Tangent vectors to a light-cone in the spacetime are not always vector timelike. The condition of tangency to a submanifolds is a linear equation in the tangent space, and this also applies to the light-cone at any point other than the vertex that is a singular point,while the equation of the light-cone in tangent space is quadratic. Considering a light-cone with vertex in $x_{o}^\mu$a: $$\eta_{\mu \nu}(x^\mu-x^\mu_{0})(x^\nu-x^\nu_{0})=0$$ Only at the vertex of the light-cone the vectors, that are tangent to the cone are all and only light-cone ones The tangent vectors to the remainder of the cone may be timelike or spacelike. Someone cuold give me an explanation?
2026-02-23 08:28:26.1771835306
Only at the vertex of the light-cone the vectors, that are tangent to the cone are all and only light-cone ones
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Consider the hypersurface $S=\{\eta_{\mu \nu} x^{\mu}x^{\nu}=0$} and a point $ P \in S$ different than the origin. A (tangent) vector $\mathbf{v}$ in $P$ is a lightlike vector if, by definition, $\eta(\mathbf{v},\mathbf{v})=0$. This is a quadratic equation. On the other hand the condition that defines the tangent space of $S$ in $P$ is linear. For simplicity, think of the lightcone as a surface of $M^3$, and suppose that $P$ lies in the future nappe of the cone. We can parametrize the cone $-x^2 +y^2 + z^2 =0$ as $$\begin{cases} x=r>0\\ y=r\cos \theta\\ z=r\sin \theta \text{.} \end{cases} $$ A basis of the tangent space is given by $\mathbf{e}_1 = (1, \cos \theta, \sin \theta)$ and $\mathbf{e}_2=(0, -r\sin \theta, r \cos \theta)$. If we consider a linear combination $\lambda \mathbf{e}_1+ \mu \mathbf{e}_2$ we get (with a straightforward calculation): $$\eta(\lambda \mathbf{e}_1+ \mu \mathbf{e}_2, \lambda \mathbf{e}_1+ \mu \mathbf{e}_2)= -\lambda^2 +\lambda^2 + r^2 \mu^2 \ge0 \text{.}$$ If, thus, $\mu \neq 0$ then the vector $\lambda \mathbf{e}_1+ \mu \mathbf{e}_2$ is spacelike. Note however that it can't be timelike. Intuitively, this is due to the fact that a cone is a convex surface. Also note that in the origin, which is a singularity, every tangent vector is lightlike (letting $r=0$).