# How do I find the maximum of this quantity?

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?

• what have you tried? Mar 1, 2020 at 16:29
• @AccidentalFourierTransform I am clueless, FindMaximum gives the maximum for the variabel r I do not know how to find the maximum in r^* Mar 1, 2020 at 16:31
• find the maximum in terms of $r$, and then plug that expression into $r^*$? Mar 1, 2020 at 16:32
• Please clarify the function definition! Mar 1, 2020 at 16:36
• @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 Mar 1, 2020 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!

• ...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! Mar 1, 2020 at 17:38
• @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. Mar 1, 2020 at 18:51
• 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... Mar 1, 2020 at 19:15
• @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 . Mar 2, 2020 at 8:03
• 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). Mar 3, 2020 at 22:37