If I just consider the two dimensional spacetime $(t,x)$ with metric $\{-1,1\}$, then the light-cone is a null surface, defined by $$f(t,x)=t-x=0.$$ If I calculate calculate the normal vector of this null surface then it's $$\nabla_\mu f(t,x)=\frac{\partial f}{\partial x^{\mu}}=(-1,-1),$$ which is inside the null-surface. Then my question is how do we get the vector $(1,-1)$ that is "normal" to the null-surface in the usual sense? Or how do I construct a basis for the original spacetime on the null-surface?

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    $\begingroup$ (1) You have only considered one brach of the light cone, but once you notice it that is fine. (2) In your case, the covariant components of the normal vector is (1,-1), and therefore, the corresponding contravariant components are (-1,-1). This is a null vector. It is not parallel to the vector lying on (tangent to) the lightcone surface, whose contravariant components read (1,1) (3) I am not sure your confusion is coming from a mix-up of covariant and contravariant components or of the normal vector of the other branch of lightcone. $\endgroup$
    – gamebm
    May 6, 2020 at 3:38

1 Answer 1


The definition is $t^2 - x^2 = 0$, though it does not make a large difference in one dimension other than making it two lines instead of one line.

You are right, the normal vector to a null surface is within the null surface. The normal plane to any world line is the surface of simultaneous events. A photon does not observe time, therefore it observes every spacetime point it passes at the same time. Also the photon cannot observe anything besides its world line, therefore you do not get any vector which is independent of the world line from that. From this physical argument there is no way to construct a basis around the null surface because the photon could not reach that anyway.


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