Editing Exact differential (section)

==Overview==

===Definition===
Even if we work in three dimensions here, the definitions of exact differentials for other dimensions are structurally similar to the three dimensional definition. In three dimensions, a form of the type
<math display="block">
 A(x, y, z) \,dx + B(x, y, z) \,dy + C(x, y, z) \,dz
</math>
is called a [[differential form]]. This form is called ''exact'' on an open domain <math>D \subset \mathbb{R}^3</math> in space if there exists some [[Differentiable function|differentiable]] [[scalar function]] <math>Q = Q(x, y, z)</math> defined on <math>D</math> such that
<math display="block">
 dQ \equiv \left(\frac{\partial Q}{\partial x}\right)_{y,z} \, dx
         + \left(\frac{\partial Q}{\partial y}\right)_{x,z} \, dy
         + \left(\frac{\partial Q}{\partial z}\right)_{x,y} \, dz,
 \quad dQ = A \, dx + B \, dy + C \, dz
</math>
throughout <math>D</math>, where <math>x, y, z</math> are [[orthogonal coordinates]] (e.g., [[Cartesian coordinate system|Cartesian]], [[cylindrical]], or [[Spherical coordinate system|spherical coordinates]]). In other words, in some open domain of a space, a differential form is an ''exact differential'' if it is equal to the general differential of a differentiable function in an orthogonal coordinate system.

The subscripts outside the parenthesis in the above mathematical expression indicate which variables are being held constant during differentiation. Due to the definition of the [[partial derivative]], these subscripts are not required, but they are explicitly shown here as reminders.

=== Integral path independence ===
The exact differential for a differentiable scalar function <math>Q</math> defined in an open domain <math>D \subset \mathbb{R}^n</math> is equal to <math>dQ = \nabla Q \cdot d \mathbf{r}</math>, where <math>\nabla Q</math> is the [[gradient]] of <math>Q</math>, <math>\cdot</math> represents the [[Dot product|scalar product]], and <math>d \mathbf{r}</math> is the general differential displacement vector, if an orthogonal coordinate system is used. If <math>    Q</math> is of differentiability class <math>C^1</math> ([[Smoothness#Multivariate differentiability classes|continuously differentiable]]), then <math>\nabla Q</math> is a [[conservative vector field]] for the corresponding potential <math>Q</math> by the definition. For three dimensional spaces, expressions such as <math>d \mathbf{r} = (dx, dy, dz)</math> and <math>\nabla Q = \left(\frac{\partial Q}{\partial x}, \frac{\partial Q}{\partial y},\frac{\partial Q}{\partial z}\right)</math> can be made.

The [[gradient theorem]] states

:<math>\int _{i}^{f} dQ = \int _{i}^{f}\nabla Q (\mathbf {r} )\cdot d \mathbf {r} = Q \left(f \right) - Q \left(i \right)</math>

that does not depend on which integral path between the given path endpoints <math>i</math> and <math>f</math> is chosen. So it is concluded that ''the integral of an exact differential is independent of the choice of an integral path between given path endpoints [[Conservative vector field#Path independence|(path independence)]].''

For three dimensional spaces, if <math>\nabla Q</math> defined on an open domain <math>D \subset \mathbb{R}^3</math> is of [[Smoothness#Multivariate differentiability classes|differentiability class]] <math>C^1</math> (equivalently <math>Q</math> is of <math>C^2</math>), then this integral path independence can also be proved by using the [[Vector calculus identity#Curl of gradient is zero|vector calculus identity]] <math>\nabla \times ( \nabla Q )  = \mathbf{0}</math> and [[Stokes' theorem]].

:<math>\oint _{\partial \Sigma }\nabla Q \cdot d \mathbf {r} = \iint _{\Sigma }(\nabla \times \nabla Q)\cdot d \mathbf {a} = 0</math>
for a simply closed loop <math>\partial \Sigma</math> with the smooth oriented surface <math>\Sigma</math> in it. If the open domain <math>D</math> is [[Simply connected space|simply connected open space]] (roughly speaking, a single piece open space without a hole within it), then any irrotational vector field (defined as a <math>C^1</math> vector field <math>\mathbf{v}</math> which curl is zero, i.e., <math>\nabla \times \mathbf{v}  = \mathbf{0}</math>) has the path independence by the Stokes' theorem, so the following statement is made; ''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.'' The equality of the path independence and conservative vector fields is shown [[Conservative vector field#Path independence and conservative vector field|here]].

==== Thermodynamic state function ====
In [[thermodynamics]], when <math>dQ</math> is exact, the function <math>Q</math> is a [[state function]] of the system: a [[Function (mathematics)|mathematical function]] which depends solely on the current [[Thermodynamic equilibrium|equilibrium state]], not on the path taken to reach that state. [[Internal energy]] <math>U</math>, [[Entropy]] <math>S</math>, [[Enthalpy]] <math>H</math>, [[Helmholtz free energy]] <math>A</math>, and [[Gibbs free energy]] <math>G</math> are [[state function]]s. Generally, neither [[Work (thermodynamics)|work]] <math>W</math> nor [[heat]] <math>Q</math> is a state function. (Note: <math>Q</math> is commonly used to represent heat in physics. It should not be confused with the use earlier in this article as the parameter of an exact differential.)

===One dimension===
In one dimension, a differential form
:<math>A(x) \, dx</math>
is exact [[if and only if]] <math>A</math> has an [[antiderivative]] (but not necessarily one in terms of elementary functions). If <math>A</math> has an antiderivative and let <math>Q</math> be an antiderivative of <math>A</math> so <math>\frac{dQ}{dx} = A</math>, then <math>A(x) \, dx</math> obviously satisfies the condition for exactness. If <math>A</math> does ''not'' have an antiderivative, then we cannot write <math>dQ = \frac{dQ}{dx}dx</math> with <math>A = \frac{dQ}{dx}</math> for a differentiable function <math>Q</math> so <math>A(x) \, dx</math> is inexact.

===Two and three dimensions===
By [[symmetry of second derivatives]], for any "well-behaved" (non-[[Pathological (mathematics)|pathological]]) function <math>Q</math>, we have

: <math>\frac{\partial ^2 Q}{\partial x \, \partial y} = \frac{\partial ^2 Q}{\partial y \, \partial x}.</math>

Hence, in a [[simply-connected]] region ''R'' of the ''xy''-plane, where <math>x,y</math> are independent,<ref name=":0">If the pair of independent variables <math>(x,y)</math> is a (locally reversible) function of dependent variables <math>(u,v)</math>, all that is needed for the following theorem to hold, is to replace the partial derivatives with respect to <math>x</math> or to <math>y</math>, by the partial derivatives with respect to <math>u</math> and to <math>v</math> involving their [[Jacobian]] components. That is: <math>A(u, v)du + B(u, v)dv,</math> is an exact differential, if and only if: <math>\frac{\partial A}{\partial u}\frac{\partial u}{\partial y} + \frac{\partial A}{\partial v}\frac{\partial v}{\partial y} = \frac{\partial B}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial B}{\partial v}\frac{\partial v}{\partial x}.</math></ref> a differential form

:<math>A(x, y)\,dx + B(x, y)\,dy</math>

is an exact differential if and only if the equation

:<math>\left( \frac{\partial A}{\partial y} \right)_x = \left( \frac{\partial B}{\partial x} \right)_y</math>

holds. If it is an exact differential so <math>A=\frac{\partial Q}{\partial x}</math> and <math>B=\frac{\partial Q}{\partial y}</math>, then <math>Q</math> is a differentiable (smoothly continuous) function along <math>x</math> and <math>y</math>, so <math>\left( \frac{\partial A}{\partial y} \right)_x = \frac{\partial ^2 Q}{\partial y \partial x} = \frac{\partial ^2 Q}{\partial x \partial y} = \left( \frac{\partial B}{\partial x} \right)_y</math>. If <math>\left( \frac{\partial A}{\partial y} \right)_x = \left( \frac{\partial B}{\partial x} \right)_y</math> holds, then <math>A</math> and <math>B</math> are differentiable (again, smoothly continuous) functions along <math>y</math> and <math>x</math> respectively, and <math>\left( \frac{\partial A}{\partial y} \right)_x = \frac{\partial ^2 Q}{\partial y \partial x} = \frac{\partial ^2 Q}{\partial x \partial y} = \left( \frac{\partial B}{\partial x} \right)_y</math> is only the case.

For three dimensions, in a simply-connected region ''R'' of the ''xyz''-coordinate system, by a similar reason, a differential

:<math>dQ = A(x, y, z) \, dx + B(x, y, z) \, dy + C(x, y, z) \, dz</math>

is an exact differential if and only if between the functions ''A'', ''B'' and ''C'' there exist the relations

:<math>\left( \frac{\partial A}{\partial y} \right)_{x,z} \!\!\!= \left( \frac{\partial B}{\partial x} \right)_{y,z}</math>''';'''&nbsp;<math>\left( \frac{\partial A}{\partial z} \right)_{x,y} \!\!\!= \left( \frac{\partial C}{\partial x} \right)_{y,z}</math>''';'''&nbsp;<math>\left( \frac{\partial B}{\partial z} \right)_{x,y} \!\!\!= \left( \frac{\partial C}{\partial y} \right)_{x,z}.</math>

These conditions are equivalent to the following sentence: If ''G'' is the graph of this vector valued function then for all tangent vectors ''X'',''Y'' of the ''surface'' ''G'' then ''s''(''X'',&nbsp;''Y'')&nbsp;=&nbsp;0 with ''s'' the [[symplectic form]].

These conditions, which are easy to generalize, arise from the independence of the order of differentiations in the calculation of the second derivatives. So, in order for a differential ''dQ'', that is a function of four variables, to be an exact differential, there are six conditions (the [[combination]] <math>C(4,2)=6</math>) to satisfy.