I am trying to understand why in the definition of a stationary spacetime the Killing vector field has to be timelike.(adsbygoogle = window.adsbygoogle || []).push({});

It is required that the metric is time independent, i.e. the time translations [itex]x^0 \to x^0 + \epsilon[/itex] leave the metric unchanged. So the Killing vector is [itex]\xi^{\mu}=\delta_{0}^{\mu}[/itex]. In other words it is required that there exists a Killing vector that has only [itex]x^{0}[/itex] component and no space components (they are 0). But in general, a timlike vector can also have non zero space components. For example, the vector [itex](1,1/2)[/itex] in two dimensional Minkowski space, is timelike because [itex](1,1/\sqrt{2})^2=-1/2<0[/itex]. This vector is a linear combination of the two independent killing vectors (1,0) (time translations) and (0,1) (space translations) of the 2-d Minkowski spacetime, so it's also a killing vector and it generates translations in time and also in space.

So, if there exists a timelike killing vector with non zero space components, then necessarily there exists the killing vector [itex]\delta_{0}^{\mu}[/itex]. So saying that there should exist timelike Killing vector is because of this? It is simply a more general statement than saying that "spacetime is stationary if it admits the Killing vector field [itex]\delta_{\mu}^{0}[/itex]"?

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# Timelike Killing vector field and stationary spacetime

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