# Del, divergence, laplacian

1. Mar 20, 2008

### Bucky

I've been reading up on these three recently, and wondered if anyone could confirm what I think they do. I'm not 100% I understand these.

del $$(\bigtriangleup)$$, when applied to a scalar, creates a vector with that scalar as each of the XYZ values. eg

$$\bigtriangleup . x = (x,x,x)$$
$$\bigtriangleup . 3 = (3,3,3)$$

divergence is applied to a vector, and sums the components of the vector into a scalar. eg

$$\bigtriangleup . (x,y,z) = x+y+z$$
$$\bigtriangleup . (1,2,3) = 1+2+3 = 6$$

finally, laplacian. This is the one I'm not as sure about. It's applied to a scalar I think?

$$\bigtriangleup ^2 = \bigtriangleup(\bigtriangleup)$$
$$\bigtriangleup ^2 . x = \bigtriangleup(\bigtriangleup . x)$$
$$= \bigtriangleup((x,x,x))$$
$$= 3x$$

That doesn't seem right (I think I'm meant to end up with a vector). Can laplacian be broken up like that or does it have a special rule?

2. Mar 20, 2008

### kdv

No.

Those are differential operators so they must applied to vector fields or scalar fields

The divergence applied to a vector field gives a number .
The gradient applied to a scala field gives a vector.
The laplacian may be applied to a scalar field or to a vector field producing something of the same nature as what it was applied to.

You can't apply those things to a single vector.

3. Mar 20, 2008

### Bucky

ok, is method right? Regardless of the size of the field, are you still doing "that" to each element?

4. Mar 20, 2008

### Tom Mattson

Staff Emeritus
Bucky,

Nothing in your original post is correct. You need to use the definition of the del operator, together with the definition of a dot product. To write a dot product of del with some scalar is nonsense.

Also, I've never seen anyone write the del operator the way that you have done it. It is always written $\nabla$. The Laplacian on the other hand can be written either as $\nabla^2$ or $\bigtriangleup$.