Supplement: exact differential
• In these units we will review some basics of
exact differentials
PHYS 4420 — FALL 2021
147
k=1
Consider
x as
a function
of zeta
two variables
y andC.21).
z. This can be w
where ζ(n)
is the
Riemann
function (eqn
.6 Partial
derivatives
x C.6
= x(y, z),
and we have
that
Partial
derivatives
# $
# $
nsider xC.6
as a function
two variables
and z. This
written
∂xyvariables
Consider
x as aoffunction
of two
y ∂x
andcan
z. be
This
can be written
Partial
derivatives
dx =
dy +
dz.
(
= x(y, z),xand
we z),
have
that
= x(y,
and
we have that∂y
∂z y
z
$
#
#
$ $
# $
Consider x as a#function
of
two
variables
y and z. This can be w
∂x
∂x ∂x
∂x
But
x(y,
to having
a function
of
dxand
= xwe=dx
dyz)
+ can lead
dz.
=
dy +
dz.z as (C.39)
(C.39
x = rearranging
x(y, z),
have
that
∂y ∂z
∂z y
z which
z y
y so that z = z(x,∂y
y), in
case
# $
# $
∂x$
∂x$
#
#
But rearranging
= x(y,
z)
can
lead
to
having
a of
function
t rearranging
x = x(y, z)xcan
lead
to
having
z
as
a
function
x and of x and
dx = ∂z
dy + ∂z z asdz.
(
that
= which
z(x, y),case
in =
which∂y
casez dx + ∂z y dy.
o that z y=soz(x,
y),z in
(
dz
∂x
∂y
y
$
$ x
#
#
$
$
#
#
Substituting C.40 into C.39 gives
But !rearranging
x = x(y,
z)∂z
can
having
z as a function of
∂z!lead" to!∂z
$
" dz
"
"
!
! = " ∂z dz#=
dx +
dx∂x
+
dy. ∂z dy.
(C.40) (C.40
∂x
∂z
∂x
∂x ∂y
∂y x dy.
y so
y),+yin which
dx
+ycase
dx
= that z = z(x,∂x
x
∂z y ∂x y
∂y# z $ ∂z y #
∂y x$
∂z
∂z
The terms multiplying dx give the
reciprocal
theorem
dx +
dy.
(
dz =
∂x y
∂y x
! "
Partial Differentials
PHYS 4420 — FALL 2021
∂x
∂z
y
1
%
= ∂z & ,
(C.41)
∂x y
and the terms multiplying dy give the reciprocity theorem
! " ! " ! "
∂x
∂y
∂z
= −1.
148
(C.42)
Substituting C.40 into C.39 gives
#! into"C.39 gives
"
! " ! " $
! " ! Substituting
C.40
#! " ∂z! " ! " $
∂x
∂z
∂x
! "
! " ∂x
dx + ∂x
dx =
∂x
∂x dy.
∂z +
∂z
dx
+
+
dy
dx
=
∂z y ∂x y
∂z ∂yy ∂y ∂zx
∂z y∂y∂x z y
∂y
g C.40 into
C.39
gives
z
y
x
Substituting C.40
into
C.39
gives
#! "
#!! reciprocal
$ theorem
" multiplying
"multiplying
"" $dx !
!terms
! terms
! "The
"
"
"
"
!
!
!
The
give
the
reciprocal
dx
give
the
theorem
∂x ∂z ∂x
∂z
∂z
∂x
∂x
∂x "
∂z
∂x
!
dx
+
+
dy.
!
"
dx
+
+
dy.
dx
=
1
∂x
∂z y ∂x y
∂y
∂z
∂y
∂z y z∂x y ∂xy ∂y xz 1 ∂z y = ∂y
% ∂z & x,
∂z
= % ∂z & ,
(C.41)
y
∂x y
∂z
multiplying
give multiplying
the reciprocal
theorem
Thedx
terms
dx give
the yreciprocal
∂x y theorem
and the
and the
! terms
" multiplying dy give the reciprocity theorem
1
∂x
" ! " ! "
1!reciprocity
∂x
terms
multiplying
theorem
∂x
∂y
∂z
= % ∂z & , dy give=the
% ∂z & , (C.41)
(C.41)
=
−1.
∂z y
" ∂z
! y " !∂x ∂y
∂x!y
y " z ∂z x ∂x y
!
"
∂x
∂y
∂z
= reciprocal
−1.
(C.42)
Reciprocal
theorem
This can
be theorem
combined
with
the
theorem to write
that
ms multiplying
dy
give multiplying
the
reciprocity
and the
terms
dy
give
the
reciprocity
theorem
Reciprocity
theorem
∂y z ∂z x ∂x
!
! "
!
"
"
y
!" " ! " ! "∂x
∂x
∂z
=
−
,
∂x
∂y
∂z ∂x
∂y
∂z
∂y=
∂ztoy write
∂y x that
= −1.
(C.42)
(C.42)
z−1.
This ∂y
can be ∂z
combined
with
the
reciprocal
theorem
∂z x! ∂x"y !
∂x!∂y
z
z
x
y
"
"
!
" !
" !
which is a very useful identity.
∂x
∂x
∂z
−write that
This
canthe
be reciprocal
combined with
the=reciprocal
theorem to, write that
e combined
with
theorem
to
! ∂z
" !y ∂y
" x
! " !!∂y""z
! "
∂x
∂x
=
−
which∂y
is a very useful
∂z y
PHYS 4420 — FALL 2021 z
C.7
∂x
∂z
Exact
∂x differentials
∂z
=, −
identity.
∂y
∂z
∂yexpression
zx
An
such
,
(C.43)
(C.43)
(C.43)
y F∂y
x dx + F2 (x, y) dy is known as
as
1 (x, y)
149 an
differential if it can be written as the differential
which
is a very useful identity.
very useful
identity.
! "
! "
∂f
∂f
df =
dx +
dy,
∂x
∂y
C.7
Exact differentials
C.7 Exact differentials
of a differentiable single-valued function f (x, y). This implies tha
Exact differentials
y
∂x y
his can be combined with the reciprocal theorem to write that
! "single-valued
! "
"
of !
a differentiable
function f (x, y). This implies that
ich is a very useful
identity.
∂x
∂x
∂z
dy
give
"
! the"reciprocity theore
= − and the terms
, !multiplying
(C.43)
∂f ! " ! "∂f! "
∂y z
∂z y ∂y x
F1 =
, ∂xF2 =∂y
(
∂z,
∂x+ F2 (x, y) dy is
∂yknown=as−1.an exac
hich is a very
identity. such as F1 (x, y) dx
Anuseful
expression
∂y
∂z
∂x y
+ F (x, y) dy is known as an exact
z
x
C.7
ls
Exact differentials
.72 Exact differentials
differential
if it form,
can beF written
the differential
or in vector
= ∇f . as
Hence
the integral of an exact differ
!
"
!
" reciprocal theorem to write
the differential
This
can
be
combined
with
the
! Exact
" such
isdifferentials
path
independent,
so
that
1known
and 2 !
arean
shorthands
for (x
∂f y)[where
∂f
C.7
expression
as
F
(x,
y)
dx
+
F
(x,
dy
is
as
exact
!
!
"
"
"
1
2
∂f
df =
dx + ∂x
dy, ∂x
(C.44
∂z
and
(x
,
y
)]
x+
dy,
(C.44)
2 written
2
∂x
∂y = −
fferential
if it can be
as
the
differential
,
∂y such'as2F1 (x, y) dx
n expression
+
F
(x,
y)
dy
is
known
as
an
! "
∂z
∂y
! "' 2 ∂y exact
2
' 2 y
z
x
∂f
∂f
differentiable
ifferential of
if ita can
be written assingle-valued
the differential function f (x, y). This implies that
! =y)
"
!dx
tion f (x, y). This implies
that
which
is"
useful
Fdf
dx+F
dy" =
F ·dr
= " df =(C.44)
f (2)−f (1), (
+ay)very
dy,identity.
1 (x,
2 (x,
!
!
∂f
∂f
∂x+
! " 1df =
∂f ∂y 1
∂f 1
dx
dy,
(C.44)
∂f
F1 ∂y
, the
F2 initial
, Path
(C.45
∂x
= only on
=
independent
and
the
answer
depends
and
final
states
of
the
sy
F
,
(C.45)
=
2
∂xf (x, y).differentials
∂y
a differentiable
single-valued
function
This
implies
that
C.7
Exact
∂y For
a differentiable
single-valued
function
f (x, y). This
implies
thattrue and knowledge of the i
"
"
!
!
an !
inexact
differential
this
is
not
" ∂f
! . "
or in vector form,
F
=
∇f
Hence∂fthe integral of an exact differentia
∂f
∂f
the integral
of
an
exact
differential
final
notFsufficient
to,Fevaluate
the
integral:
ha
y)
dxshorthands
+
F2 (x,
y) dyfor
is you
known
F1 =states
,=expression
(C.45)
=, such1as
1 (x,
Gradient
f:and
Findependent,
, An
Fis
(C.45)
2 [where
1 =
2
,
is ofpath
so
that
and
2
are
(x
1 y1
∂x
∂y
∂x
∂y
,
y
)
1 and 2 are shorthands
for
(x
differential
if it can be written as the differential
1 was
1 taken.
know which path
! "
! "
and
(x
,
y
)]
2= ∇f
2 . Hence the integral of an exact differential
rinin vector
vector form,
F
∂f
∂f loop is zero
For
an∇f
exact
differential
the integral
round
a
closed
form,
F
=
.
Hence
the
integral
of
an
exact
differential
'
'
'
dffor(
=(x1 ,2y1 ) dx +(
dy,
2
path independent,
( [where 1 and 2 are shorthands
' 2 2 so that
∂x
∂y
) (C.46
path
so
that
[where
1dyand
2 are
shorthands
forf (2)−f
(x1 , y1(1),
F
(x,
y)
dx+F
(x,
y)
=
F
·dr
=
df
=
nd (x2independent,
, y2 )]
1
2
y)
+
= F1 · function
dr = fdf
Fd' ·dr
= f (2)−fF(1),
1'(x, (C.46)
of dx
a differentiable
single-valued
(x, =
y).0,This impl(
' F2 (x,1y) dy
(x =
, y )] df
1
Exact differential
2
2
21
2
2
! "
! "
'
'
F
(x,
y)
dx+F
(x,
y)
dy
=
F
·dr
=
df
=
f
(2)−f
(1),
(C.46)
∂f
∂fthe system
2
and
the
answer
depends
2 1
2 only on the2initial and final states of
which
implies
that
∇
×
F
=
0
(by
Stokes’
theorem)
and
hence
nitial
and
final
states
of
the
system.
F1 =
, F2 =
,
1
1
1
∂x
∂y
For
an inexact
differential
this
is not
true
y)
dx+F
y)
dythe
="
·dr
df
= system.
f2and
(2)−f
(1),
!F
"=states
" knowledge
! 2(C.46)
"of the initia
!on
1 (x,
2 (x,
nd F
the
answer
only
initial
and
final
of !
the
not
true
knowledge
of∂F
the
initial
150
PHYS
4420and
— FALLdepends
2021
f
f
∂F
∂
∂
1
1
1
2
1
or=true
insufficient
form,
=
Hence
the integral
exac
and
final states
not
to
the =integral:
you
have
t(
or
an inexact
differential
this
ishave
not
and knowledge
of ∇f
the .initial
orFevaluate
. of an
evaluate
the
integral:
youis
tovector
∂xtoon
∂y
∂x∂y
∂y∂x
isthe
path
independent,
so
[where
1 and
2 are shorthand
dndthe
only
initial
and final
states
system.
finalanswer
states
isdepends
not
sufficient
evaluate
the integral:
youthat
have
toof the
know
which
path
was
taken.
andis(xnot
y2true
)] integral
which
path
was
taken.
2 , the
rnow
an
inexact
differential
this
and
knowledge
of theloop
initial
For
an
exact
differential
round
a closed
is zero:
For
thermal
physics,
a
crucial
point
to remember
is' that
functio
ral
round
a
closed
loop
is
zero:
'
'
2
2
2
(
( is zero:
(
For an exact differential
the integral round a closed loop
For
ansingle-valued
inexact
differential
thisHence
is not
tr
e-valued function f (x, y).
implies
that
or function
in vector fform,
F =
∇f .implies
the
of a This
differentiable
(x, y).
This
thatint
! "
! "
"so that [where
! is "
!sufficient
path
independent,
1 and
and final states
is
not
to evalu
∂f
∂f
∂f
∂f
and ,(x2 , F
y22)]=
, F2 =
,
=
F1(C.45)
,
(C
=
know
which
path
was
taken.
'
'
∂x
∂y
∂x
∂y
2
2
F1 the
(x, y)integral
dx+F2 (x,
=
F ·dr ro
=
or in
form,
F
∇f . Hence
ofy)andyintegral
exact
differen
Fordifferential
an= exact
differential
the
∇f . Hence the integral
of vector
an exact
1
1
(
( (x1 ,a
is path independent,
so that [where
1
and
2
are
shorthands
initial
that [where 1 and 2 are shorthands for (x1 , y1 )and the answer depends only on thefor
Exact differential
y) dy =
'
2
F ·dr =
and (x2 , y2 )]
For
andx
inexact
differential
this=
is not tru
F
(x,
y)
+
F
(x,
y)
dy
F
1
2
'
'
' 2
2
2
and final states is not sufficient to evalua
' 2 F1 (x, y) dx+F2 (x, y) dy =know which
F ·dr path
= was
df taken.
= f (2)−f (1), (C
1 = f (2)−f (1), (C.46) For
1 an exact differential
1
rou
df
which implies that
∇
×
F
= 0 the
(byintegral
Stoke
(
(
the answer depends
only on
final states of!
the syst
"the initial
!F1 (x,and
"
!Stokes
Theorem
y)and
dx +
F2 (x, y) dyof
= theF2in·
For
an
inexact
differential
this
is
not
true
knowledge
∂F
∂ f
∂F
s only on the initial and final states of the system.
2
1
and final states is not sufficient
to implies
evaluate
the
integral:
you
hav
=
or
which
that
∇
×
F
=
0
(by
Stokes’
ntial this is not true and
knowledge of the initial
∂x
" !
"
!∂x∂
"
! ∂y
know which path was taken.
2
f
∂F
∂
∂F
2
1
sufficient to evaluate theForintegral:
you have the
to integral round
an exact differential
= a closed loop
or is zero:
∂x
∂y
( For thermal physics,
( a crucial
(
point∂x∂y
to
aken.
crucial
re
dx
+ F2exact
(x, y)For
dy thermal
= F physics,
· dr = a df
= 0,point to(C
1 (x, y)is
have
differentials.
tial the integral round a closedFstate
loop
zero:
1
1 and
state have exact differentials.
(
(
which implies that ∇ × F = 0 (by Stokes’
theorem) and hence
Condition:
F2 (x, y) dy = F · dr = df! = 0," ! (C.47)
"
! 2 " ! 2 "
∂F1
∂ f
∂ f
∂F2
=
or
=
.
(C
∂xhence ∂y
∂x∂y
∂y∂x
F = 0 (by Stokes’ theorem) and
"
! 2 " For
! thermal
" physics, a crucial point to remember is that function
2
f exact differentials.
F1 PHYS 4420 — FALL 2021∂ f
∂have
151
state
or
=
.
(C.48)
y
∂x∂y
∂y∂x
crucial point to remember is that functions of
ntials.
Inexact differential đ
For an inexact differential this is not true and knowledge of the initial and final states is
not sufficient to evaluate the integral: you have to know which path was taken.
Z
f (1)
I
df = 0
f (1)
I
đf 6= 0
2
df = f (2)
1
Z
2
1
đf 6= f (2)
For thermal physics, a crucial point to remember is
that functions of state have exact differentials.
PHYS 4420 — FALL 2021
152
Example
Three possible paths
1.0
xPath
=y 1
0.8
xPath
= y 22
yPath
= x4/3
3
Problem: integrate from (0,0) to (1,1)
df = ydx + xdy
Exact
0.6
y
0.0007
0.0006
0.4
0.0005
0.0004
f
f
0.0003
0.2
0.0002
0.0001
0.0
0.0000
0.0
0.2
0.4
0.6
0.8
1.0
Path
Same surface areas under the curves!
PHYS 4420 — FALL 2021
0.0
0.2
0.4
0.6
0.8
1.0
x
153
Example
Three possible paths
1.0
xPath
=y 1
0.8
xPath
= y 22
yPath
= x4/3
3
Problem: integrate from (0,0) to (1,1)
df = ydx + x2 dy Inexact
0.6
y
0.0007
0.0006
0.4
0.0005
0.0004
f
f
0.2
0.0003
0.0002
0.0001
0.0
0.0000
0.0
0.2
0.4
0.6
Path
0.8
1.0
Different surface areas!
PHYS 4420 — FALL 2021
0.0
0.2
0.4
0.6
0.8
1.0
x
154