lecture 0 murderous mathematics
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MurderousMathematics
Reference: Chapter 1, “Introduction to Electrodynamics” by David J. Griffiths
Text: Chapter 2, “Electromagnetism” by Gerald L. Pollack and Dainel R. Stump
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“To those who do not know mathematics it is
difficult to get across a real feeling as to thebeauty, the deepest beauty, of nature. … If youwant to learn about nature, to appreciate nature, itis necessary to understand the language that shespeaks in.”
– R. P. Feynman, The Character of Physical Law
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Curvilinear coordinates
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Cartesian Coordinates
In terms of Cartesiancoordinates, the positionvector
of a point P in space, withrespect to a chosen origin:k jix ˆˆˆ z y x ++=
If the coordinate axes are rotated,
but the point P is left fixed inspace. For example, for a rotationby angle θ about :k ˆ
jiii ˆsinˆcosˆˆ θ+θ=′→
ji j j ˆcosˆsinˆˆ θ+θ−=′→
k k k ˆˆˆ
=′
→
( x , y , z )
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Cartesian Coordinates (cont’)
It follows from k jixk ji ′′+′′+′′==++ ˆˆˆˆˆˆ z y x z y x
( ) ( ) k ji ji ˆˆcosˆsinˆsinˆcos z y x ′+θ+θ−′+θ+θ′=
( ) ( ) k ji ˆˆcossinˆsincos z y x y x ′+θ′+θ′+θ′−θ′=
that or
′
′′
θθθ−θ
=
z
y
x
z
y
x
100
0cossin
0sincos
θθ−θθ
=
′
′′
z
y
x
z
y
x
100
0cossin
0sincos
RT R R =−1
The matrix R depends on the angle and axis of rotation,but not on P.
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Cylindrical Coordinates
k rx ˆˆ z r +=
In terms of cylindricalcoordinates, the positionvector of a point P in space, withrespect to a chosen origin:
z z r yr x =φ=φ= ,sin,cos
where z z y xr =+= ,22
jir ˆsinˆcosˆ φ+φ= ( ) φφ≡φ+φ−φ= ˆˆcosˆsinˆ d d d jir
It follows that k rs ˆˆˆ dz rd dr d +φφ+=
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rx ˆr =
222 z y xr ++=
Spherical Coordinates
It follows that
φφθ+θ+= ˆsinˆˆ d r rd dr d θrs
( ) ( ) jik jir ˆcosˆsinsinˆsinˆsincosˆcoscosˆ φ+φ−φθ+θ−φθ+φθθ= d d d
φθ= cossinr x
k jir ˆcosˆsinsinˆcossinˆ θ+φθ+φθ=
In terms of spherical coordinates, theposition vector of a point P in space, withrespect to a chosen origin:
φθ= sinsinr y
θ= cosr z
where
θ≡ φ≡ ˆ
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Infinitesimal displacement vector
The infinitesimal displacement vector, from ( x , y , z )to ( x + dx , y + dy , z + dz ), is
k jis ˆˆˆ dz dydxd ++=
In cylindricalcoordinates,
k rs ˆˆˆ dz rd dr d +φφ+=
φφθ+θ+=ˆ
sinˆ
ˆ d r rd dr d θrs
In sphericalcoordinates,
In general terms, let u1, u2,
u3 denote three coordinates
that specify the points inthree dimensions:
z u yu xu === 321 ,,
Cartesian coordinates
z uur u =φ==321
,,
cylindrical coordinates
φ=θ== 321 ,, uur u
spherical coordinates
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Suppose the unit vectors pointingin the directions of independent
positive displacements of u
1,u
2,u3, are respectively:,ˆ,ˆ,ˆ321eee
k e jeie ˆˆ,ˆˆ,ˆˆ321
===
k eere ˆˆ,ˆˆ,ˆˆ321
=φ== φ=== ˆˆ,ˆˆ,ˆˆ321eθere
Cartesian coordinates
cylindrical coordinates spherical coordinates
Infinitesimal displacement vector (cont’)
The infinitesimal displacement vector in space that resultsfrom changing the coordinates by du1, du2, du3 can always
be written in the form:333222111 ˆˆˆ eees duhduhduhd ++=
For Cartesian coordinates, the scale factors h1 = h2 = h3
= 1.1,,1 321 === hr hh
cylindrical coordinates
θ=== sin,,1321
r hr hh
spherical coordinates
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Vector algebra
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Definitions
A vector A is a quantity with three components
k jik jiA ˆˆˆˆˆˆ321
A A A A A A z y x ++≡++=
In suffix notation, the vector A is denoted Ai (i = 1, 2,
3).
Note 1: Ai stands for two different things: the i th
component of the vector, and the vector itself. Thecontext must be used to decide which is meant.
that transform underrotation in the sameway as the
coordinates of a point:
=
′′′
z
y
x
z
y
x
A
A
A
R
A
A
A
=
′′′
3
2
1
3
2
1
A
A
A
R
A
A
A
jiji A R A =′Einstein summation convention:
Note 2: A scalar is a quantity that does not change
under rotation of the coordinate axes.
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Definitions (cont’)
Note 3: Multiplication by a scalar & Addition of twovectors
jiji A R A =′jiji B R B =′
( ) jij jiji A R A R A α=α=′α ( ) jij jiji B R B R B β=β=′β
( ) ( ) ( ) j jij jij jijii B A R B R A R B A β+α=β+α=′β+′α
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Dot Product
The dot product, or scalar product, of two vectors
k jiA ˆˆˆ z y x A A A ++= k jiB ˆˆˆ
z y x B B B ++=
is a scalar z z y y x x B A B A B A ++≡⋅BA
∑==++=
3
1332211
iii B A B A B A B A
Einstein summation convention: jiijii B A B A δ=≡⋅BA
where the Kronecker delta tensor
≠
=
=δ ji
ji
ij if 0
if 1
It follows that
1ˆˆˆˆˆˆ =⋅=⋅=⋅ k k j jii 0ˆˆˆˆˆˆ =⋅=⋅=⋅ ik k j ji
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Dot Product (cont’)
Geometric meaning:
A dot B is the projection of A on B times the magnitude of B, or the projection of B on A times the magnitude of A.
θ=⋅ cosBABA
where θ is the angle between A and B.
A dot B is a scalar:
k jik ijk ik jijii B A R R B R A R B A ==′′=′⋅′ BA
( ) BA ⋅=δ=== k j jk k j jk
T
k jik
T
ji B A B A R R B A R R
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Cross Product
The cross product, or vector product, of two vectors
k jiA ˆˆˆ z y x A A A ++= k jiB ˆˆˆ
z y x B B B ++=
is a vector
321
321
ˆˆˆˆˆˆ
B B B
A A A
B B B
A A A
z y x
z y x
k jik ji
BA =≡×
The i th component of A x B,( )
k jijk j k
k jijk i
B A B A
ε=ε≡× ∑∑= =
3
1
3
1
BA
The Levi-Civita alternatingtensor or the completelyantisymmetric tensor for
three dimensions, ε ijk is 0
1312231123
=ε=ε=ε
1321132213 −=ε=ε=ε
It follows that jik ik jk ji ˆˆˆ,ˆˆˆ,ˆˆˆ =×=×=×
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Cross Product (cont’)
An identity of the Levi-Civita tensor
jl im jmil klmkijklmijk δδ−δδ=εε=εε
( ) t s pqr kt kr js jqipt kt s js pqr kr jqipk jijk i B A R R R R R B R A R R R R B A ε=ε=′′ε′=′×′ BA
( ) ( ) t s pqr rt
T
qs
T
ipt s pqr kt
T
rk js
T
qjip B A R R R R R B A R R R R R ε=ε=
( ) pipr q pqr ipt s pqr rt qsip R B A R B A R BA×=ε=εδδ=
A cross B is a vector:
Geometric meaning:
nBABA ˆsin θ=×
θ is the angle between A and B.
d d i i
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Vector Product Identities
Scalar triple product:
( ) ( ) ( ) ( ) CBABACACBCBA ⋅×=×⋅=×⋅=×⋅
Proof – ( ) ( ) ( ) k jiijk k jijk iii C B AC B A A ε=ε=×=×⋅ CBCBA
( ) ( ) ( )ACB ×⋅=ε=ε= ik jki jk iijk j AC BC A B
( ) ( ) ( )BAC ×⋅=ε=ε= jikijk jiijk k B AC B AC
Note – ( ) ( ) ( )ABCCABBCA ×⋅=×⋅=×⋅
V P d Id i i ( ’)
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Vector Product Identities (cont’)
Vector triple product:
( ) ( ) ( )BACCABCBA ⋅−⋅=××
Proof – ( )[ ] ( ) ( )ml klm jijk k jijk i C B A A εε=×ε=×× CBCBA
( ) ml j jl im jmil ml jklmkijml jklmijk C B AC B AC B A δδ−δδ=εε=εε=
( ) ( ) ( ) ( )BACA ⋅−⋅=δδ−δδ= iil j jl mimm j jml il C B B AC C A B
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Vector differential operators
∇ (“del”), ∇ ⋅ (“del dot”),∇ 2
(“del squared”),∇×
(“delcross”)
G di t f S l F ti
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Gradient of a Scalar Function
In Cartesian coordinates, “del f ”
k ji ˆˆˆ z
f
y
f
x
f f
∂∂+
∂∂+
∂∂=∇ i
i x
f f e
∂∂=∇
“del f ” is a vector:
( ) ( ) ( ) jij
j
jij
j ji
j
i
i f R f R x
f
x
x
x
f f ∇=∇=∂∂′∂
∂=′∂∂=∇′ ∑∑ ==
3
1
3
1
Here, ( ) ( ) ( ) ij ji
T
ji
i
j
i ji j jiji R R R x
x x R x x R x ===
′∂∂
⇒′=⇒=′ −− 11
Consider the change of f (x) resultingfrom an infinitesimal displacement:
k jix ˆˆˆ dz dydxd ++=
The change of f from x to x + d x is
( ) ( ) xxxx d f dz z
f dy
y
f dx
x
f f d f df ⋅∇=
∂∂+
∂∂+
∂∂=−+=
The “del” operator acts algebraically as a vector.
G di t f S l F ti ( t’)
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Gradient of a Scalar Function (cont’)
for any arbitrary displacement d x along the surface of constant f .
The direction of “del f ” at a point x is perpendicular tothe surface of constant f that includes the point x:
0==⋅∇ df d f x
“del f ” points in the direction of maximum increase of f:
θ∇=⋅∇= cosxx d f d f df
where θ is the angle between “del f ” and d x.
The change of f is maximum if θ = 0. The magnitude of “del f ” is the rate of change of f in that direction.
G di t f S l F ti ( t’)
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Gradient of a Scalar Function (cont’)
It follows that
( )
ii
i
u
f
h
f ∂
∂=∇1
i
ii u
f
h
f e
1
∂
∂=∇
Consider the change of f (u1, u2, u3) resulting from an
infinitesimal displacement:333222111ˆˆˆ eees duhduhduhd ++=
The change of f , df duu
f du
u
f du
u
f =∂∂+
∂∂+
∂∂
3
3
2
2
1
1
( ) ( ) ( ) 333222111 duh f duh f duh f d f ∇+∇+∇=⋅∇= s
In cylindrical coordinates, “del f ” k r ˆˆ1ˆ
z
f f
r r
f f
∂∂
+φφ∂
∂+
∂∂
=∇
In spherical coordinates, “del f ” φφ∂
∂
θ+θ∂
∂
+∂
∂
=∇ˆ
sin
1ˆ1
ˆ
f
r
f
r r
f
f θr
E l
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Example
Show that 3
1
xx
xx
xx ′−
′−=′−
∇−
Solution: ( ) ( ) ( )233
2
22
2
11x x x x x x ′−+′−+′−=′− xx
( )ii
i
x x x
′−′−
=′−∂∂
xxxx
1
It follows that i
i xe
xxxxˆ
11
′−∂
∂−=
′−∇−
i
ii
ii
x x
x exxexxxx ˆˆ
1
32 ′−
′−
=
′−∂
∂
′−=
3
1
xx
xx
xx ′−
′−=
′−−∇∴
E ample
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Example
Show that ( ) f g g f fg ∇+∇=∇
Solution: ( )[ ] ( )iii
i x f g
x g f fg
x fg
∂∂+
∂∂=
∂∂=∇
Homework: Work through Example 2 @ Page 22
Divergence of a Vector Function (or Field)
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Divergence of a Vector Function (or Field)
In Cartesian coordinates, “del dot F”
z y x F
z
F
y
F
x ∂
∂+
∂
∂+
∂
∂=⋅∇ F i
i
F
x∂
∂=⋅∇ F
“del dot F” is invariant under rotation of the coordinatesystem:
( ) k
j
ik ijk ik
ji
j
i
i
F x
R R F R x x
x F
x ∂∂=
∂∂
′∂∂
=′′∂
∂=′⋅∇′ F
( ) F⋅∇=∂∂
δ=∂∂
=∂∂
= k
j
jk k
j
jk
T
k
j
ik
T
ji F x
F x
R R F x
R R
“del dot F” is a scalar function.
The Laplacian (or “del squared”) is thedivergence of the gradient:
( ) f f 2∇=∇⋅∇
In Cartesian coordinates,
2
2
2
2
2
22
z
f
y
f
x
f
f ∂
∂
+∂
∂
+∂
∂
=∇
Divergence of a Vector Function (cont’)
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Divergence of a Vector Function (cont’)
The divergence is a measure of how the vector functionF(x) diverges, i.e., spreads out from x.
Let dV be an infinitesimal cubic volume centered at x, of size ε x ε x ε , aligned with the Cartesian directions:
∑∫ =ε ε−− ε+=⋅
3
1
2ˆ2
ˆ2i
iiiidS F F d exexAF
( )∑=
ε∂∂=
3
1
3
i
i
i
F x
x
( ) 3ε⋅∇= F
The flux of F outward through theboundary surface dS of dV ,
The divergence is equal to theflux per unit volume through an
infinitesimal closed surface: ∫⋅=⋅∇
→ S V
d
V
AFF1
lim0
Divergence of a Vector Function (cont’)
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( )21213213
333
duduhh F hh F uduu
−+ +
( ) ( ) ( ) ( )332211 duhduhduhF⋅∇
( )32321321
111
duduhh F hh F uduu
−= +
( )13132132
222
duduhh F hh F uduu
−+ +
Consider an infinitesimal cubic volumedefined by displacements du1, du2, du3:
( ) ∫ ⋅=⋅∇=dS
d dV AFF
Divergence of a Vector Function (cont’)
( ) ( ) ( ) 213213
3
132132
2
321321
1
dududuhh F u
dududuhh F u
dududuhh F u ∂
∂+∂∂+
∂∂=
( ) ( ) ( )
∂
∂
+∂
∂
+∂
∂
=⋅∇∴ 2133
1322
3211321
1hh F
uhh F
uhh F
uhhhF
Divergence of a Vector Function (cont’)
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Divergence of a Vector Function (cont’)
In cylindrical coordinates, “del dot F”
( ) z r F z F r rF r r ∂∂
+φ∂∂
+∂∂
=⋅∇ φ
11F
In spherical coordinates,
( ) ( ) φθ φ∂∂
θ+θ
θ∂∂
θ+
∂∂=⋅∇ F
r F
r F r
r r r
sin
1sin
sin
11 2
2F
Recall i
ii u
f
h f e
1
∂∂
=∇
∂∂
∂∂+
∂∂
∂∂+
∂∂
∂∂=∇⋅∇=∇
33
21
322
13
211
32
1321
2 1u f
hhh
uu f
hhh
uu f
hhh
uhhh f f
It follows that
“del squared f ” 2
2
2
2
2
2 11
z
f f
r r
f r
r r f
∂∂+
φ∂∂+
∂∂
∂∂=∇
2
2
222
2
2
2
sin
1
sinsin
11
φ∂∂
θ+
θ∂∂
θθ∂∂
θ+
∂∂
∂∂
=∇f
r
f
r r
f
r r r f
Example
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Example
Show that 012 =
′−∇−
xx
Solution: ( ) ( ) ( )2
33
2
22
2
11x x x x x x ′−+′−+′−=′− xx
( )ii
i x x x ′−′−=′−∂∂ xxxx
1
provided |x – x’| is not zero.
32
11
xxxx
xxxx ′−
′−−=
′−∂∂
′−−=
′−∂
∂ ii
ii
x x
x x
( )5
2
32
2
311
xxxxxx ′−′−+
′−−=
′−∂
∂ ii
i
x x
x
Example
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Example
Show that ( ) GGG ⋅∇+⋅∇=⋅∇ f f f
Solution: ( ) ( ) i
i
i
i
i
i
G x
f G x f G f
x f
∂∂+∂∂=∂∂=⋅∇ G
Homework: Work through Example 3 @ Page 22
Curl of a Vector Function (or Field)
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Curl of a Vector Function (or Field)
In Cartesian coordinates, “del cross F”
z y x F F F z y x ∂∂∂∂∂∂=×∇
k ji
F
ˆˆˆ
( ) k
j
ijk i F x∂∂
ε=×∇ F
“del cross F” is a vector function:
( ) ( )t kt
s j
s
pqr kr jqipk
j
ijk i F R x x
x
R R R F x ∂∂
′∂∂
ε=′′∂∂
ε′=′×∇′ F
t
s
pqr kt
T
rk js
T
qjipt
s
pqr kt kr js jqip F x
R R R R R F x
R R R R R∂∂ε=
∂∂ε=
( ) pipr
q
pqr ipt
s
pqr rt qsip R F x
R F x
R F×∇=∂∂ε=
∂∂εδδ=
The curl is a measure of vorticity, i.e., how the vectorfunction F(x) curls around the point x.
Curl of a Vector Function (cont’)
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je
ie
Let dS be an infinitesimal squarecentered at x, of size ε x ε ,aligned with the Cartesiandirections:
( )
ε+ε+
ε−ε=⋅∫ i j ji
ijdP F F dl exexF ˆ
2ˆ2
ε−ε−
ε+ε− i j ji F F exex ˆ2
ˆ2
( ) 22 ε×∇=ε
∂∂
−∂∂
= k
j
i
i
j
x
F
x
F F
The circulation of F, i.e.,the line integral of F,counterclockwise aroundthe perimeter dP (ij ) of the
square,
The curl of F is equal to thecirculation of F per unit area
around an infinitesimal loop:
( ) ∫ ⋅=×∇⋅→ C A
dl A
FFn1
limˆ0
Curl of a Vector Function (cont )
Here, C denotes a planar closed curve with area A and
normal vector .n
Curl of a Vector Function (cont’)
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Curl of a Vector Function (cont )
Consider an infinitesimalrectangular area produced bydisplacements du1, du2:
( ) ( ) ( )22113duhduhF×∇
( )11111
222
duh F h F duuu +−= ( )
22222111
duh F h F uduu
−+ +
( )
∫ ⋅=×∇=
C
dl dA FF 3
( ) ( ) 21221
12112
duduh F ududuh F u ∂∂
+∂∂
−=
332211
321
213132321 ˆˆˆ
F h F h F h
uuu
hhhhhh
∂∂∂∂∂∂=×∇∴eee
F
Curl of a Vector Function (cont’)
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Curl of a Vector Function (cont )
In cylindrical coordinates,“del cross F”:
In spherical coordinates,
“del cross F”:
z r F rF F
z r
r r
φ
∂∂φ∂∂∂∂φ
=×∇k r
F
ˆˆˆ
( ) φ∂∂−
φ∂∂=×∇∴ F
z F
r z r
1F
( ) z r F r
F z ∂∂−∂∂=×∇ φF
( ) ( ) φφ φ∂∂−
∂∂=×∇ F
r rF
r r z
11F
φθ θφ∂∂θ∂∂∂∂
φθθ=×∇
F r rF F
r
r r r
r sin
ˆsinˆsinˆ2
θr
F
( ) ( )
φ∂∂−θ
θ∂∂
θ=×∇∴ θφ F F
r r sin
sin
1F
( ) ( )
∂
∂−
φ∂
∂
θ
=×∇ φθ rF
r
F
r
r
sin
11F ( ) ( )
θ∂
∂−
∂
∂=×∇ θφ r F rF
r r
1F
Example
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Example
Show that ( ) ( ) ( ) ( )[ ]xFxxFx ′×∇=′×∇ f f
Solution: ( ) ( )[ ] ( ) ( )xxxFx ′∂∂ε=′×∇ k
j
ijk i F x
f f
( ) ( )[ ]xx ′∂∂ε= k
j
ijk F f x
( ) ( )[ ] i f xFx ′×∇=
Example
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Example
Show that
( ) ( ) ( ) ( )[ ] ( ) ( )xFxxFxxFx ′∇−′⋅∇∇=′×∇×∇ f f f 2
Solution:
( ) ( )[ ]( )
( )( )
( )xx
xx
xFx ′∂∂
∂εε=
′∂
∂ε∂∂ε=′×∇×∇ m
l j
klmkijm
l
klm
j
ijk i F x x
f F
x
f
x f
2
( ) ( ) ( )xx ′
∂∂∂δδ−δδ= m
l j
jl im jmil F x x
f 2
( )( ) ( ) ( )xxx
x ′∇−′
∂∂
∂= i j
ji
F f F
x x
f 22
Homework: Work through Example 4 @ Page 23
Homework: Work through Example 7 @ Page 35
Del Identities – Products of derivatives
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Del Identities Products of derivatives
( ) FFF2∇−⋅∇∇=×∇×∇
( ) m
l j
klmkijm
l
klm
j
ijk i F x x
F x x ∂∂
∂εε=
∂∂ε
∂∂ε=×∇×∇
2
F
( ) i j ji
ml j
jl im jmil F F
x x F
x x
222
∇−∂∂
∂
=∂∂
∂
δδ−δδ=
The curl of the gradient of a scalar function is identicallyzero:
( ) 002
=∇×∇⇒=
∂∂
∂ε=∇×∇ f
x x
f f
k j
ijk i
The divergence of the curl of a vector function isidentically zero:
( )0
2
=∂∂
∂
ε=
∂
∂
ε∂
∂
=×∇⋅∇ k ji
ijk k j
ijk i F x x F x xF
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Integral theorems
arl Friedrich Gausspril 30th 1777 – February 23rd 1855) George Gabriel Stoke(August 13th 1819 – February 1st 1903
Gauss’s Theorem (divergence theorem)
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Gauss s Theorem (divergence theorem)
The flux of a vector quantity outwardthrough a closed surface S is equal tothe integral of the divergence of the
function in the enclosed volume V ,
∫ ∫ ⋅∇=⋅V S
xd d 3FAF
The flux of a vector field F(x) through asurface S with area element dA is thesurface integral of , where is theunit normal vector at the point x on S:
( ) nxF ˆ⋅ n∫ ∫ ⋅=⋅
S S d dA AFnF ˆ
( )
3
ε⋅∇=⋅∫ FAFdS d
Stokes’s Theorem
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Stokes s Theorem
The circulation of a vector field F(x)around a loop C with length element dl is the line integral of where is theunit tangent vector at the point x on C:
( ) txF ˆ⋅ t ∫ ∫ ⋅=⋅C C
dl dl FtF ˆ
je
ie
( )∫ ∫ ⋅×∇=⋅ S C d dl AFF
( ) ( )
2
ε×∇=⋅∫k ijdP dl FF
Homework: Work throughExamples 5, 6 @ Page 27
The circulation of a vector function around a closed curveC is equal to the flux of vorticity through any surfacebounded by C,
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The Helmholtz theorem
ermann von Helmholtzugust 31st 1821 – September 8th 1894)
Preliminaries
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Preliminaries
A vector function (or field) F(x) that has zero curl:
0=×∇ F
is called irrotational.A vector function (or field) G(x) that has zero divergence:
0=⋅∇ G
is called solenoidal.
ψ −∇=F
AG ×∇=
Any vector function (or field) H(x) can be written as thesum of an irrotational function F(x) and a solenoidal
function G(x): GFH += AH ×∇+ψ −∇=
The functions F and G are not necessarily unique.However, in some cases, if suitable boundary conditions
are imposed, then the decomposition is unique.
The Helmholtz theorem
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The Helmholtz theorem
Let H(x) be differentiable at all points in space, with curland divergence:
( )xcH =×∇ ( )xH d =⋅∇
If and c(x) approach 0 faster than r –2 as , andH(x) approaches 0 faster than r –1 , then
( )xd ∞→r
AH ×∇+ψ −∇=where
( )( )
∫ ′′−π
′=ψ xd
d 3
4 xx
xx ( )
( )∫ ′
′−π′
= xd 3
4 xx
xcxA
The integration region is all of space.
The Helmholtz theorem implies that if the divergence andcurl of a vector function (field) is known, then the functioncan be determined uniquely (under the assumptions of
the theorem).
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Green’s function andthe Dirac delta function
eorge GreenJuly 14th 1793 – May 31st 1841)
Paul Adrien Maurice Dirac
Paul Adrien Maurice Dira(August 8th 1902 – October 20th 1984
The Dirac delta function
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The Dirac delta function δ ( x ) is a “generalizedfunction” or “Schwartz distribution” with the followingdefining property:
( ) ( ) ( )0 f dx x f x =δ∫ ∞
∞− ( ) 1=δ∫ ∞
∞− dx x
for every continuous function f(x).
The Dirac delta function can
be understood as the limitof a sequence of more andmore sharply peakedfunctions:
δ ( x ) is an extremely
singular function:( )
=≠
≠=δ
0when0
0allfor 0
x
x x
Exercise:
( ) ( ) ( )a f dx x f a x =−δ∫ ∞∞−
The Dirac delta function (cont’)
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( )
The 3-dimensional Dirac delta function δ 3(x) isdefined analogously:
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )0xx f f dxdydz z y x f z y x xd f ==δδδ=δ ∫ ∫ ∫ ∫ ∞
∞−
∞
∞−
∞
∞− 0,0,0,,33
for every continuous function f (x).
( )
=≠
≠=δ
0when0
0allfor 03
x
xx( ) 1
33 =δ
∫ xd x
Exercise:
( ) ( ) ( )axax f xd f =−δ∫ 33
The Green’s Function of 2∇−
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( )xx
xx′−π
=′−4
1G( ) ( )xxxx ′−δ=′−∇− 32G
along with the boundary condition that G approaches 0 atinfinity.Proof :Consider the integral∫
′−π
∇−V
xd 32
4
1
xx
in a spherical volume V of any radius around x’.
∫ ∫ ⋅
′−
∇π
−=
′−
∇π
−S V
d xd Axxxx
1
4
11
4
1 32 divergencetheorem
14
1
ˆˆ
1
4
1
ˆˆ
1
4
1 2
22 =Ωπ=Ω⋅π=⋅π= ∫ ∫ ∫ S S S d d r r dAr rrrr
Here, we recall rr
xx
xx
xxˆ
11233
r r =≡
′−
′−=
′−∇−
2