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I have the following Mathematica code:

$Assumptions = {r \[Element] Reals, r >= 0, rh \[Element] Reals, 
   rh > 0, qe \[Element] Reals, qe >= 0, qm \[Element] Reals, qm >= 0,
    z \[Element] Reals, z >= 0, zh \[Element] Reals, zh >= 0};

S[rh_] := \[Pi] rh^2

M[rh_, qe_, qm_] := 
 1/2 Sqrt[\[Pi]/S[rh]] (S[rh]^2/\[Pi]^2 + S[rh]/\[Pi]  + qe^2 + qm^2)

qe = (r* rh* \[Phi]e)/(r - rh);

f[r_, rh_, qm_, \[Phi]e_] := 
 1 + r^2/l^2 - (2 M[rh, qe, qm])/r + (qe^2 + qm^2)/r^2

Veff[r_] = f[r, rh, qm, \[Phi]e]*(L^2/r^2);

r0 = Solve[Simplify[Veff'[r] == 0], r, Reals][[2]];

\[Lambda][rh_, qm_, \[Phi]e_, l_] = 
  1/Sqrt[2] Sqrt[-((r^2 f[r, rh, qm, \[Phi]e])/L^2) Veff''[r]] /. r0 //
    Simplify;

When I try to plot $\lambda$ v/s $r_h$ for specific values of $q_m$ and $\phi_e$, I find that $\lambda$ saturates to the value $\frac{1}{l}$ when I try using different values of $l$ say 1,2,3 etc. To make it concrete, I also try to find the relation of $\lambda$ v/s $l$ graphically by using the following line of code:

\[Lambda]sat[q_, \[Phi]_, l_] = \[Lambda][100, q, \[Phi], l];

Plot[{\[Lambda]sat[1, 1, l], l^-1}, {l, 0, 100}]

As you can visually check that the relation holds from the plot. My only desire is to prove this analytically. I tried using limits and series method on Mathematica but in vain. How to use Mathematica to prove the relation $\lambda_{sat}=1/l$ analytically ?

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  • $\begingroup$ Maybe using Limit? I leave it to someone else to show how, for some easy rep. $\endgroup$
    – Hans Olo
    Commented Aug 29, 2023 at 7:24
  • $\begingroup$ I tried using that but somehow Mathematica isn't able to give a solution for Limit $rh \rightarrow \infty$ . $\endgroup$
    – codebpr
    Commented Aug 29, 2023 at 8:34
  • $\begingroup$ If you think my answer is useful please consider upvoting and accepting it. If not, let me know if something more is needed :-) $\endgroup$
    – Hans Olo
    Commented Aug 29, 2023 at 19:52
  • $\begingroup$ Sorry for the delay! I just saw your answer! Let me make some useful comments there! $\endgroup$
    – codebpr
    Commented Aug 30, 2023 at 4:46

1 Answer 1

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Printing the expression we see that it is of the form c1 Sqrt[1 + c2/l^2], where c1,c2 are appropriate constants.

Thus, doing some series expansions at small and high l we easily find:

Series[\[Lambda]sat[1, 1, l], {l, 0, 1}]//N//PowerExpand//Rationalize
(* 1./l+7.40445*10^-14 l+O[l]^3 *)

and

Series[\[Lambda]sat[1, 1,l],{l,\[Infinity],2}]//N//PowerExpand//Rationalize
(* 3.84823*10^-7+1.2993*10^6/l^2+O[1/l]^3 *)

In other words, the expression has two regimes:

  1. At low-l \[Lambda]sat[1, 1, l]~1/l
  2. At high-l \[Lambda]sat[1, 1, l]~b1+b2/l^2, where b1,b2 are some constants given in the series expansion above.

A plot confirms this and shows the transition between the regimes happens at l~10^6:

enter image description here

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  • $\begingroup$ Your analysis is spot on but it is for $r_h=100$. I basically need the $\lambda_{sat}$ at $r_h \rightarrow \infty$, so let's say we take $r_h = 10^{10}$, would you find the same results or different ? $\endgroup$
    – codebpr
    Commented Aug 30, 2023 at 5:00
  • $\begingroup$ The functional form is the same yes, as the general expression is c1 Sqrt[1 + c2/l^2], which has the low-l and high-l limits mentioned above ;-) $\endgroup$
    – Hans Olo
    Commented Aug 30, 2023 at 8:05
  • $\begingroup$ So is there a relation between $r_h$ and $l$ like how big can we go? $\endgroup$
    – codebpr
    Commented Aug 30, 2023 at 12:49
  • $\begingroup$ The relation is analytic, so you can go as high up as you want! Please play with the Series[] function ;-) $\endgroup$
    – Hans Olo
    Commented Aug 30, 2023 at 13:08
  • $\begingroup$ Let me provide you some context. Here l represent the inverse of intrinsic curvature so basically at very large values of l, the system is practically flat. So, if we think about your analysis it basically gives us a limit of curvature. I was thinking of finding some analytical proof but still don't know how to proceed. What if I use this code: Limit[\[Lambda][rh, 1, 1, l], rh -> \[Infinity]], will it work ? $\endgroup$
    – codebpr
    Commented Aug 30, 2023 at 13:28

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