Null Killing vectors

1. Jan 22, 2014

pervect

Staff Emeritus
If space-like Killing vectors are associated with a conserved momentum, and timelike Killing vectors are associated with a conserved energy, what is the conserved quantity associated with a null Killing vector?

For instance, "v" in ingoing Eddington Finklestein coordinates.

The line element is:

−(1−2m/r) dv^2+2 dv dr

None of the metric coefficient's are a function of v (and GR Tensor's KillingTest confirms that $\partial / \partial v$ is a Killing vector).

2. Jan 23, 2014

George Jones

Staff Emeritus
$\partial / \partial v$ is not lightlike.

A related post:

3. Jan 23, 2014

Staff: Mentor

Except at $r = 2M$, correct? If I'm reading this right, the ingoing E-F $v$ coordinate is timelike for $r > 2M$, null at $r = 2M$, and spacelike for $r < 2M$, just like the Painleve $T$ coordinate.

Also, for extra fun, if I'm reading this right, since ingoing radial null geodesics are curves of constant $v$, in ingoing E-F coordinates, $\partial / \partial r$ is null! (Whereas in Painleve coordinates, $\partial / \partial r$ is spacelike, and in Schwarzschild coordinates, $\partial / \partial r$ is spacelike for $r > 2M$, null at $r = 2M$, and timelike for $r < 2M$.)

I think recognizing all this also puts the question in the OP in a new light. Consider, for example, $\partial / \partial T$ in Painleve coordinates (or $\partial / \partial t$ in Schwarzschild coordinates, since the integral curves of both are the same). This is a KVF, and it is timelike outside the horizon, null on the horizon, and spacelike inside the horizon. The constant of motion associated with it is usually called "energy at infinity". Of course that interpretation is usually considered to be problematic on or inside the horizon; but does that mean that the same constant of the motion (which is constant all along an infalling geodesic, so the energy at infinity of an infalling object is the same all the way from infinity down to the singularity at $r = 0$) somehow becomes a "momentum" inside the horizon because the associated KVF is now spacelike?

Last edited: Jan 23, 2014
4. Jan 23, 2014

pervect

Staff Emeritus
Ah-ha! Thanks, George - I'm familiar with the second fundamental confusion you describe, but apparently not familiar enough to avoid falling victim to it again.