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my approach

$$\nabla \vec{r} = \left(\frac{\partial }{\partial x}\vec {i} + \frac{\partial }{\partial y}\vec {j} + \frac{\partial }{\partial z}\vec {k}\right)(x\vec i + y\vec j+z \vec k) = \sum\frac{\partial }{\partial x}\vec {i}(x\vec i+y\vec j+ z\vec k) = \vec i \vec i + \vec j \vec j+ \vec k \vec k$$

What is this $\vec i \vec i$ expression? is this correct?

$$A\cdot\nabla \vec r = (A_1 \vec i + A_2 \vec j + A_3 \vec k)\cdot (\vec i \vec i + \vec j \vec j+\vec k \vec k) = \Sigma A_1 (\vec i\cdot \vec i \vec i) = \sum A_1 \vec i = \vec A $$

is this correct??? Please clarify

Otherway I tried

$$A\cdot \nabla \vec r = (A\cdot \nabla) \vec r = \cdots = \vec A~~\text{by exapanding}~~ (A\cdot\nabla)$$

caverac
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Magneto
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1 Answers1

0

$\newcommand{\vect}[1]{{\bf #1}}$ $\newcommand{\uvect}[1]{\hat{\bf #1}}$

Since $\vect{r}$ is a vector already, when you take its gradient you'd expect to get something different from a vector: a tensor in this case. If $\uvect{e}_i$ represents the $i$-th direction, then $\vect{A} = \sum_i A_i\uvect{e}_i$, and

\begin{eqnarray} \vect{A} \cdot \nabla\vect{r} &=& \left(\sum_i A_i\uvect{e}_i\right)\cdot \left(\sum_{jk}\uvect{e}_j\partial_j (\uvect{e}_k x_k)\right) \\ &=& \sum_{ijk} \uvect{e}_i\cdot(\uvect{e}_j\uvect{e}_k)A_i \partial_j x_k\\ &=& \sum_{ijk} \uvect{e}_i\cdot(\uvect{e}_j\uvect{e}_k)A_i (\underbrace{\partial_j x_k}_{\delta_{jk}}) \\ &=& \sum_{ij}(\underbrace{\uvect{e}_i\cdot \uvect{e}_j}_{\delta_{ij}})\uvect{e}_j A_i \\ &=& \sum_i \uvect{e}_i A_i = \vect{A} \end{eqnarray}

caverac
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