# Easy curl/conservative question

1. Jan 26, 2014

### joshmccraney

hey pf!

can someone comment on this statement i believe to be true: a simply connected, curl-free vector field is conservative (path independent).

thanks!

2. Jan 26, 2014

### dextercioby

What comment to give? It's an immediate result of Stokes' theorem (integral formula) and is very useful in physics (electrodynamics).

3. Jan 26, 2014

### joshmccraney

thanks for confirming this!

4. Jan 27, 2014

### FactChecker

I am used to incompressible, irrotational fluid flow being a requirement for path independence. What is the vector field interpretation of "incompressible"?

5. Jan 27, 2014

### joshmccraney

vector notation for incompressible?? no idea. but if a fluid is incompressible then density is constant.

6. Jan 27, 2014

### FactChecker

A vector field is an incompressible flow if the total net flux through any closed surface is zero.

7. Jan 28, 2014

### WannabeNewton

If $\vec{v}$ is the velocity field of the fluid then $\vec{\nabla} \cdot \vec{v} = \frac{1}{V}\vec{\nabla}_{\vec{v}} V$ where $V$ is a volume element carried along the fluid flow. Do you know the definition of an incompressible fluid in terms of volume elements? Can this volume element change along the flow if the fluid is incompressible? What does that tell you about $\vec{\nabla} \cdot \vec{v}$?

8. Jan 29, 2014

### FactChecker

I always thought that incompressible was also required for path independence, but it is not clear to me now. Obviously irrotational is required since otherwise some circles would have different clockwise and counterclockwise integrals.

9. Feb 1, 2014

### vanhees71

The path-independence of line integrals, i.e., their only dependence of initial and final points of the paths but not their shapes implies that the line integral along all closed lines is zero. In a simply connected region of space and if the vector field in question is continuously differentiable in this region that follows from the vanishing of its curl (Poincare's Lemma)
$$\vec{\nabla} \times \vec{V}=0 \; \Leftrightarrow \; \int_{\mathcal{C}} \mathrm{d} \vec{r} \cdot \vec{V} \; \Leftrightarrow \; \vec{V}=-\vec{\nabla} \Phi.$$
In the same way you can show that a vector field whose integral over any closed surface vanishes in a region, where each such surface can be continuously contracted to a point is a solenoidal field and its divergence vanishes:
$$\vec{\nabla} \cdot \vec{V}=0 \; \Leftrightarrow \; \int_S \mathrm{d}^2 \vec{F} \cdot \vec{V}=0 \; \Leftrightarrow \; \vec{V}=\vec{\nabla} \times \vec{A}.$$
From this it follows the Helmholtz decomposition theorem, i.e., for any sufficiently well-behaved vector field in sufficiently nice regions of space you can write
$$\vec{V}=-\vec{\Phi}+\vec{\nabla} \times \vec{A}.$$
The scalar potential $\Phi$ is determined from its sources, i.e.,
$$\vec{\nabla} \cdot \vec{V}=-\Delta \vec{\Phi}$$
and the vector potential $\vec{A}$ from its vortices,
$$\vec{\nabla} \times \vec{V}= \vec{\nabla} \times (\vec{\nabla} \times \vec{A}).$$
Since $\vec{A}$ is only defined up to a gradient field, you can demand
$$\vec{\nabla} \cdot \vec{A}=0,$$
and then the latter equation simplifies to (in Cartesian coordinates)
$$\Delta \vec{A}=-\vec{\nabla} \times \vec{V}.$$
So you get the decomposition with help of any Green's function of the Laplace operator that satisfies the boundary conditions suitable for your problem:
$$\Phi(\vec{x})=\int \mathrm{d}^3 \vec{x}' G(\vec{x},\vec{x}') \nabla' \cdot \vec{V}(\vec{x}')$$
and
$$\vec{A}(\vec{x})=\int \mathrm{d}^3 \vec{x}' G(\vec{x},\vec{x}') \nabla' \times \vec{V}(\vec{x}').$$
The Green's function satisfies
$$\Delta_{\vec{x}} G(\vec{x},\vec{x}')=-\delta^{(3)}(\vec{x}-\vec{x}').$$
$$G(\vec{x},\vec{x}')=\frac{1}{4 \pi |\vec{x}-\vec{x}'|},$$
i.e., the electrostatic potential of a unit point charge located at $\vec{x}'$ (Coulomb potential).

10. Feb 1, 2014

### pasmith

If a fluid is incompressible then the density $\rho$ of individual fluid particles is constant, so that
$$\frac{\partial \rho}{\partial t} + \vec v \cdot \nabla \rho = 0$$
which is not the same as asserting that density is constant, ie. $\dfrac{\partial \rho}{\partial t} = 0$ and $\nabla \rho = 0$.

Conservation of mass requires that
$$\frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \vec v) = 0$$
and applying the product rule gives
$$\frac{\partial \rho}{\partial t} + \rho \nabla \cdot \vec v + \vec v \cdot \nabla \rho = 0$$
and if the fluid is incompressible then this reduces to
$$\nabla \cdot \vec v = 0.$$

The fact that a vector field is divergence-free says nothing about its path-independence.

11. Feb 1, 2014

### FactChecker

When people talk about path independence of line integrals of vector fields, is it always $$\int_{\mathcal{C}} \mathrm{d} \vec{r} \cdot \vec{V} =0 \;$$ that they are referring to?

What about $$\int_{\mathcal{C}} \mathrm{d} \vec{r} \times \vec{V} =0 ? \;$$ That would require an incompressible (solenoidal) vector field.