1. Jun 28, 2011

hoomanya

Hi,
Just a simple, quick question:
Does the gradient of a vector give a scalor or a vector?
Thanks!

2. Jun 28, 2011

tiny-tim

Last edited by a moderator: Apr 26, 2017
3. Jun 28, 2011

hoomanya

4. Jun 28, 2011

HallsofIvy

You can, of course, have
$$\nabla\cdot \vec{\phi}(x, y, z)= \frac{\partial f}{\partial x}+ \frac{\partial g}{\partial y}+ \frac{\partial h}{\partial z}$$
the "divergence" of the vector valued function, $\vec{\phi}(x, y, z)$, which is a scalar, or
$$\nabla\cdot \vec{\phi}= \left(\frac{\partial g}{\partial z}- \frac{\partial h}{\partial y}\right)\vec{i}+ \left(\frac{\partial f}{\partial z}- \frac{\partial h}{\partial x}\right)\vec{j}+ \left(\frac{\partial g}{\partial x}- \frac{\partial f}{\partial y}\right)\vec{i}$$
the "curl" of the vector valued function, $\vec{\phi}(x, y, z)$, which is a vector.

Perhaps that is what you are thinking of. There are three kinds of vector "multiplication" and so three ways we can attach the "del" operator to a function.

5. Jun 28, 2011

qbert

I'd say there's a perfectly good definition for the gradient of a vector.
it's a rank-2 cartesian tensor.

For the vector $$\vec{A} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k}$$

we have the beast $$\nabla \vec{A} =\nabla A_x \hat{i} + \nabla A_y \hat{j} + \nabla A_z \hat{k}$$

such that for any vector $\vec{v}$
$$\vec{v} \cdot \nabla \vec{A} = (\vec{v} \cdot \nabla A_x) \hat{i} + (\vec{v} \cdot \nabla A_y) \hat{j} + (\vec{v} \cdot \nabla A_z) \hat{k}$$

This is more clear in component notation
$$\vec{A} \rightarrow A^i$$
then
$$\nabla \vec{A} \rightarrow (\nabla A)^i_j = \frac{\partial A^i}{\partial x^j}.$$

where the product with the vector $\vec{v}$ above is really
contraction on the second index.
$$\vec{v} \cdot \nabla \vec{A} \rightarrow \sum_j \quad \left( v^j \frac{\partial A^i}{\partial x^j} \right)$$