Physlib.Relativity.MinkowskiMatrix

The Minkowski matrix in $d+1$ dimensions is defined as the real matrix of the form $\mathrm{diag}(1, -1, -1, \dots)$. It is represented as a square matrix over $\mathbb{R}$ indexed by the disjoint union of sets of size $1$ and $d$.

For any natural number $d$, the Minkowski matrix $\eta$ in $d+1$ dimensions is equal to the diagonal matrix $\mathrm{diag}(1, -1, \dots, -1)$, where the diagonal entry is $1$ for the first index (representing the time dimension) and $-1$ for the remaining $d$ indices (representing the spatial dimensions).

For any natural number $d$, the Minkowski matrix $\eta$ in $d+1$ dimensions can be expressed as a block matrix: $$\eta = \begin{pmatrix} 1 & 0 \\ 0 & -I_d \end{pmatrix}$$ where $1$ denotes the $1 \times 1$ identity matrix, $-I_d$ denotes the $d \times d$ negative identity matrix, and $0$ denotes the zero matrices of sizes $1 \times d$ and $d \times 1$.

The symbol $\eta$ is a notation for the Minkowski matrix.

For any natural number $d$, the $(d+1)$-dimensional Minkowski matrix $\eta$, indexed by $\text{Fin } 1 \oplus \text{Fin } d$, has its "time-time" component (the entry at row $\text{inl } 0$ and column $\text{inl } 0$) equal to $1$.

For any natural number $d$ and any spatial index $i \in \{0, \dots, d-1\}$, the diagonal entry of the $(d+1)$-dimensional Minkowski matrix $\eta$ corresponding to the spatial component $\text{inr } i$ is equal to $-1$. That is, $\eta_{\text{inr } i, \text{inr } i} = -1$.

For any indices $\mu, \nu \in \text{Fin } 1 \oplus \text{Fin } d$ such that $\mu \neq \nu$, the entry of the Minkowski matrix $\eta_{\mu\nu}$ is equal to $0$. This implies that the off-diagonal elements of the Minkowski matrix are zero.

Let $\eta$ be the $(d+1)$-dimensional Minkowski matrix. For any index $\mu$, the diagonal entry $\eta_{\mu\mu}$ is non-zero.

For any natural number $d$, the Minkowski matrix $\eta$ in $d+1$ dimensions is self-inverting, satisfying the property that the product of the matrix with itself is the identity matrix $I$: $$\eta \cdot \eta = I$$

Let $\eta$ be the $(d+1)$-dimensional Minkowski matrix. For any index $\mu$, the product of the diagonal entry $\eta_{\mu\mu}$ with itself is equal to $1$, i.e., $\eta_{\mu\mu} \cdot \eta_{\mu\mu} = 1$.

Let $\eta$ be the $(d+1)$-dimensional Minkowski matrix. For any index $\mu$ in the index set of the matrix, the square of the diagonal entry $\eta_{\mu\mu}$ is equal to $1$, i.e., $\eta_{\mu\mu}^2 = 1$.

Let $\eta$ be the $(d+1)$-dimensional Minkowski matrix, defined as the diagonal matrix $\text{diag}(1, -1, \dots, -1)$ with one time dimension and $d$ spatial dimensions. The determinant of this matrix is $\det(\eta) = (-1)^d$.

Let $\eta$ be the $(d+1)$-dimensional Minkowski matrix. For any index $\mu$ and any real numbers $x$ and $y$, the equality $\eta_{\mu\mu} x = \eta_{\mu\mu} y$ holds if and only if $x = y$.

Let $\eta$ be the Minkowski matrix in $1+d$ dimensions, defined as the diagonal matrix $\operatorname{diag}(1, -1, \dots, -1)$. For any vector $v \in \mathbb{R}^{1+d}$, the time component (the first component) of the matrix-vector product $\eta v$ is equal to the time component of $v$: $$(\eta v)_0 = v_0$$

Let $\eta$ be the Minkowski matrix in $1+d$ dimensions, which is the diagonal matrix $\operatorname{diag}(1, -1, \dots, -1)$. For any vector $v \in \mathbb{R}^{1+d}$ and any spatial index $i \in \{1, \dots, d\}$, the $i$-th component of the matrix-vector product $\eta v$ is equal to the negation of the $i$-th component of $v$: $$(\eta v)_i = -v_i$$

For a real matrix $\Lambda$ of dimension $(1+d) \times (1+d)$, its Minkowski dual is defined as the matrix product $\eta \Lambda^T \eta$, where $\eta$ is the Minkowski matrix $\text{diag}(1, -1, \dots, -1)$ and $\Lambda^T$ denotes the transpose of $\Lambda$.

For a given number of spatial dimensions $d$, let $I$ be the $(1+d) \times (1+d)$ identity matrix. The Minkowski dual of $I$ is equal to the identity matrix $I$: $$\operatorname{dual}(I) = I$$ where the Minkowski dual of a matrix $\Lambda$ is defined as $\eta \Lambda^T \eta$, and $\eta$ is the Minkowski matrix $\operatorname{diag}(1, -1, \dots, -1)$.

For any two real $(d+1) \times (d+1)$ matrices $\Lambda$ and $\Lambda'$, the Minkowski dual of their product is equal to the product of their Minkowski duals in reverse order: $$\text{dual}(\Lambda \Lambda') = \text{dual}(\Lambda') \cdot \text{dual}(\Lambda)$$ where the Minkowski dual of a matrix $M$ is defined as $\text{dual}(M) = \eta M^T \eta$, with $\eta$ being the Minkowski matrix $\text{diag}(1, -1, \dots, -1)$. This property shows that the Minkowski dual operation is contravariant with respect to matrix multiplication.

For any real matrix $\Lambda$ of dimension $(d+1) \times (d+1)$, the Minkowski dual operation is involutive. That is, applying the dual operation twice results in the original matrix: $$\text{dual}(\text{dual}(\Lambda)) = \Lambda$$ where the Minkowski dual is defined as $\text{dual}(\Lambda) = \eta \Lambda^T \eta$, with $\eta$ being the Minkowski matrix $\text{diag}(1, -1, \dots, -1)$.

For any real matrix $\Lambda$ of dimension $(d+1) \times (d+1)$, the Minkowski dual of the transpose of $\Lambda$ is equal to the transpose of the Minkowski dual of $\Lambda$, i.e., $\text{dual}(\Lambda^T) = (\text{dual}(\Lambda))^T$. This demonstrates that the Minkowski dual operation commutes with the matrix transpose operation.

Let $\eta$ be the $(d+1) \times (d+1)$ Minkowski matrix, defined as $\mathrm{diag}(1, -1, \dots, -1)$. For any real $(d+1) \times (d+1)$ matrix $\Lambda$, let its Minkowski dual be defined as $\text{dual}(\Lambda) = \eta \Lambda^T \eta$. Then the Minkowski dual of the Minkowski matrix is the Minkowski matrix itself, i.e., $$\text{dual}(\eta) = \eta$$

For any $(d+1) \times (d+1)$ real matrix $\Lambda$, the determinant of its Minkowski dual is equal to the determinant of $\Lambda$. That is, $$\det(\text{dual } \Lambda) = \det(\Lambda)$$ where the Minkowski dual is defined as $\text{dual}(\Lambda) = \eta \Lambda^T \eta$, and $\eta$ is the Minkowski matrix $\text{diag}(1, -1, \dots, -1)$.

For a $(1+d) \times (1+d)$ real matrix $\Lambda$, the components of its Minkowski dual, defined as $\text{dual } \Lambda = \eta \Lambda^T \eta$, are given by $$(\text{dual } \Lambda)_{\mu\nu} = \eta_{\mu\mu} \Lambda_{\nu\mu} \eta_{\nu\nu}$$ where $\eta$ is the Minkowski matrix $\mathrm{diag}(1, -1, \dots, -1)$, and $\mu, \nu$ are indices in the set $\{0, 1, \dots, d\}$.

Let $\Lambda$ be a real matrix of size $(d+1) \times (d+1)$ and $\eta$ be the Minkowski matrix $\text{diag}(1, -1, \dots, -1)$. Let $\text{dual } \Lambda = \eta \Lambda^T \eta$ denote the Minkowski dual of $\Lambda$. For any indices $\mu$ and $\nu$ in $\{0, 1, \dots, d\}$, the components of the dual matrix and the original matrix satisfy the identity: $$(\text{dual } \Lambda)_{\mu\nu} \eta_{\nu\nu} = \eta_{\mu\mu} \Lambda_{\nu\mu}$$ where $(\text{dual } \Lambda)_{\mu\nu}$ is the entry in the $\mu$-th row and $\nu$-th column of the dual matrix, and $\eta_{\mu\mu}$ and $\eta_{\nu\nu}$ are the diagonal components of the Minkowski matrix.