# Help with intuition of divergence, gradient, and curl

1. Apr 5, 2014

### joshmccraney

hey pf!

i have a few question about the physical intuition for divergence, gradient, and curl. before asking, i'll define these as i have seen them (an intuitive definition).

$$\text{Divergence} \:\: \nabla \cdot \vec{v} \equiv \lim_{V \to 0} \frac{1}{V} \oint_A \hat{n} \cdot \vec{v} da$$

$$\text{Curl} \:\: \nabla \times \vec{v} \equiv \lim_{V \to 0} \frac{1}{V} \oint_A \hat{n} \times \vec{v} da$$

$$\text{Gradient} \:\: \nabla \vec{v} \equiv \lim_{V \to 0} \frac{1}{V} \oint_A \hat{n} \vec{v} da$$

where $V$ is the volume, $\vec{v}$ is a vector field and the rest (i think) is evident.

as for the divergence, i understand that $\hat{n} \cdot \vec{v} da$ is a volumetric flow rate through a differential surface element. thus the integral is the total volumetric flow rate, and dividing by volume and the limit gives us a nice understanding that divergence is expansion/contraction of a vector field.

my understanding is predicated on the understanding of the dot product, namely $\hat{n} \cdot \vec{v} da$. however, the curl uses a cross product. i understand a cross product to be a vector orthogonal to two given vectors. thus, $\hat{n} \times \vec{v} da$ seems to be some new vector in a field such that it is tangent to the surface at any point. if we add all these vectors $\hat{n} \times \vec{v} da$ up we should get some kind of body rotation about a point, which i think parallels the general understanding of curl.

however, what on earth do we do about the dyadic product $\hat{n} \vec{v}$ embedded in the definition of gradient? i really don't have a physical interpretation of what is happening here, and thus i really don't physically understand $\nabla \vec{v}$.

2. Apr 6, 2014

### Simon Bridge

The curl is the degree of "circulation" in the vector field.
The gradient is the "slope".

You should get your intuition from using the definitions and not so much from the definitions themselves.

3. Apr 7, 2014

### joshmccraney

but the gradient of a vector is a tensor. is slope still an appropriate description?

thanks for the response.

4. Apr 7, 2014

5. Apr 7, 2014

### joshmccraney

i think you're referring to the gradient of a scalar, $\nabla f$, which returns a vector. but what i am referring to is the gradient of a vector $\nabla \vec{f}$, which returns a 2nd rank tensor (a matrix).

6. Apr 7, 2014

### electricspit

You're right my mistake, sorry I didn't see the vector symbol there!

7. Apr 7, 2014

### joshmccraney

no need to be sorry. i appreciate your interest.

8. Apr 7, 2014

### Simon Bridge

I thought you were after a conceptual understanding ... you can think of grad as an extension of the concept of "slope".

9. Apr 7, 2014

### joshmccraney

so if $\vec{v}$ is a velocity field would $\nabla \vec{v}$ be some sort velocity flux? but then what is $\hat{n} \vec{v}$? is it the same?