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The Schwarzschild horizon $r=2GM$ should have normal vectors proportional to $$\nabla^\mu r=g^{\mu \nu}\delta^r_\nu$$ Isn't it? Then I fail to understand how could one have Killing vector field $\partial_t$ being normal to the surface.
More generally, given a surface $$\Sigma:f(x)=0,$$ how could we tell if there is a Killing vector field which is normal to the surface at every spacetime point and find out that Killing vector field?

Qmechanic
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Rescy_
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1 Answers1

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I just realised this is another 'confuse yourself' question which requires better understanding of the definition. In case someone in the same scenario as I did, so the definition of a Killing Horizon is a null hypersurface whose normal vectors are Killing vectors.

Now we are given the Killing vector $\partial_t$ so we compute its norm which is simply $g_{tt}$ of our usual Schwarzschild metric $$g_{tt}=-\left(1-\frac{2GM}{r}\right)$$

The requirement of this Killing vector field being null corresponds to $r=2GM$ which is our event horizon.

Rescy_
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