# Gradient of cross product of two vectors (where first is constant)

In some book about continuum mechanics I read that from principle of virtual work follows balance of rotational momentum when $$\delta \boldsymbol{r} = \boldsymbol{\delta \varphi} \times \boldsymbol{r}, \; \boldsymbol{\delta \varphi} = \boldsymbol{\mathsf{const}}$$ ($$\boldsymbol{r}$$ is location vector, $$\delta \boldsymbol{r}$$ is its variation, $$\boldsymbol{\delta \varphi}$$ is not variation, just denoted as it for some reason like being small enough for infinitesimal $$\delta \boldsymbol{r}$$). Then there is written without any explaination $$\boldsymbol{\nabla} \delta \boldsymbol{r} = - \boldsymbol{E} \times \boldsymbol{\delta \varphi}$$. I know that $$\boldsymbol{E}$$ is bivalent “metric unit identity” tensor (the one which is neutral to dot product operation), and that $$\boldsymbol{\nabla} \boldsymbol{r} = \boldsymbol{E}$$. And that $$\boldsymbol{a} \times \boldsymbol{E} = \boldsymbol{E} \times \boldsymbol{a} \:\: \forall\boldsymbol{a}$$, no minus here. To get minus, transposing is needed: $$\left( \boldsymbol{E} \times \boldsymbol{\delta \varphi} \right)^{\mathsf{T}} \! = - \boldsymbol{E} \times \boldsymbol{\delta \varphi}$$. Thus I can’t get why $$\boldsymbol{\nabla} \delta \boldsymbol{r} = - \boldsymbol{E} \times \boldsymbol{\delta \varphi}$$ has minus sign.

For constant $$\boldsymbol{\delta \varphi}$$, $$\boldsymbol{\nabla} \boldsymbol{\delta \varphi} = {^2\boldsymbol{0}}$$ (bivalent zero tensor). Isn’t it true that $$\boldsymbol{\nabla} \! \left( \boldsymbol{\delta \varphi} \times \boldsymbol{r} \right) = \boldsymbol{\delta \varphi} \times \boldsymbol{\nabla} \boldsymbol{r} = \boldsymbol{\delta \varphi} \times \boldsymbol{E} = \boldsymbol{E} \times \boldsymbol{\delta \varphi}$$? Searching for how to get gradient of cross product of two vectors gives gradient of dot product, divergence ($$\boldsymbol{\nabla} \cdot$$) of cross product, and many other relations. But no gradient of cross product $$\boldsymbol{\nabla} \! \left( \boldsymbol{a} \times \boldsymbol{b} \right) = \ldots$$ Is it impossible or unknown how to find it? At least for the case when first vector is constant.

update

As “gradient” I mean tensor product with “nabla” $$\boldsymbol{\nabla}$$: $$\operatorname{^{+1}grad} \boldsymbol{A} \equiv \boldsymbol{\nabla} \! \boldsymbol{A}$$, here $$\boldsymbol{A}$$ may be tensor of any valence (and I don’t use “$$\otimes$$” or any other symbol for tensor product). Nabla (differential Hamilton’s operator) is $$\boldsymbol{\nabla} \equiv (\sum_i)\, \boldsymbol{r}^i \partial_i$$, $$\:(\sum_i)\, \boldsymbol{r}^i \boldsymbol{r}_i = \boldsymbol{E} \,\Leftrightarrow\, \boldsymbol{r}^i \cdot \boldsymbol{r}_j = \delta^{i}_{j}$$ (Kronecker’s delta), $$\,\boldsymbol{r}_i \equiv \partial_i \boldsymbol{r}$$ (basis vectors), $$\,\partial_i \equiv \frac{\partial}{\partial q^i}$$, $$\:\boldsymbol{r}(q^i)$$ is location vector, and $$q^i$$ $$(i = 1, 2, 3)$$ are coordinates.

• The gradient of the cross product is not defined (usually). The gradient is only defined for scalar-valued functions – rubikscube09 May 5 at 13:47
• @rubikscube09 As “gradient” I mean tensor product with nabla $\boldsymbol{\nabla}$: $\operatorname{grad} \boldsymbol{A} \equiv \boldsymbol{\nabla} \! \boldsymbol{A}$, here $\boldsymbol{A}$ may be tensor of any valency. Nabla (differential Hamilton’s operator) is $\boldsymbol{\nabla} \equiv \boldsymbol{r}^i \partial_i$ – Douglas Mencken May 5 at 13:58
• I see. Perhaps you are referring to this : en.wikipedia.org/wiki/Tensor_derivative_(continuum_mechanics) ? – rubikscube09 May 5 at 14:02
• @rubikscube09 Pretty like (I use $\boldsymbol{r}^i$ while there’s $\boldsymbol{g}^i$). But again, I see nothing about symbolic calculation of $\boldsymbol{\nabla} \! \left( \boldsymbol{a} \times \boldsymbol{b} \right)$ in that article. – Douglas Mencken May 5 at 14:16

Well, it’s easy to find such gradient. You mentioned almost everything you need, but these things —

• $$\boldsymbol{p} \times \boldsymbol{q} = - \, \boldsymbol{q} \times \boldsymbol{p}\,$$ for any two vectors $$\boldsymbol{p}$$ and $$\boldsymbol{q}$$

• partial derivative of any vector with respect to scalar, like coordinate, isn’t some more complex tensor, it’s a vector too

• for differentiation of a product “$$\circ$$” of two multipliers, the famous “product rule” https://en.wikipedia.org/wiki/Product_rule applies: $$\frac{\partial}{\partial q^i} \left( u \circ v \right) = \left( \frac{\partial}{\partial q^i} u \right) \! \circ v \, + \, u \circ \! \left( \frac{\partial}{\partial q^i} v \right)$$

So you have

$${\boldsymbol{\nabla} \! \left( \boldsymbol{a} \times \boldsymbol{b} \right)} = {\boldsymbol{r}^i \partial_i \! \left( \boldsymbol{a} \times \boldsymbol{b} \right)} = {\boldsymbol{r}^i \! \left( \partial_i \boldsymbol{a} \times \boldsymbol{b} + \boldsymbol{a} \times \partial_i \boldsymbol{b} \right)} = {\boldsymbol{r}^i \! \left( \partial_i \boldsymbol{a} \times \boldsymbol{b} - \partial_i \boldsymbol{b} \times \boldsymbol{a} \right)} = {\boldsymbol{r}^i \partial_i \boldsymbol{a} \times \boldsymbol{b} - \boldsymbol{r}^i \partial_i \boldsymbol{b} \times \boldsymbol{a}} = {\boldsymbol{\nabla} \boldsymbol{a} \times \boldsymbol{b} - \! \boldsymbol{\nabla} \boldsymbol{b} \times \boldsymbol{a}}$$

Minus sign before the second addend appears here, because you need to swap multipliers to get differentiation of the first multiplier to have full $$\boldsymbol{\nabla}$$

When a first multiplier, say $$\boldsymbol{\phi}$$, is constant in space, that is it doesn’t change with coordinates, then it becomes

$${\boldsymbol{\nabla} \! \left( \boldsymbol{\phi} \times \boldsymbol{b} \right)} = {\boldsymbol{\nabla} \! \boldsymbol{\phi} \times \boldsymbol{b} - \! \boldsymbol{\nabla} \boldsymbol{b} \times \boldsymbol{\phi}} = {{^2{\boldsymbol{0}}} \times \boldsymbol{b} - \! \boldsymbol{\nabla} \boldsymbol{b} \times \boldsymbol{\phi}} = {{^2{\boldsymbol{0}}} - \! \boldsymbol{\nabla} \boldsymbol{b} \times \boldsymbol{\phi}} = {- \boldsymbol{\nabla} \boldsymbol{b} \times \boldsymbol{\phi}}$$

That’s why there’s a minus sign