# Conservative vector field

(Redirected from Irrotational vector field)

In vector calculus a conservative vector field is a vector field that is the gradient of some function, known in this context as a scalar potential.[1] Conservative vector fields have the property that the line integral is path independent, i.e. the choice of integration path between any point and another does not change the result. Path independence of a line integral is equivalent to the vector field being conservative. A conservative vector field is also irrotational; in three dimensions this means that it has vanishing curl. An irrotational vector field is necessarily conservative provided that a certain condition on the geometry of the domain holds, i.e. the domain is simply connected.

Conservative vector fields appear naturally in mechanics: they are vector fields representing forces of physical systems in which energy is conserved.[2] For a conservative system, the work done in moving along a path in configuration space depends only on the endpoints of the path, so it is possible to define a potential energy independently of the path taken.

## Intuitive explanation

M. C. Escher's lithograph Ascending and Descending

M. C. Escher's painting Ascending and Descending illustrates a non-conservative vector field, impossibly made to appear to be the gradient of the varying height above ground as one moves along the staircase. It is "rotational" in that one can keep getting higher or keep getting lower while going around in circles. It is non-conservative in that one can return to one's starting point while ascending more than one descends or vice-versa. On a real staircase the height above the ground is a scalar potential field: if one returns to the same place, one goes upward exactly as much as one goes downward. Its gradient would be a conservative vector field, and is irrotational. The situation depicted in the painting is impossible.

## Definition

A vector field $\mathbf{v}$ is said to be conservative if there exists a scalar field $\varphi$ such that

$\mathbf{v}=\nabla\varphi.$

Here $\nabla\varphi$ denotes the gradient of $\varphi$. When the above equation holds, $\varphi$ is called a scalar potential for $\mathbf{v}$.

The fundamental theorem of vector calculus states that any vector field can be expressed as the sum of a conservative vector field and a solenoidal field.

## Path independence

A key property of a conservative vector field is that its integral along a path depends only on the endpoints of that path, not the particular route taken. Suppose that $S\subseteq\mathbb{R}^3$ is a region of three-dimensional space, and that $P$ is a rectifiable path in $S$ with start point $A$ and end point $B$. If $\mathbf{v}=\nabla\varphi$ is a conservative vector field then the gradient theorem states that

$\int_P \mathbf{v}\cdot d\mathbf{r}=\varphi(B)-\varphi(A).$

This holds as a consequence of the chain rule and the fundamental theorem of calculus.

An equivalent formulation of this is to say that

$\oint \mathbf{v}\cdot d\mathbf{r}=0$

for every closed loop in S. The converse of this statement is also true: if the circulation of v around every closed loop in an open set S is zero, then v is a conservative vector field.

## Irrotational vector fields

The above field v = (−y/(x2 + y2), +x/(x2 + y2), 0) includes a vortex at its center, so it is non-irrotational; it is neither conservative, nor does it have path independence. However, any simply connected subset that excludes the vortex line (0,0,z) will have zero curl, ∇ × v = 0. Such vortex-free regions are examples of irrotational vector fields.

A vector field $\mathbf{v}$ is said to be irrotational if its curl is zero. That is, if

$\nabla\times\mathbf{v} = \mathbf{0}.$

For this reason, such vector fields are sometimes referred to as curl free field (curl-free vector field) or curl-less vector fields.

It is an identity of vector calculus that for any scalar field $\varphi$:

$\nabla \times \nabla \varphi=\mathbf{0}.$

Therefore every conservative vector field is also an irrotational vector field.

Provided that $S$ is a simply connected region, the converse of this is true: every irrotational vector field is also a conservative vector field.

The above statement is not true if $S$ is not simply connected. Let $S$ be the usual 3-dimensional space, except with the $z$-axis removed; that is $S=\mathbb{R}^3\setminus\{(0,0,z)~|~z\in\mathbb{R}\}$. Now define a vector field by

$\mathbf{v}= \left( \frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2}, 0 \right).$

Then $\mathbf{v}$ exists and has zero curl at every point in $S$; that is $\mathbf{v}$ is irrotational. However the circulation of $\mathbf{v}$ around the unit circle in the $x,y$-plane is equal to $2\pi$. Indeed we note that in polar coordinates $\mathbf{v}= \mathbf{e_{\phi}}/r$, so the integral over the unit circle is equal $\int \mathbf{v}\mathbf{e_{\phi}}\mathrm{d}\phi=2\pi$. Therefore $\mathbf{v}$ does not have the path independence property discussed above, and is not conservative. (However, in any simply connected subregion of S, it is still true that it is conservative. In fact, the field above is the gradient of $\arg (x + iy)$. As we know from complex analysis, this is a multi-valued function which requires a branch cut from the origin to infinity to be defined in a continuous way; hence, in a region that does not go around the z-axis, its gradient is conservative.)

In a simply connected region an irrotational vector field has the path independence property. This can be seen by noting that in such a region an irrotational vector field is conservative, and conservative vector fields have the path independence property. The result can also be proved directly by using Stokes' theorem. In a connected region any vector field which has the path independence property must also be irrotational.

More abstractly, a conservative vector field is an exact 1-form. That is, it is a 1-form equal to the exterior derivative of some 0-form (scalar field) $\phi$. An irrotational vector field is a closed 1-form. Since d2 = 0, any exact form is closed, so any conservative vector field is irrotational. The domain is simply connected if and only if its first homology group is 0, which is equivalent to its first cohomology group being 0. The first de Rham cohomology group $H_{\mathrm{dR}}^{1}$ is 0 if and only if all closed 1-forms are exact.

## Irrotational flows

The flow velocity $\mathbf{v}$ of a fluid is a vector field, and the vorticity $\boldsymbol{\omega}$ of the flow can be defined by

$\boldsymbol{\omega}=\nabla\times\mathbf{v}.$

A common alternative notation for vorticity is $\zeta\;$.[3]

If $\mathbf{v}$ is irrotational, with $\nabla\times\mathbf{v} =\mathbf{0}$, then the flow is said to be an irrotational flow. The vorticity of an irrotational flow is zero.[4]

Kelvin's circulation theorem states that a fluid that is irrotational in an inviscid flow will remain irrotational. This result can be derived from the vorticity transport equation, obtained by taking the curl of the Navier-stokes equations.

For a two-dimensional flow the vorticity acts as a measure of the local rotation of fluid elements. Note that the vorticity does not imply anything about the global behaviour of a fluid. It is possible for a fluid traveling in a straight line to have vorticity, and it is possible for a fluid which moves in a circle to be irrotational.

## Conservative forces

Examples of potential and gradient fields in physics
Scalar fields (scalar potentials) (yellow): VG - gravitational potential; Wpot - potential energy; VC - Coulomb potential; Vector fields (gradient fields) (cyan): aG - gravitational acceleration; F - force; E - electric field strength

If the vector field associated to a force $\mathbf{F}$ is conservative then the force is said to be a conservative force.

The most prominent examples of conservative forces are the force of gravity and the electric field associated to a static charge. According to Newton's law of gravitation, the gravitational force, $\mathbf{F}_G$, acting on a mass $m$, due to a mass $M$ which is a distance $r$ away, obeys the equation

$\mathbf{F}_G=-\frac{GmM\hat{\mathbf{r}}}{r^2},$

where $G$ is the gravitational constant and $\hat{\mathbf{r}}$ is a unit vector pointing from $M$ towards $m$. The force of gravity is conservative because $\mathbf{F}_G=-\nabla\Phi_G$, where

$\Phi_G=-\frac{GmM}{r}$

For conservative forces, path independence can be interpreted to mean that the work done in going from a point $A$ to a point $B$ is independent of the path chosen, and that the work W done in going around a closed loop is zero:

$W=\oint \mathbf{F}\cdot d\mathbf{r}=0.$

The total energy of a particle moving under the influence of conservative forces is conserved, in the sense that a loss of potential energy is converted to an equal quantity of kinetic energy or vice versa.