Why must this expression for the curl be wrong?

1. May 4, 2017

scotty_le_b

1. The problem statement, all variables and given/known data

Without explicit calculation, argue why the following expression cannot be correct: $$\nabla \times (\mathbf{c} \times \mathbf{r}) = c_{2}\mathbf{e_{1}}+c_{1}\mathbf{e_{2}}+c_{3}\mathbf{e_{3}}$$ where $\mathbf{c}$ is a constant vector and $\mathbf{r}$ is the position vector.

2. Relevant equations

3. The attempt at a solution

So I can do the explicit calculation to see that in fact the curl should be parallel to the vector $\mathbf{c}$ but then I struggle to provide an argument for why this should be so without the calculation.

I think that the incorrect solution has flipped the vector $\mathbf{c}$ in the x-y plane but left the z component unchanged. The position vector treats all directions equally so it seems strange that the z-component of $\mathbf{c}$ should be unchanged by this operation. However, I am unable to explain why this solution can't be true.

2. May 4, 2017

BvU

Hi,
Can you show us in detail ? What does 'parallel' mean to you ?

Are you allowed to use / familiar with the triple product expansion ?

3. May 4, 2017

scotty_le_b

Hi,

So I used the formula $\nabla \times(\mathbf{c}\times\mathbf{r}) = (\nabla \cdot \mathbf{r})\mathbf{c}+(\mathbf{r}\cdot\nabla)\mathbf{c}-(\nabla\cdot\mathbf{c})\mathbf{r}-(\mathbf{c}\cdot\nabla)\mathbf{r}$. Then the terms where $\nabla$ acts on $\mathbf{c}$ will be zero since $\mathbf{c}$ is constant. Also $\nabla \cdot \mathbf{r}=3$ and $(\mathbf{c}\cdot\nabla)\mathbf{r}=\mathbf{c}$ so the whole expression reduces to $3\mathbf{c}-\mathbf{c}=2\mathbf{c}$ which is why I though that the answer should then be parallel to $\mathbf{c}$.

However, I think the point of the question was to justify this intuitively without explicitly doing the calculation above. And that is where I'm unsure.

Thanks

4. May 4, 2017

BvU

I see. 'Parallel' in the sense of 'linearly dependent'.
I was under the impresssion you worked out the components of $\vec c \times\vec r$ and then applied the $\vec \nabla \times$ to the result. That, to me, is an explicit calculation. I tried it and I think it yields $2\,\vec c$ as you found.

So you are fine.

However, with the triple product expansion expression in the link I gave, I managed to confuse myself: the Lagrange formula reads $${\bf a}\times\left ( {\bf b} \times {\bf c} \right ) = {\bf b} \left ( {\bf a} \cdot {\bf c} \right ) - {\bf c} \left ( {\bf a} \cdot {\bf b} \right )$$so that $$\nabla \times(\mathbf{c}\times\mathbf{r}) = \mathbf{c} (\nabla \cdot \mathbf{r}) - \mathbf{r} (\nabla\cdot \mathbf{c}\ ) \ ,$$ only two terms, and yielding $3\bf c$....

Perhaps some math expert can put me right ?