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?
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Sign up to join this communityThe 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!
r*=r+2Log[r-2m], m>0
shows, that real r*
only exist if r>2m
. Thus the singularity V[0+]
is excluded .
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Mar 2, 2020 at 8:03
FindMaximum
gives the maximum for the variabelr
I do not know how to find the maximum inr^*
$\endgroup$V
defined in terms of the coordinater', but I need the maximum of the potential in terms of the variable
r*` whch depends onr
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