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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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    The normal to a $f=$constant surface is $n_\mu = \partial_\mu f$ (lowered index, not raised). $\partial_t$ is not supposed to normal to the surface. It generates the null geodesics (so it is null) and also happens to be a Killing vector field. – Prahar Apr 24 '23 at 07:12

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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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  • I think you have answered my recent question "Does condition $g_{00}(r_{0})=0$ define the event horizon on $r_{0}$?" (https://physics.stackexchange.com/q/759098/281096), too. Would you mind to write your answer there? – JanG Apr 24 '23 at 14:24