In multivariate calculus, a differential or differential form is said to be exact or perfect (exact differential), as contrasted with an inexact differential, if it is equal to the general differential
dQ
Q
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 path-independent, and this fact is used to identify state functions in thermodynamics.
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
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 an open domain
D\subsetR3
Q=Q(x,y,z)
D
dQ\equiv\left(
\partialQ | |
\partialx |
\right)y,zdx+\left(
\partialQ | |
\partialy |
\right)x,zdy+\left(
\partialQ | |
\partialz |
\right)x,ydz,
dQ=Adx+Bdy+Cdz
throughout
D
x,y,z
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 explicitly shown here as reminders.
The exact differential for a differentiable scalar function
Q
D\subsetRn
dQ=\nablaQ ⋅ dr
\nablaQ
Q
⋅
dr
Q
C1
\nablaQ
Q
dr=(dx,dy,dz)
\nablaQ=(
\partialQ | |
\partialx |
,
\partialQ | , | |
\partialy |
\partialQ | |
\partialz |
)
The gradient theorem states
\int
f | |
i |
dQ=\int
f | |
i |
\nablaQ(r) ⋅ dr=Q\left(f\right)-Q\left(i\right)
that does not depend on which integral path between the given path endpoints
i
f
For three dimensional spaces, if
\nablaQ
D\subsetR3
C1
Q
C2
\nabla x (\nablaQ)=0
\oint\partial\nablaQ ⋅ dr=\iint\Sigma(\nabla x \nablaQ) ⋅ da=0
\partial\Sigma
\Sigma
D
C1
v
\nabla x v=0
C1
In thermodynamics, when
dQ
Q
U
S
H
A
G
W
Q
Q
In one dimension, a differential form
A(x)dx
A
A
Q
A
dQ | |
dx |
=A
A(x)dx
A
dQ=
dQ | |
dx |
dx
A=
dQ | |
dx |
Q
A(x)dx
By symmetry of second derivatives, for any "well-behaved" (non-pathological) function
Q
\partial2Q | |
\partialx\partialy |
=
\partial2Q | |
\partialy\partialx |
.
Hence, in a simply-connected region R of the xy-plane, where
x,y
A(x,y)dx+B(x,y)dy
is an exact differential if and only if the equation
\left(
\partialA | |
\partialy |
\right)x=\left(
\partialB | |
\partialx |
\right)y
holds. If it is an exact differential so
A= | \partialQ |
\partialx |
B= | \partialQ |
\partialy |
Q
x
y
\left(
\partialA | |
\partialy |
\right)x=
\partial2Q | |
\partialy\partialx |
=
\partial2Q | |
\partialx\partialy |
=\left(
\partialB | |
\partialx |
\right)y
\left(
\partialA | |
\partialy |
\right)x=\left(
\partialB | |
\partialx |
\right)y
A
B
y
x
\left(
\partialA | |
\partialy |
\right)x=
\partial2Q | |
\partialy\partialx |
=
\partial2Q | |
\partialx\partialy |
=\left(
\partialB | |
\partialx |
\right)y
For three dimensions, in a simply-connected region R of the xyz-coordinate system, by a similar reason, a differential
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
\left(
\partialA | |
\partialy |
\right)x,z=\left(
\partialB | |
\partialx |
\right)y,z
\left(
\partialA | |
\partialz |
\right)x,y=\left(
\partialC | |
\partialx |
\right)y,z
\left(
\partialB | |
\partialz |
\right)x,y=\left(
\partialC | |
\partialy |
\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.
C(4,2)=6
z(x,y)
z(x,y)
x
y
(x,y)
x(y,z)
dx={\left(
\partialx | |
\partialy |
\right)}zdy+{\left(
\partialx | |
\partialz |
\right)}ydz
dz={\left(
\partialz | |
\partialx |
\right)}ydx+{\left(
\partialz | |
\partialy |
\right)}xdy.
Substituting the first equation into the second and rearranging, we obtain
dz={\left(
\partialz | |
\partialx |
\right)}y\left[{\left(
\partialx | |
\partialy |
\right)}zdy+{\left(
\partialx | |
\partialz |
\right)}ydz\right]+{\left(
\partialz | |
\partialy |
\right)}xdy,
dz=\left[{\left(
\partialz | |
\partialx |
\right)}y{\left(
\partialx | |
\partialy |
\right)}z+{\left(
\partialz | |
\partialy |
\right)}x\right]dy+{\left(
\partialz | |
\partialx |
\right)}y{\left(
\partialx | |
\partialz |
\right)}ydz,
\left[1-{\left(
\partialz | |
\partialx |
\right)}y{\left(
\partialx | |
\partialz |
\right)}y\right]dz=\left[{\left(
\partialz | |
\partialx |
\right)}y{\left(
\partialx | |
\partialy |
\right)}z+{\left(
\partialz | |
\partialy |
\right)}x\right]dy.
y
z
dy
dz
Setting the first term in brackets equal to zero yields
{\left(
\partialz | |
\partialx |
\right)}y{\left(
\partialx | |
\partialz |
\right)}y=1.
A slight rearrangement gives a reciprocity relation,
{\left(
\partialz | |
\partialx |
\right)}y=
1 | |||||
|
y}.
There are two more permutations of the foregoing derivation that give a total of three reciprocity relations between
x
y
z
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
{\left(
\partialz | |
\partialx |
\right)}y{\left(
\partialx | |
\partialy |
\right)}z=-{\left(
\partialz | |
\partialy |
\right)}x.
Using a reciprocity relation for
\tfrac{\partialz}{\partialy}
{\left(
\partialx | |
\partialy |
\right)}z{\left(
\partialy | |
\partialz |
\right)}x{\left(
\partialz | |
\partialx |
\right)}y=-1.
If, instead, reciprocity relations for
\tfrac{\partialx}{\partialy}
\tfrac{\partialy}{\partialz}
{\left(
\partialy | |
\partialx |
\right)}z=-
| ||||||
y |
}{{\left(
\partialz | |
\partialy |
\right)}x}.
(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
v
but also by the chain rule:
and
so that (by substituting (2) and (3) into (1)):
which implies that (by comparing (4) with (1)):
Letting
v=y
Letting
u=y
Letting
u=y
v=z
using (
\partiala/\partialb)c=1/(\partialb/\partiala)c
(x,y)
(u,v)
x
y
u
v
A(u,v)du+B(u,v)dv,
\partialA | |
\partialu |
\partialu | |
\partialy |
+
\partialA | |
\partialv |
\partialv | |
\partialy |
=
\partialB | |
\partialu |
\partialu | |
\partialx |
+
\partialB | |
\partialv |
\partialv | |
\partialx |
.