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.

  • 3
    $\begingroup$ $g$ is not the divergence, it's the metric tensor. $\endgroup$
    – Javier
    Commented Dec 16, 2016 at 17:54

2 Answers 2


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.

  • $\begingroup$ Basically the diagonal terms are zero. Say $u=r-t$, $w=t+t$ yields metric tensor $dudw$ up to a sign, therefore both $u$ and $v$ are null. Does this make sense? $\endgroup$ Commented Mar 25, 2020 at 14:03
  • $\begingroup$ @AlexanderCska yes, that's correct. $\endgroup$
    – Javier
    Commented Mar 25, 2020 at 14:28

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.

  • $\begingroup$ well if I understand correctly, the null coordinates are geodesics in Minkowski space. I guess that in flat space-time we should just have $\frac{d^2x^{\mu}}{d\tau^2}=0$ therefore the transformation $u=x+t,v=x-t$ should satisfy this condition. (All Christoffel symbols are zero.) However, how does one show that explicitly. Where is the affine parameter $\tau$ here? Do we regard $u$ and $v$ as affine parameters and differentiate with respect to them? $\endgroup$ Commented May 2, 2019 at 9:40
  • $\begingroup$ @AlexanderCska: I don't know what to tell you other than to pick some parameterization of a curve $u(\tau), v(\tau)$, and compute everything. in this very simple case, that is very easy. $\endgroup$ Commented May 2, 2019 at 14:23
  • $\begingroup$ I don't understand your remark . Could you show how the calculation goes precisely. $\endgroup$ Commented May 20, 2019 at 15:10
  • $\begingroup$ @AlexanderCska: take the metric above. You are correct that $\Gamma_{ab}{}^{c} = 0$. Then, the geodesic equation becomes $0 = {\ddot x}^{a}$, and any curve of the form $(a\tau + b, c\tau + d)$ will satisfy this. $\endgroup$ Commented May 20, 2019 at 15:14
  • $\begingroup$ @JerrySchrimer that is clear. What I don't understand is how you get the $u=x+t$ and $v=x-t$. In flat spacetime the null condition reads $g_{\mu\nu}\frac{dx^{\mu}}{ds}\frac{dx^{\nu}}{ds}=\left(\frac{dt}{ds}\right)^2-\left(\frac{dx}{ds}\right)^2=0$ . This is solved by separation of variables to give $x=t=\pm C\cdot s$. I don't see how ones links the solutions of $g_{\mu\nu}\frac{dx^{\mu}}{ds}\frac{dx^{\nu}}{ds}$ and $\frac{d^2x^{\mu}}{d^2s}$ together, and arrives at $u=x+t$ and $v=x-t$. $\endgroup$ Commented May 21, 2019 at 15:18

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