Editing Conservative vector field

{{short description|Vector field that is the gradient of some function}}
{{More footnotes|date=May 2009}}
In [[vector calculus]], a '''conservative vector field''' is a [[vector field]] that is the [[gradient]] of some [[function (mathematics)|function]].<ref>{{cite book|title = Vector calculus|first1 = Jerrold|last1 = Marsden|author-link1 = Jerrold Marsden |first2 = Anthony|last2 = Tromba|publisher = W.H.Freedman and Company|edition = Fifth|year = 2003|pages = 550&ndash;561}}</ref> A conservative vector field has the property that its [[line integral]] is path independent; the choice of path between two points does not change the value of the line integral. Path independence of the line integral is equivalent to the vector field under the line integral being conservative. A conservative vector field is also '''irrotational'''; in three dimensions, this means that it has vanishing [[curl (mathematics)|curl]]. An irrotational vector field is necessarily conservative provided that the domain is [[simply connected]].

Conservative vector fields appear naturally in [[mechanics]]: They are vector fields representing [[force]]s of [[physical system]]s in which [[energy]] is [[conservation of energy|conserved]].<ref>George B. Arfken and Hans J. Weber, ''Mathematical Methods for Physicists'', 6th edition, Elsevier Academic Press (2005)</ref> For a conservative system, the [[work (physics)|work]] done in moving along a path in a configuration space depends on only the endpoints of the path, so it is possible to define [[potential energy]] that is independent of the actual path taken.

==Informal treatment==

In a two- and three-dimensional space, there is an ambiguity in taking an integral between two points as there are infinitely many paths between the two points—apart from the straight line formed between the two points, one could choose a curved path of greater length as shown in the figure. Therefore, in general, the value of the integral depends on the path taken. However, in the special case of a conservative vector field, the value of the integral is independent of the path taken, which can be thought of as a large-scale cancellation of all elements <math>d{R}</math> that do not have a component along the straight line between the two points. To visualize this, imagine two people climbing a cliff; one decides to scale the cliff by going vertically up it, and the second decides to walk along a winding path that is longer in length than the height of the cliff, but at only a small angle to the horizontal. Although the two hikers have taken different routes to get up to the top of the cliff, at the top, they will have both gained the same amount of gravitational potential energy. This is because a gravitational field is conservative.

[[File:Pathdependence.png|thumbnail|Depiction of two possible paths to integrate. In green is the simplest possible path; blue shows a more convoluted curve]]

==Intuitive explanation==

[[M. C. Escher|M. C. Escher's]] lithograph print ''[[Ascending and Descending]]'' illustrates a non-conservative vector field, impossibly made to appear to be the gradient of the varying height above ground (gravitational potential) as one moves along the staircase. The force field experienced by the one moving on the staircase is non-conservative in that one can return to the starting point while ascending more than one descends or vice versa, resulting in nonzero work done by gravity. On a real staircase, the height above the ground is a scalar potential field: one has to go upward exactly as much as one goes downward in order to return to the same place, in which case the work by gravity totals to zero. This suggests path-independence of work done on the staircase; equivalently, the force field experienced is conservative (see the later section: [[Conservative vector field#Path_independence_and_conservative_vector_field|Path independence and conservative vector field]]). The situation depicted in the print is impossible.

==Definition==

A [[vector field]] <math>\mathbf{v}: U \to \R^n</math>, where <math>U</math> is an open subset of <math>\R^n</math>, is said to be conservative if  there exists a <math>C^1</math> ([[Smoothness#Multivariate differentiability classes|continuously differentiable]]) [[scalar field]] <math>\varphi</math><ref name=":1">For <math>\mathbf{v} = \nabla \varphi</math> to be [[Conservative vector field#Path independence|path-independent]], <math>\varphi</math> is not necessarily continuously differentiable, the condition of being differentiable is enough, since the [[Gradient theorem]], that proves the path independence of <math>\nabla \varphi</math>, does not require <math>\varphi</math> to be continuously differentiable.

There must be a reason for the definition of conservative vector fields to require <math>\varphi</math> to be [[Smoothness#Multivariate differentiability classes|continuously differentiable]].</ref> on <math>U</math> such that

<math display="block">\mathbf{v} = \nabla \varphi.</math>

Here, <math>\nabla \varphi</math> denotes the [[gradient]] of <math>\varphi</math>. Since <math>\varphi</math> is continuously differentiable, <math>\mathbf{v}</math> is continuous. When the equation above holds, <math>\varphi</math> is called a [[scalar potential]] for <math>\mathbf{v}</math>.

The [[Helmholtz decomposition|fundamental theorem of vector calculus]] states that, under some regularity conditions, any vector field can be expressed as the sum of a conservative vector field and a [[solenoidal field]].

==Path independence and conservative vector field==
{{main article|Gradient theorem}}

=== Path independence ===
A line integral of a vector field <math>\mathbf{v}</math> is said to be path-independent if it depends on only two integral path endpoints regardless of which path between them is chosen:<ref name=":0">{{Cite book |last=Stewart |first=James |title=Calculus |publisher=Cengage Learning |year=2015 |isbn=978-1-285-74062-1 |edition=8th |pages=1127–1134 |language=English |chapter=16.3 The Fundamental Theorem of Line Integrals"}}</ref> 
<math display="block">\int_{P_1} \mathbf{v} \cdot d \mathbf{r} = \int_{P_2} \mathbf{v} \cdot d \mathbf{r}</math>

for any pair of integral paths <math>P_1</math> and <math>P_2</math> between a given pair of path endpoints in <math>U</math>.

The path independence is also equivalently expressed as

<math display="block">\int_{P_c} \mathbf{v} \cdot d \mathbf{r} = 0</math>
for any [[piecewise]] smooth closed path <math>P_c</math> in <math>U</math> where the two endpoints are coincident. Two expressions are equivalent since any closed path <math>P_c</math> can be made by two path; <math>P_1</math> from an endpoint <math>A</math> to another endpoint <math>B</math>, and <math>P_2</math> from <math>B</math> to <math>A</math>, so
<math display="block">\int_{P_c} \mathbf{v} \cdot d \mathbf{r} = \int_{P_1} \mathbf{v} \cdot d \mathbf{r} + \int_{P_2} \mathbf{v} \cdot d \mathbf{r} = \int_{P_1} \mathbf{v} \cdot d \mathbf{r} - \int_{-P_2} \mathbf{v} \cdot d \mathbf{r} = 0</math>
where <math>-P_2</math> is the reverse of <math>P_2</math> and the last equality holds due to the path independence <math display="inline">\displaystyle \int_{P_1} \mathbf{v} \cdot d \mathbf{r} = \int_{-P_2} \mathbf{v} \cdot d \mathbf{r}.</math>

=== Conservative vector field ===
A key property of a conservative vector field <math>\mathbf{v}</math> is that its integral along a path depends on only the endpoints of that path, not the particular route taken. In other words, ''if it is a conservative vector field, then its line integral is path-independent.'' Suppose that <math>\mathbf{v} = \nabla \varphi</math> for some <math>C^1</math> ([[Smoothness#Multivariate differentiability classes|continuously differentiable]]) scalar field <math>\varphi</math><ref name=":1" /> over <math>U</math> as an open subset of <math>\R^n</math> (so <math>\mathbf{v}</math> is a conservative vector field that is continuous) and <math>P</math> is a differentiable path (i.e., it can be parameterized by a [[differentiable function]]) in <math>U</math> with an initial point <math>A</math> and a terminal point <math>B</math>. Then the [[gradient theorem]] (also called ''fundamental theorem of calculus for line integrals'') states that
<math display="block">\int_{P} \mathbf{v} \cdot d{\mathbf{r}} = \varphi(B) - \varphi(A).</math>

This holds as a consequence of the [[Line integral#Line integral of a vector field|definition of a line integral]], the [[chain rule]], and the [[fundamental theorem of calculus|second fundamental theorem of calculus]]. <math>\mathbf{v} \cdot d\mathbf{r} = \nabla{\varphi} \cdot d\mathbf{r}</math> in the line integral is an [[exact differential]] for an orthogonal coordinate system (e.g., [[Cartesian coordinate system|Cartesian]], [[cylindrical]], or [[Spherical coordinate system|spherical coordinates]]). Since the gradient theorem is applicable for a differentiable path, the path independence of a conservative vector field over piecewise-differential curves is also proved by the proof per differentiable curve component.<ref>Need to verify if exact differentials also exist for non-orthogonal coordinate systems.</ref>

So far it has been proven that a conservative vector field <math>\mathbf{v}</math> is line integral path-independent. Conversely, ''if a continuous vector field <math>\mathbf{v}</math> is (line integral) path-independent, then it is a conservative vector field'', so the following [[Logical biconditional|biconditional]] statement holds:<ref name=":0" />
{{block indent | em = 1.5 | text = For a continuous [[vector field]] <math>\mathbf{v}: U \to \R^n</math>, where <math>U</math> is an open subset of <math>\R^n</math>, it is conservative if and only if its line integral along a path in <math>U</math> is path-independent, meaning that the line integral depends on only both path endpoints regardless of which path between them is chosen.}}
The proof of this converse statement is the following.
[[File:Line Integral paths to prove the Relation between Path Independence and Conservative Vector Field, 2022-03-13.png|thumb|Line integral paths used to prove the following statement: if the line integral of a vector field is path-independent, then the vector field is a conservative vector field.]]
<math>\mathbf{v}</math> is a continuous vector field which line integral is path-independent. Then, let's make a function <math>\varphi</math> defined as
<math display="block">\varphi(x,y) = \int_{a,b}^{x,y} \mathbf{v} \cdot d{\mathbf{r}}</math>
over an arbitrary path between a chosen starting point <math>(a,b)</math> and an arbitrary point <math>(x,y)</math>. Since it is path-independent, it depends on only <math>(a,b)</math> and <math>(x,y)</math> regardless of which path between these points is chosen.

Let's choose the path shown in the left of the right figure where a 2-dimensional [[Cartesian coordinate system]] is used. The second segment of this path is parallel to the <math>x</math> axis so there is no change along the <math>y</math> axis. The line integral along this path is
<math display="block">\int_{a,b}^{x,y} \mathbf{v} \cdot d{\mathbf{r}} = \int_{a,b}^{x_1, y} \mathbf{v} \cdot d{\mathbf{r}} + \int_{x_1, y}^{x,y} \mathbf{v} \cdot d{\mathbf{r}}.</math>
By the path independence, its [[partial derivative]] with respect to <math>x</math> (for <math>\varphi</math> to have partial derivatives, <math>\mathbf{v}</math> needs to be continuous.) is
<math display="block">\frac{\partial \varphi}{\partial x} = \frac{\partial}{\partial x} \int_{a,b}^{x,y} \mathbf{v} \cdot d{\mathbf{r}} = \frac{\partial}{\partial x} \int_{a,b}^{x_1,y} \mathbf{v} \cdot d{\mathbf{r}} + \frac{\partial}{\partial x} \int_{x_1,y}^{x,y} \mathbf{v} \cdot d{\mathbf{r}} = 0 + \frac{\partial}{\partial x} \int_{x_1,y}^{x,y} \mathbf{v} \cdot d{\mathbf{r}}</math>
since <math>x_1</math> and <math>x</math> are independent to each other. Let's express <math>\mathbf{v}</math> as <math>{\displaystyle \mathbf {v} } = P(x,y) \mathbf{i} + Q(x,y) \mathbf{j}</math> where <math>\mathbf{i}</math> and <math>\mathbf{j}</math> are unit vectors along the <math>x</math> and <math>y</math> axes respectively, then, since <math>d \mathbf{r} = dx \mathbf{i} + dy \mathbf{j}</math>,
<math display="block">\frac{\partial}{\partial x} \varphi (x,y) = \frac{\partial}{\partial x} \int_{x_1,y}^{x,y} \mathbf{v} \cdot d\mathbf{r} = \frac{\partial}{\partial x} \int_{x_1,y}^{x,y} P(t,y) dt = P(x,y) </math>
where the last equality is from the [[Fundamental theorem of calculus|second fundamental theorem of calculus]].

A similar approach for the line integral path shown in the right of the right figure results in <math display="inline">\frac{\partial}{\partial y} \varphi (x,y) = Q(x,y) </math> so
<math display="block">\mathbf{v} = P(x,y) \mathbf{i}+ Q(x,y) \mathbf{j} = \frac{\partial \varphi}{\partial x} \mathbf{i} + \frac{\partial \varphi}{\partial y} \mathbf{j} = \nabla \varphi</math>
is proved for the 2-dimensional [[Cartesian coordinate system]]. This proof method can be straightforwardly expanded to a higher dimensional orthogonal coordinate system (e.g., a 3-dimensional [[spherical coordinate system]]) so the converse statement is proved. Another proof is found [[Gradient theorem|here]] as the converse of the gradient theorem.

==Irrotational vector fields==

[[File:Irrotational vector field.svg|300px|thumbnail|right|The above vector field <math>\mathbf{v} = \left( - \frac{y}{x^2 + y^2},\frac{x}{x^2 + y^2},0 \right)</math> defined on <math>U = \R^3 \setminus \{ (0,0,z) \mid z \in \R \}</math>, i.e., <math>\R^3</math> with removing all coordinates on the <math>z</math>-axis (so not a simply connected space), has zero curl in <math>U</math> and is thus irrotational. However, it is not conservative and does not have path independence.]]

Let <math>n = 3</math> (3-dimensional space), and let <math>\mathbf{v}: U \to \R^3</math> be a <math>C^1</math> ([[Smoothness#Multivariate differentiability classes|continuously differentiable]]) vector field, with an open subset <math>U</math> of <math>\R^n</math>. Then <math>\mathbf{v}</math> is called irrotational if  its [[Curl (mathematics)|curl]] is <math>\mathbf{0}</math> everywhere in <math>U</math>, i.e., if
<math display="block">\nabla \times \mathbf{v} \equiv \mathbf{0}.</math>

For this reason, such vector fields are sometimes referred to as curl-free vector fields or curl-less vector fields. They are also referred to as [[Helmholtz decomposition#Longitudinal and transverse fields |longitudinal vector fields]].

It is an [[Vector calculus identities#Curl of gradient is zero|identity of vector calculus]] that for any <math>C^2</math> ([[Smoothness#Multivariate differentiability classes|continuously differentiable up to the 2nd derivative]]) scalar field <math>\varphi</math> on <math>U</math>, we have
<math display="block">\nabla \times (\nabla \varphi) \equiv \mathbf{0}.</math>

Therefore, ''every <math>C^1</math> conservative vector field in <math>U</math> is also an irrotational vector field in <math>U</math>''. This result can be easily proved by expressing <math>\nabla \times (\nabla \varphi)</math> in a [[Cartesian coordinate system]] with [[Symmetry of second derivatives#Schwarz's theorem|Schwarz's theorem]] (also called Clairaut's theorem on equality of mixed partials).

Provided that <math>U</math> is a [[simply connected space|simply connected open space]] (roughly speaking, a single piece open space without a hole within it), the converse of this is also true: ''Every irrotational vector field in a simply connected open space <math>U</math> is a <math>C^1</math> conservative vector field in <math>U</math>''.

The above statement is ''not'' true in general if <math>U</math> is not simply connected. Let <math>U</math> be <math>\R^3</math> with removing all coordinates on the <math>z</math>-axis (so not a simply connected space), i.e., <math>U = \R^3 \setminus \{ (0,0,z) \mid z \in \R \}</math>. Now, define a vector field <math>\mathbf{v}</math> on <math>U</math> by
<math display="block">\mathbf{v}(x,y,z) ~ \stackrel{\text{def}}{=} ~ \left( - \frac{y}{x^2 + y^2},\frac{x}{x^2 + y^2},0 \right).</math>

Then <math>\mathbf{v}</math> has zero curl everywhere in <math>U</math> (<math>\nabla \times \mathbf{v} \equiv \mathbf{0}</math> at everywhere in <math>U</math>), i.e., <math>\mathbf{v}</math> is irrotational. However, the [[Circulation (physics)|circulation]] of <math>\mathbf{v}</math> around the [[unit circle]] in the <math>xy</math>-plane is <math>2 \pi</math>; in [[polar coordinates]], <math>\mathbf{v} = \mathbf{e}_{\phi} / r</math>, so the integral over the unit circle is
<math display="block">\oint_{C} \mathbf{v} \cdot \mathbf{e}_{\phi} ~ d{\phi} = 2 \pi.</math>

Therefore, <math>\mathbf{v}</math> does not have the path-independence property discussed above so is not conservative even if <math>\nabla \times \mathbf{v} \equiv \mathbf{0}</math> since <math>U</math> where <math>\mathbf{v}</math> is defined is not a simply connected open space. 

Say again, in a simply connected open region, an irrotational vector field <math>\mathbf{v}</math> has the path-independence property (so <math>\mathbf{v}</math> as conservative). This can be proved directly by using [[Stokes' theorem]],<math display="block">\oint _{P_c} \mathbf{v} \cdot d \mathbf {r} = \iint _{A}(\nabla \times \mathbf{v})\cdot d \mathbf {a} = 0</math>for any smooth oriented surface <math>A</math> which boundary is a simple closed path <math>P_c</math>. So, it is concluded that ''In a simply connected open region, any'' <math>C^1</math> ''vector field that has the path-independence property (so it is a conservative vector field.) must also be irrotational and vice versa.''

=== Abstraction ===
More abstractly, in the presence of a [[Riemannian metric]], vector fields correspond to [[differential form|differential {{nowrap|<math>1</math>-forms}}]]. The conservative vector fields correspond to the [[closed and exact differential forms|exact]] {{nowrap|<math>1</math>-forms}}, that is, to the forms which are the [[exterior derivative]] <math>d\phi</math> of a function (scalar field) <math>\phi</math> on <math>U</math>. The irrotational vector fields correspond to the [[closed and exact differential forms|closed]] {{nowrap|<math>1</math>-forms}}, that is, to the {{nowrap|<math>1</math>-forms}} <math>\omega</math> such that <math>d\omega = 0</math>. As {{nowrap|<math>d^2 = 0</math>,}} any exact form is closed, so any conservative vector field is irrotational. Conversely, all closed {{nowrap|<math>1</math>-forms}} are exact if <math>U</math> is [[simply connected]].

=== Vorticity ===
{{main article|Vorticity}}

The [[vorticity]] <math>\boldsymbol{\omega}</math> of a vector field can be defined by:
<math display="block">\boldsymbol{\omega} ~ \stackrel{\text{def}}{=} ~ \nabla \times \mathbf{v}.</math>

The vorticity of an irrotational field is zero everywhere.<ref>{{citation|title = Elements of Gas Dynamics|first1 = H.W.|last1 = Liepmann|author-link1 = Hans W. Liepmann|first2 = A.|last2 = Roshko|author-link2 = Anatol Roshko|publisher = Courier Dover Publications|year = 1993|orig-year = 1957|isbn = 0-486-41963-0}}, pp. 194–196.</ref> [[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 field, the vorticity acts as a measure of the ''local'' rotation of fluid elements. The vorticity does ''not'' imply anything about the global behavior of a fluid. It is possible for a fluid that travels in a straight line to have vorticity, and it is possible for a fluid that moves in a circle to be irrotational.

==Conservative forces==

[[File:Conservative fields.png|thumb|upright=1.5|Examples of potential and gradient fields in physics:{{unbulleted list
| {{color box|yellow}} Scalar fields, scalar potentials:{{unbulleted list
 | style=margin-left:1.6em;
 | '''V<sub>G</sub>''', gravitational potential
 | '''W<sub>pot</sub>''', (gravitational or electrostatic) potential energy
 | '''V<sub>C</sub>''', Coulomb potential
 }}| {{color box|cyan}} Vector fields, gradient fields:{{unbulleted  list
 | style=margin-left:1.6em;
 | '''a<sub>G</sub>''', gravitational acceleration
 | '''F''', (gravitational or electrostatic) force
 | '''E''', electric field strength
 }}}}
]]

If the vector field associated to a force <math>\mathbf{F}</math> is conservative, then the force is said to be a [[conservative force]].

The most prominent examples of conservative forces are gravitational force (associated with a gravitational field) and electric force (associated with an electrostatic field). According to [[Newton's law of universal gravitation|Newton's law of gravitation]], a [[gravitational force]] <math>\mathbf{F}_{G}</math> acting on a mass <math>m</math> due to a mass <math>M</math> located at a distance <math>r</math> from <math>m</math>, obeys the equation

<math display="block">\mathbf{F}_{G} = - \frac{G m M}{r^2} \hat{\mathbf{r}},</math>

where <math>G</math> is the [[gravitational constant]] and <math>\hat{\mathbf{r}}</math> is a ''unit'' vector pointing from <math>M</math> toward <math>m</math>. The force of gravity is conservative because <math>\mathbf{F}_{G} = - \nabla \Phi_{G}</math>, where

<math display="block">\Phi_{G} ~ \stackrel{\text{def}}{=} - \frac{G m M}{r}</math>

is the [[gravitational potential energy]]. In other words, the gravitation field <math>\frac{\mathbf{F}_{G}}{m}</math> associated with the gravitational force <math>\mathbf{F}_{G}</math> is the [[gradient]] of the gravitation potential <math>\frac{\Phi_{G}}{m}</math> associated with the gravitational potential energy <math>\Phi_{G}</math>. It can be shown that any vector field of the form <math>\mathbf{F}=F(r) \hat{\mathbf{r}}</math> is conservative, provided that <math>F(r)</math> is integrable.

For [[conservative force]]s, ''path independence'' can be interpreted to mean that the [[work done]] in going from a point <math>A</math> to a point <math>B</math> is independent of the moving path chosen (dependent on only the points <math>A</math> and <math>B</math>), and that the work <math>W</math> done in going around a simple closed loop <math>C</math> is <math>0</math>:

<math display="block">W = \oint_{C} \mathbf{F} \cdot d{\mathbf{r}} = 0.</math>

The total [[conservation of energy|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 the equal quantity of kinetic energy, or vice versa.

==See also==
* [[Beltrami vector field]]
* [[Conservative force]]
* [[Conservative system]]
* [[Complex lamellar vector field]]
* [[Helmholtz decomposition]]
* [[Laplacian vector field]]
* [[Longitudinal and transverse vector fields]]
* [[Solenoidal vector field]]

==References==
{{reflist}}

==Further reading==
* {{cite book |first= D. J. |last= Acheson |title= Elementary Fluid Dynamics |publisher= Oxford University Press |date= 1990 |isbn= 0198596790 }}

[[Category:Vector calculus]]
[[Category:Force]]