Derivative of cross product w.r.t. a vector

Question

How to compute the derivative of $\vec{a}\times\vec{b}$ w.r.t. $\vec{c}$, all of which are 3D vectors for simplicity? Here $\vec{a}(\vec{c})$ and $\vec{b}(\vec{c})$ are both dependent on $\vec{c}$. I know that the result should be a $3\times3$ matrix. And we cannot simply apply the chain rule as

$$\frac{\partial(\vec{a}\times\vec{b})}{\partial \vec{c}} = \frac{\partial\vec{a}}{\partial\vec{c}} \times \vec{b} + \vec{a} \times \frac{\partial\vec{b}}{\partial\vec{c}}$$

as $\frac{\partial\vec{a}}{\partial\vec{c}}$ would be a matrix then. Is there any analytical expression for the result? For example, $$\frac{\partial\vec{x}}{\partial\vec{x}} = \left[\array{1&0&0\\0&1&0\\0&0&1}\right]$$

Answer 1

In tensorial form, we can write $$ (a\times b)_i=\epsilon_{ijk}a^jb^k, $$ where $\epsilon_{ijk}$ is the Levi-Civita's symbol. Therefore, what we are looking for, can be expressed as $$ \frac{\partial (a\times b)_i}{\partial c^\mu} {= \frac{\partial\epsilon_{ijk}a^jb^k}{\partial c^\mu} \\= \epsilon_{ijk}\frac{\partial a^jb^k}{\partial c^\mu} \\= \epsilon_{ijk}\frac{\partial a^j}{\partial c^\mu}b^k+\epsilon_{ijk}a^j\frac{\partial b^k}{\partial c^\mu} \\\equiv \begin{bmatrix} \frac{\partial a}{\partial c^1}\times b+a\times\frac{\partial b}{\partial c^1}& \frac{\partial a}{\partial c^2}\times b+a\times\frac{\partial b}{\partial c^2}& \frac{\partial a}{\partial c^3}\times b+a\times\frac{\partial b}{\partial c^3} \end{bmatrix}. } $$

Answer 2

$ \def\LR#1{\left(#1\right)} \def\qiq{\quad\implies\quad} \def\p{\partial} \def\grad#1#2{\frac{\p #1}{\p #2}} \def\m#1{\left[\begin{array}{r}#1\end{array}\right]} \def\gradLR#1#2{\LR{\grad{#1}{#2}}} $Use an uppercase letter to denote the $\sf Cross$-$\sf Product\;Matrix$ of a vector, i.e. $$\eqalign{ b = \m{b_1\\b_2\\b_3} \qquad B = \m{ 0\; & -b_3 & b_2\; \\ b_3 & 0\; & -b_1\; \\ -b_2 & b_1 & 0\;\; \\ } \;=\; -B^T }$$ Use this to rewrite the desired cross product (in several different ways) $$\eqalign{ p &= \LR{a\times b} \;&=\; -\LR{b\times a} \\ &= Ab \;&=\; -Ba \;\;\;=\; B^Ta \\ }$$ Then calculate its differential and gradient $$\eqalign{ dp &= A\:db + B^Tda \qiq \grad{p}{c} \:=\: A\gradLR{b}{c} \;+\; B^T\!\gradLR{a}{c} \\ }$$