Penrose Diagrams Null Coordinates I am an engineering student interested in astrophysics. I started to watch a video about Penrose diagrams, since I wanted to find out how they work. Around the seventh minute the lecturer in the video said that in order to create a Penrose Diagram "non-compact"coordinates (where at least one goes against infinity) are replaced by null coordinates (which are still non-compact). He then went ahead and defined a coordinate function $u$ as null coordinate if its $$g(\frac{\partial}{\partial u},\frac{\partial}{\partial u})=0.$$ He also noted that such coordinates are "light-like". Well, I have a hard time to understand what he means. First, I am not familiar with his notation: I assume $g$ is the divergence? But the divergence of what field? Why does light have the property of divergence equals zero? I would be grateful for some clarification.
 A: Expanding on my answer: $g$ is the metric tensor, i.e. the scalar product; we have $g(\vec{U},\vec{V}) = \vec{U} \cdot \vec{V} = g_{\mu\nu}U^\mu V^\nu$. So $u$ being a null coordinate means that the vector $\partial/\partial u$ is null, that is, the vector $\partial/\partial u$ is a possible tangent vector for a light ray.
A: Let's do this in 1 + 1 special relativity.  In this case, the metric tensor is simple:
$$g_{ab}ds^{a}ds^{b} = - dt^{2} + dx^{2}$$
This is this metric expressed in time and space coordinates.  If, instead, we did a transformation:
$$\begin{align}
u &= \frac{1}{\sqrt{2}}\left(t + x\right)\\
v &= \frac{1}{\sqrt{2}}\left(t - x\right)
\end{align}$$
Then, it should be pretty apparent than $du$ and $dv$ are null.  Transforming the metric tensor, we get:
$$g_{ab}ds^{a}ds^{b} = -2 du\,dv$$
Which reflects the fact that $du$ and $dv$ are both null, but that their inner product is nonzero.  Having the metric in a form like this simplifies a lot of things, and this procedure can be generalized to arbitrary metrics.  But the simple thing is "null coordinates" just means that we're doing a generalization of this $(x,t) \rightarrow (u,v)$ transformation.
