Physlib.Relativity.LorentzGroup.ToVector

For an element $\Lambda$ of the Lorentz group $\mathcal{L}_d$ with $d$ spatial dimensions, this function defines a Lorentz vector $v \in \text{Vector}_d$ as the result of the group action of $\Lambda$ on the first basis vector $e_0$ (the basis vector corresponding to the temporal component). In terms of matrix components, this corresponds to selecting the first column of the matrix $\Lambda$.

For any spatial dimension $d \in \mathbb{N}$ and any two elements $\Lambda_1, \Lambda_2$ of the Lorentz group $\mathcal{L}_d$, the Lorentz vector obtained from the product of the two matrices $\Lambda_1 \Lambda_2$ is equal to the Lorentz group action of $\Lambda_1$ on the Lorentz vector obtained from $\Lambda_2$. That is, \[ \text{toVector}(\Lambda_1 \Lambda_2) = \Lambda_1 \cdot \text{toVector}(\Lambda_2) \] where $\text{toVector}(\Lambda)$ denotes the Lorentz vector formed by the first column of the matrix $\Lambda$, and $\cdot$ denotes the standard action of the Lorentz group on Lorentz vectors.

For a given number of spatial dimensions $d \in \mathbb{N}$ and an element $\Lambda$ of the Lorentz group $\mathcal{L}_d$, let $\text{toVector}(\Lambda)$ be the Lorentz vector defined as the first column of the matrix $\Lambda$ (representing the action of $\Lambda$ on the temporal basis vector $e_0$). Then, the vector obtained from the negation of the Lorentz transformation, $-\Lambda$, is equal to the negation of the vector obtained from $\Lambda$: $$\text{toVector}(-\Lambda) = -\text{toVector}(\Lambda)$$

For a spacetime with $d$ spatial dimensions, let $\Lambda$ be an element of the Lorentz group $\mathcal{L}_d$. For any index $i \in \text{Fin } 1 \oplus \text{Fin } d$, the $i$-th component of the Lorentz vector $\text{toVector}(\Lambda)$ is equal to the matrix entry $\Lambda_{i, 0}$ (the entry in the $i$-th row and the first column, corresponding to the temporal index).

For a spacetime with $d$ spatial dimensions, let $\Lambda$ be an element of the Lorentz group $\mathcal{L}_d$. The Lorentz vector $\text{toVector}(\Lambda)$ is defined as the function that maps each index $i \in \text{Fin } 1 \oplus \text{Fin } d$ to the matrix entry $\Lambda_{i, 0}$ (the entry in the $i$-th row and the first column of the matrix $\Lambda$).

Let $d$ be a natural number representing the number of spatial dimensions. The function $\text{toVector} : \mathcal{L}_d \to \text{Vector}_d$, which maps an element $\Lambda$ of the Lorentz group to its first column as a Lorentz vector, is continuous.

For an element $\Lambda$ of the Lorentz group $\mathcal{L}_d$ with $d$ spatial dimensions, let $v = \text{toVector}(\Lambda)$ be the Lorentz vector defined as the first column of the matrix $\Lambda$. The time component of this vector, $v^0$, is equal to the top-left entry of the matrix, $\Lambda_{0,0}$.

For any spatial dimension $d \in \mathbb{N}$ and any element $\Lambda$ of the Lorentz group $\mathcal{L}_d$, let $v = \text{toVector}(\Lambda)$ be the Lorentz vector defined as the first column of the matrix $\Lambda$. The Minkowski product of $v$ with itself is equal to 1: $$\langle v, v \rangle_m = 1$$

For any spatial dimension $d \in \mathbb{N}$ and any element $\Lambda$ of the Lorentz group $\mathcal{L}_d$, the absolute value of the top-left entry of its matrix representation (the temporal-temporal component $\Lambda_{0,0}$) is greater than or equal to 1: $$1 \leq |\Lambda_{0,0}|$$ where $\Lambda_{0,0}$ corresponds to the component indexed by the temporal dimension in both row and column.

For any spatial dimension $d \in \mathbb{N}$ and any element $\Lambda$ of the Lorentz group $\mathcal{L}_d$, let $v = \text{toVector}(\Lambda)$ be the Lorentz vector defined by the first column of the matrix $\Lambda$. Then $v$ is equal to the standard temporal basis vector $e_0$ if and only if the top-left entry of the matrix $\Lambda$ (the component $\Lambda_{0,0}$) is equal to 1.

For a spacetime with $d$ spatial dimensions, let $\Lambda$ be an element of the Lorentz group $\mathcal{L}_d$ and $v$ be a Lorentz vector. The time component of the transformed vector $\Lambda \cdot v$ is equal to the Minkowski product of the vector corresponding to the first column of the inverse transformation $\Lambda^{-1}$ and the vector $v$: $$(\Lambda \cdot v)^0 = \langle \text{toVector}(\Lambda^{-1}), v \rangle_m$$ where $(\Lambda \cdot v)^0$ denotes the $0$-th (temporal) component of the vector resulting from the action of $\Lambda$ on $v$, $\text{toVector}(\Lambda^{-1})$ is the Lorentz vector defined by the first column of the matrix $\Lambda^{-1}$, and $\langle \cdot, \cdot \rangle_m$ is the Minkowski product.