Cheng; 3/2/2009; 2-1 Chapter 2. Vector Analysis

2.1 Overview

At a given position and time a scalar function → a magnitude, a vector function → a magnitude and a direction

Function conversion between different coordinates Physical laws should be independent of the coordinates. Coordinate system is chosen by convenience

Three main topics (1) Vector algebra : addition, subtraction, multiplication (2) Orthogonal coordinate system : Cartesian, Cylindrical, Spherical (3) Vector calculus : differentiation, integration(gradient, divergence, curl)

2.2 Vector Addition and Subtraction A vector has a magnitude and a direction AAa= A ↑ ↑ magnitude, A A unit vector, A

Graphical representation

Two vectors are equal if they have the same magnitude and direction, even though they may be displaced in space.

• Vector addition, CA= + B Two vectors, AB and , form a plane

Parallelogram rule : is the diagonal of the parallelogram C Head-To-Tail rule : The head of A touches the tail of B . C is drawn from the tail of A to the head of B .

Cheng; 3/2/2009; 2-2

Note C =+=+A B B A Commutative law ABF++=++()() ABF Associative law

• Vector subtraction ˆ AB−=+− Aej B, where −=−BaB()B

2.3 Vector Multiplication Multiplication of a vector by a positive scalar kA= bg kA a A

A. Scalar or Dot Product AB•≡ ABcosθAB

Note that •≤ (1) A B AB (2) •≤ •≥ A B 00 or A B •=× (3) A B A Projection of B onto A =B × Projection of A onto B (4) A •=B 0 when A and B are perpendicular to each other

Note that A •=A A2 → AAA=•

A •=•B B A : Commutative law

ABF•+=•+•() ABAF : Distributive law

Cheng; 3/2/2009; 2-3 Example Law of Cosine Use vectors to prove CAB222=+−2 ABcos α

CBA=− CCCBABA2 =•⇒( −) •( −) ⇒•+•−•−•BB AAAB BA ⇒+−AB222cos AB α

B. Vector or Cross Product AB×≡ aABnABsin θ ↑ unit vector normal to A and B (the right hand rule)

AB× = Area of the parallelogram formed by A and B

Note that AB×=−× BA NOT Commutative ABF××≠××()() ABF NOT Associative ABF×+=×+×() ABAF Distributive law

C. Product of Three Vectors Scalar triple product ABC•×=•ejej AaBCn sin α ↑ ↑ Area of parallelogram Height of the parallelepiped

→ Volume of the parallelepiped

Vector triple product ABCBACCAB××=( ) ( •−) ( •) back-cab rule

Cheng; 3/2/2009; 2-4 2.4 Orthogonal Coordinate Systems

A point in space is represented by three surfaces, u1 = const., u2 = const. and u3 = const. If they are mutually perpendicular, they form an orthogonal coordinate system. y Unit vectors along uuu,, → Base vectors, aaaˆˆˆ, , 123 uu123 u A vector in (uuu123,,) coordinate system → A =+Aauu11ˆˆˆ Aa u 2 u 2 + Aa u 3 u 3

y Differential change dui → Differential length change dliii= h du ↑

metric coefficient hi

Vector differential length dl=+ aˆˆuu111( h du) a 2( h 2 du 2) + a ˆ u 3( h 3 du 3)

Differential volume

dv= ( h11 du) X( h 2 du 2) X( h 3 du 3)

Vector differential area ds= ds aˆn , aˆn : surface normal

Note

ds11 aˆˆuu= h 232 h du du 31 a

ds22 aˆˆuu= h 131 h du du 32 a

ds33 aˆˆuu= h 121 h du du 23 a

A. Cartesian Coordinate System

bgbguuu123,,= xyz ,,

A point bgxyz111,, : The intersection of three planes xx= 1 , yy= 1 , zz= 1

Base vectors : aaa xyz, ,

Properties of the base vectors

aaxy×= a z,

aayz×= a x,

aazx×= a y

aaxy•=•=•= aa yz aa xz0

aaxx•=•=•= aa yy aa zz1

The position vector to the point

Pxbg111,, y z → → OP=++ x111 axyz y a z a

A vector A in Cartesian coord. AAaAaAa=++xx yy zz

Cheng; 3/2/2009; 2-5 y Vector differential length → dl=++ dx axyz dy a dz a

Differential area

dsx = dydz

dsy = dxdz

dsz = dxdy

Differential volume dv= dxdydz y The dot product A•= Bdidi Aaxx + Aa yy + Aa zz • Ba xx + Ba yy + Ba zz ⇒ AB x x + AB y y + AB z z

The cross product A×= Bdidi Aaxx + Aa yy + Aa zz × Ba xx + Ba yy + Ba zz

aaa xyz AB×=AAAxyz ⇒−diAByz AB zy a x +−bg AB zx AB xz a y +−di AB xy AB yx a z BBBxyz

Example

A straight line L1 is given by 2xy+= 4 .

(a) Find the unit normal to L1 starting from the origin

(b) Fine the normal line to L1 passing through P(0,2)

Solution:

(a) A vector along L1 : −+24xyˆˆ (1)

A vector from the origin to a point Q on L1 : xxˆˆ+−(42 x) y (2)

From (1) and (2) (−+24xyxxˆˆˆ)i( +−(42 xy) ˆ) =0 → Q(1.6, 0.8)

1 The unit normal is ()2xyˆˆ+ 5

1 (b) The straight line parallel to the unit normal is yxc= + 2 1 This line should pass through P(0,2) → yx= + 2 2

Cheng; 3/2/2009; 2-6 B. Cylindrical coordinates

bgbguuu123,,=φ r ,, z

A point Prbg111,,φ z : a cylindrical surface with radius of r1 , a half-plane rotated by φ1 from x-axis, a plane cutting z-axis at z1 .

Three base vectors :

aaarz, φ , ↑ ↑ directions change with P

aarz×=φ a ,

aaφ ×=zr a ,

aaazr×=φ

y Differential length dl=+ dr arz rdφ aφ + dz a ↑

Differential areas

dsr = rdφ dz

dsφ = drdz

dsz = rdrdφ

Differential volume dv= rdrdφ dz

Cheng; 3/2/2009; 2-7 y A vector in cylindrical coords. AAaAaAa=++rrφφ zz Transformation of A into Cartesian coord. AAaAaaAaaAaaxxrrxxzzxr=•=•+•+•⇒φφ Acosφφ − A φ sin ↑ ↑ ↑ =0 = cos φ = − sin φ

Similarly using aary• =φsin , aaφ • y = cos φ

AAyr=+sinφφ Aφ cos

In matrix form MLAx PO MLcosφφ− sin 0POMLAr PO MAy P = Msinφφ cos 0PMAφ P NMAz QP NM 001QPNMAz QP

y A point in cylindrical coord is transformed into Cartesian coord. x = r cos φ yr= sin φ zz=

Conversion from Cartesian to cylindrical coords. rxy=+22 y φ = tan−1 x zz=

→ Exercise Convert a vector in Cartesian coords., OQ=++345 axyz a a , to cylindrical coords. → OQa=dididi345xyzrrxyz ++• a aaaa + 345 ++• a aaaaφφ + 345 xyzzz ++• a aaa ↑ ↑ ↑ 34cosφφ+ sin 34bg−+sinφφ cos 5

From Cartesian Coords. we know that 3 4 cos φ= and sin φ= 5 5

Therefore in cylindrical Coords. → OQ=+55 arz a

Can you convert this vector back to Cartesian Coords.? Cheng; 3/2/2009; 2-8 C. Spherical Coordinates

bgbguuu123,,=θφ R ,,

A point PRbg111,,θφ : a sphere with radius R1

a cone with a half-angle of θ1

a half-plane rotated by φ1 from x-axis

z φφ= 1 plane θθ= 1 cone

θ1

aR P a φ a θ

φ1 y x RR= 1 sphere

Three base vectors

aaaR ×=θφ,

aaθφ×= a R ,

aaφθ×=R a

A vector in spherical coord AAaAaAa=++RRθθ φφ y Vector differential length dl=+ dR aR Rdθθφ aθφ + Rsin d a ↑ ↑

Differential areas 2 dsR = Rsinθ dθφ d

dsθ = Rsinθ dRdφ

dsφ = RdRdθ

Differential volume dv= R2 sinθ dRdθφ d

Cheng; 3/2/2009; 2-9

• Conversion of a point xR= sinθφ cos yR= sinθφ sin zR= cos θ

Conversion from Cartesian to spherical coords.

Rxyz=++222 θ =+tan−122⎡⎤xyz / ⎣⎦ y φ = tan−1 x

Integrals Containing Vector Functions Fdv Fxˆˆˆ++ Fy Fz dv ∫V ∫V ( xyz) Vdl V() xdxˆˆˆ++ ydy zdz ∫C ∫C → Fdli Fdli ∫C ∫C Adsi Adsi ∫S ∫S

Sign convention for the surface integral

Cheng; 3/2/2009; 2-10

Cheng; 3/2/2009; 2-11 2.5 Gradient of a Scalar Field

A scalar field Vtbg,,, u123 u u

Move from VV= 1 surface to VV=+1 dV surface PP12→ → dn PP13→ → dl ↑ ↑

same dV different distances

The directional derivative dV : The space rate of change of V along l dl

The gradient of a scalar function V is defined as dV grad V≡∇ V ≡ a : The maximum directional derivative of V dn n

(an is perpendicular to V=const. plane)

• The directional derivative dV dV dn dV dV =⇒cos α ⇒•⇒∇•aanlbg Va l : projection of gradV onto al direction dl dn dl dn dn → dV=∇bg V • dl , (1)

In Cartesian coord. ∂ ∂ ∂ ∂ ∂ ∂ V V V F V V V I dV =++⇒dx dy dz axyz ++a adxadyadza•++di xyz ∂x ∂y ∂z HG ∂x ∂y ∂z KJ ↑ (2) = dl From (1) and (2) ∂V ∂V ∂V F ∂ ∂ ∂ I ∇=V a +a +a ⇒a +a + aV ∂x xyz∂y ∂z HG ∂x xyz∂y ∂z KJ ↑ ≡ ∇ , “del” operator

In bguuu123,, coordinates 11∂ ∂ 1∂ ∇=a +a + a uu123 u hu11∂ hu 22∂ hu 3∂ 3

Cheng; 3/2/2009; 2-12 2.6 Divergence of a Vector Field Flux is defined as a quantity per unit area per unit time

A vector field is represented by flux lines Its direction → Direction of the flux line Its magnitude → Density of the flux lines

• Divergence of A The net outward flux of A per unit volume AdS• zS divA ≡ lim : dS= dS an ΔV →0 ΔV

Find divA at Pxbgooo,, y z

For a small volume ΔxΔΔy z A•= dS A •+ dS A •+ dS A •+ dS A •+ dS A •+ dS A • dS zSz frontz backz rightz leftz topz botom

(1) On the front surface 1 AdSA•=front •ΔΔΔΔ S front ⇒ Ax x(,,) o +2 xy o z o yz zfront ↑ ↑ ΔxA∂ x bgΔΔyzax , Axxo(,,) y o z o+ , Taylor series 2 ∂x bgxyzooo,, (2) On the back surface 1 A•= dSAback •ΔΔΔΔΔΔ S back ⇒ A back •−bg yza x ⇒−− A x(,,) x o2 x y o z o yz zback ↑ ΔxA∂ Ax(,,) y z − x xo o o 2 ∂x bgxyzooo,, Combining ∂A AdS•+ AdS •=x ΔΔΔxyz (1) zfrontz back ∂x bgxyzooo,,

Cheng; 3/2/2009; 2-13 Similarly, combining contributions from the right and left surfaces ∂A AdS•+ AdS •=y ΔΔΔxyz (2) zrightz left ∂y bgxyzooo,,

Similarly, combining contributions from the top and bottom surfaces ∂A AdS•+ AdS •=z ΔΔΔxyz (3) ztopz bottom ∂z bgxyzooo,,

From (1), (2) and (3)

F ∂A ∂Ay ∂A I AdS•=x + +z ΔΔΔxyz zS HG ∂x ∂y ∂z KJ bgxyzooo,,

∂A ∂Ay ∂A → divA =++x z ∂x ∂y ∂z

We define ∇•AdivA ≡

In general 1 L ∂ ∂ ∂ O ∇•A =M bgbgbghhA231 +hh13 A 2 + hh12 A 3P hhh123N∂ u 1 ∂u2 ∂u3 Q

2.7 Divergence Theorem The volume integral of the divergence of a vector field equals the total outward flux of the vector through the surface

∇•AdV = A • dS zzVS

Proof:

Consider a differential volume ΔV j bounded by S j From the definition of divA ej∇•AVΔ j = AdS • j zs j

Add all ΔV j ’s L N O L N O limM ∇•AVΔ j P =• lim M AdSAdSP ⇒• ∑ejj ∑ ΔΔV jj→0M P V →0M zs j P zS N j =1 Q N j =1 Q ↑ ↑ ↑ Integral at the external surface ∇•Adv An internal surface is shared by two adjacent volumes. zV ej (Opposite surface normals)

• ∇•A ≠0 means the existence of flow sources in the given volume

Cheng; 3/2/2009; 2-14 2.8 Curl of a Vector Field Divergence measures flow source Curl measures vortex source

The circulation of a vector field around a contour C Adl• zC

The definition of curl : The maximum circulation of A per unit area with the direction normal to the loop area (Right hand rule) 1 L O curlA=∇× A ≡lim Adl • an Δs→0 Δs NMzC QPmax

y The component of curl A in the direction of au 1 ej∇×Aa =u • ej ∇× A =lim Adl • u Δsu →0 Δs zcu u

y The x-component of ∇×A 1 L O ej∇×A =lim Adl • x ΔΔyz→0 ΔΔyzNMzsides,,1234 , , QP

At side 1: L O Δy ∂Az Adl•=M Axyzzooobg,, + + ...PΔz zside1 M 2 ∂y P N bgxyzooo,, Q

At side 3 : L O Δy ∂Az Adl•=M Axyzzooobg,, − + ...P bg−Δz zside3 M 2 ∂y P N bgxyzooo,, Q

Combine the two L∂A O Adl•=M z +...PΔΔyz, (1) zside13, M ∂y P N bgxyzooo,, Q

Similarly, combining contributions from sides 2 and 4 L O ∂Ay Adl•=−M +...PΔΔyz (2) zsides 24, M ∂z P N bgxyzooo,, Q

From (1) and (2)

1 L O ∂A ∂A ∇×A =lim M Adl • P ⇒−z y : Note the cyclic order in x, y, and z ejx ΔΔyz→0 ΔΔyz ∂y ∂z NMzsides,,1234 , , QP

Cheng; 3/2/2009; 2-15 In Cartesian coordinates ∂A ∂A ∂A ∂A ∂A ∂A F z y I F xzI F y x I ∇×A = − ax +−G Jay +−az HG ∂y ∂z KJ H ∂z ∂x K HG ∂x ∂y KJ

aaaxyz ∂ ∂ ∂ ∇×A = ∂xyz∂ ∂ AAAxyz

In general ah ah ah uu12123 u 3 1 ∂ ∂ ∂ ∇×A = hhh123 ∂uuu123∂ ∂ hA123xyz hA hA

If ∇×A =0, A is called an irrotational field, or a conservative field

2.9 Stokes’s Theorem

Consider a differential area ΔS j bounded by a contour C j

From the definition of ∇×A ∇×ASaAdl •Δ jj = • ejj di zC j

Adding all the contributions The left side: N lim ∇×ASa •Δ jj ⇒ ∇× AdS • ∑ejj di ej ΔS j →0 zS j =1

The right side: N F I lim Adl• ⇒• Adl Δ → ∑G J S j 0 zC j zC j =1 H K ↑ External contour C bounding S

Stokes’s theorem is defined ej∇×AdSAdl • = • : Right-hand rule for dS and dl zzSC

→ The surface integral of the curl of a vector over a surface is equal to the closed line integral of the vector along the bounding contour.

Note ej∇×AdS • =0 zS

Cheng; 3/2/2009; 2-16 2.10 Two Null Identities Identity I The curl of the gradient of a scalar field is always zero ∇×bg ∇V =0

Proof: (1) Direct operations of the gradient and curl (2) Using Stokes’s theorem ∇×bg ∇VdSVdldV • = ∇ • ⇒ ⇒0 zzzSCC

• The converse statement If a vector field is curl-free, then it can be expressed as the gradient of a scalar field ∇×E =0 → EV=−∇

Identity 2 The divergence of the curl of any vector field is identically zero ∇•ej ∇×A =0

Using divergence theorem ∇•ej ∇×AdV = ej ∇× A • dS zzVS

Split the surface, SS= 12+ S and apply Stokes’s theorem The right side ej∇×AdS • ⇒ ej ∇× AadS •n12 + ej ∇× AadS •n ⇒0 zzS S12 zS ↑ ↑ Adl• Adl• zC1 zC2

• The converse statement If a vector field is divergenceless, then it can be expressed as the curl of another vector field ∇•B =0 → BA=∇×

2.11 Helmholtz’s Theorem

1. A static electric field in a charge-free region ∇•EE =00, and ∇× = : solenoidal, irrotational

2. A steady magnetic field in a current-carrying conductor ∇•HH =00, and ∇× ≠ : solenoidal, not irrotational

3. A static electric field in a charged region ∇•EE ≠00, and ∇× = : not solenoidal, irrotational

4. An electric field in a charge medium with a time-varying magnetic field ∇•EE ≠00, and ∇× ≠ : not solenoidal, not irrotational

Helmholtz’s Theorem A vector field is determined if both its divergence and its curl are specified everywhere