Conservative Vector Fields

Stokes' Theorem

Stokes' Theorem is effectively a generalization of Green's theorem to 3 dimensions, and the "curl" is a generalization of the quantity ${\frac {\partial F_{y}}{\partial x}}-{\frac {\partial F_{x}}{\partial y}}$ to 3 dimensions. An arbitrary oriented surface $\sigma$ can be articulated into a family of infinitesimal surfaces, some parallel to the xy-plane, others parallel to the zx-plane, and the remainder parallel to the yz-plane. Let $\mathbf {F}$ denote an arbitrary vector field.

Let $\sigma$ be a surface that is parallel to the yz-plane with counter-clockwise oriented boundary $C$ . Green's theorem gives:

$\int _{\mathbf {q} \in C}\mathbf {F} (\mathbf {q} )\cdot d\mathbf {q} =\iint _{\mathbf {q} \in \sigma }\left({\frac {\partial F_{z}}{\partial y}}-{\frac {\partial F_{y}}{\partial z}}\right)(\mathbf {dS} \cdot \mathbf {i} )=\iint _{\mathbf {q} \in \sigma }\left(\left({\frac {\partial F_{z}}{\partial y}}-{\frac {\partial F_{y}}{\partial z}}\right)\mathbf {i} \right)\cdot \mathbf {dS}$

$\mathbf {dS} \cdot \mathbf {i}$ is positive if the normal direction to $\sigma$ points in the positive x direction and is negative if otherwise. If the normal direction to $\sigma$ points in the negative x direction, then $C$ is oriented clockwise instead of counter-clockwise in the yz-plane.

Decomposing a 3D loop into an ensemble of infinitesimal loops that are parallel to the yz, zx, or xy planes.

Repeating this argument for $\sigma$ being parallel to the zx-plane and xy-plane respectively gives:

$\int _{\mathbf {q} \in C}\mathbf {F} (\mathbf {q} )\cdot d\mathbf {q} =\iint _{\mathbf {q} \in \sigma }\left({\frac {\partial F_{x}}{\partial z}}-{\frac {\partial F_{z}}{\partial x}}\right)(\mathbf {dS} \cdot \mathbf {j} )=\iint _{\mathbf {q} \in \sigma }\left(\left({\frac {\partial F_{x}}{\partial z}}-{\frac {\partial F_{z}}{\partial x}}\right)\mathbf {j} \right)\cdot \mathbf {dS}$

and

$\int _{\mathbf {q} \in C}\mathbf {F} (\mathbf {q} )\cdot d\mathbf {q} =\iint _{\mathbf {q} \in \sigma }\left({\frac {\partial F_{y}}{\partial x}}-{\frac {\partial F_{x}}{\partial y}}\right)(\mathbf {dS} \cdot \mathbf {k} )=\iint _{\mathbf {q} \in \sigma }\left(\left({\frac {\partial F_{y}}{\partial x}}-{\frac {\partial F_{x}}{\partial y}}\right)\mathbf {k} \right)\cdot \mathbf {dS}$

Treating $\sigma$ as an ensemble of infinitesimal surfaces parallel to the yz-plane, zx-plane, or xy-plane gives:

$\int _{\mathbf {q} \in C}\mathbf {F} (\mathbf {q} )\cdot d\mathbf {q} =\iint _{\mathbf {q} \in \sigma }\left(\left({\frac {\partial F_{z}}{\partial y}}-{\frac {\partial F_{y}}{\partial z}}\right)\mathbf {i} +\left({\frac {\partial F_{x}}{\partial z}}-{\frac {\partial F_{z}}{\partial x}}\right)\mathbf {j} +\left({\frac {\partial F_{y}}{\partial x}}-{\frac {\partial F_{x}}{\partial y}}\right)\mathbf {k} \right)\cdot \mathbf {dS}$

This is Stokes' theorem, and $\nabla \times \mathbf {F} =\left({\frac {\partial F_{z}}{\partial y}}-{\frac {\partial F_{y}}{\partial z}}\right)\mathbf {i} +\left({\frac {\partial F_{x}}{\partial z}}-{\frac {\partial F_{z}}{\partial x}}\right)\mathbf {j} +\left({\frac {\partial F_{y}}{\partial x}}-{\frac {\partial F_{x}}{\partial y}}\right)\mathbf {k}$ is the "curl" of $\mathbf {F}$ which generalizes the "circulation density" to 3 dimensions.

The direction of $\nabla \times \mathbf {F}$ at $\mathbf {q}$ is effectively an "axis of rotation" around which the counterclockwise circulation density in a plane whose normal is parallel to $\nabla \times \mathbf {F}$ is $|\nabla \times \mathbf {F} |$ . Out of all planes that pass through $\mathbf {q}$ , the plane whose normal is parallel to $\nabla \times \mathbf {F}$ has the largest counterclockwise circulation density at $\mathbf {q}$ which is $|\nabla \times \mathbf {F} |$ .

An arbitrary vector field $\mathbf {F}$ that is differentiable everywhere is considered to be "irrotational" or "conservative" if $\nabla \times \mathbf {F} =\mathbf {0}$ everywhere, or equivalently that $\int _{\mathbf {q} \in C}\mathbf {F} (\mathbf {q} )\cdot d\mathbf {q} =0$ for all continuous closed curves $C$

Conservative vector fields

A vector field $\mathbf {F}$ for which $\nabla \times \mathbf {F} =\mathbf {0}$ at all points is an "conservative" vector field. $\mathbf {F}$ can also be referred to as being "irrotational" since the gain around any closed curve is always 0.

A key property of a conservative vector field $\mathbf {F}$ is that the gain of $\mathbf {F}$ along a continuous curve is purely a function of the curve's end points. If $C_{1}$ and $C_{2}$ are two continuous curves which share the same starting point $\mathbf {q} _{0}$ and end point $\mathbf {q} _{1}$ , then $\int _{\mathbf {q} \in C_{1}}\mathbf {F} (\mathbf {q} )\cdot d\mathbf {q} =\int _{\mathbf {q} \in C_{2}}\mathbf {F} (\mathbf {q} )\cdot d\mathbf {q}$ . In other words, the gain is purely a function of $\mathbf {q} _{0}$ and $\mathbf {q} _{1}$ . This property can be derived from Stokes' theorem as follows:

Invert the orientation of $C_{2}$ to get $-C_{2}$ and combine $C_{1}$ and $-C_{2}$ to get a continuous closed curve $C_{3}=C_{1}-C_{2}$ , linking the curves together at the endpoints $\mathbf {q} _{0}$ and $\mathbf {q} _{1}$ . Let $\sigma$ denote a surface for which $C_{3}$ is the counterclockwise oriented boundary.

Stokes' theorem states that $\int _{\mathbf {q} \in C_{3}}\mathbf {F} (\mathbf {q} )\cdot d\mathbf {q} =\iint _{\mathbf {q} \in \sigma }(\nabla \times \mathbf {F} )|_{\mathbf {q} }\cdot \mathbf {dS} =0$ . The gain around $C_{3}$ is the gain along $C_{1}$ minus the gain along $C_{2}$ : $\int _{\mathbf {q} \in C_{3}}\mathbf {F} (\mathbf {q} )\cdot d\mathbf {q} =\int _{\mathbf {q} \in C_{1}}\mathbf {F} (\mathbf {q} )\cdot d\mathbf {q} +\int _{\mathbf {q} \in -C_{2}}\mathbf {F} (\mathbf {q} )\cdot d\mathbf {q}$ $=\int _{\mathbf {q} \in C_{1}}\mathbf {F} (\mathbf {q} )\cdot d\mathbf {q} -\int _{\mathbf {q} \in C_{2}}\mathbf {F} (\mathbf {q} )\cdot d\mathbf {q}$ . Therefore:

$\int _{\mathbf {q} \in C_{3}}\mathbf {F} (\mathbf {q} )\cdot d\mathbf {q} =0$ $\implies \int _{\mathbf {q} \in C_{1}}\mathbf {F} (\mathbf {q} )\cdot d\mathbf {q} =\int _{\mathbf {q} \in C_{2}}\mathbf {F} (\mathbf {q} )\cdot d\mathbf {q}$