Dirk Van de moortel wrote:
> "Ikke" <ikkepoker@hotmail.com> wrote in message
> news:1156243687.592259.294440@b28g2000cwb.googlegroups.com...
> > Hello,
> >
> > I really hope I could get an answer here, because I couldnt find it on
> > the internet and cant contact my prof atm.
> >
> > I have to calculate the curvature scalar for spherical coordinates
> > (theta, phi). So the metric tensor is:
> >
> > g_00 = 1 , g_11 = sin^2(theta), g_01 = g_10 = 0
> >
> >>From my calculations I get:
> >
> > R=-1/tan^2(theta)
>
> The Ricci scalar is -2.
> You probably made a little mistake somewhere.

Ricci Tensor is zero in free space even under the curvature of
spacetime.  Thus, you definitely made a mistake somewhere.  Ricci
scalar is ZERO for flat spacetime (in this case) with the following
metric.

**  g_00 = g_11 = 1
**  g_22 = r^2 cos^2(phi)
**  g_33 = r^2
**  Others = 0

Where

**  phi = latitude

Ricci scalar is a very poor gauge to measure the curvature of
spacetime.  Hilbert's Lagrangian that leads to Einstein Field
Equations is totally unqualified and unjustified.