-1
$\begingroup$

I have the following:

\[CapitalSigma] = r^2 + a^2 Cos[\[Theta]]^2;
\[CapitalDelta] = r^2 - 2 M r + a^2 - k/3 r^2 (r^2 + a^2);
grr = \[CapitalSigma]/\[CapitalDelta];

and I want to obtain the expansion of grr up to linear terms in $M/r$ and $k r^2$. What I have done is to employ Series but I could not get the right answer which is $$1+\frac{2M}{r}-\frac{kr^2}{3}$$ Does anyone have an idea as to what I am missing here? Thanks

Improve this question
$\endgroup$
5
  • 2
    $\begingroup$ Does this answer your question? Multivariable Taylor expansion does not work as expected $\endgroup$
    – Carl Woll
    Commented Aug 16, 2021 at 16:30
  • $\begingroup$ Why is there no dependence on a? $\endgroup$
    – Carl Woll
    Commented Aug 16, 2021 at 16:34
  • $\begingroup$ That also bothers me @CarlWoll, although the instruction is to expand up to linear terms in $M/r$ and $kr^2$ only $\endgroup$
    – user583893
    Commented Aug 16, 2021 at 16:41
  • $\begingroup$ @CarlWoll, I am only expanding with respect to $r$ for large $r$. $\endgroup$
    – user583893
    Commented Aug 16, 2021 at 16:48
  • $\begingroup$ Is the required true? Limit[grr, r -> Infinity] produces 0. $\endgroup$
    – user64494
    Commented Aug 16, 2021 at 16:50

1 Answer 1

Reset to default
0
$\begingroup$

Assuming k r^2, M/r, a/r to be small of same order, expansion gives nearly the result you're expecting:

grr /. {M -> r eps M/r , k -> 1/r^2 eps k r^2,a -> r  eps a /r } // Simplify  
Normal[Series[%, {eps, 0, 1}]] /. eps -> 1
(*1 + (2 M)/r + (k r^2)/3*)
Improve this answer
$\endgroup$
4
  • $\begingroup$ Pay your attention to the minus sign - k r^2/3 in the required expression. $\endgroup$
    – user64494
    Commented Aug 16, 2021 at 18:50
  • $\begingroup$ What a pity, thanks. $\endgroup$
    – Ulrich Neumann
    Commented Aug 16, 2021 at 19:01
  • $\begingroup$ Your solution helps @UlrichNeumann, although the sign of the third term "should" be negative. $\endgroup$
    – user583893
    Commented Aug 17, 2021 at 1:40
  • $\begingroup$ Ok , I think you need to define the order of parameter a too. Otherwice it isn't possible to do the series expansion. $\endgroup$
    – Ulrich Neumann
    Commented Aug 17, 2021 at 6:26

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