# Normal vector to a null surface?

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?

• (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. May 6, 2020 at 3:38

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.