0
$\begingroup$

from here: $dl^2 = dr^2 + \frac{r^2 dr^2}{\kappa^{-1}R^2 -r^2}$ i got $dl^2 = dr^2 -\frac{(dr)^2}{1-\frac {R^2}{\kappa r^2}}$

but here is different: $dl^2 = \frac{dr^2}{1 - \kappa\frac{r^2}{R^2}} + r^2d\theta^2 + r^2\sin^2\theta d\phi^2,$ which one is correct $(1- \frac {R^2}{\kappa r^2}),$

${1 - \kappa\frac{r^2}{R^2}}$

and why?

  • 0
    @wisefool en.wikipedia.org/wiki/Curved_space2012-12-17

1 Answers 1

2

Ok, I think I understood your question. You started from $dl^2=dx^2+dy^2+dz^w+\frac{(xdx+ydy+zdz)^2}{\kappa^{-1}R^2-x^2-y^2-z^2}$ and tried to pass to spherical coordinates, as it is done on the wiki page.

Set $x=r\sin\theta\cos\phi$, $y=r\sin\theta\sin\phi$, $z=r\cos\theta$, then the Jacobian matrix is $J=\begin{pmatrix}\sin\theta\cos\phi&r\cos\theta\cos\phi&-r\sin\theta\sin\phi\\ \sin\theta\sin\phi&r\cos\theta\sin\phi&r\sin\theta\cos\phi\\\cos\theta&-r\sin\theta&0\end{pmatrix}$ so the metric $dx^2+dy^2+dz^2$ in these new coordinates is given by $g=J^tJ=\begin{pmatrix}1&0&0\\0&r^2&0\\0&0&r^2\sin^2\theta\end{pmatrix}$ i.e. $dr^2+r^2d\theta^2+r^2\sin^2\theta d\phi^2$.

So $dl^2=dr^2+r^2d\theta^2+r^2\sin^2\theta d\phi^2+\frac{r^2dr^2}{\kappa^{-1}R^2-r^2}$ from which $dl^2=dr^2\left(1+\frac{1}{\frac{R^2}{\kappa r^2}-1}\right)+r^2d\theta^2+r^2\sin^2\theta d\phi^2$ and $1+\frac{1}{\frac{R^2}{\kappa r^2}-1}=\frac{\frac{R^2}{\kappa r^2}}{\frac{R^2}{\kappa r^2}-1}=\frac{1}{1-\frac{\kappa r^2}{R^2}}$ so, plugging this into the previous, you obtain $dl^2=\frac{dr^2}{1-\frac{\kappa r^2}{R^2}}+r^2d\theta^2+r^2\sin^2\theta d\phi^2\;.$ Does this answer to your doubt?

  • 0
    ok tanQ its cooooool2012-12-17