$A(\textbf{r})= \textbf{r}\times \nabla\phi(\textbf{r})$ is orthogonal to $\textbf{r}$ and $\nabla\phi(\textbf{r})$

Question

Let ${\textbf{r}} = (x,y,z)$ be the position vector. Show that the vector field $A(\textbf{r})= \textbf{r}\times \nabla\phi(\textbf{r})$ is orthogonal to $\textbf{r}$ and $\nabla\phi(\textbf{r})$, that is, the following expressions are true : $A(\textbf{r})\cdot \textbf{r} = 0$ and $A(\textbf{r})\cdot \nabla\phi(\textbf{r}) = 0$.

I managed to prove this by direct application of the definitions in cartesian coordinates.

$\textbf{Now I am trying to do it using index notation since my professor required it }$:

$A(\textbf{r})\cdot \textbf{r} = (\textbf{r}\times \nabla\phi(\textbf{r})) \cdot \textbf{r} = \epsilon_{ijk}\textbf{r}_j\partial_k\phi\textbf{r}_i$ and $A(\textbf{r})\cdot \nabla\phi(\textbf{r}) =(\textbf{r}\times \nabla\phi(\textbf{r}))\cdot \nabla\phi(\textbf{r}) = \epsilon_{ijk}\textbf{r}_j\partial_k\phi\partial_i\phi $.

But I am not sure how to proceed from here, I can't see how these two expressions are going to be zero.

Any help will be appreciated . Thank you

Answer 1

Using the triple product (which can be proved using the Levi Civita Symbol, see here ) we have that :

$ \textbf{1 -}A(\textbf{r})\cdot \nabla\phi(\textbf{r}) = (\textbf{r} \times \nabla\phi(\textbf{r}))\cdot \nabla\phi(\textbf{r}) = (\nabla\phi(\textbf{r})\times \nabla\phi(\textbf{r}))\cdot\textbf{r} = 0\cdot\textbf{r} = 0 $

Similarly:

$ \textbf{2-} A(\textbf{r})\cdot \textbf{r} = (\textbf{r}\times \nabla\phi(\textbf{r}))\cdot \textbf{r} = (\textbf{r}\times\textbf{r})\cdot \nabla\phi(\textbf{r}) = 0 \cdot \nabla\phi(\textbf{r}) = 0 $

From $\textbf{1}$ and $\textbf{2}$ we have the orthogonality.

Answer 2

There is no need for you to do any of that, since the cross product $v\times w$ of two vectors $v$ and $w$ is always orthogonal to both $v$ and to $w$.