# Nabla operator

1. Jun 8, 2007

### f(x)

Could some one explain what does Nabla operator actually signify ? I understand that the various products with nabla are used to find curl,divergence,gradient in EM, but what does Nabla represent in itself ? A more basic question would be, what does del operator(partial derivative) represent , in a 3d system ?

2. Jun 8, 2007

### chroot

Staff Emeritus
It's a generalization of the derivative operator to multiple dimensions. That's all.

In one dimension, say along the x-axis, the derivative operator looks like this:

$\frac{d} {{dx}} = \frac{\partial } {{\partial x}} = \vec i \frac{\partial } {{\partial x}}$

Since there's only one dimension, the "normal" derivative and partial derivative are the same. Also, there's only way way to take a derivative in one dimension -- along that dimension. Thus, the $\vec i[/tex] is implied. In multiple dimensions, say x, y and z, it looks like: [itex] \nabla = \vec i \frac{\partial } {{\partial x}} + \vec j \frac{\partial } {{\partial y}} + \vec k \frac{\partial } {{\partial z}}$

Same thing, just with more dimensions.

- Warren

3. Jun 8, 2007

### arildno

I would disagree there, chroot.
The one-dimensional analogue of the partial derivative at a point, is the derivative with respect to the elements of some particular sequence converging to that point.

Remember that existence of all partial derivatives does not guarantee differentiability at that point; some similar restriction ought to be provable for "sequential" derivatives in the one-dimensional case.

4. Jun 8, 2007

### chroot

Staff Emeritus
So you're saying that del doesn't reduce to the standard one-dimensional derivative d/dx? Can you explain a bit more?

- Warren

5. Jun 9, 2007

### f(x)

What I was referring to was, what does a partial derivative represent (in multiple dimensions) like it represents slope in one dimension co-ordinate system ?

6. Jun 9, 2007

### FunkyDwarf

a gradient vector. the direction in which the function is changing most rapidly, and the magnitude is the amount its changing

7. Jun 9, 2007

### neutrino

If it acts on a scalar field.

8. Jun 9, 2007

### arildno

Hmm..what I meant is that a partial derivative is the derivative with respect to some proper subset of arguments in the vicinity of the point.
For example along the x-axis (or some line parallell to that) in 2-D, or along the rationals in the 1-D case.

Last edited: Jun 9, 2007
9. Jun 9, 2007

### cepheid

Staff Emeritus
It's still a vector operator...so I would say that in one dimension:

$$\nabla = \hat{x}\frac{d}{dx}$$

I'm sure Arildno's answer was better, but I didn't follow what he was saying.