Solving GR:Tetrad Problem in Schwarzschild Spacetime

[Mentz114]

Homework Statement

I'm trying to find the tetrad and dual tetrad described by E.Poisson here,
http://www.livingreviews.org/lrr-2004-6
in the section on Fermi coordinates 3.2.2 ( equations 113, 114, 115), for a specific radial geodesic in the Schwarzschild spacetime.

Homework Equations

The geodesic is given by the four-velocity
U^\mu= \left( \frac{r}{r-2M}, \sqrt{\frac{2M}{r}},0,0 \right)
U_\mu= \left( -1, \frac{r}{r-2M} \sqrt{\frac{2M}{r}},0,0 \right)

The Attempt at a Solution

The tetrad is U^{\bar{\mu}},e^{\bar{\mu}}_a where the tetrad index 'a' runs from 1 to 3. I take this to mean that the time-like basis vector is just U^\mu

\vec{e}_0= \frac{r}{r-2M}\partial_t+\sqrt{\frac{2M}{r}}\partial_r

and in order to satisfy g_{\mu\nu}e^{\mu}_a e^{\nu}_b=\delta_{ab} I calculated the other basis vectors,

\vec{e}_1= \frac{\sqrt{2Mr}}{r-2M}\partial_t+\partial_r,\ \ \vec{e}_2=\partial_{\theta}/r,\ \ \ \vec{e}_3=\partial_{\phi}/(r\sin(\theta))<br />

The orthonormality condition is satisfied by these basis vectors, and U^\mu is a normalised tangent vector, but it comes apart because this equation (115)

g^{\mu\nu}=-U^\mu U^\nu + \delta^{ab}e^\mu_a e^\nu_b is not satisfied.

What I get doesn't look like the inverse of any form of the Schwarzschild metric I know.

<br /> \left[ \begin{array}{cccc}<br /> -\frac{{r}^{2}}{{\left( 2\,M-r\right) }^{2}} &amp; \frac{\sqrt{2}\,\sqrt{r}\,\sqrt{M}}{2\,M-r} &amp; 0 &amp; 0\\\<br /> \frac{\sqrt{2}\,\sqrt{r}\,\sqrt{M}}{2\,M-r} &amp; -\frac{2\,M-r}{r} &amp; 0 &amp; 0\\\<br /> 0 &amp; 0 &amp; \frac{1}{{r}^{2}} &amp; 0\\\<br /> 0 &amp; 0 &amp; 0 &amp; \frac{1}{{r}^{2}\,{sin\left( \theta\right) }^{2}}<br /> \end{array} \right]<br />

although the spatial parts are correct. I don't think I've made a calculation error, but maybe misunderstood something ? Any help would be much appreciated.

 

[Mentz114]

The metric from the inverse above is

<br /> \left[ \begin{array}{cccc}<br /> <br /> \frac{{\left( 2\,M-r\right) }^{3}}{{r}^{3}} &amp; \sqrt{ \frac{2M}{r} }\left( \frac{2M-r}{r} \right)<br /> &amp; 0 &amp; 0\\\<br /> \sqrt{ \frac{2M}{r} }\left( \frac{2M-r}{r} \right)<br /> &amp; 1 &amp; 0 &amp; 0\\\<br /> 0 &amp; 0 &amp; {r}^{2} &amp; 0\\\<br /> 0 &amp; 0 &amp; 0 &amp; {r}^{2}\,{sin\left( \theta\right) }^{2}<br /> \end{array} \right]<br />

Which, with the transformation

dT=(1-2M/r)dt

is the Schwarzschild metric in Gull-Painleve coords . I don't know why I didn't see that straight away ...