In multivariate calculus , a differential or differential form is said to be exact or perfect (so called an exact differential ), as contrasted with an inexact differential , if it is equal to the general differential dQ for some differentiable function Q .
An exact differential is sometimes also called a total differential , or a full differential , or, in the study of differential geometry , it is termed an exact form .
The integral of an exact differential over any integral path is integral path-independent , and this fact is used to identity state functions in thermodynamics .
Overview
Definition
Even if we work in three dimensions here, the definitions of exact differentials for other dimensions are similar to the three dimensional definition. In three dimensions, a form of the type
A
(
x
,
y
,
z
)
d
x
+
B
(
x
,
y
,
z
)
d
y
+
C
(
x
,
y
,
z
)
d
z
{\displaystyle A(x,y,z)\,dx+B(x,y,z)\,dy+C(x,y,z)\,dz}
is called a differential form . This form is called exact on a domain
D
⊂
R
3
{\displaystyle D\subset \mathbb {R} ^{3}}
differentiable scalar function
Q
=
Q
(
x
,
y
,
z
)
{\displaystyle Q=Q(x,y,z)}
D
{\displaystyle D}
d
Q
≡
(
∂
Q
∂
x
)
y
,
z
d
x
+
(
∂
Q
∂
y
)
x
,
z
d
y
+
(
∂
Q
∂
z
)
x
,
y
d
z
,
{\displaystyle 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,}
d
Q
=
A
d
x
+
B
d
y
+
C
d
z
{\displaystyle dQ=A\,dx+B\,dy+C\,dz}
throughout
D
{\displaystyle D}
exact differential if it is equal to the general differential of a differentiable function.
Note: In this mathematical expression, the subscripts outside the parenthesis 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 included as a reminder.
Integral path independence
The exact differential for a differentiable function
Q
{\displaystyle Q}
d
Q
=
∇
Q
⋅
d
r
{\displaystyle dQ=\nabla Q\cdot d\mathbf {r} }
scalar product between the conservative vector field
(
A
,
B
,
C
)
=
∇
Q
{\displaystyle (A,B,C)=\nabla Q}
gradient of
Q
{\displaystyle Q}
Q
{\displaystyle Q}
(
d
x
,
d
y
,
d
z
)
=
d
r
{\displaystyle (dx,dy,dz)=d\mathbf {r} }
The gradient theorem states
∫
i
f
d
Q
=
∫
i
f
∇
Q
(
r
)
⋅
d
r
=
Q
(
f
)
−
Q
(
i
)
{\displaystyle \int _{i}^{f}dQ=\int _{i}^{f}\nabla Q(\mathbf {r} )\cdot d\mathbf {r} =Q\left(f\right)-Q\left(i\right)}
i and f is chosen (integral path-independent ).For three dimensional spaces, this integral path independence can also be proved by using the vector calculus identity
∇
×
(
∇
Q
)
=
0
{\displaystyle \nabla \times (\nabla Q)=\mathbf {0} }
Stokes' theorem .
∮
∂
Σ
∇
Q
⋅
d
r
=
∬
Σ
(
∇
×
∇
Q
)
⋅
d
a
=
0
{\displaystyle \oint _{\partial \Sigma }\nabla Q\cdot d\mathbf {r} =\iint _{\Sigma }(\nabla \times \nabla Q)\cdot d\mathbf {a} =0}
∂
Σ
{\displaystyle \partial \Sigma }
Σ
{\displaystyle \Sigma }
In thermodynamics , when dQ is exact, the function Q is a state function (a mathematical function depending only on the current equilibrium state , not depending on the system path taken to reach the equilibrium state) of the system. The thermodynamic functions U (internal energy ), S (entropy ), H (enthalpy ), A (Helmholtz free energy ), and G (Gibbs free energy ) are state functions . Generally, neither work nor heat is a state function.
One dimension
In one dimension, a differential form
A
(
x
)
d
x
{\displaystyle A(x)\,dx}
is exact if and only if
A
{\displaystyle A}
antiderivative (but not necessarily one in terms of elementary functions). If
A
{\displaystyle A}
Q
{\displaystyle Q}
A
{\displaystyle A}
Q
{\displaystyle Q}
A
{\displaystyle A}
not have an antiderivative, we cannot write
d
Q
=
A
(
x
)
d
x
{\displaystyle dQ=A(x)\,dx}
Two and three dimensions
By symmetry of second derivatives , for any "well-behaved" (non-pathological ) function
Q
{\displaystyle Q}
∂
2
Q
∂
x
∂
y
=
∂
2
Q
∂
y
∂
x
.
{\displaystyle {\frac {\partial ^{2}Q}{\partial x\,\partial y}}={\frac {\partial ^{2}Q}{\partial y\,\partial x}}.}
Hence, in a simply-connected region R of the xy -plane, a differential form
A
(
x
,
y
)
d
x
+
B
(
x
,
y
)
d
y
{\displaystyle A(x,y)\,dx+B(x,y)\,dy}
is an exact differential if and only if the equation
(
∂
A
∂
y
)
x
=
(
∂
B
∂
x
)
y
{\displaystyle \left({\frac {\partial A}{\partial y}}\right)_{x}=\left({\frac {\partial B}{\partial x}}\right)_{y}}
holds. If it is an exact differential so
A
=
∂
Q
∂
x
{\displaystyle A={\frac {\partial Q}{\partial x}}}
B
=
∂
Q
∂
y
{\displaystyle B={\frac {\partial Q}{\partial y}}}
Q
{\displaystyle Q}
x
{\displaystyle x}
y
{\displaystyle y}
(
∂
A
∂
y
)
x
=
∂
2
Q
∂
y
∂
x
=
∂
2
Q
∂
x
∂
y
=
(
∂
B
∂
x
)
y
{\displaystyle \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}}
(
∂
A
∂
y
)
x
=
(
∂
B
∂
x
)
y
{\displaystyle \left({\frac {\partial A}{\partial y}}\right)_{x}=\left({\frac {\partial B}{\partial x}}\right)_{y}}
A
{\displaystyle A}
B
{\displaystyle B}
y
{\displaystyle y}
x
{\displaystyle x}
(
∂
A
∂
y
)
x
=
∂
2
Q
∂
y
∂
x
=
∂
2
Q
∂
x
∂
y
=
(
∂
B
∂
x
)
y
{\displaystyle \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}}
For three dimensions, in a simply-connected region R of the xyz -coordinate system, by a similar reason, a differential
d
Q
=
A
(
x
,
y
,
z
)
d
x
+
B
(
x
,
y
,
z
)
d
y
+
C
(
x
,
y
,
z
)
d
z
{\displaystyle dQ=A(x,y,z)\,dx+B(x,y,z)\,dy+C(x,y,z)\,dz}
is an exact differential if and only if between the functions A , B and C there exist the relations
(
∂
A
∂
y
)
x
,
z
=
(
∂
B
∂
x
)
y
,
z
{\displaystyle \left({\frac {\partial A}{\partial y}}\right)_{x,z}\!\!\!=\left({\frac {\partial B}{\partial x}}\right)_{y,z}}
;
(
∂
A
∂
z
)
x
,
y
=
(
∂
C
∂
x
)
y
,
z
{\displaystyle \left({\frac {\partial A}{\partial z}}\right)_{x,y}\!\!\!=\left({\frac {\partial C}{\partial x}}\right)_{y,z}}
;
(
∂
B
∂
z
)
x
,
y
=
(
∂
C
∂
y
)
x
,
z
.
{\displaystyle \left({\frac {\partial B}{\partial z}}\right)_{x,y}\!\!\!=\left({\frac {\partial C}{\partial y}}\right)_{x,z}.}
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 , Y ) = 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
C
(
4
,
2
)
=
6
{\displaystyle C(4,2)=6}
Partial differential relations
If three variables,
x
{\displaystyle x}
y
{\displaystyle y}
z
{\displaystyle z}
F
(
x
,
y
,
z
)
=
constant
{\displaystyle F(x,y,z)={\text{constant}}}
F
(
x
,
y
,
z
)
{\displaystyle F(x,y,z)}
total differentials exist[ 1] : 667&669
d
x
=
(
∂
x
∂
y
)
z
d
y
+
(
∂
x
∂
z
)
y
d
z
{\displaystyle dx={\left({\frac {\partial x}{\partial y}}\right)}_{z}\,dy+{\left({\frac {\partial x}{\partial z}}\right)}_{y}\,dz}
d
z
=
(
∂
z
∂
x
)
y
d
x
+
(
∂
z
∂
y
)
x
d
y
.
{\displaystyle dz={\left({\frac {\partial z}{\partial x}}\right)}_{y}\,dx+{\left({\frac {\partial z}{\partial y}}\right)}_{x}\,dy.}
Substituting the first equation into the second and rearranging, we obtain[ 1] : 669
d
z
=
(
∂
z
∂
x
)
y
[
(
∂
x
∂
y
)
z
d
y
+
(
∂
x
∂
z
)
y
d
z
]
+
(
∂
z
∂
y
)
x
d
y
,
{\displaystyle dz={\left({\frac {\partial z}{\partial x}}\right)}_{y}\left[{\left({\frac {\partial x}{\partial y}}\right)}_{z}dy+{\left({\frac {\partial x}{\partial z}}\right)}_{y}dz\right]+{\left({\frac {\partial z}{\partial y}}\right)}_{x}dy,}
d
z
=
[
(
∂
z
∂
x
)
y
(
∂
x
∂
y
)
z
+
(
∂
z
∂
y
)
x
]
d
y
+
(
∂
z
∂
x
)
y
(
∂
x
∂
z
)
y
d
z
,
{\displaystyle dz=\left[{\left({\frac {\partial z}{\partial x}}\right)}_{y}{\left({\frac {\partial x}{\partial y}}\right)}_{z}+{\left({\frac {\partial z}{\partial y}}\right)}_{x}\right]dy+{\left({\frac {\partial z}{\partial x}}\right)}_{y}{\left({\frac {\partial x}{\partial z}}\right)}_{y}dz,}
[
1
−
(
∂
z
∂
x
)
y
(
∂
x
∂
z
)
y
]
d
z
=
[
(
∂
z
∂
x
)
y
(
∂
x
∂
y
)
z
+
(
∂
z
∂
y
)
x
]
d
y
.
{\displaystyle \left[1-{\left({\frac {\partial z}{\partial x}}\right)}_{y}{\left({\frac {\partial x}{\partial z}}\right)}_{y}\right]dz=\left[{\left({\frac {\partial z}{\partial x}}\right)}_{y}{\left({\frac {\partial x}{\partial y}}\right)}_{z}+{\left({\frac {\partial z}{\partial y}}\right)}_{x}\right]dy.}
Since
y
{\displaystyle y}
z
{\displaystyle z}
d
y
{\displaystyle dy}
d
z
{\displaystyle dz}
[ 1] : 669
Reciprocity relation
Setting the first term in brackets equal to zero yields[ 1]
(
∂
z
∂
x
)
y
(
∂
x
∂
z
)
y
=
1.
{\displaystyle {\left({\frac {\partial z}{\partial x}}\right)}_{y}{\left({\frac {\partial x}{\partial z}}\right)}_{y}=1.}
A slight rearrangement gives a reciprocity relation,[ 1] : 670
(
∂
z
∂
x
)
y
=
1
(
∂
x
∂
z
)
y
.
{\displaystyle {\left({\frac {\partial z}{\partial x}}\right)}_{y}={\frac {1}{{\left({\frac {\partial x}{\partial z}}\right)}_{y}}}.}
There are two more permutations of the foregoing derivation that give a total of three reciprocity relations between
x
{\displaystyle x}
y
{\displaystyle y}
z
{\displaystyle z}
Reciprocity relations show that the inverse of a partial derivative is equal to its reciprocal.
Cyclic relation
The cyclic relation is also known as the cyclic rule or the Triple product rule . Setting the second term in brackets equal to zero yields[ 1] : 670
(
∂
z
∂
x
)
y
(
∂
x
∂
y
)
z
=
−
(
∂
z
∂
y
)
x
.
{\displaystyle {\left({\frac {\partial z}{\partial x}}\right)}_{y}{\left({\frac {\partial x}{\partial y}}\right)}_{z}=-{\left({\frac {\partial z}{\partial y}}\right)}_{x}.}
Using a reciprocity relation for
∂
z
∂
y
{\displaystyle {\tfrac {\partial z}{\partial y}}}
triple product rule ),[ 1] : 670
(
∂
x
∂
y
)
z
(
∂
y
∂
z
)
x
(
∂
z
∂
x
)
y
=
−
1.
{\displaystyle {\left({\frac {\partial x}{\partial y}}\right)}_{z}{\left({\frac {\partial y}{\partial z}}\right)}_{x}{\left({\frac {\partial z}{\partial x}}\right)}_{y}=-1.}
If, instead , a reciprocity relation for
∂
x
∂
y
{\displaystyle {\tfrac {\partial x}{\partial y}}}
standard form for implicit differentiation is obtained:
(
∂
y
∂
x
)
z
=
−
(
∂
z
∂
x
)
y
(
∂
z
∂
y
)
x
.
{\displaystyle {\left({\frac {\partial y}{\partial x}}\right)}_{z}=-{\frac {{\left({\frac {\partial z}{\partial x}}\right)}_{y}}{{\left({\frac {\partial z}{\partial y}}\right)}_{x}}}.}
Some useful equations derived from exact differentials in two dimensions
(See also Bridgman's thermodynamic equations for the use of exact differentials in the theory of thermodynamic equations )
Suppose we have five state functions
z
,
x
,
y
,
u
{\displaystyle z,x,y,u}
v
{\displaystyle v}
chain rule
d
z
=
(
∂
z
∂
x
)
y
d
x
+
(
∂
z
∂
y
)
x
d
y
=
(
∂
z
∂
u
)
v
d
u
+
(
∂
z
∂
v
)
u
d
v
{\displaystyle dz=\left({\frac {\partial z}{\partial x}}\right)_{y}dx+\left({\frac {\partial z}{\partial y}}\right)_{x}dy=\left({\frac {\partial z}{\partial u}}\right)_{v}du+\left({\frac {\partial z}{\partial v}}\right)_{u}dv}
1
but also by the chain rule:
d
x
=
(
∂
x
∂
u
)
v
d
u
+
(
∂
x
∂
v
)
u
d
v
{\displaystyle dx=\left({\frac {\partial x}{\partial u}}\right)_{v}du+\left({\frac {\partial x}{\partial v}}\right)_{u}dv}
2
and
d
y
=
(
∂
y
∂
u
)
v
d
u
+
(
∂
y
∂
v
)
u
d
v
{\displaystyle dy=\left({\frac {\partial y}{\partial u}}\right)_{v}du+\left({\frac {\partial y}{\partial v}}\right)_{u}dv}
3
so that:
d
z
=
[
(
∂
z
∂
x
)
y
(
∂
x
∂
u
)
v
+
(
∂
z
∂
y
)
x
(
∂
y
∂
u
)
v
]
d
u
+
[
(
∂
z
∂
x
)
y
(
∂
x
∂
v
)
u
+
(
∂
z
∂
y
)
x
(
∂
y
∂
v
)
u
]
d
v
{\displaystyle {\begin{aligned}dz=&\left[\left({\frac {\partial z}{\partial x}}\right)_{y}\left({\frac {\partial x}{\partial u}}\right)_{v}+\left({\frac {\partial z}{\partial y}}\right)_{x}\left({\frac {\partial y}{\partial u}}\right)_{v}\right]du\\+&\left[\left({\frac {\partial z}{\partial x}}\right)_{y}\left({\frac {\partial x}{\partial v}}\right)_{u}+\left({\frac {\partial z}{\partial y}}\right)_{x}\left({\frac {\partial y}{\partial v}}\right)_{u}\right]dv\end{aligned}}}
4
which implies that:
(
∂
z
∂
u
)
v
=
(
∂
z
∂
x
)
y
(
∂
x
∂
u
)
v
+
(
∂
z
∂
y
)
x
(
∂
y
∂
u
)
v
{\displaystyle \left({\frac {\partial z}{\partial u}}\right)_{v}=\left({\frac {\partial z}{\partial x}}\right)_{y}\left({\frac {\partial x}{\partial u}}\right)_{v}+\left({\frac {\partial z}{\partial y}}\right)_{x}\left({\frac {\partial y}{\partial u}}\right)_{v}}
5
Letting
v
=
y
{\displaystyle v=y}
(
∂
z
∂
u
)
y
=
(
∂
z
∂
x
)
y
(
∂
x
∂
u
)
y
{\displaystyle \left({\frac {\partial z}{\partial u}}\right)_{y}=\left({\frac {\partial z}{\partial x}}\right)_{y}\left({\frac {\partial x}{\partial u}}\right)_{y}}
6
Letting
u
=
y
{\displaystyle u=y}
(
∂
z
∂
y
)
v
=
(
∂
z
∂
y
)
x
+
(
∂
z
∂
x
)
y
(
∂
x
∂
y
)
v
{\displaystyle \left({\frac {\partial z}{\partial y}}\right)_{v}=\left({\frac {\partial z}{\partial y}}\right)_{x}+\left({\frac {\partial z}{\partial x}}\right)_{y}\left({\frac {\partial x}{\partial y}}\right)_{v}}
7
Letting
u
=
y
{\displaystyle u=y}
v
=
z
{\displaystyle v=z}
(
∂
z
∂
y
)
x
=
−
(
∂
z
∂
x
)
y
(
∂
x
∂
y
)
z
{\displaystyle \left({\frac {\partial z}{\partial y}}\right)_{x}=-\left({\frac {\partial z}{\partial x}}\right)_{y}\left({\frac {\partial x}{\partial y}}\right)_{z}}
8
using (
∂
a
/
∂
b
)
c
=
1
/
(
∂
b
/
∂
a
)
c
{\displaystyle \partial a/\partial b)_{c}=1/(\partial b/\partial a)_{c}}
triple product rule :
(
∂
z
∂
x
)
y
(
∂
x
∂
y
)
z
(
∂
y
∂
z
)
x
=
−
1
{\displaystyle \left({\frac {\partial z}{\partial x}}\right)_{y}\left({\frac {\partial x}{\partial y}}\right)_{z}\left({\frac {\partial y}{\partial z}}\right)_{x}=-1}
9
See also
References
^ a b c d e f g Çengel, Yunus A.; Boles, Michael A. (1998) [1989]. "Thermodynamics Property Relations". Thermodynamics - An Engineering Approach . McGraw-Hill Series in Mechanical Engineering (3rd ed.). Boston, MA.: McGraw-Hill. ISBN 0-07-011927-9
External links
![{\displaystyle dz={\left({\frac {\partial z}{\partial x}}\right)}_{y}\left[{\left({\frac {\partial x}{\partial y}}\right)}_{z}dy+{\left({\frac {\partial x}{\partial z}}\right)}_{y}dz\right]+{\left({\frac {\partial z}{\partial y}}\right)}_{x}dy,}](https://wikimedia.org/enwiki/api/rest_v1/media/math/render/svg/770ab72f3175cec4682bf55b2a60449158c232a3)
![{\displaystyle dz=\left[{\left({\frac {\partial z}{\partial x}}\right)}_{y}{\left({\frac {\partial x}{\partial y}}\right)}_{z}+{\left({\frac {\partial z}{\partial y}}\right)}_{x}\right]dy+{\left({\frac {\partial z}{\partial x}}\right)}_{y}{\left({\frac {\partial x}{\partial z}}\right)}_{y}dz,}](https://wikimedia.org/enwiki/api/rest_v1/media/math/render/svg/39499fa9a236408664263ae32ff4becb8668435f)
![{\displaystyle \left[1-{\left({\frac {\partial z}{\partial x}}\right)}_{y}{\left({\frac {\partial x}{\partial z}}\right)}_{y}\right]dz=\left[{\left({\frac {\partial z}{\partial x}}\right)}_{y}{\left({\frac {\partial x}{\partial y}}\right)}_{z}+{\left({\frac {\partial z}{\partial y}}\right)}_{x}\right]dy.}](https://wikimedia.org/enwiki/api/rest_v1/media/math/render/svg/34f0e189b4df0b982460664880de0506cd6107a6)
![{\displaystyle {\begin{aligned}dz=&\left[\left({\frac {\partial z}{\partial x}}\right)_{y}\left({\frac {\partial x}{\partial u}}\right)_{v}+\left({\frac {\partial z}{\partial y}}\right)_{x}\left({\frac {\partial y}{\partial u}}\right)_{v}\right]du\\+&\left[\left({\frac {\partial z}{\partial x}}\right)_{y}\left({\frac {\partial x}{\partial v}}\right)_{u}+\left({\frac {\partial z}{\partial y}}\right)_{x}\left({\frac {\partial y}{\partial v}}\right)_{u}\right]dv\end{aligned}}}](https://wikimedia.org/enwiki/api/rest_v1/media/math/render/svg/45b74975dee47742cb368c261912a75590a8a8e8)
