I am looking for the maximum of the function

$$V(r^*)=\left( 1-\dfrac{2m}{r} \right)\left( \dfrac{l(l+1)}{r^2}-\dfrac{6m}{r^3} \right)$$

in the coordinate $r^*$ for given $l$ and $m$, where $r^*=r+2m \ln(r-2m)$.

How can I do it?

  • $\begingroup$ what have you tried? $\endgroup$ – AccidentalFourierTransform Mar 1 at 16:29
  • $\begingroup$ @AccidentalFourierTransform I am clueless, FindMaximum gives the maximum for the variabel r I do not know how to find the maximum in r^* $\endgroup$ – mattiav27 Mar 1 at 16:31
  • 1
    $\begingroup$ find the maximum in terms of $r$, and then plug that expression into $r^*$? $\endgroup$ – AccidentalFourierTransform Mar 1 at 16:32
  • $\begingroup$ Please clarify the function definition! $\endgroup$ – Ulrich Neumann Mar 1 at 16:36
  • $\begingroup$ @UlrichNeumann I have the potential V defined in terms of the coordinate r', but I need the maximum of the potential in terms of the variable r*` whch depends on r $\endgroup$ – mattiav27 Mar 1 at 16:39

The necessary condition to maximize is D[V,r]/D[r*,r] which can be solved analytically

sol=Solve[0 == D[(1 - 2 m/r) ((l (l + 1))/r^2 - (6 m)/r^3), r]/D[r + 2 m Log[r - 2 m], r], r]
(* {{r -> 2 m}, 
{r -> (9 m + 3 l m + 3 l^2 m - 
Sqrt[-96 (l + l^2) m^2 + (-9 m - 3 l m - 3 l^2 m)^2])/(2 (l + l^2))}, 
{r -> (9 m + 3 l m + 3 l^2 m + 
Sqrt[-96 (l + l^2) m^2 + (-9 m - 3 l m - 3 l^2 m)^2])/(2 (l + l^2))}}*)

Knowing the parameters l,m you might check the result to be a maximum!

| improve this answer | |
  • $\begingroup$ ...i'm kinda confused. The function $V(z)$ clearly diverges (to $\infty$) at $z\to0^+$, so how can it have a maximum at all? Your solutions seem to be minima (or local maxima), not a global maximum! $\endgroup$ – AccidentalFourierTransform Mar 1 at 17:38
  • $\begingroup$ @AccidentalFourierTransform It's only a necessary condition for an extremum. I didn't check for maximum! The influence of the constraint shoulden'd be disregared. $\endgroup$ – Ulrich Neumann Mar 1 at 18:51
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    $\begingroup$ I'm not sure what you mean by "the constraint", but the function is unbounded from above for all $m,l$, so the problem has no solution... $\endgroup$ – AccidentalFourierTransform Mar 1 at 19:15
  • $\begingroup$ @AccidentalFourierTransform The constraint (sorry for my wording) r*=r+2Log[r-2m], m>0 shows, that real r* only exist if r>2m. Thus the singularity V[0+] is excluded . $\endgroup$ – Ulrich Neumann Mar 2 at 8:03
  • $\begingroup$ Ah, I see, that makes sense! (But do note that $r=2m$ is a boundary, so the maximum could in principle be located there. Strictly speaking, one should test that value too, together with the three you write in your answer). $\endgroup$ – AccidentalFourierTransform Mar 3 at 22:37

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