Physlib.Relativity.Tensors.RealTensor.Vector.Causality.TimeLike

For any vectors $v$ and $w$ in a $d$-dimensional Lorentz vector space, if their time components (denoted by $v^0$ and $w^0$) are both negative, then the product of these components is positive, i.e., $v^0 w^0 > 0$.

Let $p$ be a Lorentz vector in a Minkowski space with $d$ spatial dimensions. The causal character of $p$ is classified as timelike if and only if its Minkowski inner product with itself is strictly positive, i.e., $\langle p, p \rangle_m > 0$.

For any timelike Lorentz vector $v$ in a Minkowski space with $d$ spatial dimensions, the Euclidean inner product of its spatial part with itself is strictly less than the square of its time component: $$\langle \mathbf{v}_{\text{spatial}}, \mathbf{v}_{\text{spatial}} \rangle_{\mathbb{R}} < (v^0)^2$$ where $\mathbf{v}_{\text{spatial}}$ denotes the spatial part of $v$ and $v^0$ denotes its temporal component.

Let $v$ be a Lorentz vector in a Minkowski space with $d$ spatial dimensions. If $v$ is timelike (meaning its causal character is classified as timelike, which implies $\langle v, v \rangle_m > 0$), then its temporal component $v^0$ (the component indexed by the first index of the spacetime decomposition) is non-zero, i.e., $v^0 \neq 0$.

For any Lorentz vector $v$ in a Minkowski space with $d$ spatial dimensions, if $v$ is timelike (meaning its causal character is classified as timelike), then its temporal component $v^0$ is non-zero, i.e., $v^0 \neq 0$.

For a Lorentz vector $v$ in $(1+d)$-dimensional Minkowski spacetime, let $v^0$ denote its temporal component (time component) and $\mathbf{v} \in \mathbb{R}^d$ denote its spatial part. The vector $v$ is classified as timelike if and only if the squared Euclidean norm of its spatial part is strictly less than the square of its temporal component: $$\|\mathbf{v}\|^2 < (v^0)^2$$ where $\|\mathbf{v}\|^2 = \sum_{i=1}^d (v^i)^2$ is the sum of the squares of the spatial components.

Let $v$ be a Lorentz vector in a Minkowski space with $d$ spatial dimensions. If $v$ is timelike (meaning its causal character is classified as timelike, which implies $\langle v, v \rangle_m > 0$), then the square of its temporal component $v^0$ is strictly positive, i.e., $(v^0)^2 > 0$.

For any Lorentz vector $v$ in a Minkowski space with $d$ spatial dimensions, if $v$ is timelike, then the squared Euclidean norm of its spatial part $\|\mathbf{v}\|^2$ (given by the inner product $\langle \text{spatialPart } v, \text{spatialPart } v \rangle_{\mathbb{R}}$) is strictly less than the square of its temporal component $(v^0)^2$.