# F = ∇·F

1. Jun 13, 2014

### Jhenrique

A scalar field can be the exact form of a vector field (potential form)? It's make sense?

2. Jun 13, 2014

### Matterwave

Nope, doesn't make sense. A divergence is not an exterior derivative. A scalar field is a 0-form. It can't be an exact form because an exact n-form must be the exterior derivative of of a n-1 form. There are no -1 forms, so a 0-form cannot be considered exact.

3. Jun 13, 2014

### Jhenrique

You have a lot of knowledge in several areas of science, impressive!

4. Jun 14, 2014

### Jhenrique

But the operation $\vec{\nabla} \cdot \vec{F} = F$ exist! But still so $\vec{F}$ isn't a potential form and $F$ isn't an exact form?

5. Jun 14, 2014

### Matterwave

You can't call it that. Those terms you used have very specific meanings. You call $\vec{F}$ the vector field and $F$ its divergence.

6. Jun 14, 2014

### Jhenrique

7. Jun 14, 2014

### Matterwave

The analogy this article is talking about is if $\vec{\nabla}\cdot\vec{F}=0$ we call $\vec{F}$ incompressible analogous to closed. If $\vec{F}=\vec{\nabla}\times\vec{A}$ then we call $\vec{F}$ solenoidal analogous to exact. The final analogy is that exact implies closed $\vec{\nabla}\cdot(\vec{\nabla}\times\vec{A})=0$.

But these are analogies. You can NOT call $\vec{F}$ a (closed or exact) form. A form has a specific mathematical definition.

Give you an example. Say I have a bicycle and a car. If I put a motor on my bicycle, it turns into a motorized vehicle which is somewhat analogous to a car. But I DON'T call my motorized bicycle a car. I call it a motorized bicycle, or a motorcycle. In the same way F above is analogous to a closed or exact form but I CANNOT call F a closed or exact form. I call it an incompressible or solenoidal vector field.

8. Jun 26, 2014

### Geometry_dude

Maybe what OP was asking for is the following identities for the Cartan derivatives $d$ of $k$-forms in $\mathbb R ^3$ with canonical metric:
$$0) \, d \phi = (\nabla \phi)^T \cdot d \bar x$$
$$1) \, d A = (\nabla \times \bar A)^T \cdot \star d \bar x \quad, A:= \bar A^T \cdot d \bar x$$
$$2) \, d B = (\nabla \cdot \bar B) \star 1 \quad, B:= \bar B^T \cdot \star d \bar x$$
Here ${}^T$ is the transpose and $\star$ the Hodge-operator.

Last edited: Jun 26, 2014