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Given a Killing Horizon $\mathcal{N}$ of Killing vector field $\xi$, one can prove that the surface gravity $\kappa$ is constant on orbits of $\xi$, i.e $$\xi\cdot\partial\kappa^2 = 0.$$ What is the meaning of orbit of Killing vector field? Do they cover all of the Killing Horizon? (this second question, is because $\xi$ is supposed to be normal and tangent to the hypersurface $\mathcal{N}$, and therefore it generates the latter.)

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  • $\begingroup$ Do you know what the orbit of a vector field (or rather the orbit of its flow) is in general? $\endgroup$
    – ACuriousMind
    May 22, 2020 at 22:18
  • $\begingroup$ @ACuriousMind No, I am only familiar with orbits that relate to motion in physics (planets for instance). $\endgroup$
    – devCharaf
    May 23, 2020 at 8:12

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Since the Lie bracket of two Killing fields is a Killing field, the Killing fields form a Lie subalgebra of the vector fields - which is the flow or the orbit of the Killing field.

For instance, in Minkowski space-time, with the line element $$ds^2=-dt^2+dx^2+dy^2+dz^2,$$ there are $10$ Killing fields, namely, $$\partial_t,\partial_x,\partial_y,\partial_z,$$ $$-y\partial_x + x \partial_y,z\partial_y+y\partial_z,z\partial_x-x\partial_z,$$ $$x\partial_t + ct \partial_x, y\partial_t + ct \partial_y, z\partial_t + ct \partial_z.$$

which are the infinitesimal vector generators for translations, Lorentz rotations and Lorentz boosts, respectively.

The Killing Horizon is where the magnitude of the Killing vector vanishes which creates null hypersurfaces similar to the light cone.

As an example, let $\xi=x\partial_t + ct \partial_x$ be the Killing vector for the Lorentz boost in the $x$-$t$ plane. Then setting the square of the magnitude of $\xi$ to $0$

$$g(\xi,\xi)=x^2-(ct)^2=(x-ct)(x+ct)=0$$

where the Killing Horizon is $(x-ct)=0$ and $(x+ct)=0$ taken together.

And $$\xi\cdot\partial\kappa^2 = 0.$$

However, in Minkowski space-time $\kappa=0$ so the Killing Horizon is considered degenerate.

See "https://en.wikipedia.org/wiki/Killing_horizon"

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  • $\begingroup$ So if the Killing Horizon in the Minkowski place is degenerate (one point $x=ct=0$ ?) why do we get, when solving the equation $g(\xi,\xi)=0$, a light cone ? $\endgroup$
    – devCharaf
    May 26, 2020 at 9:27
  • $\begingroup$ The Killing Horizon is where the magnitude of the Killing vector vanishes, i.e., $g(\xi,\xi)=0$ - which indicates a null hypersurfaces similar to a light cone. In the case of Minkowski space-time, the null hypersurface is the light cone - hence it's consider to be a degenerate Killing Horizon. In some black holes, there are non-generate Killing Horizons where the Killing vectors outside the horizon are time-like, but inside the horizon they are space-like. $\endgroup$ May 26, 2020 at 20:53

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