Definition of orbits of a Killing vector field

Question

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.)

Answer 1

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"