# A GR question about null surfaces, vectors and coordinates

1. Oct 8, 2013

### qtm912

I wondered anyone can explain the significance of the above as applied to metrics in the context of general relativity. This came up when the video lecturer in GR mentioned that r for example, was null or this or that vector or surface was null, say in the context of the eddington finkelstein coordinate system. I am unable to grasp the meaning or significance of the word null. I think null geodesics are clear however but even here not sure if there is any connection with null vectors etc.

2. Oct 8, 2013

### Staff: Mentor

A null vector is a vector whose length is zero; that is, a vector $k^a$ such that $g_{ab} k^a k^b = 0$, where $g_{ab}$ is the metric. One of the things that makes spacetime different from an ordinary Riemannian space is that you can have vectors with zero length that are not the zero vector.

A null surface is a surface that has null tangent vectors; in other words, there are null vectors that lie completely within the surface. For example, the event horizon of a black hole is a null surface.

A null geodesic is just a geodesic whose tangent vector is null; for example, the worldline of a light ray traveling through a vacuum is a null geodesic. So there is indeed a close connection between null vectors and null geodesics.

3. Oct 8, 2013

### qtm912

Ok thanks I especially found helpful your comment about 4 vectors that have zero length but that are not the zero vector. A follow on question if ok : what then is a null coordinate.

4. Oct 8, 2013

### WannabeNewton

It means the vector field associated with the coordinate is null.

5. Oct 8, 2013

### qtm912

Ok many thanks to you both for clarifying.

6. Oct 8, 2013

### WannabeNewton

By the way, in case you wanted an example, consider the surface $\Sigma$ given by $r = 2M$ in Schwarzschild space-time. The vector field $\nabla r$ is a normal vector field to $\Sigma$. Now $\nabla r = g^{\mu r}\partial_{\mu} = (1 - \frac{2M}{r})\partial_r$ so $g(\nabla r, \nabla r) = (1 - \frac{2M}{r})$. On $\Sigma$ then, $g(\nabla r, \nabla r) = 0$ so $\nabla r$ is null on $\Sigma$ meaning $\Sigma$ is a null hypersurface ("hyper" because it is of codimension 1); this is of course just the event horizon of the Schwarzschild black hole.

Note that for any tangent vector $v \in T_p \Sigma$, $v \perp \nabla r$ implies that $v$ is itself null so this is equivalent to Peter's definition above.

7. Oct 9, 2013

### qtm912

Hi again, I am a bit unfamiliar with the notation here, how is g(del-r, del-r) defined?

8. Jan 13, 2015

### Abbas Sherif

That's the scalar product (the dot product) of the vector field del-r with itself.