# Can someone explain this about gradients to me?

1. Jan 29, 2010

### theneedtoknow

Say I have two 3d vector functions A and B
I would intuitively think that the xhat component of (A dot Nabla)B

would be
(dAx/dx+dAy/dy+dAz/dz)Bx

but my book gives it as
Ax*dBx/dx + Ay*dBx/dy + Az*dBx/dz

why is this so?
why isn't A dot nabla equal to (dAx/dx+dAy/dy+dAz/dz)?
My book says that the dot product of 2 vectors is commutative, and it says hat although the del operator is not exactly a vector, we can treat it as such in most cases. It never mentions that taking dot products is NOT one of those cases, yet from this it seems that
A dot nabla is NOT equal to nabla dot A

I've been trying to prove the dot product rule for gradients for 2 hours and It is not working out cause I kept doing (A dot nabla)B and (B dot nabla)A the way I thought they should be done, which gives different results from how they are apparently supposed to be done, but the proper way contradicts what I thought I knew about the del operator so clearly I have some kind of misconception that I need straightened out. Thanks!

2. Jan 29, 2010

### Ben Niehoff

The del operator is a derivative operator, so it doesn't necessarily commute with other things. It might help to do an example just with single variables, instead of vectors. Surely you would agree that

$$f \frac{d}{dx}$$

and

$$\frac{d}{dx} f$$

are two different things, despite the fact that multiplication is generally considered to be commutative. It is the same with the nabla operator, except that it's with the dot product instead of just the ordinary product.

You have to remember that nabla is an operator that acts on whatever appears to its right; so it is not an ordinary vector, even if it might sometimes be written to look like one. Hence

$$\nabla \cdot \vec A$$

is the divergence of A, but

$$\vec A \cdot \nabla$$

is a differential operator that looks like

$$A_x \frac{\partial}{\partial x} + A_y \frac{\partial}{\partial y} + A_z \frac{\partial}{\partial z}$$

3. Jan 29, 2010

### theneedtoknow

All is clear now, thank you sir!

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