Physlib.SpaceAndTime.Space.Derivatives.MatrixDiv

For a matrix-valued field $T : \text{Space } d \to \mathbb{R}^{d \times d}$, the matrix divergence $\text{matrixDiv } T$ is a vector-valued function from $\text{Space } d$ to $\mathbb{R}^d$. At any point $x \in \text{Space } d$, its $i$-th component is defined as the divergence of the $i$-th row of the matrix field $T$: $$(\text{matrixDiv } T)_i(x) = \sum_{j=0}^{d-1} \frac{\partial T_{ij}}{\partial x_j}(x)$$ where $T_{ij}(x)$ denotes the entry in the $i$-th row and $j$-th column of the matrix $T(x)$, and $\frac{\partial}{\partial x_j}$ is the partial derivative with respect to the $j$-th spatial coordinate.

Let $d$ be a natural number representing the dimension of the space. For any matrix-valued field $T: \text{Space } d \to \mathbb{R}^{d \times d}$, any point $x \in \text{Space } d$, and any index $i \in \{0, \dots, d-1\}$, the $i$-th component of the matrix divergence evaluated at $x$ is equal to the sum of the partial derivatives of the $i$-th row entries with respect to their corresponding column indices: $$(\text{matrixDiv } T)_i(x) = \sum_{j=0}^{d-1} \frac{\partial T_{ij}}{\partial x_j}(x)$$ where $T_{ij}$ is the function $x \mapsto T(x)_{ij}$ and $\frac{\partial}{\partial x_j}$ denotes the partial derivative with respect to the $j$-th spatial coordinate.

For any dimension $d \in \mathbb{N}$, the matrix divergence of the zero matrix-valued function $T : \text{Space } d \to \mathbb{R}^{d \times d}$ (defined such that $T(x) = 0$ for all $x \in \text{Space } d$) is the zero vector field.

For any dimension $d \in \mathbb{N}$ and any constant matrix $T \in \mathbb{R}^{d \times d}$, the matrix divergence of the constant matrix-valued function $f(x) = T$ is the zero vector field: $$\text{matrixDiv}(x \mapsto T) = 0.$$ Here, $x \mapsto T$ represents a function from $\text{Space } d$ that takes any point $x$ to the constant matrix $T$.

For a natural number $d$, let $T_1, T_2 : \text{Space } d \to \mathbb{R}^{d \times d}$ be two differentiable matrix-valued functions. Then the matrix divergence of their sum is equal to the sum of their individual matrix divergences: $$\text{matrixDiv}(T_1 + T_2) = \text{matrixDiv } T_1 + \text{matrixDiv } T_2$$

Let $d$ be a natural number and $\text{Space } d$ be the $d$-dimensional flat Euclidean space. For any real scalar $k$ and any differentiable matrix-valued field $T : \text{Space } d \to \mathbb{R}^{d \times d}$, the matrix divergence of the scalar multiple $k \cdot T$ is equal to $k$ times the matrix divergence of $T$: $$\text{matrixDiv}(k \cdot T) = k \cdot \text{matrixDiv}(T)$$ where $\text{matrixDiv}$ is the operator that maps a matrix field to a vector field whose $i$-th component is the sum of the partial derivatives of the $i$-th row entries, i.e., $(\text{matrixDiv } T)_i = \sum_{j=0}^{d-1} \partial_j T_{ij}$.