0
$\begingroup$

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?

$\endgroup$
9
  • $\begingroup$ what have you tried? $\endgroup$ Mar 1, 2020 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, 2020 at 16:31
  • 1
    $\begingroup$ find the maximum in terms of $r$, and then plug that expression into $r^*$? $\endgroup$ Mar 1, 2020 at 16:32
  • $\begingroup$ Please clarify the function definition! $\endgroup$ Mar 1, 2020 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, 2020 at 16:39

1 Answer 1

2
$\begingroup$

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!

$\endgroup$
5
  • $\begingroup$ ...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! $\endgroup$ Mar 1, 2020 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. $\endgroup$ Mar 1, 2020 at 18:51
  • 1
    $\begingroup$ 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... $\endgroup$ Mar 1, 2020 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 . $\endgroup$ Mar 2, 2020 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). $\endgroup$ Mar 3, 2020 at 22:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.