Physlib.Relativity.Tensors.RealTensor.Vector.Causality.LightLike

For any Lorentz vector $p$ in a $(1+d)$-dimensional spacetime, the causal character of $p$ is light-like if and only if its Minkowski inner product with itself, denoted by $\langle p, p \rangle_m$, is equal to zero.

For any spatial dimension $d$, the causal character of the zero vector $0$ in the Lorentz vector space $\text{Vector}_d$ is light-like.

For any spatial dimension $d \in \mathbb{N}$ and any Lorentz vector $p$ in a $(1+d)$-dimensional spacetime, the relation of causal precedence is reflexive. That is, $p$ causally precedes itself.

In a $(1+d)$-dimensional Minkowski spacetime, let $v$ and $w$ be two light-like vectors. If their temporal components are equal ($v^0 = w^0$), then the Euclidean inner products of their spatial parts with themselves are equal, i.e., $\langle \vec{v}, \vec{v} \rangle = \langle \vec{w}, \vec{w} \rangle$. This implies that the Euclidean norms of their spatial parts are also equal: $\|\vec{v}\| = \|\vec{w}\|$.

For any two Lorentz vectors $v$ and $w$ in a $(1+d)$-dimensional spacetime that have a light-like causal character, if $v$ is proportional to $w$ (i.e., there exists a scalar $r \in \mathbb{R}$ such that $v = r w$), then the absolute values of their temporal components $v^0$ and $w^0$ (the components at index 0) are also proportional, satisfying $|v^0| = |r| |w^0|$.