Lecture 0 Murderous Mathematics

Murderous Mathematics Text: Chapter 2, “Electromagnetism” by Gerald L. Pollack and Dainel R. Stump Reference: Chapter 1, “Introduction to Electrodynamics” by David J. Griffiths

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“To those who do not know mathematics it is difficult to get across a real feeling as to the beauty, the deepest beauty, of nature. … If you want to learn about nature, to appreciate nature, it is necessary to understand the language that she speaks in.” –

R. P. Feynman, The Character of Physical Law

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Curvilinear coordinates

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Cartesian Coordinates In terms of Cartesian coordinates, the position vector of a point P in space, with respect to a ˆchosen origin: ˆ ˆ x = xi + yj + zk

(x, y, z)

If the coordinate axes are rotated, but the point P is left fixed in space. For example, for a rotation kˆ θ about : by angle ˆi → ˆi ′ = cos θˆi + sin θˆj ˆj → ˆj′ = − sin θˆi + cos θˆj kˆ → kˆ ′ = kˆ

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Cartesian Coordinates (cont’) It follows from

xˆi + yˆj + zkˆ = x = x′ˆi ′ + y′ˆj′ + z ′kˆ ′

= x′( cos θˆi + sin θˆj) + y′( − sin θˆi + cos θˆj) + z ′kˆ = ( x′ cos θ − y′ sin θ) ˆi + ( x′ sin θ + y′ cos θ) ˆj + z ′kˆ that  x   cos θ − sin θ 0 x′        y  =  sin θ cos θ 0 y′  z  0 0 1 z ′    

or

 x′   cos θ sin θ 0 x        y′  =  − sin θ cos θ 0 y   z′   0 0 1 z    

R R −1 = R T The matrix R depends on the angle and axis of rotation, but not on P. 5

Cylindrical Coordinates In terms of cylindrical coordinates, the position vector of a point P in space, with respect to a chosen origin: x = rrˆ + zkˆ where

r = x2 + y2 , z = z

x = r cos φ,

y = r sin φ, z = z

rˆ = cos φˆi + sin φˆj It follows that

drˆ = dφ( − sin φˆi + cos φˆj) ≡ dφφˆ

ds = drrˆ + rdφφˆ + dzkˆ 6

Spherical Coordinates In terms of spherical coordinates, the position vector of a point P in space, with respect to a chosen origin:

x = rrˆ

x = r sin θ cos φ

2 2 2 r = x + y + z where

y = r sin θ sin φ

rˆ = sin θ cos φˆi + sin θ sin φˆj + cos θkˆ It follows that

z = r cos θ

ds = drrˆ + rdθθˆ + r sin θdφφˆ

drˆ = dθ( cos θ cos φˆi + cos θ sin φˆj − sin θkˆ ) + sin θdφ( − sin φˆi + cos φˆj) ≡ θˆ

≡ φˆ

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Infinitesimal displacement vector The infinitesimal displacement vector, from (x, y, z) to (x + dx, y + dy, z + dz), is ds = dxˆi + dyˆj + dzkˆ In cylindrical coordinates,

ds = drrˆ + rdφφˆ + dzkˆ

In spherical coordinates,

ds = drrˆ + rdθθˆ + r sin θdφφˆ

In general terms, let u1, u2, u3 denote three coordinates that specify the points in three dimensions:

u1 = x, u2 = y, u3 = z Cartesian coordinates

u1 = r , u2 = φ, u3 = z

u1 = r , u2 = θ, u3 = φ

cylindrical coordinates

8 spherical coordinates

Infinitesimal displacement vector (cont’) Suppose the unit vectors pointing eˆ1 = ˆi , eˆ 2 = ˆj, eˆ 3 = kˆ in the directions of independent positive displacements of u1, u2, Cartesian coordinates e ˆ , e ˆ , e ˆ , u3, are respectively: 1 2 3 eˆ1 = rˆ , eˆ 2 = φˆ , eˆ 3 = kˆ

eˆ1 = rˆ , eˆ 2 = θˆ , eˆ 3 = φˆ

cylindrical coordinates spherical coordinates The infinitesimal displacement vector in space that results from changing the coordinates by du1, du2, du3 can always be written in the form: ds = h1du1eˆ1+ h2du2eˆ 2 + h3du3eˆ 3 For Cartesian coordinates, the scale factors h1 = h2 = h3 = 1. h1 = 1, h2 = r , h3 = r sin θ h1 = 1, h2 = r , h3 = 1 cylindrical coordinates

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spherical coordinates

Vector algebra

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Definitions A vector A is a quantity with three components A = Ax ˆi + Ay ˆj + Az kˆ ≡ A1ˆi + A2ˆj + A3kˆ that transform under  Ax′   Ax   A1′   A1         rotation in the same  A′y  = R Ay   A2′  = R A2 way as the  A′  A   A′  A   z  z  3  3 coordinates of a point: In suffix notation, the vector A is denoted Ai (i = 1, 2, 3). Ai′ = Rij A j Einstein summation convention: Note 1: Ai stands for two different things: the i th component of the vector, and the vector itself. The context must be used to decide which is meant. 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 two vectors Ai′ = Rij A j Bi′ = Rij B j αAi′ = αRij A j = Rij ( αA j )

β Bi′ = β Rij B j = Rij ( β B j )

αAi′ + β Bi′ = Rij ( αA j ) + Rij ( βB j ) = Rij ( αA j + β B j )

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Dot Product The dot product, or scalar product, of two vectors A = Ax ˆi + Ay ˆj + Az kˆ is a scalar

B = Bx ˆi + B y ˆj + Bz kˆ

A ⋅ B ≡ Ax Bx + Ay B y + Az Bz 3

= A1B1+ A2B2 + A3B3 = ∑ Ai Bi i =1

It follows that ˆi ⋅ ˆi = ˆj ⋅ ˆj = kˆ ⋅ kˆ = 1

ˆi ⋅ ˆj = ˆj ⋅ kˆ = kˆ ⋅ ˆi = 0

Einstein summation convention:

A ⋅ B ≡ Ai Bi = δij Ai B j

where the Kronecker delta tensor

1if i = j δij =  0if i ≠ j

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Dot Product (cont’) A dot B is a scalar: A′ ⋅ B′ = Ai′Bi′ = Rij A j Rik Bk = Rij Rik A j Bk = R Tji Rik A j Bk = ( R T R ) jk A j Bk = δ jk A j Bk = A ⋅ B 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. A ⋅ B = A B cos θ where θ is the angle between A and B.

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Cross Product The cross product, or vector product, of two vectors A = Ax ˆi + Ay ˆj + Az kˆ

is a vector

It follows that

ˆi A × B ≡ Ax Bx

B = Bx ˆi + B y ˆj + Bz kˆ ˆj Ay By

ˆi ˆj kˆ kˆ Az = A1 A2 A3 Bz B1 B2 B3

ˆi × ˆj = kˆ , ˆj × kˆ = ˆi , kˆ × ˆi = ˆj 3

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( A × B ) i ≡ ∑∑ εijk A j Bk = εijk A j Bk The i th component of A x B, j =1 k =1

The Levi-Civita alternating tensor or the completely antisymmetric tensor for three dimensions, ε ijk is 0

ε123 = ε 231 = ε 312 = 1 ε 213 = ε132 = ε 321 = −15 1

Cross Product (cont’) An identity of the Levi-Civita tensor ε ijk ε klm = ε kij ε klm = δil δ jm − δim δ jl A cross B is a vector:

( A′ × B′) i = ε′ijk A′j Bk′ = Rip R jq Rkr ε pqr R js As Rkt Bt = Rip R jq R js Rkr Rkt ε pqr As Bt = Rip RqjT R js RrkT Rkt ε pqr As Bt = Rip ( R T R ) qs ( R T R ) rt ε pqr As Bt = Rip δ qs δ rt ε pqr As Bt = Rip ε pqr Aq Br = Rip ( A × B ) p Geometric meaning: A × B = A B sin θnˆ θ is the angle between A and B.

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Vector Product Identities Scalar triple product: A ⋅ ( B × C) = B ⋅ ( C × A ) = C ⋅ ( A × B ) = ( A × B ) ⋅ C Proof – A ⋅ ( B × C) = Ai ( B × C ) i = Ai ( ε ijk B j Ck ) = ε ijk Ai B j Ck = B j ( ε ijk Ai Ck ) = B j ( ε jki Ck Ai ) = B ⋅ ( C × A ) = Ck ( ε ijk Ai B j ) = Ck ( ε kij Ai B j ) = C ⋅ ( A × B ) Note –

A ⋅ ( C × B ) = B ⋅ ( A × C) = C ⋅ ( B × A )

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Vector Product Identities (cont’) Vector triple product: A × ( B × C) = B( A ⋅ C) − C( A ⋅ B ) Proof –

[ A × ( B × C) ] i = εijk A j ( B × C) k

= ε ijk A j ( ε klm Bl Cm )

= ε ijk ε klm A j Bl Cm = ε kij ε klm A j Bl Cm = ( δil δ jm − δim δ jl ) A j Bl Cm = δil Bl ( δ jm A j Cm ) − δimCm ( δ jl A j Bl ) = Bi ( A ⋅ C) − Ci ( A ⋅ B )

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Vector differential operators ∇ (“del”), ∇ ⋅ (“del dot”), ∇ 2 (“del squared”), ∇ × (“del cross”) 19

Gradient of a Scalar Function Consider the change of f (x) resulting from an infinitesimal displacement:

dx = dxˆi + dyˆj + dzkˆ

The change of f from x to x + dx is ∂f ∂f ∂f df = f ( x + dx ) − f ( x ) = dx + dy + dz = ∇f ⋅ dx ∂x ∂y ∂z In Cartesian coordinates, “del f ” ∂f ˆ ∂f ˆ ∂f ˆ ∇f = i + j + k ∂x ∂y ∂z

∂f ∇f = eˆ i ∂xi

“del f ” is a vector: 3 ∂x 3 ∂f j ∂f ( ∇′f ) i = = ∑ = ∑ Rij ( ∇f ) j = Rij ( ∇f ) j ∂xi′ j =1 ∂xi′ ∂x j j =1 ∂x j = ( R −1) ji = ( R T ) ji = Rij Here, xi′ = Rij x j ⇒ x j = ( R ) ji xi′ ⇒ ∂xi′ −1

The “del” operator acts algebraically as a vector.

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Gradient of a Scalar Function (cont’) The direction of “del f ” at a point x is perpendicular to the surface of constant f that includes the point x: ∇f ⋅ dx = df = 0 for any arbitrary displacement dx along the surface of constant f. “del f ” points in the direction of maximum increase of f: df = ∇f ⋅ dx = ∇f dx cos θ where θ is the angle between “del f ” and dx. The change of f is maximum if θ = 0. The magnitude of “del f ” is the rate of change of f in that direction. 21

Gradient of a Scalar Function (cont’) Consider the change of f (u1, u2, u3) resulting from an infinitesimal displacement: ds = h1du1eˆ1+ h2du2eˆ 2 + h3du3eˆ 3 ∂f ∂f ∂f du1+ du2 + du3 = df The change of f, ∂u1 ∂u2 ∂u3 = ∇f ⋅ ds = ( ∇f ) 1h1du1+ ( ∇f ) 2h2du2 + ( ∇f ) 3h3du3 It follows that

( ∇f ) i

1 ∂f = hi ∂ui

1 ∂f ∇f = eˆ i hi ∂ui

∂f 1∂f ˆ ∂f ˆ φ+ k In cylindrical coordinates, “del f ” ∇f = rˆ + ∂r r ∂φ ∂z ∂f 1∂f ˆ 1 ∂f ˆ θ+ 22φ In spherical coordinates, “del f ”∇f = rˆ + ∂r r ∂θ r sin θ ∂φ

Example 1 x − x′ −∇ = x − x′ x − x′ 3

Show that Solution:

x − x′ =

( x1− x1′ ) 2 + ( x2 − x′2) 2 + ( x3− x3′ ) 2 ∂ 1 x − x′ = ( xi − xi′ ) ∂xi x − x′

It follows that

1 ∂  1   eˆ i −∇ =− x − x′ ∂xi  x − x′   1  ∂ xi − xi′  = x − x′ eˆ i = eˆ 2 3 i x − x′  ∂xi x − x′  1 x − x′ ∴ −∇ = x − x′ x − x′ 3 23

Example Show that Solution:

∇( fg ) = f ∇g + g ∇f

[ ∇( fg ) ] i

∂ ∂g ∂f = ( fg ) = f + g ∂xi ∂xi ∂xi

Homework: Work through Example 2 @ Page 22

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Divergence of a Vector Function (or Field) In Cartesian coordinates, “del dot F” ∂ ∂ ∂ ∂ ∇⋅F = Fi ∇ ⋅ F = Fx + Fy + Fz ∂xi ∂x ∂y ∂z “del dot F” is invariant under rotation of the coordinate system: ∂x j ∂ ∂ ∂ ∇′ ⋅ F ′ = Fi′ = ( Rik Fk ) = Rij Rik Fk ∂xi′ ∂xi′ ∂x j ∂x j ∂ ∂ ∂ T = R Rik Fk = ( R R ) jk Fk = δ jk Fk = ∇ ⋅ F ∂x j ∂x j ∂x j T ji

“del dot F” is a scalar function. The Laplacian (or “del squared”) is the divergence of the gradient:

∇ ⋅ ( ∇f ) = ∇ 2f

In Cartesian coordinates, ∂ 2f ∂ 2f ∂ 2f ∇ f = 2+ 2+ 2 ∂x ∂y ∂z 2

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Divergence of a Vector Function (cont’) The divergence is a measure of how the vector function F(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: The flux of F outward through the boundary surface dS of dV, 3 ε  ε  2    ∫dSF ⋅ dA = ∑  Fi  x + 2 eˆ i  − Fi  x − 2 eˆ i  ε i =1 3 ∂ =∑ Fi ( x ) ε 3 i =1 ∂xi = ( ∇ ⋅ F ) ε3 The divergence is equal to the flux per unit volume through an infinitesimal closed surface:

1 ∇ ⋅ F = lim ∫ F ⋅ dA V →0 V S 26

Divergence of a Vector Function (cont’) Consider an infinitesimal cubic volume defined by displacements du1, du2, du3:

( ∇ ⋅ F )( h1du1 )( h2 du2 )( h3du3 ) = ( ∇ ⋅ F ) dV = ∫ F ⋅ dA dS

(

)

(

)

= F1h2 h3 u1 + du1 − F1h2 h3 u1 du2 du3 + F2 h3h1 u2 + du2 − F2 h3 h1 u2 du3 du1

(

)

+ F3 h1h2 u3 + du3 − F3h1h2 u3 du1du2

∂ ∂ ∂ = ( F1h2 h3 ) du1du2 du3 + ( F2 h3h1 ) du2 du3du1 + ( F3h1h2 ) du3du1du2 ∂u1 ∂u2 ∂u3 ∴∇ ⋅ F =

 1  ∂ ∂ ∂ ( F h h ) + ( F h h ) + ( F h h ) 1 2 3 2 3 1 3 1 2  h1h2 h3  ∂u1 ∂u2 ∂u3 

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Divergence of a Vector Function (cont’) 1 ∂f ∇f = eˆ i Recall It follows that hi ∂ui 1  ∂  h2 h3 ∂f  ∂  h3 h1 ∂f  ∂  h1h2 ∂f    ∇ f = ∇ ⋅ ∇f =  +  +  h1h2 h3  ∂u1  h1 ∂u1  ∂u2  h2 ∂u2  ∂u3  h3 ∂u3  2

In cylindrical coordinates, “del dot F” 1∂ 1 ∂ ∂ ∇⋅F = ( rFr ) + Fφ + Fz r ∂r r ∂φ ∂z 1 ∂  ∂f  1 ∂ 2 f ∂ 2 f “del squared f ”∇ f = r  + 2 2 + 2 r ∂r  ∂r  r ∂φ ∂z 2

In spherical coordinates, 1 ∂ 2 1 ∂ 1 ∂ ∇ ⋅ F = 2 ( r Fr ) + ( sin θFθ ) + Fφ r ∂r r sin θ ∂θ r sin θ ∂φ 1 ∂  2 ∂f  1 ∂  ∂f  1 ∂2 f ∇ f = 2 r + 2  sin θ  + 2 2 r ∂r  ∂r  r sin θ ∂θ  ∂θ  r sin θ ∂φ 2 2

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Example 1 −∇ =0 x − x′ 2

Show that

provided |x – x’| is not zero. Solution:

x − x′ =

( x1 − x1′ ) 2 + ( x2 − x′2 ) 2 + ( x3 − x3′ ) 2 ∂ 1 ′ x−x = ( xi − xi′ ) ∂xi x − x′

 ∂  1  1  ∂ xi − xi′   = −  x − x′  = − 2  3 ∂xi  x − x′  ′ x − x′  ∂xi x − x  2 ( xi − xi′ ) ∂2  1  1   = − +3 3 5 2  ∂xi  x − x′  x − x′ x − x′

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Example Show that

∇ ⋅ ( f G ) = ∇f ⋅ G + f ∇ ⋅ G

Solution:

∂ ∂f ∂ ∇ ⋅( f G) = ( f Gi ) = Gi + f Gi ∂xi ∂xi ∂xi

Homework: Work through Example 3 @ Page 22

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Curl of a Vector Function (or Field) In Cartesian coordinates, “del cross F” ˆi ˆj kˆ ∂ ( ∇ × F ) i = εijk Fk ∇ × F = ∂ ∂x ∂ ∂y ∂ ∂z ∂x j Fx Fy Fz “del cross F” is a vector function: ∂x ( ∇′ × F′) i = ε′ijk ∂ Fk′ = Rip R jq Rkr ε pqr s ∂ ( Rkt Ft ) ∂x′j ∂x′j ∂xs = Rip R jq R js Rkr Rkt ε pqr = Rip δ qs δ rt ε pqr

∂ ∂ Ft = Rip RqjT R js RrkT Rkt ε pqr Ft ∂xs ∂xs

∂ ∂ Ft = Rip ε pqr Fr = Rip ( ∇ × F ) p ∂xs ∂xq

The curl is a measure of vorticity, i.e., how the vector function F(x) curls around the point x. 31

Curl of a Vector Function (cont’) Let dS be an infinitesimal squareeˆ j centered at x, of size ε x ε , aligned with the Cartesian directions: The circulation of F, i.e., the line integral of F, counterclockwise around the perimeter dP (ij ) of the square, ε  ε    eˆ i ∫dPF( ij⋅ )dl = εFi  x − 2eˆ j  + εF j  x + 2eˆ i  The curl of F is equal to the ε ε  − εFi  x + eˆ j  − εF j  x − eˆ i circulation of F per unit area  2  2 around    an infinitesimal loop:  ∂F j ∂Fi  2 1 ε = ( ∇ × F ) k ε 2 n ˆ ⋅ ( ∇ × F ) = lim F ⋅ dl =  − ∫  A→0 A C  ∂xi ∂x j  Here, C denotes a planar closed curve with area A and32 nˆ vector . normal

Curl of a Vector Function (cont’) Consider an infinitesimal rectangular area produced by displacements du1, du2:

( ∇ × F ) 3( h1du1)( h2du2) = ( ∇ × F ) 3dA = ∫ F ⋅ dl C

(

)

= F1h1u2 − F1h1u2+ du2 du1

(

)

+ F2h2u1+ du1 − F2h2u1 du2

∂ ∂ =− ( F1h1) du2du1+ ( F2h2) du1du2 ∂u2 ∂u1 eˆ 1 h2h3 eˆ 2 h3h1 eˆ 3 h1h2 ∴ ∇ × F = ∂ ∂u1 ∂ ∂u2 ∂ ∂u3 h1F1 h2F2 h3F3

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Curl of a Vector Function (cont’) In cylindrical coordinates, “del cross F”: rˆ r φˆ kˆ r ∇ × F = ∂ ∂r ∂ ∂φ ∂ ∂z Fr rFφ Fz

1∂ ∂ ∴( ∇ × F) r = Fz − Fφ r ∂φ ∂z

( ∇ × F) φ

∂ ∂ = Fr − Fz ∂z ∂r

1∂ 1∂ ( rFφ ) − Fφ ( ∇ × F) z = r ∂r r ∂φ

In spherical coordinates, “del cross F”: rˆ r 2sin θ θˆ r sin θ φˆ r ∇ × F = ∂ ∂r ∂ ∂θ ∂ ∂φ Fr rFθ r sin θFφ

1 ∂ ∂  ( sin θFφ ) − Fθ  ∴( ∇ × F) r =  r sin θ  ∂θ ∂φ 

( ∇ × F) θ

1 1 ∂ ∂  =  Fr − ( rFφ )  r  sin θ ∂φ ∂r 

1 ∂ ∂ ( ∇ × F ) φ =  ( rFθ ) − Fr  r  ∂r ∂θ 34

Example Show that

∇f ( x ) × F( x′) = ∇ × [ f ( x ) F( x′) ]

Solution:

[ ∇f ( x ) × F( x′) ] i = εijk ∂f ( x ) Fk ( x′) ∂x j

∂ [ f ( x ) Fk ( x′) ] = ε ijk ∂x j = { ∇ × [ f ( x ) F( x′) ]} i

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Example Show that ∇ × ∇f ( x ) × F( x′) = ∇{ ∇ ⋅ [ f ( x ) F( x′) ]} − ∇ 2f ( x ) F( x′) Solution:

[ ∇ × ∇f ( x ) × F( x′) ] i

∂ = ε ijk ∂x j

  ∂f ( x ) ∂ 2f ( x ) ε klm ∂x Fm ( x′)  = ε kij ε klm ∂x ∂x Fm ( x′)   l j l

∂ 2f ( x ) = ( δil δ jm − δim δ jl ) Fm ( x′) ∂x j ∂xl ∂ 2f ( x ) = F j ( x′) − ∇ 2f ( x ) Fi ( x′) ∂xi ∂x j

Homework: Work through Example 4 @ Page 23 Homework: Work through Example 7 @ Page 35

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Del Identities – Products of derivatives The curl of the gradient of a scalar function is identically zero: ∂ 2f ( ∇ × ∇f ) i = εijk = 0⇒ ∇ × ∇f = 0 ∂x j ∂xk The divergence of the curl of a vector function is 2 identically zero:  ∂  ∂ ∂ ∇ ⋅ ( ∇ × F) = ε ijk Fk  = ε ijk Fk = 0  ∂xi  ∂x j  ∂xi ∂x j ∇ × ∇ × F = ∇( ∇ ⋅ F ) − ∇ 2F  ∂  ∂ ∂2 ( ∇ × ∇ × F ) i = εijk  ε klm Fm  = ε kij ε klm Fm ∂x j  ∂xl ∂x j ∂xl  ∂2 ∂2 = ( δil δ jm − δim δ jl ) Fm = F j − ∇ 2Fi ∂x j ∂xl ∂xi ∂x j

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Integral theorems

Carl Friedrich Gauss April 30th 1777 – February 23rd 1855)

George Gabriel38Stoke (August 13th 1819 – February 1st 1903)

Gauss’s Theorem (divergence theorem) The flux of a vector field F(x) through a surface S with area element dA is the F( x ) ⋅ nˆ of nˆ , where is the surface integral unit normal vector at the point x on S: The flux of a vector quantity outward through a closed surface S is equal to the integral of the divergence of the function in the enclosed volume V,

3 F ⋅ d A = ( ∇ ⋅ F ) ε ∫ dS

∫ F ⋅ nˆ dA = ∫ F ⋅ dA S

S

3 F ⋅ d A = ∇ ⋅ F d x ∫ ∫ S

V

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Stokes’s Theorem The circulation of a vector field F(x) around a loop C with length element dl ˆdl = F ⋅ dl F ⋅ t ∫C ∫C ( x ) ⋅ tˆ is the lineFintegral of tˆ where is the unit tangent vector at the point x on C: The circulation of a vector function around a closed curve C is equal to the flux of vorticity through any surface eˆ j bounded by C, ∫ F ⋅ dl = ∫ ( ∇ × F ) ⋅ dA C

eˆ i Homework: Work through Examples 5, 6 @ Page 27

S

2 F ⋅ dl = ( ∇ × F ) ε k ∫ dP ( ij )

40

The Helmholtz theorem

Hermann von Helmholtz August 31st 1821 – September 8th 1894)

41

Preliminaries A vector function (or field) F(x) that has zero curl: ∇×F = 0

F = −∇ ψ

is called irrotational. A vector function (or field) G(x) that has zero divergence: ∇ ⋅G = 0

G = ∇×A

is called solenoidal. Any vector function (or field) H(x) can be written as the sum of an irrotational function F(x) and a solenoidal function G(x): H = F+G H = −∇ ψ + ∇ × A The functions F and G are not necessarily unique. However, in some cases, if suitable boundary conditions 42 are imposed, then the decomposition is unique.

The Helmholtz theorem Let H(x) be differentiable at all points in space, with curl and divergence: ∇ × H = c( x )

∇ ⋅ H = d ( x)

→∞ If d ( x ) and c(x) approach 0 faster than r –2r as H(x) approaches 0 faster than r –1 , then

, and

H = −∇ ψ + ∇ × A where

d ( x′) ψ( x ) = ∫ d 3 x′ 4π x − x′

c( x′) A( x ) = ∫ d 3 x′ 4π x − x′

The integration region is all of space. The Helmholtz theorem implies that if the divergence and curl of a vector function (field) is known, then the function can be determined uniquely (under the assumptions of43 the theorem).

Green’s function and the Dirac delta function Paul Adrien Maurice Dirac

George Green (July 14th 1793 – May 31st 1841)

44 Dira Paul Adrien Maurice (August 8th 1902 – October 20th 1984)

The Dirac delta function The Dirac delta function δ (x) is a “generalized function” or “Schwartz distribution” with the following defining ∞ property: ∞ ∫ δ( x ) f ( x ) dx = f ( 0) ∫ δ( x ) dx = 1 −∞

−∞

for every continuous function f(x). The Dirac delta function can δ (x) is an extremely be understood as the limit singular function:  = 0 for all x ≠ 0 of a sequence of more and δ( x )  more sharply peaked ≠ 0 when x = 0 functions: Exercise: ∞

∫ δ( x − a ) f ( x ) dx = f ( a ) −∞

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The Dirac delta function (cont’) The 3-dimensional Dirac delta function δ 3(x) is defined analogously:

∫ δ ( x) f ( x) d 3

3

x=∫

∞

∞

∞

∫ ∫ δ( x ) δ( y ) δ( z ) f ( x, y, z ) dxdydz = f ( 0,0,0) = f ( 0)

−∞ −∞ −∞

for every continuous function f (x).

∫ δ ( x) d 3

3

x =1

 = 0 for all x ≠ 0 δ ( x)  ≠ 0 when x = 0 3

Exercise: 3 3 δ ( x − a ) f ( x ) d x = f ( a) ∫

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∇2 The Green’s Function −of − ∇ G ( x − x′) = δ ( x − x′) 2

3

G ( x − x′) =

1 4π x − x′

along with the boundary condition that G approaches 0 at infinity. Proof:  3 1 2 d x Consider the integral∫V − ∇   4π x − x′  in a spherical volume V of any radius around x’.  1  1 1  3 1 2 divergen d x = − ∫ ∇  ⋅ dA − ∫ ∇  4π V  x − x′  4π S  x − x′  ce theorem 1 1 1 1 1 2 = r ˆ ⋅ dA r ˆ = r ˆ ⋅ r dΩrˆ = dΩ = 1 2 2 ∫ ∫ ∫ S S S 4π r 4π r 4π 1 x − x′ r 1 = ≡ 3 = 2 rˆ Here, we recall − ∇ 3 x − x′ x − x′ r r ∇ 2 (1 x − x′ ) = 0 In addition, when |x – x’| ≠ 0, then

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