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I am trying to find the value of $r_0$ using $L^2=\frac{{r_0}^3 f'({r_0})}{2 f({r_0})-{r_0} f'({r_0})}$ and making a 3d plot with the help of the code given below:

$Assumptions = Thread[{l, L, Q, r0, rp} > 0];
f[r_] := 1 - (2 M)/r + Q^2/r^2 + r^2/l^2;
M = rp/2 (1 + Q^2/rp^2 + rp^2/l^2);
Veff[r_] = f[r] (L^2/r^2 + 1)
eqn = L^2 == (r0^3 f'[r0])/(2 f[r0] - r0 f'[r0]) // Simplify;
sol = Reduce[eqn, r0][[2, 2, 1]] // ToRules;
\[Lambda][Q_, rp_] = 
  Block[{l = 1, L = 20}, 
    1/2 Sqrt[(r0 f'[r0] - 2 f[r0]) Veff''[r0]] /. sol] // Simplify;
Plot3D[Log[100, (\[Lambda][Q, rp] + 1)], {Q, 0, 1.2}, {rp, 0, 1}]

which gives me root objects as output instead of algebraic expressions.
I tried ToRadicalsas discussed in other related answers but it didn't work for me.
The resulting 3d plot is also blank.
How to solve this problem?
Reference paper (page 8)

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  • $\begingroup$ The Root objects are roots of 6th order polynomials, where no general solution in terms of radicals exists $\endgroup$
    – Lukas Lang
    Sep 6, 2022 at 14:34
  • $\begingroup$ Thanks for the clarification, but then how can I express $r_0$ in terms of $L$ as stated in the reference paper? Is there any possibility to do that? $\endgroup$
    – codebpr
    Sep 6, 2022 at 14:41
  • $\begingroup$ In the paper they say, just above (3.10), that (3.9) "can be used to express $r_0$ in terms of $L$", but they don't give any expression or even imply it's analytic in terms of radicals ;-) $\endgroup$
    – Hans Olo
    Sep 6, 2022 at 14:52
  • $\begingroup$ @HansOlo that's what troubles me. Even if I use the root object as input, I don't get the desired 3D plot as given in Figure 6, for the massive particles, instead I get a blank plot. $\endgroup$
    – codebpr
    Sep 6, 2022 at 15:00
  • $\begingroup$ All of your variables have positive values. This knowledge should be shared with Mathematica. By using $Assumptions = Thread[{l, L, Q, r0, rp} > 0]; the assumptions will automatically be available to any function that uses the option Assumptions. After defining $Assumptions, simplify the equation, eqn = L^2 == (r0^3 f'[r0])/(2 f[r0]-r0 f'[r0])//Simplify;If you are having trouble plotting, show the code that you used for the plot. $\endgroup$
    – Bob Hanlon
    Sep 6, 2022 at 15:37

1 Answer 1

Reset to default
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Clear["Global`*"]

$Assumptions = Thread[{l, L, Q, r0, rp} > 0];

f[r_] := 1 - (2 M)/r + Q^2/r^2 + r^2/l^2;
M = rp/2 (1 + Q^2/rp^2 + rp^2/l^2);
Veff[r_] = f[r] (L^2/r^2 + 1);
eqn = L^2 == (r0^3 f'[r0])/(2 f[r0] - r0 f'[r0]) //
   Simplify;

sol = {Reduce[eqn, r0] // Simplify // ToRules}[[2 ;; 3]];

λ[Q_, rp_] = Block[{l = 1, L = 20},
    1/2 Sqrt[(r0 f'[r0] - 2 f[r0]) Veff''[r0]] /. 
     sol] // Simplify;

Column[
 Plot3D[Evaluate@Log[100, (# + 1)],
    {Q, 0, 6/5}, {rp, 0, 1},
    PlotPoints -> 60,
    MaxRecursion -> 3,
    ClippingStyle -> None,
    WorkingPrecision -> 15,
    AxesLabel ->
     {Q, rp, 
      Rotate[HoldForm@Log[100, λ[Q, rp] + 1], 90 Degree]},
    ImageSize -> 480] & /@
  λ[Q, rp]]

enter image description here

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  • $\begingroup$ Thank you for the illuminating answer. It helped me getting started. The plot is very close to the Figure 6 on page 8 of the reference paper. There are minor differences, which maybe due to the difference in the plot range. $\endgroup$
    – codebpr
    Sep 6, 2022 at 18:43

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