A.2 General w case
A.2.4 Recovering the w = 1 result
Here, we show that when substituting w = 1 on the general solution, given by fullsimplify22, we recover the matter as a scalar field solution, given by fullsimplify1, and its corresponding series expansion.
Expand[fullsimplify1/.{ → 1}] == Expand[fullsimplify22/.{w → 1, → 1}]//FullSimplify Expand[fullsimplify1/.{ → 1}] == Expand[fullsimplify22/.{w → 1, → 1}]//FullSimplify Expand[fullsimplify1/.{ → 1}] == Expand[fullsimplify22/.{w → 1, → 1}]//FullSimplify True
Normal converts the series into a normal expression, i.e. it removes O from the series. When comparing expressions on Wolfram Mathematica we sometimes have to add an extra FullSimplify in
order to get a True or False output, otherwise it can give an expression equal to another expression as an output.
Normal[series1] == Normal[series22/.{w → 1, → 1}]//FullSimplify Normal[series1] == Normal[series22/.{w → 1, → 1}]//FullSimplify Normal[series1] == Normal[series22/.{w → 1, → 1}]//FullSimplify True
Appendix B
Mathematica codes for the b + branch mLQC-I
In this appendix we provide a detailed explanation of the Wolfram Mathematica notebook used to de-termine the effective actions for the b+branch of mLQC-I. Moreover, this appendix is divided in two sections, one for the case of matter as a scalar field and one for the general w case. First of all, we define the modified Friedmann equation corresponding to the b+ branch of mLQC-I, given by eq. (3.38), as H2mLQCIbPlus, and Ψ (ρ) as Psirho2, which is generically defined by eq. (5.3) and, in this case, is given by eq. (5.34).
B.1 Matter as a scalar field
B.1.1 Solving the differential equation
We define the modified Friedmann eq. (5.6), as H2flatZeroΛΛΛ. We choose to write the general expression and substitute w by 1. This substitution is indicated by “/.” as follows.
H2flatZeroΛ =
In this case, we want to solve eq. (5.36), which we define as diffEq3. Notice, below, that we substitute the energy density ρ, in the whole equation, by its corresponding expression in terms of the lower order Ricci scalar, using eq. (5.8).
diffEq3 = (H2flatZeroΛ −κρ3 ) == Psirho2/.{w → 1, ρ → R/(−2κ)} Using DSolve[eqn,u,x], which solves a differential equation for the function u, with independent variable x, we are able to compute the solution of diffEq3, which we will denote by sol3. In this step, we provide information about the constants α, ρΛ(ρΛ on the code), κ, ρIc (ρc1 on the code) and γ, as
well as information about the variable R. The presence of Clear[R,φ3,sol3] before DSolve[eqn,u,x]
is important, since we have to ensure Wolfram Mathematica solves the differential equations properly.
In that regard, we have experienced that the software can, sometimes, give the wrong solutions, and attempting Clear[R,φ3,sol3] seems to solve this issue. Moreover, we may also quit the kernel, which leaves the notebook intact, but seems to forget what was done before, including definitions.
Clear[φ3, R, sol3]
The result is given as a pure function, meaning it is given in the format φ3 → Function[{R}, ...]. This is achived by writing DSolve[eqn,φ3, Rφ3, Rφ3, R] instead of DSolve[eqn,φ3[R], Rφ3[R], Rφ3[R], R], which is also possible. Then, we use Simplify, in order to obtain a first simplication of the solution, where we provide the same as-sumptions as before. In order to obtain a final simplification, we use FullSimplify while setting constants c1 and c2 to zero. Recall that removing the terms of the solution that are multiplied by c1 and c2, only
removes the terms that correspond to eq. (5.15). Writting sol3[[1]]sol3[[1]]sol3[[1]] allows us to select the first component of sol3sol3sol3, which is technically a list, although it only contains one component. Since sol3[[1]]sol3[[1]]sol3[[1]] is itself a list with one component, writting sol3[[1]][[1]]sol3[[1]][[1]]sol3[[1]][[1]] allows us to sellect φ3 → Function[{R}, ...], which has two components, {φ3, Function[{R}, ...]}. Writting sol3[[1]][[1]][[2]]sol3[[1]][[1]][[2]]sol3[[1]][[1]][[2]] sellects its second component which is Function[{R}, ...], which also has two components, {{R}, ...]}. Finally, writting sol3[[1]][[1]][[2]][[2]]sol3[[1]][[1]][[2]][[2]]sol3[[1]][[1]][[2]][[2]], selects the actual solution.
simplify3 = simplify3 = simplify3 =
Simplify[sol3[[1]][[1]][[2]][[2]], Assumptions → { ∈ Reals, κ ∈ Reals, κ > 0, ρc1 ∈ Reals, ρc1 > 0, Simplify[sol3[[1]][[1]][[2]][[2]], Assumptions → { ∈ Reals, κ ∈ Reals, κ > 0, ρc1 ∈ Reals, ρc1 > 0, Simplify[sol3[[1]][[1]][[2]][[2]], Assumptions → { ∈ Reals, κ ∈ Reals, κ > 0, ρc1 ∈ Reals, ρc1 > 0, γ ∈ Reals, γ > 0, α ∈ Reals, α > 0, ρΛ ∈ Reals, ρΛ > 0, R ∈ Reals, R ≤ 0}];
γ ∈ Reals, γ > 0, α ∈ Reals, α > 0, ρΛ ∈ Reals, ρΛ > 0, R ∈ Reals, R ≤ 0}];
γ ∈ Reals, γ > 0, α ∈ Reals, α > 0, ρΛ ∈ Reals, ρΛ > 0, R ∈ Reals, R ≤ 0}];
fullsimplify3 = FullSimplify[sol3[[1]][[1]][[2]][[2]]/.{C[1] → 0, C[2] → 0, → 1}, fullsimplify3 = FullSimplify[sol3[[1]][[1]][[2]][[2]]/.{C[1] → 0, C[2] → 0, → 1}, fullsimplify3 = FullSimplify[sol3[[1]][[1]][[2]][[2]]/.{C[1] → 0, C[2] → 0, → 1},
Assumptions → { ∈ Reals, κ ∈ Reals, κ > 0, ρc1 ∈ Reals, ρc1 > 0, γ ∈ Reals, γ > 0, α ∈ Reals, Assumptions → { ∈ Reals, κ ∈ Reals, κ > 0, ρc1 ∈ Reals, ρc1 > 0, γ ∈ Reals, γ > 0, α ∈ Reals, Assumptions → { ∈ Reals, κ ∈ Reals, κ > 0, ρc1 ∈ Reals, ρc1 > 0, γ ∈ Reals, γ > 0, α ∈ Reals, α > 0, ρΛ ∈ Reals, ρΛ > 0, R ∈ Reals, R ≤ 0}]
α > 0, ρΛ ∈ Reals, ρΛ > 0, R ∈ Reals, R ≤ 0}]
α > 0, ρΛ ∈ Reals, ρΛ > 0, R ∈ Reals, R ≤ 0}]
R − ακρΛ +Rα(R+18(−1+2γ2)κρc1)ρΛ 72γ2κρc12
+20ρc11 αρΛ
− 3(R − 2(−2 +√
10)κρc1) Hypergeometric2F1[−12,16(−4 −√
10),16(2 −√
10), −2κρc1R ]
− 3(R + 2(2 +√
10)κρc1)Hypergeometric2F1[−12,16(−4 +√
10),16(2 +√
10), −2κρc1R ] + (R + 2κρc1)
− (−1 +√
10)Hypergeometric2F1[12,16(−4 −√
10),16(2 −√
10), −2κρc1R ] + (1 +√
10)Hypergeometric2F1[12,16(−4 +√
10),16(2 +√
10), −2κρc1R ]
B.1.3 Obtaining a series expansion
In order to obtain a Taylor expansion of the solution, we use Series[f, {x, x[f, {x, x[f, {x, x000, n}], n}], n}], which generates a power series expansion for f about the point x = x0to order (x − x0)n, where n is an explicit integer.
FullSimplify is then used to further simplify each term. We denote the series, for this case, as series2.
series3 = Series[fullsimplify3, {R, 0, 3}]//FullSimplify series3 = Series[fullsimplify3, {R, 0, 3}]//FullSimplify series3 = Series[fullsimplify3, {R, 0, 3}]//FullSimplify
−2(ακρΛ) + (1 + α(−1+5γ4γ2ρc12)ρΛ)R + α(4−3γ288γ22κρc1)ρΛR2 2 +3968καρΛR2ρc13 3 + O[R]4
B.2 General w case
In order to obtain the general result, we solve the differential equation for the two possible intervals of w, separatly, given the fact that eq. (5.1) does not depend smoothly on w, becoming a first order equation for w = 1/3. In that regard, we solve the equation, first, for −1 < w < 1/3 and, then, for 1/3 < w < 1.
B.2.1 Solving the differential equation
−1 < w < 1/3
−1 < w < 1/3
−1 < w < 1/3
We define the modified Friedmann eq. (5.1), as H2flatZeroΛΛΛw1 for this branch of w.
H2flatZeroΛw1 = (1/3)κρ − (/3)((1/2)(3w + 1)φ410[R]κρ + (1/2)φ41[R] − 3(1 + w)(1 − 3w)
In this case, we want to solve eq. (5.73) for φ41[R], which we define as diffEq41. Notice, below, that we substitute the energy density ρ, in the whole equation, by its corresponding expression in terms of the lower order Ricci scalar, using eq. (5.2).
diffEq41 = (H2flatZeroΛw1 −κρ3 ) == Psirho2/.{ρ → R/(−κ(3w − 1))}
Using DSolve[eqn,u,x], we are able to compute the solution of diffEq41, which we will denote by sol41. In this step, we provide information about the constants α, ρΛ(ρΛ on the code), κ, ρIc(ρc1 on the code) and γ, as well as information about the variable R and possible values for w.
Clear[R, φ41]
Here, we define eq. (5.1) as H2flatZeroΛΛΛw2, with φ42[R] instead of φ41[R] so we can consider the second branch of values of w, while not mixing the branches of w.
H2flatZeroΛw2 = (1/3)κρ − (/3)((1/2)(3w + 1)φ420[R]κρ + (1/2)φ42[R] − 3(1 + w)(1 − 3w)
In this case, we define the differential eq. (5.73) as diffEq42 and, once again, substitute the energy density ρ by its corresponding expression in terms of the lower order Ricci scalar, using eq. (5.2).
diffEq42 = (H2flatZeroΛw2 −κρ3 ) == Psirho2/.{ρ → R/(−κ(3w − 1))}
Using DSolve[eqn,u,x], we are able to compute the solution of diffEq42, which we will denote by sol42. In this step, we provide information about the constants α, ρΛ(ρΛ on the code), κ, ρIc(ρc1 on the code) and γ, as well as information about the variable R and possible values for w.
Clear[R, φ42]
Clear[R, φ42]
Clear[R, φ42]
sol42 = DSolve[diffEq42, φ42, R, sol42 = DSolve[diffEq42, φ42, R, sol42 = DSolve[diffEq42, φ42, R,
Assumptions → { ∈ Reals, κ ∈ Reals, κ > 0, ρc1 ∈ Reals, ρc1 > 0, γ ∈ Reals, γ > 0, α ∈ Reals, Assumptions → { ∈ Reals, κ ∈ Reals, κ > 0, ρc1 ∈ Reals, ρc1 > 0, γ ∈ Reals, γ > 0, α ∈ Reals, Assumptions → { ∈ Reals, κ ∈ Reals, κ > 0, ρc1 ∈ Reals, ρc1 > 0, γ ∈ Reals, γ > 0, α ∈ Reals, α > 0, ρΛ ∈ Reals, ρΛ > 0, R ∈ Reals, w ∈ Reals, 1/3 < w < 1}];
α > 0, ρΛ ∈ Reals, ρΛ > 0, R ∈ Reals, w ∈ Reals, 1/3 < w < 1}];
α > 0, ρΛ ∈ Reals, ρΛ > 0, R ∈ Reals, w ∈ Reals, 1/3 < w < 1}];
B.2.2 Simplifying the solution
Once again, the results sol41 and sol42 are given as pure functions, meaning they are given in the format φ41 → Function[{R}, ...] and φ42 → Function[{R}, ...]. We use Simplify, while setting constants c1
and c2 to zero, in order to obtain a simplication of the solutions, where we provide the relevant assump-tions. Attempting to use FullSimplify does not provide a more simplified version of the soluassump-tions. Once again, writting sol41[[1]][[1]][[2]][[2]]sol41[[1]][[1]][[2]][[2]]sol41[[1]][[1]][[2]][[2]] or sol42[[1]][[1]][[2]][[2]]sol42[[1]][[1]][[2]][[2]]sol42[[1]][[1]][[2]][[2]], selects the solution. For more details, we refer to the section B.1.2.
−1 < w < 1/3
−1 < w < 1/3
−1 < w < 1/3
simplify41 = Simplify[sol41[[1]][[1]][[2]][[2]]/.{C[1] → 0, C[2] → 0, → 1}, simplify41 = Simplify[sol41[[1]][[1]][[2]][[2]]/.{C[1] → 0, C[2] → 0, → 1}, simplify41 = Simplify[sol41[[1]][[1]][[2]][[2]]/.{C[1] → 0, C[2] → 0, → 1},
Assumptions → { ∈ Reals, κ ∈ Reals, κ > 0, ρc1 ∈ Reals, ρc1 > 0, γ ∈ Reals, γ > 0, w ∈ Reals, Assumptions → { ∈ Reals, κ ∈ Reals, κ > 0, ρc1 ∈ Reals, ρc1 > 0, γ ∈ Reals, γ > 0, w ∈ Reals, Assumptions → { ∈ Reals, κ ∈ Reals, κ > 0, ρc1 ∈ Reals, ρc1 > 0, γ ∈ Reals, γ > 0, w ∈ Reals,
−1 < w < 1/3, α ∈ Reals, α > 0, ρΛ ∈ Reals, ρΛ > 0, R ∈ Reals}]
−1 < w < 1/3, α ∈ Reals, α > 0, ρΛ ∈ Reals, ρΛ > 0, R ∈ Reals}]
−1 < w < 1/3, α ∈ Reals, α > 0, ρΛ ∈ Reals, ρΛ > 0, R ∈ Reals}];
1/3 < w < 1 1/3 < w < 1 1/3 < w < 1
simplify42 = Simplify[sol42[[1]][[1]][[2]][[2]]/.{C[1] → 0, C[2] → 0, → 1}, simplify42 = Simplify[sol42[[1]][[1]][[2]][[2]]/.{C[1] → 0, C[2] → 0, → 1}, simplify42 = Simplify[sol42[[1]][[1]][[2]][[2]]/.{C[1] → 0, C[2] → 0, → 1},
Assumptions → { ∈ Reals, κ ∈ Reals, κ > 0, ρc1 ∈ Reals, ρc1 > 0, γ ∈ Reals, γ > 0, α ∈ Reals, Assumptions → { ∈ Reals, κ ∈ Reals, κ > 0, ρc1 ∈ Reals, ρc1 > 0, γ ∈ Reals, γ > 0, α ∈ Reals, Assumptions → { ∈ Reals, κ ∈ Reals, κ > 0, ρc1 ∈ Reals, ρc1 > 0, γ ∈ Reals, γ > 0, α ∈ Reals, α > 0, ρΛ ∈ Reals, ρΛ > 0, R ∈ Reals, w ∈ Reals, 1/3 < w < 1}];
α > 0, ρΛ ∈ Reals, ρΛ > 0, R ∈ Reals, w ∈ Reals, 1/3 < w < 1}];
α > 0, ρΛ ∈ Reals, ρΛ > 0, R ∈ Reals, w ∈ Reals, 1/3 < w < 1}];
In this case were not able to verify if the two solutions are the same, since the software is not able to give a True or False output, for the command
Expand[simplify41] == Expand[simplify42]
Expand[simplify41] == Expand[simplify42]
Expand[simplify41] == Expand[simplify42]
B.2.3 Obtaining a series expansion
Since we cannot confirm if the two solutions are the same, we compute the Taylor expansion for the 1/3 < w < 1 case, in order to compare with the series we have obtained for the matter as scalar field case, given by series3. For that we use Series[f, {x, x[f, {x, x[f, {x, x000, n}], n}], n}], which generates a power series expansion for f about the point x = x0to order (x − x0)n, where n is an explicit integer. We only calculate the 0th order term, for it is sufficient to show that it is not the series expansion we have obtained before.
series42 = Series[simplify42, {R, 0, 0}];
series42 = Series[simplify42, {R, 0, 0}];
series42 = Series[simplify42, {R, 0, 0}];
B.2.4 Recovering the w = 1 result
When attempting to compare the general solution, given by simplify42 while substituting w = 1, with the solution previously obtained, given by fullsimplify3, we are also not able to get a True or False answer.
Expand[fullsimplify3/.{ → 1}] == Expand[simplify42/.{w → 1, → 1}]
Expand[fullsimplify3/.{ → 1}] == Expand[simplify42/.{w → 1, → 1}]
Expand[fullsimplify3/.{ → 1}] == Expand[simplify42/.{w → 1, → 1}]
As such, we show the expression we get when we make this substitution, as well as the corresponding series expansion.
3 (R + 2κρc1)2ρΛHypergeometric2F1h
5
+
It is clear we do not recover the particular case of w = 1.
Appendix C
Mathematica codes for mLQC-II
In this appendix we provide a detailed explanation of the Wolfram Mathematica notebook used to deter-mine the effective actions for mLQC-II. Moreover, this appendix is divided in two sections, one for the case of matter as a scalar field and one for the general w case. First of all, we define the modified Fried-mann equation corresponding to mLQC-II, given by eq. (3.51), as H2mLQCII, and Ψ (ρ) as Psirho3, which is generically defined by eq. (5.3) and, in this case, is given by eq. (5.41)
H2mLQCII = 2κρ3
C.1 Matter as a scalar field
C.1.1 Solving the differential equation
We define the modified Friedmann eq. (5.6), as H2flatZeroΛΛΛ. We choose to write the general expression and substitute w by 1. This substitution is indicated by “/.” as follows.
H2flatZeroΛ = (1/3)κρ − (/3)
In this case, we want to solve eq. (5.78), which we define as diffEq5. Notice below how we substitute the energy density ρ, in the whole equation, by its corresponding expression in terms of the lower order Ricci scalar, using eq. (5.8).
diffEq5 = H2flatZeroΛ −κρ3 ==Psirho3/.{ρ → R/(−2κ)}
Using DSolve[eqn,u,x], which solves a differential equation for the function u, with independent variable x, we are able to compute the solution of diffEq5, which we will denote by sol5. In this step, we provide information about the constants κ, ρIIc (ρc2 on the code) and γ, as well as information about the variable R. The presence of Clear[R,φ5,sol5] before DSolve[eqn,u,x] is important, since we have to ensure Wolfram Mathematica solves the differential equations properly. In that regard, we have
experienced that the software can, sometimes, give the wrong solutions, and attempting Clear[R,φ5,sol5]
seems to solve this issue. Moreover, we may also quit the kernel, which leaves the notebook intact, but seems to forget what was done before, including definitions.
Clear[R, φ5, sol5] + 36κ2ρc22Hypergeometric2F1
10κ2ρc22Hypergeometric2F1
− 36κ2ρc22Hypergeometric2F1
10κ2ρc22Hypergeometric2F1
The result is given as a pure function, meaning it is given in the format φ5 → Function[{R}, ...]. This is achived by writing DSolve[eqn,φ5, Rφ5, Rφ5, R] instead of DSolve[eqn,φ5[R], Rφ5[R], Rφ5[R], R], which is also possible. Then, we use Simplify, in order to obtain a first simplication of the solution, where we provide the same as-sumptions as before. In order to obtain a final simplification, we use FullSimplify while setting constants c1 and c2 to zero. Recall that removing the terms of the solution that are multiplied by c1 and c2, only removes the terms that correspond to eq. (5.15). Writting sol5[[1]]sol5[[1]]sol5[[1]] allows us to select the first component of sol5sol5sol5, which is technically a list, although it only contains one component. Since sol5[[1]]sol5[[1]]sol5[[1]] is itself a list with one component, writting sol5[[1]][[1]]sol5[[1]][[1]]sol5[[1]][[1]] allows us to sellect φ5 → Function[{R}, ...], which has two
components, {φ5, Function[{R}, ...]}. Writting sol5[[1]][[1]][[2]]sol5[[1]][[1]][[2]]sol5[[1]][[1]][[2]] sellects its second component which is Function[{R}, ...], which also has two components, {{R}, ...}. Finally, writting sol5[[1]][[1]][[2]][[2]]sol5[[1]][[1]][[2]][[2]]sol5[[1]][[1]][[2]][[2]], selects the actual solution.
fullsimplify5 = FullSimplify[sol5[[1]][[1]][[2]][[2]]/.{C[1] → 0, C[2] → 0, → 1}, fullsimplify5 = FullSimplify[sol5[[1]][[1]][[2]][[2]]/.{C[1] → 0, C[2] → 0, → 1}, fullsimplify5 = FullSimplify[sol5[[1]][[1]][[2]][[2]]/.{C[1] → 0, C[2] → 0, → 1}, Assumptions → {κ ∈ Reals, κ > 0, ρc2 ∈ Reals, ρc2 > 0, γ ∈ Reals, γ > 0, R ∈ Reals}]
C.1.3 Obtaining a series expansion
In order to obtain a Taylor expansion of the solution, we use Series[f, {x, x[f, {x, x[f, {x, x000, n}], n}], n}], which generates a power series expansion for f about the point x = x0to order (x − x0)n, where n is an explicit integer.
FullSimplifyFullSimplifyFullSimplify is then used to further simplify each term. We denote the series, for this specific case, as series3, and we will, later on, compare with the series expansion for the general case.
series5 = Series[fullsimplify5, {R, 0, 3}]//FullSimplify series5 = Series[fullsimplify5, {R, 0, 3}]//FullSimplify series5 = Series[fullsimplify5, {R, 0, 3}]//FullSimplify
(1−2γ2−3γ4)R2
In order to obtain the general result, we solve the differential equation for the two possible intervals of w, separatly, given the fact that eq. (5.1) does not depend smoothly on w, becoming a first order equation for w = 1/3. In that regard, we solve the equation, first, for −1 < w < 1/3 and, then, for 1/3 < w < 1.
We latter show, that the solutions are the same.
C.2.1 Solving the differential equation
−1 < w < 1/3
−1 < w < 1/3
−1 < w < 1/3
We define the modified Friedmann eq. (5.1), as H2flatZeroΛΛΛw1 for this branch of w.
H2flatZeroΛw1 = (1/3)κρ − (/3)
(1/2)(3w + 1)φ610[R]κρ + (1/2)φ61[R] − 3(1 + w)(1 − 3w) H2flatZeroΛw1 = (1/3)κρ − (/3)
In this case, we want to solve eq. (5.78) for φ61[R], which we define as diffEq61. Notice, below, that we substitute the energy density ρ, in the whole equation, by its corresponding expression in terms of the lower order Ricci scalar, using eq. (5.2).
diffEq61 = H2flatZeroΛw1 −κρ3 == Psirho3/.{ρ → R/(−κ(3w − 1))}
Using DSolve[eqn,u,x], we are able to compute the solution of diffEq61, which we will denote by sol61. In this step, we provide information about the constants κ, ρIIc (ρc2 on the code) and γ, as well as information about the variable R and possible values for w.
Clear[φ61, R, sol61]
Here, we define eq. (5.1) as H2flatZeroΛΛΛw2, with φ62[R] instead of φ61[R] so we can consider the second branch of values of w, while not mixing the branches of w.
H2flatZeroΛw2 = (1/3)κρ − (/3)
(1/2)(3w + 1)φ620[R]κρ + (1/2)φ62[R] − 3(1 + w)(1 − 3w) H2flatZeroΛw2 = (1/3)κρ − (/3)
(1/2)(3w + 1)φ620[R]κρ + (1/2)φ62[R] − 3(1 + w)(1 − 3w) H2flatZeroΛw2 = (1/3)κρ − (/3)
In this case, we define the differential eq. (5.78) as diffEq62 and, once again, substitute the energy density ρ by its corresponding expression in terms of the lower order Ricci scalar, using eq. (5.2).
diffEq62 = H2flatZeroΛw2 −κρ3 == Psirho3/.{ρ → R/(−κ(3w − 1))}
Using DSolve[eqn,u,x], we are able to compute the solution of diffEq62, which we will denote by sol22. In this step, we provide information about the constants κ, ρIIc (ρc2 on the code) and γ, as well as information about the variable R and possible values for w.
Clear[sol62, φ62, R]
Once again, the results sol61 and sol62 are given as pure functions, meaning they are given in the for-mat φ61 → Function[{R}, ...] and φ62 → Function[{R}, ...], except, this time, we do not show them explicitly. We use Simplify, in order to obtain a first simplication of the solutions, where we provide the relevant assumptions. In order to obtain a final simplification, for both branches of w, we use FullSim-plify while setting constants c1 and c2 to zero. Recall that removing the terms of the solution that are multiplied by c1and c2, is only removing the solution to the homogeneous eq. (5.15). As before, we also provide the assumptions about the constants and variable.
−1 < w < 1/3
−1 < w < 1/3
−1 < w < 1/3
Once again, writting sol61[[1]][[1]][[2]][[2]]sol61[[1]][[1]][[2]][[2]]sol61[[1]][[1]][[2]][[2]], selects the solution. For more details, we refer to the section C.1.2.
fullsimplify61 = FullSimplify[sol61[[1]][[1]][[2]][[2]]/.{C[1] → 0, C[2] → 0}, fullsimplify61 = FullSimplify[sol61[[1]][[1]][[2]][[2]]/.{C[1] → 0, C[2] → 0}, fullsimplify61 = FullSimplify[sol61[[1]][[1]][[2]][[2]]/.{C[1] → 0, C[2] → 0},
Assumptions → { ∈ Reals, κ ∈ Reals, κ > 0, ρc2 ∈ Reals, ρc2 > 0, γ ∈ Reals, γ > 0, R ∈ Reals,
18(1 + w)(−1 + 3w)pR (73 + 78w + 9w2)γ4κρc2
1/3 < w < 1 1/3 < w < 1 1/3 < w < 1
Once again, writting sol62[[1]][[1]][[2]][[2]]sol62[[1]][[1]][[2]][[2]]sol62[[1]][[1]][[2]][[2]], selects the solution. For more details, we refer to the section C.1.2.
fullsimplify62 = FullSimplify[sol62[[1]][[1]][[2]][[2]]/.{C[1] → 0, C[2] → 0}, fullsimplify62 = FullSimplify[sol62[[1]][[1]][[2]][[2]]/.{C[1] → 0, C[2] → 0}, fullsimplify62 = FullSimplify[sol62[[1]][[1]][[2]][[2]]/.{C[1] → 0, C[2] → 0},
Assumptions → { ∈ Reals, κ ∈ Reals, κ > 0, ρc2 ∈ Reals, ρc2 > 0, γ ∈ Reals, γ > 0,
Finally we are able to verify that, in actuality, the two solutions are the same. We use Expand so that the software is able to compare all the terms, individually.
Expand[fullsimplify61] == Expand[fullsimplify62]
Expand[fullsimplify61] == Expand[fullsimplify62]
Expand[fullsimplify61] == Expand[fullsimplify62]
True
C.2.3 Obtaining a series expansion
Since the two solutions are the same, we compute the Taylor expansion of one of them. For that we use Series[f, {x, x[f, {x, x[f, {x, x000, n}], n}], n}], which generates a power series expansion for f about the point x = x0 to order
(x − x0)n, where n is an explicit integer. FullSimplifyFullSimplifyFullSimplify is then used to further simplify each term.
series62 = Series[fullsimplify62, {R, 0, 3}]//FullSimplify series62 = Series[fullsimplify62, {R, 0, 3}]//FullSimplify series62 = Series[fullsimplify62, {R, 0, 3}]//FullSimplify
− (2−4γ2−6γ4)R2
9κρc2−18wκρc2−27w2κρc2 − 2
γ2(1+γ2)3R3
(1−3w)2(16+15w)κ2ρc22 + O[R]7/2
C.2.4 Recovering the w = 1 result
Here, we show that when substituting w = 1 on the general solution, given by fullsimplify62, we recover the matter as a scalar field solution, given by fullsimplify5 and its corresponding series expansion.
Expand[fullsimplify5/.{ → 1}] == Expand[fullsimplify62/.{ → 1, w → 1}]//FullSimplify Expand[fullsimplify5/.{ → 1}] == Expand[fullsimplify62/.{ → 1, w → 1}]//FullSimplify Expand[fullsimplify5/.{ → 1}] == Expand[fullsimplify62/.{ → 1, w → 1}]//FullSimplify True
series62/.{w → 1}
series62/.{w → 1}
series62/.{w → 1}
(2−4γ2−6γ4)R2
36κρc2 −
γ2(1+γ2)3R3
62(κ2ρc22) + O[R]7/2
Normal converts the series into a normal expression, i.e. it removes O from the series. When comparing expressions on Wolfram Mathematica we sometimes have to add an extra FullSimplify in order to get a True or False output, otherwise it can give an expression equal to another expression as an output.
Normal[series5] == Normal[series62/.{w → 1, → 1}]//FullSimplify Normal[series5] == Normal[series62/.{w → 1, → 1}]//FullSimplify Normal[series5] == Normal[series62/.{w → 1, → 1}]//FullSimplify True