[(r*R*)1]B[1(r*R*)]

 

=

π(π+B)[1(r*R*)].


[12 January 2019]: Here's what appears to work pretty well, empirically:

xP

=

1η2{1ηcot[η(π0.8)]}.

η

=

r*R*(π0.6π).


Let's work through the analytic derivatives again. Keeping in mind that,

ddη[cot(ηB)]

=

[1+cot2(ηB)],

and starting with the guess,

xP

=

bη2[1ηcot(ηB)],

we have,

dxPdη

=

2bη3[1ηcot(ηB)]bη2{cot(ηB)η[1+cot2(ηB)]}

(η3b)dxPdη

=

[22ηcot(ηB)]{ηcot(ηB)η2[1+cot2(ηB)]}

 

=

η22+ηcot(ηB)+η2cot2(ηB).

The second derivative then gives,

d2xPdη2

=

ddη{bη3[η22+ηcot(ηB)+η2cot2(ηB)]}

 

=

3bη4[η22+ηcot(ηB)+η2cot2(ηB)]

 

 

+bη3{2η+cot(ηB)+2ηcot2(ηB)+ηddη[cot(ηB)]+2η2cot(ηB)ddη[cot(ηB)]}

d2xPdη2

=

bη4[63η23ηcot(ηB)3η2cot2(ηB)]

 

 

+bη4{2η2+ηcot(ηB)+2η2cot2(ηB)η2[1+cot2(ηB)]2η3cot(ηB)[1+cot2(ηB)]}

η4bd2xPdη2

=

63η23ηcot(ηB)3η2cot2(ηB)

 

 

+2η2+ηcot(ηB)+2η2cot2(ηB)η2η2cot2(ηB)2η3cot(ηB)2η3cot3(ηB)

 

=

2[3η2(η+η3)cot(ηB)η2cot2(ηB)η3cot3(ηB)].

Recalling that,

Q=[1ηcot(ηB)],

plugging these expressions into the relevant envelope LAWE gives,

LAWE

=

d2xdη2+{42Q}1ηdxdη2Qxη2

 

=

d2xdη2+{42[1ηcot(ηB)]}1ηdxdη[1ηcot(ηB)]2xη2

 

=

bη4{η4bd2xdη2+[1+ηcot(ηB)]2η3bdxdη[1ηcot(ηB)]2η2xb}

 

=

2bη4{3η2(η+η3)cot(ηB)η2cot2(ηB)η3cot3(ηB)

 

 

+[1+ηcot(ηB)][η22+ηcot(ηB)+η2cot2(ηB)][1ηcot(ηB)][1ηcot(ηB)]}

 

=

2bη4{3η2(η+η3)cot(ηB)η2cot2(ηB)η3cot3(ηB)+[η22+ηcot(ηB)+η2cot2(ηB)][1ηcot(ηB)]

 

 

+ηcot(ηB)[η22+ηcot(ηB)+η2cot2(ηB)]+ηcot(ηB)[1ηcot(ηB)]}

 

=

2bη4{(η+η3)cot(ηB)η3cot3(ηB)+2ηcot(ηB)

 

 

+η3cot(ηB)2ηcot(ηB)+η2cot2(ηB)+η3cot3(ηB)+ηcot(ηB)η2cot2(ηB)}

 

=

2bη4{[ηcot(ηB)ηcot(ηB)+2ηcot(ηB)]+[η3cot(ηB)η3cot(ηB)]

 

 

+[η2cot2(ηB)η2cot2(ηB)]+[η3cot3(ηB)η3cot3(ηB)]}

 

=

0.

Okay. Now let's determine at what value of η the logarithmic derivative of xP goes to negative one.

dlnxPdlnη=ηxPdxPdη

=

η3b[1ηcot(ηB)]1dxPdη

 

=

[1ηcot(ηB)]1[η22+ηcot(ηB)+η2cot2(ηB)].

Setting this to negative one, we have,

[1ηcot(ηB)]

=

[η22+ηcot(ηB)+η2cot2(ηB)]

1

=

η2[1+cot2(ηB)]

 

=

η2[1sin2(ηB)]

1

=

η2sin2(ηB).

And this occurs when,

(Aϕ)2=1.

Third Attempt[edit]

Prior to the Brute-Force Trial Fit[edit]

Let's work through the analytic derivatives again. Keeping in mind that,

ddη[cot(ηC)]

=

[1+cot2(ηC)],

and starting with the guess,

xP

=

bη2[1ηcot(ηC)],

we have,

(η3b)dxPdη

=

η22+ηcot(ηC)+η2cot2(ηC),

and,

η4bd2xPdη2

=

2[3η2(η+η3)cot(ηC)η2cot2(ηC)η3cot3(ηC)].


Note that the relevant logarithmic derivative is,

dlnxPdlnη

=

(bη2)[η22+ηcot(ηC)+η2cot2(ηC)]xP1

 

=

[η22+ηcot(ηC)+η2cot2(ηC)][1ηcot(ηC)]1

If we know the logarithmic slope and the value of η at the interface, then we can solve for

yiηicot(ηiC),

via the quadratic relation,

(1yi)[dlnxPdlnη]i

=

ηi22+yi+yi2

0

=

ηi22+yi+yi2(1yi)[dlnxPdlnη]i

 

=

yi2+yi{1+[dlnxPdlnη]i}+{ηi22[dlnxPdlnη]i}.

(In practice it appears as though the "plus" solution to this quadratic equation is desired if the quantity inside the last set of curly braces is positive; and the "minus" solution is desired if this quantity is negative.) Once the value of yi is known, we can solve for the key coefficient, C, via the relation,

tan(ηiC)

=

ηiyi

C

=

ηitan1(ηiyi).


Recalling that,

Q=[1ηcot(ηB)],

plugging these expressions into the relevant envelope LAWE gives,

LAWE

=

d2xdη2+{42Q}1ηdxdη2Qxη2

 

=

d2xdη2+{42[1ηcot(ηB)]}1ηdxdη[1ηcot(ηB)]2xη2

 

=

bη4{η4bd2xdη2+[1+ηcot(ηB)]2η3bdxdη[1ηcot(ηB)]2η2xb}

 

=

2bη4{3η2(η+η3)cot(ηC)η2cot2(ηC)η3cot3(ηC)

 

 

+[1+ηcot(ηB)][η22+ηcot(ηC)+η2cot2(ηC)][1ηcot(ηB)][1ηcot(ηC)]}

 

=

2bη4{3η2(η+η3)cot(ηC)η2cot2(ηC)η3cot3(ηC)+[η22+ηcot(ηC)+η2cot2(ηC)][1ηcot(ηC)]

 

 

+ηcot(ηB)[η22+ηcot(ηC)+η2cot2(ηC)]+ηcot(ηB)[1ηcot(ηC)]}

 

=

2bη4{(ηη3)cot(ηC)η3cot3(ηC)+ηcot(ηB)[η21+η2cot2(ηC)]}

 

=

2bη4{(ηη3)[cot(ηC)cot(ηB)]+η3cot2(ηC)[cot(ηB)cot(ηC)]}

 

=

2bη4[cot(ηC)cot(ηB)][ηη3η3cot2(ηC)]

 

=

2bη3[cot(ηC)cot(ηB)]{1η2[1+cot2(ηC)]}.

This will go to zero if C=(B2mπ), where m is a positive integer. When m=1, for example,

cot(ηC)

=

cot[η(B2π)]=cot(ηB).


Okay. Now let's determine at what value of η the logarithmic derivative of xP goes to negative one.

Brute-Force Trial Fit[edit]

Photo of white board with steps showing development of trial eigenfunction. This should be paired with an Excel spreadsheet.

Using a couple of separate Excel spreadsheets — FaulknerBipolytrope2.xlsx/mu100Mode0 and AnalyticTrialBipolytropeA.xlsx/Sheet2, both stored in a DropBox account under the folder Wiki_edits/Bipolytrope/LinearPerturbation — we used an inelegant and inefficient trial & error technique in search of an eigenfunction that had the same analytic form as the one represented above for xP, but that, when plotted, appeared to qualitatively match the numerically determined envelope eigenfunction. Then, on a whiteboard — see the photo, here on the right — we formulated a concise expression for a trial function that seemed to work pretty well. Our primary finding was that α, appearing as the argument to the tanα function, needed to be shifted by something like 3π/4.

 

THIS SPACE

INTENTIONALLY

LEFT BLANK

Following Up on the Brute-Force Trial Fit[edit]

In an accompanying discussion — see especially Attempt #2 — we have determined by visual inspection that a decent fit to the envelope's eigenfunction is given by the expression,

xtrial

=

b0Λ2{1Λ[tan(ηiΛ3π/4)+fα1fαtan(ηiΛ3π/4)]}a0

 

=

b0Λ2{1Λcot(ΛE)}a0,

Limiting Parameter Values
  min max α=αs
ηF ηi ηs 8π(ηsηi)2+2ηsηi
α π2 5π8 ηiηs3π4
Λ ηiπ4 ηiπ8 ηs

where, over the range, ηiηηs,

E

ηi5π4+tan1fα,

Λ(η)

ηi+gF[ηi2ηs+η]=Λ0+gFη,

1fα=tan(αs)

tan[(ηsηi+3π4)],

gF

π8(ηsηi).



Here, we reference a separate discussion of the bipolytrope's underlying equilibrium structure

B=ηiπ2+tan1f E=ηi5π4+tan1fα

cot(ηiB)

=

tan[π2(ηiB)]

 

=

tan[π2(π2tan1f)]

 

=

f

f

=

tan(B+π2ηi)

cot(ηiE)

=

tan[π2(ηiE)]

 

=

tan[π2(5π4tan1fα)]

 

=

tan(tan1fα3π4)

 

=

tan(3π4tan1fα)

 

=

cot(tan1fα)

 

=

1fα

Hence …     cot(ηB)

=

tan[π2(ηB)]

 

=

tan[π2η+ηiπ2+tan1f]

 

=

tan[ηiη+tan1f]

 

=

tan(ηiη)+f1ftan(ηiη)

Hence …     cot(ΛE)

=

tan[π2(ΛE)]

 

=

tan[ηiΛ3π4+tan1fα]

 

=

tan(ηiΛ3π4)+fα1fαtan(ηiΛ3π4)

Also …    B=ηsπ

f=cot(ηiB)

=

cot(ηiηs+π)

1f

=

tan(ηiηs+π)



Let's examine the first and second derivatives of this trial eigenfunction, recognizing that,

dxtrialdη=dΛdηdxtrialdΛ=gFdxtrialdΛ

      and      

d2xtrialdη2=dΛdηddΛ[gFdxtrialdΛ]=gF2d2xtrialdΛ2.

and drawing from the derivative expressions already derived, above. For the first derivative, we have,

dxtrialdη

=

gF(b0Λ3)[Λ22+Λcot(ΛE)+Λ2cot2(ΛE)].

And the second derivative gives,

d2xtrialdη2

=

gF2(2b0Λ4)[3Λ2(Λ+Λ3)cot(ΛE)Λ2cot2(ΛE)Λ3cot3(ΛE)].

Hence,

LAWE

=

d2xtrialdη2+{42Q}1ηdxtrialdη2Qxtrialη2

 

=

d2xtrialdη2+{42[1ηcot(ηB)]}1ηdxtrialdη[1ηcot(ηB)]2xtrialη2

 

=

b0η4{η4b0d2xtrialdη2+[1+ηcot(ηB)]2η3b0dxtrialdη[1ηcot(ηB)]2η2xtrialb0}

 

=

b0η4{η4b0gF2(2b0Λ4)[3Λ2(Λ+Λ3)cot(ΛE)Λ2cot2(ΛE)Λ3cot3(ΛE)]

 

 

+[1+ηcot(ηB)]2η3b0gF(b0Λ3)[Λ22+Λcot(ΛE)+Λ2cot2(ΛE)]

 

 

[1ηcot(ηB)]2η2b0[b0Λ2{1Λcot(ΛE)}a0]}

 

=

b0η4{gF2(2η4Λ4)[3Λ2(Λ+Λ3)cot(ΛE)Λ2cot2(ΛE)Λ3cot3(ΛE)]

 

 

+[1+ηcot(ηB)]gF(2η3Λ3)[Λ22+Λcot(ΛE)+Λ2cot2(ΛE)]

 

 

[1ηcot(ηB)][2η2Λ2[1Λcot(ΛE)]2η2a0b0]}

 

=

2b0Λ4η2{gF2η2[3Λ2(Λ+Λ3)cot(ΛE)Λ2cot2(ΛE)Λ3cot3(ΛE)]

 

 

+[1+ηcot(ηB)]gFΛη[Λ22+Λcot(ΛE)+Λ2cot2(ΛE)]

 

 

[1ηcot(ηB)][Λ2[1Λcot(ΛE)]a0Λ4b0]}

(Λ42b0)  LAWE

=

gF2[3Λ2(Λ+Λ3)cot(ΛE)Λ2cot2(ΛE)Λ3cot3(ΛE)]

 

 

+gFΛη[Λ22+Λcot(ΛE)+Λ2cot2(ΛE)](Λη)2[1Λcot(ΛE)a0Λ2b0]

 

 

+[ηcot(ηB)]{gFΛη[Λ22+Λcot(ΛE)+Λ2cot2(ΛE)]+(Λη)2[1Λcot(ΛE)a0Λ2b0]}


Fourth Attempt[edit]

XXXX[edit]

If we assume that, αe=(34/2)=1 and σc2=0, then the relevant envelope LAWE is,

0

=

d2xdη2+{42Q}1ηdxdη[2Q]xη2,

where,

Qdlnϕdlnη

=

[1ηcot(ηB0)].


Let's work through the analytic derivatives again. Keeping in mind that,

ddη[cot(ηB)]

=

[1+cot2(ηB)];

and that the,

Precise Solution to the Polytropic LAWE

xP

=

b(n1)2n[1+(n3n1)(1ηϕn)dϕdη]

 

=

b[(1ηϕ)dϕdη]

 

=

bη2[dlnϕdlnη]

 

=

bQη2

xP

=

bη2[1ηcot(ηB0)].

As we have already tried once, above, let's try a more general form of this expression, namely,

xQ

=

A+C(ηF)2[1(ηD)cot(ηB)].

Hence,

dxQdη

=

[1(ηD)cot(ηB)]ddη[C(ηF)2]C(ηF)2ddη[(ηD)cot(ηB)]

 

=

[1(ηD)cot(ηB)][2C(ηF)3]C(ηF)2[cot(ηB)]+C(ηD)(ηF)2[1+cot2(ηB)]

 

=

C(ηF)3{2+2(ηD)cot(ηB)(ηF)[cot(ηB)]+(ηD)(ηF)[1+cot2(ηB)]}

 

=

C(ηF)3{[(ηD)(ηF)2]+(η2D+F)cot(ηB)+(ηD)(ηF)cot2(ηB)}.

And,

d2xQdη2

=

{[(ηD)(ηF)2]+(η2D+F)cot(ηB)+(ηD)(ηF)cot2(ηB)}ddη[C(ηF)3]

 

 

+C(ηF)3ddη{[(ηD)(ηF)2]+(η2D+F)cot(ηB)+(ηD)(ηF)cot2(ηB)}

 

=

{[(ηD)(ηF)2]+(η2D+F)cot(ηB)+(ηD)(ηF)cot2(ηB)}[3C(ηF)4]

 

 

+C(ηF)3{[2η(D+F)]+cot(ηB)(η2D+F)[1+cot2(ηB)]+[2η(D+F)]cot2(ηB)2[η2η(D+F)+DF]cot(ηB)[1+cot2(ηB)]}

 

=

C(ηF)4{3[(ηD)(ηF)2]3(η2D+F)cot(ηB)3(ηD)(ηF)cot2(ηB)}

 

 

+C(ηF)3{[2η(D+F)](η2D+F)+cot(ηB)(η2D+F)cot2(ηB)+[2η(D+F)]cot2(ηB)2[η2η(D+F)+DF]cot(ηB)2[η2η(D+F)+DF]cot3(ηB)}

YYYY[edit]

And,

d2xQdη2

=

{[(ηD)(ηF)2]+(η2D+F)cot(ηB)+(ηD)(ηF)cot2(ηB)}ddη[C(ηF)3]

 

 

+C(ηF)3ddη{[(ηD)(ηF)2]+(η2D+F)cot(ηB)+(ηD)(ηF)cot2(ηB)}

 

=

{[(ηD)(ηF)2]+(η2D+F)cot(ηB)+(ηD)(ηF)cot2(ηB)}[3C(ηF)4]

 

 

+C(ηF)3{ddη[(η2D+F)cot(ηB)]+ddη[(ηD)(ηF)cot2(ηB)]}

 

=

C(ηF)4{3[(ηD)(ηF)2]3(η2D+F)cot(ηB)3(ηD)(ηF)cot2(ηB)

 

 

+(ηF)ddη[(η2D+F)cot(ηB)]+(ηF)ddη[(ηD)(ηF)cot2(ηB)]}

 

=

C(ηF)4{3[(ηD)(ηF)2]3(η2D+F)cot(ηB)3(ηD)(ηF)cot2(ηB)

 

 

+(ηF)cot(ηB)ddη[(η2D+F)]+(ηF)(η2D+F)ddη[cot(ηB)]

 

 

+(ηF)cot2(ηB)ddη[η2η(D+F)+DF]+(ηF)(ηD)(ηF)ddη[cot2(ηB)]}

 

=

C(ηF)4{3[(ηD)(ηF)2]3(η2D+F)cot(ηB)3(ηD)(ηF)cot2(ηB)

 

 

+(ηF)cot(ηB)(ηF)(η2D+F)[1+cot2(ηB)]

 

 

+(ηF)cot2(ηB)[2η(D+F)]2(ηF)(ηD)(ηF)cot(ηB)[1+cot2(ηB)]}

 

=

C(ηF)4{3[(ηD)(ηF)2](ηF)(η2D+F)

 

 

+[(ηF)3(η2D+F)2(ηF)(ηD)(ηF)]cot(ηB)

 

 

+[(ηF)[2η(D+F)]3(ηD)(ηF)2(ηF)(ηD)(ηF)cot(ηB)(ηF)(η2D+F)]cot2(ηB)}

So the envelope LAWE becomes,

(ηF)4CLAWE

=

(ηF)4Cd2xQdη2+(ηF)4C[1+ηcot(ηB0)]2ηdxQdη(ηF)4C[1ηcot(ηB0)]2xQη2

 

=

{3[(ηD)(ηF)2](ηF)(η2D+F)+[(ηF)3(η2D+F)2(ηF)(ηD)(ηF)]cot(ηB)

 

 

+[(ηF)[2η(D+F)]3(ηD)(ηF)2(ηF)(ηD)(ηF)cot(ηB)(ηF)(η2D+F)]cot2(ηB)}

 

 

+(ηF)[1+ηcot(ηB0)]2η{[(ηD)(ηF)2]+(η2D+F)cot(ηB)+(ηD)(ηF)cot2(ηB)}

 

 

(ηF)4C[1ηcot(ηB0)]2η2{A+C(ηF)2[1(ηD)cot(ηB)]}

 

=

3[(ηD)(ηF)2](ηF)(η2D+F)+[(ηF)3(η2D+F)2(ηF)(ηD)(ηF)]cot(ηB)

 

 

+(ηF)[1+ηcot(ηB0)]2η{[(ηD)(ηF)2]+(η2D+F)cot(ηB)}

 

 

+[(ηF)[2η(D+F)]3(ηD)(ηF)2(ηF)(ηD)(ηF)cot(ηB)(ηF)(η2D+F)]cot2(ηB)

 

 

+(ηF)[1+ηcot(ηB0)]2η[(ηD)(ηF)cot2(ηB)](ηF)2[1ηcot(ηB0)]2η2[1(ηD)cot(ηB)]

 

 

(ηF)4C[1ηcot(ηB0)]2Aη2.

What does this reduce to if A=D=F=0.

η4CLAWE

=

3[(η)(η)2](η)(η)+[(η)3(η)2(η)(η)(η)]cot(ηB)

 

 

+(η)[1+ηcot(ηB0)]2η{[(η)(η)2]+(η)cot(ηB)}

 

 

+[(η)[2η]3(η)(η)2(η)(η)(η)cot(ηB)(η)(η)]cot2(ηB)

 

 

+(η)[1+ηcot(ηB0)]2η[(η)(η)cot2(ηB)](η)2[1ηcot(ηB0)]2η2[1(η)cot(ηB)]

 

 

(η)4C[1ηcot(ηB0)]2Aη2.

 

=

64η22(η+η3)cot(ηB)+2[1+ηcot(ηB0)][η22+ηcot(ηB)]2[η2+η3cot(ηB)]cot2(ηB)

 

 

+2[1+ηcot(ηB0)][η2cot2(ηB)]2[1ηcot(ηB0)][1ηcot(ηB)]2Aη2C[1ηcot(ηB0)]

 

=

64η22(η+η3)cot(ηB)+2η24+2ηcot(ηB)+2η3cot(ηB0)4ηcot(ηB0)+2η2cot(ηB0)cot(ηB)2η2cot2(ηB)2η3cot3(ηB)

 

 

+2η2cot2(ηB)+2η3cot(ηB0)cot2(ηB)2+4ηcot(ηB0)2η2cot2(ηB0)2Aη2C[1ηcot(ηB0)]

 

=

2η22η3[cot(ηB)+cot3(ηB)]+cot(ηB0)[2η3+2η2cot(ηB)+2η3cot2(ηB)]2η2cot2(ηB0)2Aη2C[1ηcot(ηB0)]

 

=

2η2+2η3[cot(ηB0)cot(ηB)][1+cot2(ηB)]+2η2cot(ηB0)[cot(ηB)cot(ηB0)]2Aη2C[1ηcot(ηB0)].

See Also[edit]

  • In an accompanying Chapter within our "Ramblings" Appendix, we have played with the adiabatic wave equation for polytropes, examining its form when the primary perturbation variable is an enthalpy-like quantity, rather than the radial displacement of a spherical mass shell. This was done in an effort to mimic the approach that has been taken in studies of the stability of Papaloizou-Pringle tori.
  • n=32 … D. Lucas (1953, Bul. Soc. Roy. Sci. Liege, 25, 585) … Citation obtained from the Prasad & Gurm (1961) article.
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