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In the Kruskal extension of the Schwarzschild metric, where the metric is now transformed into the new coordinates [tex](T,X,\theta,\phi)[/tex], the line element is

[tex]ds^2=-\frac{32M^3e^{-r/2m} }{r}(-dT^2+dX^2)+r^2(d\theta^2+\sin^2(\theta)d\phi^2),[/tex]

with

[tex]\left (r/2m-1 \right ) e^{r/2m}=X^2-T^2 [/tex] (1)

and [tex]t/2m=2\tanh^{-1}(T/X).[/tex] (2)

From the equation (1) we see that [tex]\nabla_{\alpha}r=0[/tex] at [tex]X=T=0[/tex] and this is obvious. Okay, but we know that the static Killing field [tex]\xi^{\alpha}[/tex] becomes collinear with [tex]\nabla_{\alpha}r=0[/tex] thus requiring [tex]\xi^{\alpha}[/tex] to also vanish there. My question is that how is this possible? Looking at the static Killing vectors of the Schwarzschild metric in a general Cartesian-like coordinate system [tex]x^{\mu}[/tex],

[tex]\xi^0=0[/tex] and [tex]{\xi}^{i}={\epsilon}^{{ik}}{x}^{k}[/tex]

where

[tex]\epsilon^{ik}=-\epsilon^{ki} =\left[ \begin {array}{ccc} 0&a&-b\\ \noalign{\medskip}-a&0&c

\\ \noalign{\medskip}b&-c&0\end {array} \right], [/tex]

with [tex]a,b,c[/tex] being all arbitrary constants, how come the above requirement [tex](X=T=0)[/tex] gets all Killing vectors to vanish? My own view on the problem is that we must find the explicit expressions for [tex]r[/tex] and [tex]t[/tex] from (1) and (2), respectively, and then calculate the Killing vectors and put [tex]X=T=0.[/tex] Am I on the right track or what?

Thanks in advance

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# The static Killing vectors in Kruskal coordinates

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