I am trying to reproduce Figure 4 from this paper (page 7), for which I need to use the line $\tilde{T_p}$ from this figure:
in the output of this figure:
Now by varying the value of $\tilde{Q}$, I need to find the first and last intersection point that the line $\tilde{T_p}$ makes with the above curve(in the blue and green region only) for all the generated plots and then make a plot showing the difference of the intersection points v/s $t$ as shown below:
The partial code that I used to generate the intersection points is given below:
T[rp_, Q_] := 1/(4 \[Pi] rp) (1 - Q^2/rp^2 + (3 rp^2)/l^2)
M[rp_, Q_] := rp/2 (1 + Q^2/rp^2 + rp^2/l^2)
S[rp_] := \[Pi] rp^2
\[Phi][rp_, Q_] := Q/rp
F[rp_, Q_] := Simplify[M[rp, Q] - (T[rp, Q] S[rp])]
Q = Qt l;
rp = rpt l;
Tt[rpt_, Qt_] = T[rp, Q] l;
Ft[rpt_, Qt_] = F[rp, Q]/l;
Mt[rpt_, Qt_] = M[rp, Q]/l;
r = rt l;
{rptc, Qtc} =
SolveValues[{D[Tt[rpt, Qt], {rpt, 2}] == 0,
D[Tt[rpt, Qt], rpt] == 0}, {rpt, Qt}][[4]];
Ttc = Tt[rpt, Qt] /. rpt -> rptc /. Qt -> Qtc;
\[Phi]c = \[Phi][rp, Q] /. Qt -> Qtc /. rpt -> rptc;
f[r_, rpt_, Qt_] := 1 - (2 Mt[rpt, Qt])/r + Q^2/r^2 + r^2/l^2
pplt[Qt_] :=
ParametricPlot[Evaluate@{Tt[rpt, Qt], Ft[rpt, Qt]}, {rpt, 0.01, 3},
PlotRange -> {{0.2, 3.2}, {0, 0.15}}, AspectRatio -> 1]
ip[Qt_] := Graphics`Mesh`FindIntersections[pplt[Qt]][[1]]
Tp = Table[ip[Qt][[1]], {Qt, 0.01, 0.166, 0.001}];
ClearAll[l, L, \[Delta]1, Qt, r0];
Veff[r_] := f[r, rpt, Qt] (L^2/r^2 + \[Delta]1)
eqn = L^2 == (r0^3 D[f[r0, rpt, Qt], r0])/(2 f[r0, rpt, Qt] -
r0 D[f[r0, rpt, Qt], r0]) // Simplify;
sol = {Reduce[eqn, r0] // Simplify // ToRules}[[5]];
l = 1; \[Delta]1 = 1; L = 20 l;
\[Lambda][rpt_, Qt_] =
1/2 Sqrt[(r0 D[f[r0, rpt, Qt], r0] - 2 f[r0, rpt, Qt]) Veff''[r0]] /.
sol // Simplify;
Qtt = Table[Qt, {Qt, 0.01, 0.166, 0.001}];
pplt1[Qt_] :=
ParametricPlot[{Tt[rpt, Qt], \[Lambda][rpt, Qt]}, {rpt, 0.01, 2},
PlotRange -> {{0.0, 2}, {0, 26.00}}, AspectRatio -> 1,
AxesLabel -> {"T", "\[Lambda]"}, PlotPoints -> 500]
also,
\[Lambda]c = \[Lambda][rptc, Qtc]
t = Tp/Ttc;
Any help in this regard would be truly beneficial!