topic 02 (vectoranalysis)
DESCRIPTION
Topic 02 (VectorAnalysis)TRANSCRIPT
Vector Analysis
EE 141 Lecture NotesTopic 2
Professor K. E. OughstunSchool of Engineering
College of Engineering & Mathematical SciencesUniversity of Vermont
2014
Motivation
First published in 1901.
Gradient of a Scalar Field
Let F (P) = F (r) = F (x , y , z) be a scalar function of position (orpoint function) with value F (P1) at the point P1 = (x , y , z) andvalue F (P2) at a neighboring point P2 = (x + dx , y + dy , z + dz).
The two points are separated by the differential distance vector
d ~̀= 1̂xdx + 1̂ydy + 1̂zdz (1)
where dx = 1̂x · d ~̀, dy = 1̂y · d ~̀, and dz = 1̂z · d ~̀.
Gradient of a Scalar Field
The Total Differential of F (P)
dF =∂F
∂xdx +
∂F
∂ydy +
∂F
∂zdz (2)
is then given by
dF =∂F
∂x1̂x · d ~̀+
∂F
∂y1̂y · d ~̀+
∂F
∂z1̂z · d ~̀
=
(1̂x∂F
∂x+ 1̂y
∂F
∂y+ 1̂z
∂F
∂z
)· d ~̀ (3)
The Gradient of the scalar field F (r) is then defined as the vector field
grad F (r) ≡ ∇F (r) ≡ 1̂x∂F
∂x+ 1̂y
∂F
∂y+ 1̂z
∂F
∂z(4)
so that Eq. (3) for the total differential may then be expressed as
dF = ∇F · d ~̀ (5)
Vector Differential Operator ∇
The vector differential operator ∇ or “del” is defined as
∇ ≡ 1̂x∂
∂x+ 1̂y
∂
∂y+ 1̂z
∂
∂z(6)
Notice that this is not a vector! It is an operator.
Explicit expressions for the various differential operations (gradient,divergence, curl, Laplacian) need to be derived in the variousorthogonal curvilinear coordinate systems that naturally arise inengineering problems with specific geometric symmetries, e.g.:
cylindrical polar coordinates with axial symmetry
spherical polar coordinates with point symmetry.
Directional Derivative
The directional derivative of the scalar field F (r) along the direction
specified by the unit vector 1̂` with d ~̀= 1̂`d` then follows from therelation (5), viz.
dF = ∇F · d ~̀= ∇F · 1̂`d`
asdF
d`= 1̂` · ∇F . (7)
It then follows that dF/d` has its maximum value when 1̂` is alongthe direction specified by ∇F . The streamlines of the vector field ∇Fare then orthogonal to the isotimic surfaces F (r) = constant.
The path integral of Eq. (5) then gives
F (P2)− F (P1) =
∫ P2
P1
∇F · d ~̀ (8)
independent of the path taken.
Properties of the Gradient Operator
For any two scalar point functions U(r) and V (r),
Distributive Law over Addition:
∇(U(r) + V (r)
)= ∇U(r) +∇V (r) (9)
Product Rule:
∇(U(r)V (r)
)= U(r)∇V (r) + V (r)∇U(r) (10)
Power Rule: For any n,
∇V n(r) = nV n−1(r)∇V (r) (11)
Problem 2: Derive each of these properties in rectangular coordinates.
Example: Gradient of the Plane Wave Propagation
Factor e jk·r
Consider the plane wave propagation factor U(r) = e jk·r with fixedwave vector k = 1̂xkx + 1̂yky + 1̂zkz , where r = 1̂xx + 1̂yy + 1̂zz isthe variable position vector. Then
∇e jk·r = ∇e j(kxx+kyy+kzz)
=
(1̂x
∂
∂x+ 1̂y
∂
∂y+ 1̂z
∂
∂z
)e j(kxx+kyy+kzz)
=(1̂x jkx + 1̂y jky + 1̂z jkz
)e j(kxx+kyy+kzz)
= jke jk·r
Note thatd
dte jωt = jωe jωt
Hence, just as one has one-dimensional Fourier analysis in ω-space,one also has three-dimensional Fourier analysis in k-space.
Divergence of a Vector Field
A vector field F(r) = F(x , y , z) may be described graphically by itsfield lines, flux lines, or streamlines.
The flux density of F(r) passing through a differential element ofsurface ds = n̂ds whose directed orientation in space is specified bythe unit surface normal n̂, is defined as the amount of flux crossing aunit surface element of area, viz.
flux density ≡ F · n̂ds
ds= F · n̂ (12)
Divergence of a Vector Field
The total flux crossing a closed surface S with outward unit normalvector n̂ is then given by the surface integral
total flux =
∮SF · ds =
∮SF · n̂ds (13)
A positive value means that more flux leaves the region enclosedby the surface S than enters, indicating that S encloses a source.
A negative value means that more flux enters the regionenclosed by S than exits, indicating that S encloses a sink.
Divergence of a Vector Field
Consider determining the total outward flux of the vector field
F(r) = 1̂xFx(x , y , z) + 1̂yFy (x , y , z) + 1̂zFz(x , y , z) (14)
passing through the closed surface S comprised of the rectangularsurface elements Sj with outward unit normal vectors n̂j parallel tothe Cartesian coordinate axes.
Divergence of a Vector Field
The outward flux F1 through the surface element S1 with outwardunit normal vector n̂1 = −1̂x is given by
F1 =
∫∫S1
F · n̂ds
=
∫∫S1
(1̂xFx + 1̂yFy + 1̂zFz
)·(−1̂x
)dydz = −Fx(r1)∆y∆z ,
where Fx(r1) is the value of Fx at the center of surface element S1.
Similarly, the outward flux F2 through the surface element S2 withoutward unit normal vector n̂2 = 1̂x is given by
F2 =
∫∫S2
F · n̂ds
=
∫∫S2
(1̂xFx + 1̂yFy + 1̂zFz
)·(1̂x
)dydz = Fx(r2)∆y∆z ,
where Fx(r2) is the value of Fx at the center of surface element S2.
Divergence of a Vector Field
Because r2 = r1 + 1̂x∆x , then one has the Taylor series expansion
Fx(r2) = Fx(r1) +∂Fx
∂x∆x +O
{(∆x)2
}.
With this substitution, the expression for the outward flux F2 throughthe surface element S2 becomes
F2 =
[Fx(r1) +
∂Fx
∂x∆x
]∆y∆z ,
plus higher-order terms that are of the order of O{(∆x)2∆y∆z} as∆x → 0, ∆y → 0, ∆z → 0.
Notation: f (x) = O{
g(x)}
as x → x0 means that |f (x)/g(x)| ≤ Kas x → x0, where K > 0 is some positive constant.
Divergence of a Vector Field
The sum of the outward fluxes of the vector field F(r) through thetwo surface elements S1 and S2 is then given by
F1 + F2 =∂Fx
∂x∆x∆y∆z .
Similarly, the sum of the outward fluxes of F(r) through the twosurface elements S3 and S4 is given by
F3 + F4 =∂Fy
∂y∆x∆y∆z ,
and the sum of the outward fluxes of F(r) through the two surfaceelements S5 and S6 is given by
F5 + F6 =∂Fz
∂z∆x∆y∆z ,
plus higher-order terms of order O{(∆x)2∆y∆z}, O{∆x(∆y)2∆z}or O{∆x∆y(∆z)2}, respectively.
Divergence of a Vector Field
The total outward flux of the vector field F(r) through the closedsurface S is then given by
∑6j=1 Fj , so that∮
SF · ds =
(∂Fx
∂x+∂Fy
∂y+∂Fz
∂z
)∆V , (15)
where ∆V = ∆x∆y∆z is the element of volume enclosed by S.
The divergence of a vector field is then defined as the scalar field
divF(r) ≡ lim∆V→0
{1
∆V
∮SF · ds
}(16)
where ∆V is the volume element enclosed by the simply-connectedsurface S containing the point r, and where ds = n̂ds with outwardunit normal n̂ to the surface S.
Divergence of a Vector Field
Because ∇ ≡ 1̂x∂∂x
+ 1̂y∂∂y
+ 1̂z∂∂z
, then the divergence operation inrectangular coordinates is given by
divF(r) = ∇ · F(r) =∂Fx
∂x+∂Fy
∂y+∂Fz
∂z(17)
The vector field F(r) has a positive divergence at a point r if thenet flux out of an infinitesimally small surface S surroundingthat point is positive; ∆V then contains a flux source.
The vector field F(r) has a negative divergence at a point r ifthe net flux out of an infinitesimally small surface S surroundingthat point is negative; ∆V then contains a flux sink.
The vector field F(r) has a zero divergence at a point r if thenet flux out of an infinitesimally small surface S surroundingthat point vanishes; the field is then said to be divergenceless orsolenoidal at that point.
Properties of the Divergence Operator
Solenoidal Vector Field ⇐⇒ ∇ · F(r) = 0 ⇐⇒∮S F · ds = 0.
For any two vector functions F(r) and G(r) and scalar function φ(r),
Distributive Law over Vector Addition:
∇ ·(F(r) + G(r)
)= ∇ · F(r) +∇ · G(r) (18)
Product Rule for Scalar Multiplication:
∇ ·(φ(r)F(r)
)= φ(r)∇ · F(r) + F(r) · ∇φ(r) (19)
Problem 3: Derive each of these properties in rectangular coordinates.
Problem 4: Show that ∇ · r = 3, where r ≡ 1̂xx + 1̂yy + 1̂zz is theposition vector.
Divergence Theorem
From Eq. (15), for each volume element ∆Vj with closed surface SJof a simply connected region V =
∑j Vj ,(
∇ · F(r))∆Vj =
∮SjF(r) · ds.
Summing over all of the volume elements ∆Vj in V and taking thelimit as ∆Vj → 0, this becomes∑
j
(∇ · F(r)
)∆Vj =
∑j
∮SjF(r) · ds =
∮SF(r) · ds,
where S is the surface enclosing V . In this limit as ∆Vj → 0, thesummation on the left goes over to the volume integral over V ,resulting in the Divergence Theorem∫∫∫
V
(∇ · F(r)
)d3r =
∮SF(r) · ds (20)
Example: Divergence of the Vector Plane Wave
Propagation Factor Ae jk·r
Consider the vector plane wave propagation factor Ae jk·r with fixedwave vector k = 1̂xkx + 1̂yky + 1̂zkz and variable position vector
r = 1̂xx + 1̂yy + 1̂zz . Here A is a constant vector which describesthe fixed orientation of the vector wave field in space (Polarization).Then
∇ · Ae jk·r =∂
∂x
(Axe jk·r)+
∂
∂y
(Aye jk·r)+
∂
∂z
(Aze jk·r)
= jkxAxe jk·r + jkyAye jk·r + jkzAze jk·r
= jk · Ae jk·r,
or, using the product rule (19) for scalar multiplication and the factthat ∇ · A = 0,
∇ · Ae jk·r = (∇ · A) e jk·r + A · ∇e jk·r = jk · Ae jk·r.
Circulation of a Vector Field around a Closed
Contour
The circulation of a vector field F(r) around a closed contour C isdefined by the contour integral
Circulation ≡∮CF(r) · d ~̀ (21)
where d ~̀ denotes the differential element of arc length tangent to thecontour C at the point r.
For example, the circulation of a uniform vector field F(r) = K iszero, as ∮
CK · d ~̀= K ·
∮C
d ~̀= 0
for any closed contour C.
Green’s Theorem for Regular Regions
Consider evaluating the surface integral∫∫ ∂u(x ,y)
∂ydxdy over a plane
region in the xy -plane that is bounded below and above by theregular arcs AB and CD, respectively, and on the sides by straightlines parallel to the y -axis, as illustrated.
This surface integral may be directly evaluated by integrating firstwith respect to y between the two points P ′(x , y) and P ′′(x , y) onthe lower and upper curves, respectively, and then with respect to xbetween the extreme limits a and b, as follows:
Green’s Theorem for Regular Regions
∫∫∂u(x , y)
∂ydxdy =
∫ b
a
[u(x , y ′′)− u(x , y ′)
]dx
=
∫ C
D
udx −∫ B
A
udx = −∫ D
C
udx −∫ B
A
udx .
Along the straight line segments BC and DA, x is constant so that∫ C
B
udx =
∫ D
A
udx = 0.
Adding these two path integrals to the above expression then gives∫∫∂u(x , y)
∂ydxdy = −
∮u(x , y)dx ,
the closed contour integral on the right hand side being taken in thecounterclockwise direction.
Green’s Theorem for Regular Regions
If a plane region S can be partitioned into a finite number of parts,each of the type just considered, then an equation of the above typeis valid for each of them. It then follows that∫∫
S
∂u(x , y)
∂ydxdy = −
∮C
u(x , y)dx , (22)
where C is the entire boundary of S.In a similar manner, if the plane region S can be partitioned into afinite number of subregions, each bounded on the sides by regulararcs with respect to the y -axis and above and below by straight linesegments parallel to the x-axis, then∫∫
S
∂v(x , y)
∂xdxdy =
∮C
v(x , y)dy . (23)
Green’s Theorem for Regular Regions
The sum of Eqs. (22) and (23) then gives Green’s Theorem forRegular Regions∫∫
S
(∂v
∂x− ∂u
∂y
)dxdy =
∮C
(udx + vdy
)(24)
Notice that if the positive unit normal vector n̂ to the plane surfaceS in the xy -plane is taken along the 1̂z direction, then the positivesense of integration around the contour C enclosing the region S isdetermined by the right-hand rule.
Unless otherwise specifically stated, this sign convention determinedby the right-hand rule is adopted here.
George Green (1793 – 1841)
Curl of a Vector Field
In Cartesian coordinates, the curl of a continuous vector field
F(r) = 1̂xu(x , y , z) + 1̂yv(x , y , z) + 1̂zw(x , y , z)
with continuous first partial derivatives is defined as
∇× F(r) =
(1̂x
∂
∂x+ 1̂y
∂
∂y+ 1̂z
∂
∂z
)×(1̂xu(x , y , z) + 1̂yv(x , y , z) + 1̂zw(x , y , z)
)=
∣∣∣∣∣∣1̂x 1̂y 1̂z
∂/∂x ∂/∂y ∂/∂zu(x , y , z) v(x , y , z) w(x , y , z)
∣∣∣∣∣∣= 1̂x
(∂w
∂y− ∂v
∂z
)+ 1̂y
(∂u
∂z− ∂w
∂x
)+ 1̂z
(∂v
∂x− ∂u
∂y
).
(25)
Curl of a Vector Field, Green’s Theorem
With this definition, the integrand appearing on the left-hand side ofEq. (24) is seen to be the z-component of the curl of F(r) and theintegrand appearing on the right-hand side of that equation is seen tobe F · dr in the xy -plane. Consequently, Green’s theorem for regularregions in the plane (24) is equivalent to∫∫
S∇× F · d~a =
∮CF · dr (26)
with d~a = 1̂zdxdy and dr = 1̂xdx + 1̂ydy taken to be tangent to thecontour C, the positive direction of integration about the contour Cbeing determined by the right-hand rule. Analogous expressions areobtained in the xz- and yz-planes.
Curl of a Vector Field, Stokes’ Theorem
The generalization of Green’s theorem (26) to R3 then leads toStokes’ Theorem ∫∫
S∇× F · n̂d2r =
∮CF · dr (27)
where n̂ denotes the unit normal vector to the surface S, the positivedirection of integration about the contour C being determined by theright-hand rule with respect to n̂.
G. G. Stokes, 1st Baronet (1819–1903)
Curl of a Vector Field
Application of Stokes’ theorem to a regular surface element ∆Scontaining the point P in its interior gives∫∫
∆S(∇× F(r)) · n̂d2r︸ ︷︷ ︸(∇×F(r))·n̂∆S
=
∮CF(r) · dr,
where, by the mean value theorem for integrals, (∇× F(r)) · n̂ issome intermediate value between the maximum and minimum valuesof the quantity (∇× F(r)) · n̂ on ∆S. Hence, in the limit as ∆Sshrinks to the point P ,
(∇× F(P)) · n̂ = lim∆S→0
1
∆S
∮CF(r) · dr (28)
This Integral Definition of the Curl shows that the normal componentof the curl of a vector field at a point P is given by the circulationper unit area about that point in the plane orthogonal to n̂.
Properties of the Curl Operator
Irrotational Vector Field ⇐⇒ ∇× F(r) = 0 ⇐⇒∮C F · d ~̀= 0.
For any two vector functions F(r) and G(r) and scalar function φ(r),Distributive Law over Vector Addition:
∇×(F(r) + G(r)
)= ∇× F(r) +∇× G(r) (29)
Product Rule for Scalar Multiplication:
∇×(φ(r)F(r)
)= φ(r)∇× F(r) +∇φ(r)× F(r) (30)
Product Rule for Vector Multiplication
∇× (F× G) = G · ∇F− F · ∇G + F(∇ · G)− G(∇ · F) (31)
The curl of the gradient of a scalar field vanishes
∇× (∇φ) = 0 (32)
The divergence of the curl of a vector field vanishes
∇ · (∇× F) = 0 (33)
Additional Vector Differentiation Properties
Problem 5: Derive Properties (32) and (33).
Problem 6: Show that ∇× r = 0,where r = 1̂xx + 1̂yy + 1̂zz .
Problem 7: Show that
F(r) · ∇r = F(r) (34)
For any two vector functions F(r) and G(r),
∇(F · G) = F · ∇G + G · ∇F + F× (∇× G) + G× (∇× F) (35)
Notice that, for example
F · ∇G = (F · ∇)G,
where
F · ∇ = Fx∂
∂x+ Fy
∂
∂y+ Fz
∂
∂z(36)
is the projection of the vector differential operator ∇ onto vector F.
Example: Curl of the Vector Plane Wave
Propagation Factor Ae jk·r
Consider the vector plane wave propagation factor Ae jk·r with fixedwave vector k = 1̂xkx + 1̂yky + 1̂zkz and A a constant vector. Then
∇× Ae jk·r = 1̂x
[Az
∂
∂y
(e jk·r)− Ay
∂
∂z
(e jk·r)]
+1̂y
[Ax
∂
∂z
(e jk·r)− Az
∂
∂x
(e jk·r)]
+1̂z
[Ay
∂
∂x
(e jk·r)− Ax
∂
∂y
(e jk·r)]
= j[1̂x(Azky − Aykz) + 1̂y (Axkz − Azkx)
+1̂z(Aykx − Axky )]e jk·r = jk× Ae jk·r,
or, using the product rule (30) with ∇× A = 0,
∇× Ae jk·r = (∇× A) e jk·r +∇e jk·r × A = jk× Ae jk·r.
The Laplacian Operator
Consider a scalar field φ(r) = φ(x , y , z) with gradient
∇φ(r) = 1̂x∂φ(r)
∂x+ 1̂y
∂φ(r)
∂y+ 1̂z
∂φ(r)
∂z,
which in turn is a vector field with divergence
∇ ·(∇φ(r)
)=
(1̂x
∂
∂x+ 1̂y
∂
∂y+ 1̂z
∂
∂z
)·(1̂x∂φ
∂x+ 1̂y
∂φ
∂y+ 1̂z
∂φ
∂z
)=
∂2φ
∂x2+∂2φ
∂y 2+∂2φ
∂z2.
The Laplacian operator is then defined as
∇2 ≡ ∇ · ∇ =∂2
∂x2+
∂2
∂y 2+
∂2
∂z2(37)
The Laplacian & CurlCurl Operators
The Laplacian of a vector field F(r) = 1̂xFx(r) + 1̂yFy (r) + 1̂zFz(r) isgiven by
∇2F(r) = 1̂x∇2Fx + 1̂y∇2Fy + 1̂z∇2Fz . (38)
Through direct substitution, it is found that the curlcurl operator isgiven by
∇× (∇× F) = ∇(∇ · F)−∇2F (39)
when operating on a continuous vector field F with continuous first-and second-order partial derivatives.
Problem 8: Verify Eq. (39) for the curlcurl operator in Cartesiancoordinates.
Pierre-Simon Laplace (1749 – 1827)
Green’s Integral Identities
Begin with the divergence theorem [see Eq. (20)]∮SF · ds =
∫∫∫V
(∇ · F
)d3r .
With F(r) = φ(r)∇ψ(r), one obtains Green’s First Integral Identity∮S
(φ∇ψ
)· ds =
∫∫∫V
(φ∇2ψ +∇φ · ∇ψ
)d3r (40)
Interchanging φ and ψ in this equation gives∮S
(ψ∇φ
)· ds =
∫∫∫V
(ψ∇2φ +∇ψ · ∇φ
)d3r .
Subtracting this result from the expression given in Eq. (40) yieldsGreen’s Second Integral Identity∮
S
(φ∇ψ − ψ∇φ
)· ds =
∫∫∫V
(φ∇2ψ − ψ∇2φ
)d3r (41)
Problem 9
Beginning with the divergence theorem (∮S F · n̂d2r =
∫V ∇ · Fd3r),
where the closed surface S forms the complete boundary of theregion V with outward unit normal vector n̂, show that, for anysufficiently continuous vector field G = G(r) and scalar field f = f (r):∮
Sf (r)n̂d2r =
∫V
(∇f (r)
)d3r , (42)∮
Sn̂× G(r)d2r =
∫V
(∇× G(r)
)d3r . (43)
Problem 10
Beginning with the Stokes’ theorem (∮C F · dr =
∫S(∇× F) · n̂d2r),
where the closed contour C forms the complete boundary of thesurface S with unit normal vector n̂ taken in the positive direction,show that, for any sufficiently continuous vector field G = G(r) andscalar field f = f (r):∮
Cdr × G(r) =
∫S
(n̂×∇
)× G(r)d2r , (44)∮
Cf (r)dr =
∫S
(n̂×∇f (r)
)d2r . (45)