Editing Conservative vector field (section)

==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.