IV-1

Definitions: f f(x,y,z) f(r)

dr dxi x i = dxj x j = dx x + dyy+ dz z

ds |dr| = [dxidxi]½

[dr/ds] = dx/ds x + dy/dsy+ dz/ds z = [dxi/ds]x i= cosα1 x + cosα2y+ cosα3 z= v = unit vector along dr

Chapter IV: Vector AnalysisIn this chapter we shall be working primarily in the Cartesian system. Unless stated otherwise assume the

system is Cartesian and any transformation, A, to another system is a rotation.

Recall: in Cartesian systems, g = 1 ; Fi = gx ij Fj = δij F

j.

F(r) is a "vector field" and the Fi are contravariant vector components under rotations between Cartesiansystems. Likewise the f(r) is a "scalar field" and is a zero rank tensor under rotations between Cartesian coordinatesystems. The word "field" implies that f and F depend on r and carry information about a physical quantity (say electricfield or force).

Absolute differential, dfdf = f/x dx + f/y dy + f/z dz

= dxi f/xi

= dxi i f= dr· f = dqj 'j (in a general coordinate system, qk)

Intrinsic (absolute) derivative, df/dtdf/dt = [f/x] dx/dt + [f/y] dy/dt + [f/z] dz/dt

= dxi/dt [f/xi] = [f/qk] dqk/dt (in general system)= [dr/dt·] f

Note: x, y and z each functions of one variable, t, ==> use dx/dt (not x/t).

Directional derivative, df/dsdf/ds]v = [dr/ds·] f since s replaces the parameter, t.

= v· f note: v =dr/ds must be given.

IV-2

df dr· f differential

dfdt

drdt

· f derivative

dfds v v · f directional derivative along v

One can apply all of the above formulas to F(r):

dF = dr· F = dr· Fi xi= xi dr· Fi since the xi are constant.

In the general system, the formula is different!

dF/dt = xi [dr/dt ·] Fi

dF/ds]v = xi [ v ·] Fi

Note the advantage of being in a Cartesian system: the xi are constant.

nth order derivativesd²f/ds² = d/ds [df/ds] = d/ds [dr·] f

= [(dr/ds)·]2 f

dnf/dsn = [(dr/ds)·]n f

Similarly,

dnF/dsn = xi [(dr/ds)·]n Fi

Example:Find the directional derivative of f(r) = e²/|a-r| along v = [ x - y]/2.

1. First note: |a-r| = [(a-r)·(a-r)]1/2 =[a² - 2a·r +r²]1/2

2. df/ds]v = [ x - y]/2 · e²/|a-r|

3. evaluate e²/|a-r| (there is an easy way to do this).

First we look at some shortcuts.

IV-3

Learn to recognize and write down immediately:

1. xixi = r·r = r2

2. x i i =

3. iFi = ·F

4. xk/x = δk

= qk/q (general system)

5. δii = 3

6. f/qj qj/xi = f/xi

7. C x D = εijkx iCjDk (Cartesian system!)

8. W·(C x D) = εijk WiCjDk (Cartesian system!)

9. r/xi = i [r ·r]1/2 = i [xjxj]1/2

= ½ [xjxj]-1/2 [xjixj + xjixj]

= ½ [1/r] [xjδij + xj

i δjkxk]= ½ [1/r] [xi + xj δjk δi

k]= ½ [1/r] [xi + xj δji ]= ½ [1/r] [xi + xi ]= xi/r

10. ·r = 3

11. dxidxi = dr·dr = ds2

Exercise: /xi r2 = _________________________________ (write down answer).

Exercise: /xi ln(r·r) = _____________________________.

Exercise: Find ²r/xixk .

IV-4

Replace /xi with i:1. if(|r|) = if(r) = [df/dr] i r = f'xi/r, where f' df/dr.

2. iF(r) = xk iFk

3. ixk = δik ****

4. i r = i |r| = xi/r

5. i [1/r] = -[1/r²]xi /r = -xi/r3

6. i rn = nrn-1 ir = n rn-1 xi /r = n rn-2 xi

7. i lnr = [1/r]i r = [1/r] xi /r = xi /r²

8. ir = xk i xk = xk δik = x i

9. i (r·a) = i (xkak) = ak δik = ai

10. i (r-a) = xk i (xk-ak) = xk δik = x i

11. i (r-a)·b = i (xk- ak)bk = bk δik = bi

12. i sin[(r-a)·r] = cos[(r-a)·r] i (xk-ak)xk = cos[(r-a)·r] [xk i(xk-ak) + (xk-ak)iδkjxj]= cos[(r-a)·r] [xk δi

k + (xk-ak)δkj δij] = cos[(r-a)·r] [xi + (xi-ai)]

13. i|r-a| = i [(r-a)·(r-a)]1/2 = ½[(r-a)·(r-a)]-1/2 i (xk-ak)(xk-ak)

= ½[1/|r-a|][(xk-ak)i(xk-ak) + (xk-ak)i δkj (xj-aj) ]= ½[1/|r-a|][(xk-ak)δi

k + (xk-ak)δkj i(xj-aj) ]= ½[1/|r-a|][(xk-ak)δi

k + (xk-ak)δkj δij]

= ½[1/|r-a|][(xi-ai) + (xk-ak)δki]= ½[1/|r-a|][(xi-ai) + (xi-ai)]= (xi-ai)/|r-a|

14. i [1/|r-a|] = [-1/|r-a|²] i |r-a| = -(xi-ai)/|r-a|3

15. i ln|r-a| = [1/|r-a|] (xi-ai)/|r-a| = (xi-ai)/|r-a|²

16. i|r-a|n = n|r-a|n-1 (xi-ai)/|r-a| = n(xi-ai)|r-a|n-2

17. i(rxa)·b = iεjkxjakb = εjkδij akb = εikakb = (a x b)i

IV-5

Finally, manipulate with : Some general expressions for acting on functions:1. f(r) = x i

i f(r)2. f(|r|) = x i

i f(r) = x i[df/dr]ir = df/dr r3. F(r) = x i

i xk Fk(r) = xk Fk(r)-----------------------------------------------------------------------------------------------------------------------These occur often and you should learn to do them quickly:

4. r = x i i r = x i xi /r = r/r = r5. rn = nrn-1

r = nrn-1 r6. ln(r)= [1/r]r = r/r7. |r| = r

8. r = x i i xk xk = x ixk δik = x ix i note sum, product

9. · r = i xi = δii = 3

10. |r-a|= x i i |r-a| = x i (xi-ai)/|r-a|= (r-a)/|r-a|

11.[1/|r-a|]= -[1/|r-a|²]|r-a| = -[1/|r-a|²](r-a)/|r-a| = -(r-a)/|r-a|3[1/r] = -r/r3

12.|r-a|n = n|r-a|n-1|r-a| = n|r-a|n-2(r-a)

13.(r·a) = x ii xkak = x iδi

kak = x iai = a

14.[1/(r·a)] = [-1/(r·a)²](r·a) = -a/(r·a)²

15. exp[i(r·a)]= exp[i(r·a)]i(r·a) = ia exp(ir·a)

16. sin(r·a)= cos(r·a)(r·a) = a cos(r·a)----------------------------------------------------------------------------------------------------------------------These are more complicated: but they are important results.

17. ²exp[i(r·a)]= · exp[i(r·a)]= · ia exp[i(r·a)] = ia· ia exp[i(r·a)] = -a²exp[i(r·a)]

18. ²r = ·r = · r/r = [1/r]· r + r·[1/r] = 3/r + r·[-1/r²]r= 3/r + r·[-1/r²]r/r= 3/r + r²[-1/r²]/r= 3/r + [-1/r] = 2/r if r 0

19. ²[1/r] = · [1/r] = ·[-r/r3] = -[1/r3]·r - r·[1/r3]= -[1/r3] 3 - r·[-3/r4] r= - 3/r3 + r·[3/r4] r/r = 0 if r 0.

Note that ²[1/r] [-1/r²]²r !

IV-6

Back to the Example on page IV-2:Find the directional derivative of f(r) = e²/|a-r| along v = [x - y]/2.df/ds]v = [x - y]/2 · e²/|a-r|

= [x - y]/2 · [-e²(r-a)/|a-r|3]= [-e²/2](x - y)·(r-a)/|a-r|3= [-e²/2](x - y)·(r-a)/|a-r|3

Exercises:1. d/ds)v r =

2. d/ds)v [1/r] =

3. d/ds)v r =

4. d/ds)v r·r =

5. d/ds)v r·a =

6. d/ds)v [r x a] =

IV-7

Example: Find the directional derivative of f(r) = e² cos[r·a/a²]/|r-a| along v = [x + y]/2.

df/ds)v = [x + y]/2· e² cos[r·a/a²]/|r-a|= e²[x + y]/2· -[sin[r·a/a²]/|r-a|] r·a/a² + cos[r·a/a²][1/|r-a|] = e²[x + y]/2· -[sin[r·a/a²]/|r-a|] a/a² + cos[r·a/a²][-(r-a)/|r-a|3] = -[e²/(2|r-a|)] [x + y]·[a/a²]sin[r·a/a²] + [x + y]·[(r-a)/|r-a|2] cos[r·a/a²]

check: are the units correct?a. assume r and a have length units.b. has units "like" 1/r; [x + y]/2 has no units.c. f has units "like" 1/r, so df/ds must have units "like" 1/r². Both terms satisfy this simple test.

Exercises:

1. Find df/ds)v for f(r)= ln[|r-a|/r]; leave answer in terms of v .

2. d/ds)v rln(r/a) =

IV-8

3. d/ds)v [1/r] =

4. d²/ds²)v (r·r) =

5. Find df if dr is given as 10-12 a/a and f(r) = | 105r+ a|3. Estimate df at r·a = πa².

IV-9

FORMULAS FOR OPERATIONS INVOLVING These expressions are assumed to be in the Cartesian system. f, g, F and G are functions of r.Square brackets [ ] ==> operators act ONLY on functions within the bracket; parenthesis () ==>operators act on all functions which follow. But note that sometimes these expressions are "operatorexpressions" intended to operate on whatever function follows the expression.

1. (f + g) = f + g

2. ·(F + G) = ·F + ·G

3. x (F + G) = x F + x G

4. (fg) = [f]g + fg (when possible, maintain order of functions)

5. ·(fF) = f(·F) + F·f (assuming that F and f commute)

6. · (F x G) = [ x F]·G - F· ( x G)

7. G·(fF) = F(G·f) + f(G·)F

8. x (fF) = f(xF) + (f)xF

9. x (F x G) = (G·)F - (F·)G + F(·G) - G(·F)

10. (F·G) = (F·)G + (G·)F + Fx(xG) + Gx(xF)

11. (G·)F = ½ x(FxG) + (F·G) - F(·G) + G(·F) - Fx(xG) - Gx(xF)

12. · f = ²f

13. (·) F = xk ²Fi

14. (·F) = ²F + x ( x F)

15. x ( x F) = (·F) - ²F

16. x (f) = 0 (unless followed by a function)

17. · ( x F) = 0 (unless followed by a function)

IV-10

When f or F are functions of (r-r'), denoted by f(r-r') and F(r-r') then:

f(r-r') = -'f(r-r') F(r-r') = -'F(r-r')

where ' xk (/x'k).

Note that if f is a function of (r-r') every time r appears in the function, it must be followed by (-r').When used in the integrand of an integral over r' this trick often allows one to do the integration.

Derivations of some of the formulas on page IV-9

5. ·(fF) = x i · [i f xkFk]= x i ·xk [i f Fk]= δi

k [fi Fk + Fkif]

= [fi Fi + Fiif] (assuming that Fi and f commute)

= f(·F) + F·f

16. x (f) = 0 (unless followed by a function) = εijk x ijkf (Cartesian system!)= -εikj x ikjf since the partial derivatives commute= 0

Exercise: Derive the final result for 6.:

· (F x G) =

IV-11

f(r%Δr) ' f(r) % dfds

/000 rΔr % d 2f

ds 2/000 r

(Δr)2

2!%

d 3fds 3

/000 r(Δr)3

3!%...

'[ 1 % (Δr ·L) % (Δr ·L)2

2!% ... | f(r)

' e Δr ·L f(r)

similarly, F(r%Δr) ' e Δr·LF(r)

if all derivatives of f and Fk exist andif the series converges to the function indicated

TAYLOR SERIES EXPANSION

Δr / |Δr|Note that [d²f/ds²]|Δr|² = (v ·L)²f |Δr|² = (|Δr|v ·L)²f = (Δr·L)²f where v is by definition a unit vectoralong Δr. In using this formalism you must carefully determine at which point, r, you want toevaluate the function and what will be chosen for Δr. All derivatives must be taken before theevaluation dnf/dsn|r is made. Often it is less confusing to define the point at which the function is tobe evaluated as ro and the shift as Δro so that one finds f(ro+Δro) and the general formula in the boxdoesn't lead to confusion about the variable r and the special point of evaluation, ro.

Note also what a mess the Taylor series expansion is if one writes this as explicit functions of x,y,z,M/Mx, M/My, etc.

(Δr·L)² = (Δx M/Mx + Δy M/My + Δz M/Mz)²= Δx²(M²/Mx²) + 2ΔxΔy(M²/MxMy) + 2ΔxΔz(M²/MxMz) +

2ΔyΔz(M²/MyMz) + Δy2(M²/My²) + Δz2(M²/Mz²)

To indicate that the terms remaining are of order of |Δr|3 you can add the expression:

+O(|Δr|3) to your answer.

Warning: Usually when you expand a function about a point you intend to use the expansion onlyin region where |Δr| is "small". In this case you should use just enough terms to guarantee that theremainder is negligible (or cancels out in the limit as |Δr| --> 0).

IV-12

Operators which generate translations in functionsThe three dimensional Taylor series expansion provides a compact expression for "space

translations", "rotations" and "time translations" operators which act on functions. Recall that indealing with rotations, we considered only the R(ψ,θ,φ) matrix and its effect on the componentsof the position vector, r. Now we can talk about the operator which acts on functions of x,y,andz.

f(x + ∆x) = e∆x /x f(x)= [1 + ∆x(/x) + ∆x²(/x)²/2! + ...] f(x)

This works for any function:

h(w + ∆w) = e∆w /w h(w)

Example: time translations in quantum mechanics.

H ψ(r,t) = ih/(/t)ψ(r,t), where H -h/²²/(2m) + V(r)

Using -iH/h/ = /t as the "generator" of time translations:

ψ(r,t + ∆t) = e∆t/t ψ(r,t)

= e-i∆tH/h/ ψ(r,t)

If the wave functions are eigenfunctions of H (they usually are) this expression is very convenient.

Example: space translations in quantum mechanics.

ψ(r+∆r,t) = e∆r· ψ(r,t)

= ei∆r·P/h/ ψ(r,t) where P [h//i]

The momentum operator, P [h//i], "generates" the space translations.

IV-13

Example: rotations about the z axis by φ.

ψ(r,θ,φ+∆φ) = e∆φ /φ ψ(r,θ,φ)

= ei∆φ Lz/h/ ψ(r,θ,φ) where Lz [h//i]/φ

The n projection of the angular momentum operator, n· L n·(r x P) = n· (r x [h//i]) is the

generator of rotations about the n axis. When n = z, this operator is Lz [h//i]/φ. The latteroperator is the "generator" of rotations about the z axis.

Similarly, one can generate the rotations φ+∆φ and θ+∆θ:

ψ(r,θ+∆θ,φ+∆φ) = ei[∆φLz+∆θLx']/h/ ψ(r,θ,φ) where x 'is in rotated frame.

Examples using the Taylor series:

1. Evaluate sin[(r+ε)·a] to first order in |ε| = ε << r.

f(r+∆r) = [1 + (ε·)] sin[r·a] + O(ε²)

= sin(r·a) + ε·a cos(r·a) + O(ε²)

*Note that these units are the same as for f.

2. Evaluate rln(r·a) at r = ro + ε to first order in ε.

F(r+∆r) = [1 + (ε·)] r ln(r·a)|ro + O(ε²)

= roln(ro·a) + ε·xkk [rln(r·a)]|ro + O(ε²)

= roln(ro·a) + ε·xk[ xk ln(r·a) + r[1/r·a]ak]|ro + O(ε²)

= roln(ro·a) + εk [xk ln(r·a) + r[1/(r·a)]ak]|ro + O(ε²)

= roln(ro·a) + ε ln(ro·a) + ro ε·a [1/(ro·a)] + O(ε²)

= (ro+ε)ln(ro·a) + ro[ε·a /(ro·a)]+ O(ε²)

Exercise: Estimate (rxa)exp[sin(π(k·r)/k)] at r = 10.03 k/k. Note: k |k|.

To show that

z r p z r i ∇ i∂∂

See Chapter II page 11:

z rcos − sin

r i ∇ 1r2 sin

u1 u2 u3

u1 r u2 r u3 r∇r ∇ ∇

u1 g11u1 ru2 g22u2 r2 g22 r u3 g33u3 r2 sin2 g33 r sin

z r i ∇i rcos− sin 1

r2 sin

r r r sinr 0 0∇r ∇ ∇

− i 1r2

r r r sinr 0 0∇r ∇ ∇

i 1r2 r r∇

i∂∂

IV-14

Source position vector, r', and observation point, rUsually in physics as well as in engineering problems, one needs to identify two position vectors, r'and r in the samecoordinate system with basis vectors, x, y and z. The r vector is generally identified as the observation point andthe r' is assigned to the field source point; the source points are often the source points are integrated over to determinethe effect of an extended source.

The distance between the source point and the observation point is |r - r'|.

r = xi /xi and r' xj /x'j

Note that the basis vectors are the same.

If the function on which r (or r') acts is only a function of

r - r', then one can convert from r to r' or vice versa.

That is,

r f(r - r') = xi /xi f(r - r')= xi /xi f( xn[xn -x'n] )= xi /[xi-x'i] f( xn[xn -x'n] ) since -x'n always follows xn,= - xi /[x'i-xi] f( xn[xn -x'n] )= - xi /x'i f( xn[xn -x'n] )

= - r' f(r - r')Similarly,

r F(r - r') = - r' F(r - r').

Note that the minus sign comes from the conversion of the derivative and the "inner most" functional form, r-r', so that,

r f(|r - r'|) = - r' f(|r - r'|) and

r F(|r - r'|) = - r' F(|r - r'|), also.

Other variations on this "trick" for converting from one gradient to another are:

r f(|r + r'|) = r' f(|r + r'|)

r F(|r - r'|) = r' F(|r + r'|) and

r f(r + ar') = [1/a] r' f(r + ar')

r F(r + ar') = [1/a] r' F(r + ar').

IV-15

Soleniodal and Irrotational vectors

A solenoidal vector field, F(r), is one for which ·F = 0.

examples:

·(rxa) = 0 where a is a constant vector.

·B = 0 where B is a magnetic field.

·(rxp) = 0 where p is linear momentum (not an operator) and rxp = angular momentum.

Solenoidal fields have no sources or sinks and appear to "curve" in on themselves like a "solenoid" !

An irrotational vector field, F(r), is one for which x F = 0.

examples:

x r = εijk ixjxk = εijk δijxk = 0

x rf(|r|) = εijk ixjfxk = εijk [fixjxk + xjifxk] = 0 + [df/dr]rxr/r = 0

x f(r) = 0

x d/dt[mr] = md/dt[xr] = 0 where m = mass and mdr/dt = p and p is not an operator.

Irrotational fields can have sources and sinks.

IV-16

L x L = αL ==> [Lj,Lk] = αεjkL

Expressions involving the momentum, p1. x (r x p) = m x (r x dr/dt)

= m εijk xij (r x dr/dt)· xk

= m εijk xij (εnm xxn[dxm/dt] )· xk

= m εijk xij εknmxn[dxm/dt]

= m εijk εknm xij xn[dxm/dt]

= m [δinδj

m - δimδj

n] xij xn[dxm/dt] = m [ xij xi[dxj/dt] - xij xj[dxi/dt] ]

= m xi[ [dxj/dt]δji + xi[d/dt(·r)] - [dxi/dt]δj

j -xj[d/dt(δji)]

= m[dr/dt] + mr[d/dt(3)] - m[dr/dt]3- m·0= -2m[dr/dt] = -2p 0

x L = -2p

2. Now suppose that p [/i] , an operator: L x L 0!

L x L = εijk xiLj(rxp)k

= εijk xiLj εkmxpm

= εijk εkm xiLj xpm

= [δiδj

m - δimδj

] xiLj xpm

= xiLj [ xipj - xjpi]

= xiεjstxspt [ xipj - xjpi] now replace pj with [/i]j

= (/i)² xiεjst xst [ xij - xji]

= (/i)² xiεjst xs[ t xij - t xji]; recall txi = δtn nxi = δtn δn

i = δti

= (/i)² xiεjst xs[ δti

j + xitj- δtji - xjt i]

= (/i)²[ xtεjst xsj + rεjstx

stj - xsεjstδtj - εjstxsxjt ]

= (/i)²[- xtεsjt xsj + 0 - 0 - 0]

= (/i)²[- (r x )] = -i /(-1)(r x [/i])] = i (r x [/i])] = i (r x p)]

L x L = i L

So, when L is an operator, L x L 0! Another way to write this is εijk xiLjLk = i xiLi and,

xi[ εijk LjLk -iLi ] = 0 and since xi 0,

εijk LjLk = iLi ; multiply by εinm and sum over i

εijkεinmLjLk = iLiεinmLnLm - LmLn = iLiεinm note: n and m are not summed over.

[Ln,Lm] = iLiεinm

IV-17

Integrations Using Vector Notation

I. Defining Surfaces, Curves and Points:Surface: a surface is defined by f(x,y,z) = 0.

examples: r - 4 = 0 (sphere) x²+y²+z² - 16 = 0 (sphere)θ - π/4 = 0 (cone with angle π/4, vertex at origin)x - y = 0 (plane)x² + y² - a² = 0 (cylinder)x² +2y² -9 = 0 (ellipsoidal cylinder)z - 3x²= 0 (paraboloid of revolution) φ -π/2 = 0 (the yz half-plane with y > 0)

note: when one variable is "missing" it can take on any value; this helps visualize the surface.

Curve: a curve is defined by the intersection of two surfaces:f1(x,y,z) = 0f2(x,y,z) = 0

examples: r-4 = 0 and z = 2 (a circle of radius 2 %3 in z = 2 plane)

φ - π/4 = 0r -2 = 0 (a half-circle of radius 2 with x = y > 0 )

(note that tanφ = tan(π/4) = x/y = 1, so x-y = 0)

Point: a point is specified by the intersection of three surfaces:f1(x,y,z) = 0f2(x,y,z) = 0f3(x,y,z) = 0

examples: x² + y² + z² -4 = 0z -2 = 0x-y = 0 ( x = 0; y = 0, z = 2)

exercise: φ - π/4 = 0r -2 = 0θ - π/2 = 0

note: θ = π/4 ==> cos θ = z/r = 1//2 or 2z²-r² = 0

II. Line Integrals: C is a curve in 3-dimensional space

a) mC f(r) dr =xmC f(x,y,z)dx + ymC f(x,y,z)dy + z mC f(x,y,z)dz = xk mC f(r) dxk

b) mC W(r)·dr = mC W1 dx + mC W2 dy + mC W3 dz = mC Wk dxk

c) mC W x dr = xmC [W2dz - W3dy] + ymC [W3dx - W1dz] + z mC [W1dy - W2dx] = εijk x i mC Wj dxk (cartesian systems)

In each integral f(r) or W is restricted to values of x,y,and z which lie on the curve, C. the dx, dy and dz mustalso lie on the curve.

IV-18

ns = f / |f| is normal to the surface at r.

Examples of SIMPLE line integrals:

1. C dr = 0(where C is defined by r -a = 0 and z = 0; in fact for any closed curve)(You can do this the long way; but think about the integral as being a sum of vectors, dr.)

2. Cr·dr = 0 This is done by noting that r is perpendicular to dr. (for same curve as in 1)

3. Cr x dr = zC [xdy -ydx] (for same curve as in 1 and 2)

= z02π [a sinθcosφ d(a sinθsinφ) - a sinθsinφ d(a sinθcosφ)

= z 02π[a²(cos²φ + sin² φ)dφ] = z 2πa²

exercise:

4. C [1/r]·dr =(where C is the same as in the above examples)

Claim: f(r) is perpendicular to the surface f(r) = 0(An interesting way to "show" this is to use the Taylor series expansion.)

1. Assume f(r) 0 and that r + ∆r is a point on the surface so that f(r) = f(r + ∆r) = 0.

2. Assume the following converges for |∆r| "small" and ∆r/|∆r| --> δr as |∆r| -- > 0:

f(r + ∆r) = f(r) + ∆r· f(r) + (∆r·)² f(r)/2! + ...0 = 0 + ∆r· f(r) + (∆r·)² f(r)/2! + ...

divide by |∆r|: and take limit as |∆r| --> 0:

limit [∆r· f(r)/|∆r|]|∆r|-->0 = - (∆r·)² f(r)/(2!|∆r|) - (∆r·)3 f(r)/(3!|∆r| ) - ...--> 0and δr· f = 0

3. f 0 and δr is an arbitrary tangent vector to surface at r, so f must be perpendicular to δr.

IV-19

dA = g dq1dq2u3

dV = g dq1dq2dq3

III. Surface Integrals:1. First consider the simple surface q3(x,y,z) -constant = 0.

dq1 dr·u1u1 = dq1u1 = vector tangent to q2(x,y,z) = constantdq2 dr·u2u2 = dq2u2 = vector tangent to q1(x,y,z) = constant;

(both are tangent to the surface)

dA = dq1 x dq2 See page III-8 for volume element:= dq1u1 x dq2u2

= g εijk dqidqjuk δi1δj2 dV = gdq1dq2dq3

= g dq1dq2 u3

A Method for Calculating Surface Integrals:

∆A = ∆a ∆b where ∆xi = ∆a cosγ and ∆b = ∆xj

∆Aij ∆xi∆xj = ∆a ∆b cosγ = ∆A cos γ

∆A = ∆xi∆xj/|cosγ| = ∆xi∆xj/| nij· ns|

IV-20

dA = nsdxidxj/|ns·nij|

Note that the ns n = f/|f| is the normal to the surface at the point r and nij is normal to the xixj plane.

The surface area of the entire surface is approximated by summing over N small elements ∆Am:

= limitN ->,∆xi-->0 [m=1 to m = N ∆xi∆xj/| nij· ns|

S dA

= Sij dxidxj/| nij· n|

Sij the projection of S onto the xi xj plane

That is, one integrates over the PROJECTION OF S ONTO the xixj plane!

TO DETERMINE THE PROJECTION:

1. First determine the best "single valued projection" by examining the surface, S defined by f(x,y,z) = 0.

2. Determine the projection by solving simultaneously the f(x,y,z) = 0 and the boundary "surface" equations.The boundaries are usually determined by auxiliary equations, b1(x,y,z) = 0, etc.

3. Sometimes it is easier to take a sum of projections, one on the xy plane, one on the xz plane, etc.

example: find the projection of the surface r = 3, z > 2 onto the xy plane.

x² + y² + z² = 9z = 2

x² + y² + 4 = 9 ===> x² + y² = 5 is the projection on z = 0

Other integrals which can be done this way (providing f(x,y,z) and S are well behaved):

a) S dA = Sij f dxidxj/|f· nij|(Note that here dA has a direction! -- each dA is along n)

b) S g(r)dA = Sij g(r) f dxidxj/|f· nij|

c) S W(r)·dA = Sij W(r)·f dxidxj/|f· nij|

d) S W(r) x dA = Sij W(r) x f dxidxj/|f· nij|

IV-21

Example: Evaluate the integral S r·dA if S is the surface x² + 4y² - z = 0 for which z 1.

1. f = 2x x + 8y y - z

2. S r·dA = Sxy [x x + y y + z z]·f dx dy /|f· z| note: we let nij = z

3. = Sxy [2x² + 8y² -z] dx dy /|[2x x + 8y y - z]· z|

= Sxy [2x² + 8y² -z] dx dy /| - z· z|

= Sx Sy [2x² + 8y² -z] dx dy

4. Now put in the value of z (the variable not integrated over) from f(x,y,z) = 0.

S r·dA = Sx Sy [2x² + 8y² -(x² + 4y²)] dx dy

= Sx Sy [x² + 4y²] dx dy

5. Determine the projection onto the xy plane and the limits on the integrals over x and y. We choose to do they integral first.

projection: solve for the intersection of x² + 4y² - z = 0 and z = 1

x² + 4y² = 1 ==> -1 x 1 and -y1 y y1where y1 [(1-x²)/4]1/2

6. S r·dA = -1 to +1 dx -y1 to y1 [x² + 4y²] dy

= -1 to +1 dx [x²y + 4y3/3]|y1-y1

= -1 to +1 [x²(1-x²) + (4·2/3) (1-x²)3/2/23] dx

= -1 to +1 (1-x²)[ 2x²+ 1]/3 dx

= (2/3)[ -x(1-x²)3/2/4 + x(1-x²)/8 + (1/8)sin-1x] + (1/6)x

(1-x²) + (1/6)sin-1x |-1 to +1

= (2/3)·(1/8)·[π/2 - (-π/2)] + (1/6)·[π/2 -(-π/2)]

= π/4

IV-22

Special Problem: intersecting cylindersFind the surface area of x² + z² = a² which is enclosed by x² + y² = a².

f(x,y,z) = x² + z² - a² = 0

f = 2x x + 2z zmake the projection on the xy plane

intersection: x² + y² = a² sincethis is the cross section of the vertical cylinder!

Area = 2· Sxy dxdy/| z·f/|f| | where z 0;

= 2· Sxy |f|dxdy/| z·(2x x+2z z)|

|f| = 2[x²+z²]1/2 = 2a

Area = 2· Sxy 2a dxdy/|2z| note z must be on f(x,y,z)!

= 2a· -a to a dx -(a²-x²) to (a²-x) [a² -x²]-1/2 dy

= 2a· 20 to a [a² -x²]-1/2 dx 20 to (a²-x) dy

= 8a· 0 to a [a² -x²]-1/2 dx (a²-x)

= 8a· 0 to a dx

= 8a²

IV-23

Solid Angle, Ω

Ω = ± S [r·ns]/r3 dA

Solid Angles:

The solid angle, Ω projection of a surface, S, onto a sphere with radius = 1. The viewing point is at the center of thesphere. If S is inside the unit sphere, use picture on above right.

Ω = projection of S onto unit sphere sinθ dθ dφ dΩ

In order to derive the expression for Ω, we first consider the following with F r/r3 and the volume, V, enclosed byS, its projection onto the unit sphere, and the radius vectors tracing out the boundary of S (see figures above and notedirection of outer normal, r, to that part of the enclosing surface on the unit sphere ):

V ·F dV = S+Ω+sides F·dA

V [(-3/r4)r· r+3/r3] dV = S r· ns/r3 dA + Ω(r/r3)·( r)r²dΩ + sides(r/r3)·( nsides ) dA

0 = S r· ns/r3 dA + Ω + 0

Ω = ± S r· ns/r3 dA ( take sign which makes Ω 0)

Note: if the surface, S, is outside (inside) the unit sphere take +1 (-1) so that Ω is positive.

If the viewing point is not at the origin, but at ro:

Ω = ± S [(r-ro)/|r-ro|3]·dA

where, r = point on the surface S (defined by f(x,y,z)=0) and dA ns dA.

IV-24

Example: Find the solid angle subtended by r = a; z 3a/4 when viewed from the origin.(Cartesian coordinates make the integral difficult to do, but illustrate the method.)

f = 2xi xi = 2r; |f| = 2r; ns = f/|f|.

Ω = Sxy r · dA /r3

= Sxy[(r · ns)/r3] dxdy/| z· ns |

= Sxy[(r · f)/r3] dxdy/| z·f |

= Sxy[(r · 2r)/r3] dxdy/| z·2r |

= Sxy[(r · 2r)/r3] dxdy/| 2z |

= Sxy[1/(rz)] dxdy note that r = a on the surface.

= Sxy 1/[a(a² - x² - y²)1/2] dxdy keep z on the surface f = 0.

To do this integral, we can change the (xy only!) to cylindrical coordinates, ρ and φ':

= Sρφ' 1/[a(a² - x² - y²)1/2] ρdρdφ'

= 0 to 2π dφ' 0 to ρmax 1/[a(a² - ρ²)1/2] ½d[ρ²]; let t (a² - ρ²)1/2/a;

= 2π -1 to -tmax d[-t] where tmax = z/r = 3/4;

= 2π·[ -3/4 + 1]

= π/2

Exercise: Find the solid angle subtended by the surface |r| = a when viewed from the point, ro = 20a y. Hint: to determine that part of the surface visible from ro you need to find the intersection of the line ofsight from ro to the "edge" of the surface. Think about this as "something like" a view of the moon.

IV-25

V F(r) dV = x i V Fi dV

V F·G dV = V Fi(r)Gi(r) dV

V x F dV = εijk x i V j FkdV

IV. Volume IntegralsRecall that in general: dV = gdq1dq2dq3

whenever possible(and especially if the integration is complicated) try using Cartesian coordinates.

IV-26

Divergence TheoremIf F(r) is differentiable with continuous iF in the volume, V

enclosed by surface, S, and if the integrals on S exist,

V ·F dV = S F·dA

where dA = nsdA

Stokes theorem

S [ x F]·dA = C F·dr

where the "direction" of the curve, C, is determined fromthe right hand rule, with the thumb along the outer normal to V on S.

Note that ns is the outer normal to S.

Stokes Theorem

If F(r) is single valued, differentiable and has continuous partial derivatives in V;V is a finite region containing S;S is simply connected (no "holes");C is regular (is composed of a finite number of arcs; every point of each arc must be continuous and

such that x(t), y(t), z(t) which determine C have unique partial continuous derivatives notall = 0)

(that is, C is piecewise smooth).then

Other integral theorems

A. If f(r) and g(r) are twice differentiable with continuous second partials on V and f and g are differentiable withcontinuous partial derivatives on S:

V[ f²g - g²f] dV = S [fg - gf]· ns dA

Derivation:1. · [fg] = f·g + f²g

· [gf] = f·g + g²f; subtract and integrate over the closed volume, V:

IV-27

Let F(r) ρ(r)v(r) where ρ is the density of a quantity (mass, charge, etc.) and v is the velocity of the"flow" of the quantity at r. If there are no sources or sinks of the quantity, the net change of thequantity within an infinitesimal dV is related to the flux of F across the surface dS enclosing dV:

ρ/t dV = - F(r)·dA or,

ρ/t dV + F(r)·dA = 0Using the divergence theorem:

ρ/t dV + ·F(r) dV = 0.

If dV is constant, one obtains the continuity equation:

ρ/t + ·F(r) = 0.

ρ/t + ·(ρv) = 0.

2. V [ · [fg] - · [gf] ]dV = V [f²g - g²f] dV; use divergence th. on left side;

3. S [fg - gf]· ns dA = V [f²g - g²f] dV....

B. With the same conditions as in A:

V ·[fg] dV = S fg· ns dA

V ²f dV = S f· ns dA

V[ f·f + f²f]dV = S ff· ns dA

Theorems related to Stokes' Theorem:

S[ n x ] x F dA = -C F x dr

S[ n x f] dA = C f(r) dr .

Continuity Equation

Note that dρ/dt = ρ/t + [dr/dt]· ρ, and ·[vρ] = v·ρ + ρ·v

so that another form of the continuity equation is:

dρ/dt + ρ·v = 0

IV-28

Theorem: If (1) x W(r) = 0 inside V;

(2) W(r) satisfies the conditions of Stokes theorem inside V;

(3) V is finite and simply-connected (no holes);

thenW(r) = f(r) inside V.

proof:

1. Consider an arbitrary, simply connected surface, S, inside V bounded by a closed, piecewise-smooth curve, C. Then for all such S and C:

S [ x W] · dA = C W·dr = 0, since x W = 0.

2. Now break C into two parts, C1 and C2, as shown: a and b are any two points on C.

0 = C1 W·dr + C2 W·dr

= a to b (along C1) W·dr + b to a (along C2) W·dr

and

a to b (along C1) W·dr = - b to a (along C2) W·dr

= a to b (along -C2) W·dr

Thus for any a and b in V and for any C1 and C2 between a and b a to b (along C1) W·dr = a to b (along -C2) W·dr

3. But a line integral is dependent only on its endpoints iff the integrand is a perfect differential. Thus,

a to b (along any C) W·dr = a to b df = f(b) - f(a).

4. df = (dr·)f (see page IV-2)

= f· dr where acts only on f.

5. Thus, W·dr = f· dr and W(r) = f(r).

........

IV-29

Theorem I:

If (a) ·F(r) = ρ(r) in volume, V, enclosed by S;

(b) x F(r) = J(r);

(c) F·ns = α(rs) is specified on S;

(d) V is simply-connected and S is closed;

then F(r) is uniquely specified in V.

proof:

1. Assume G(r) and F(r) both satisfy conditions (a), (b), and (c), and define

W(r) F(r) - G(r).

2. Then W(r) satisfies:

(a') ·W(r) = ρ(r) - ρ(r) = 0;(b') x W(r) = J(r) - J(r) = 0;(c') W(r)· ns = α(r) - α(r) = 0;

3. Using the theorem on page IV-28:

x W(r) = 0 ===> W(r) = Φ(r); and·W(r) = 0 ===> ²Φ(r) = 0.

4. Now we consider the following integral:

S Φ(r) W(r)· ns dA = V ·[Φ(r)W(r)] dV

0 = V ·[Φ(r)Φ(r)] dV

= V [Φ(r)·Φ(r) + Φ(r) ²Φ(r)] dV

= V |Φ(r)|² dV + V [Φ(r) ²Φ(r)] dV

= V |Φ(r)|² dV + V [Φ(r) · 0 ] dV

= V |Φ(r)|² dV

= V |W(r)|² dV.

Thus, |W(r)|² = 0 for all r in V and W(r) = 0. Hence F(r) = G(r) for all r.(See page I-2 under the properties of a unitary vector space: |x|² = (x,x) = 0 iff x = 0. )

...

Dirac Delta Function

In one dimension, *(x-xo) is defined to be such that:+

*0 if xo is not in [a,b].ma to b f(x) *(x-xo)dx / *½f(xo) if xo = a or b;*f(xo) if xo , (a,b)..

The Dirac Delta Function, *(x-xo) IV-30

Properties of *(x-xo): (you should know those marked with *)

*1. *(x-xo) = 0 if x … xo

*2. m-4 to +4 *(x)dx = 1

3. *(ax) = *(x)/|a|*4. *(-x) = *(x) 5. *(x²-a²) = [*(x-a) + *(x+a)]/(2a); a $ 0

6. m-4 to +4 *(x-a)*(x-b)dx = *(a-b)

*7. *(g(x)) = 3i *(x-xoi)/|dg/dx|x=xoi

where g(xoi) = 0 and dg/dx exists at and in a region around xoi.

8. f(x)*(x-a) = f(a)*(x-a)

9. *(x) is a "symbolic" function which provides convenient notation for many mathematical expressions. Often one "uses" *(x) in expressions which are not integrated over. However, it is understood that eventually theseexpressions will be integrated over so that the definition of * (box above) applies.

10. No ordinary function having exactly the properties of *(x) exists. However, one can approximate *(x) by thelimit of a sequence of (non-unique) functions, *n(x). Some examples of *n(x) which work are given below. In all these cases,

m-4 to +4 *n(x)dx = 1 œ n and limitn --> 4 m-4 to +4 *n(x-xo)f(x)dx = f(xo).

+

*0 for x < -1/(2n)(a) *n(x) / *n for -1/(2n) # x # 1/(2n)

*0 for x > 1/(2n).

(b) *n(x) / n//B exp[-n²x²]

(c) *n(x) / (n//B)A 1/(1 + n²x²)

(d) *n(x) / sin(nx)/Bx = [1/(2B)]m-n to n exp(ixt)dt

fxgxdg fxx − x0dx fx0

fxgxdg/dxdx fxx − x0dx fx0

gxdg/dx x − x0

gx x − x0/dg/dx

IV-31

important expressions involving δ(x-xo)

13. δ(x-xo) = [1/2π] - to + eik(x-xo)dk

14. δ(r-ro) δ(x-xo)δ(y-yo)δ(z-zo)= [1/2π]3all k space exp[ik·(r-ro)]dkxdkydkz

15. δ(g(x)) = n δ(x-βn)/|dg/dx|x = βn where g(βn) = 0.

16. δ(r-ro) = δ(q1-q1o)δ(q2-q2

o)δ(q3-q3o)/g in general system.

+ (-1)r drf/dxr|xo if xo ε (a,b)11. a to b f(x) dr/dxr δ(x-xo) dx = *½[(-1)rdrf/dxr|xo if xo = a or b

.0 otherwisef(x) is arbitrary, continuous function at x = xo

12. a to b xr dr/dxr δ(x-xo)dx = a to b (-1)rr! δ(x-xo) dx where xo ε (a,b).

Dirac Delta Function in 3 Dimensions: δ (r - ro) δ(x-xo)δ(y-yo)δ(z-zo)

17. δ(k - ko) = [1/2π]3 - to +- to +- to + exp[ir·(k - ko)] dxdydz

18. δ(r - ro) = δ(q1-q1o)δ(q2-q2

o)δ(q3-q3o)/g

derivation: δ(r - ro)f(r) d3x = f(ro) = δ(q1-q1o)δ(q2-q2

o)δ(q3-q3o)/g· f(r(qi)) gdq1dq2dq3

19. δ(r - ro)spherical coordinates = δ(r-ro)δ(φ-φo)δ(θ-θo)/(r²sinθ)

20. ½[δ(x-a)+δ(x+a)] = [1/π] 0 to cos(ka) cos(kx) dk Exercises: Evaluate the following integrals.

a) -1 to 5 δ(2x-π) exp[ sin3(x-π) ] dx

b) all spaceδ(r·r-a²)δ(cosθ-1/2)δ(sinφ-½)exp[ik·r]d3x

IV-32

Helmholtz TheoremIf

(a) ·F(r) = ρ(r) everywhere for finite r;(b) x F(r) = J(r) "(c) limitr-->ρ(r) = 0;(d) limitr-->|J(r)| = 0;

thenF(r) = -Φ(r) + x A(r)

where Φ and A are determined from ρ and J as shown below.

CLAIM: ² 1|rr'|

4π δ(rr')

proof:

1. Define Φ and A as follows:

Φ(r) = [1/4π]V= all space ρ(r')/|r - r'| d3x' + Φo(r), where ²Φo(r) = 0;

A(r) = [1/4π]V= all space J(r')/|r - r'| d3x' + Ao(r), where x(xAo(r)) = 0;

2. Let F(r) = -Φ(r) + xA(r); we shall show that this F satisfies the conditions (a) and (b) if (c) and (d) hold:

·F = ·[-Φ(r) + xA(r)] = -²Φ(r) x F = x[-Φ(r) + xA(r)] = x[xA(r)].

3. ·F = -²Φ(r) = -[1/4π]V= all space ρ(r')²[1/|r - r'|] d3x' + 0

4. But ²[1/|r - r'|] = ·[1/|r - r'|] = ·[-(r - r')/|r - r'|3]=-[3/|r - r'|3 + [-3/|r - r'|4](r - r')·(r - r')/|r - r'| ]=-[3/|r - r'|3 + [-3/|r - r'|3] ]= 0 if r r'

5 What happens if r = r'? We shall see that the expression --> , but with a crucial additional property!

Derivation: (a) Consider the following integral, V ·[1/r] d3x = V ²[1/r] d3x

where V = all space and V' = all space except a sphere of radius δ centered on the origin and a "funnel" extending from r = 0 to r = . See figure below.

We shall show that this integral = -4π if r = 0 is in V.Note: V contains r = 0 and V' does not.

IV-33

Use the divergence theorem:

V'-·[1/r] d3x = S1-[1/r]· r dA1 + cone sides -[1/r]· ns dA2 + small sphere less circle -[1/r]·- r dA3 0 = S1 [r/r3]· r dA1 + cone sides [r/r3]· ns dA2 + small sphere less circle [r/r3]·- r dA3

0 = S1 [r/r3]· r dA1 + 0 + small sphere less circle [r/r3]·- r dA3

(b) Now take limit as the angle of the cone, ε, ---> 0:

S1 ---> complete sphere at r ---> S;

S3 ---> complete sphere of radius δ, centered at the origin.

0 = S [r/r3]· r dA1 + - small sphere with radius δ [ r/δ²]· r δ²sinθdθdφ

0 = S-[1/r]· r dA + - small sphere with radius δ [ r/δ²]· r δ²dΩ

S [1/r]· r dA1 = - 4π independent of δ!

V·[1/r] d3x = - 4π if r = 0 is in V.= 0 otherwise.

This is just the property of the Dirac delta function! All this applies to ²[1/|r - r'|] also, so

volume containing r'=r ² [1/|r - r'|] d3x = -4π δ(r - r') d3x.

6. Thus, ·F = -[1/4π]V= all space ρ(r')²[1/|r - r'|] d3x' = ρ(r) and F satisfies (a).

7. Next, we examine x F = x[xA(r)].

= (·A) - ²A = (·A) + J(r) (note that ² acts on the A integral in the same way it operated on the Φ integral.

IV-34

8. We now show that (·A) = 0:

(·A) = [1/4π]V= all space [·J(r')/|r - r'|] d3x'

= [1/4π]V= all space [ [J(r')·[1/|r - r'|] + [1/|r - r'|]·J(r') ] d3x'; but ·J(r') =0.

= [1/4π]V= all space [J(r')·[1/|r - r'|] d3x'; let G [1/|r - r'|].

[J(r')·G(r,r')] = (J·)G + (G·)J(r') + Jx(xG) + Gx[xJ(r')] (see page IV-9)= (J·)G + 0 + Jx(xG) + 0 since J is not a function of r.= (J·)G + 0 + Jx( x [1/|r - r'|] )

= (J·)G + 0 + Jx( εijk xijk [1/|r - r'|] ) = (J·)G + 0 + Jx( 0 ).

Then,(·A) = [1/4π]V= all space[J(r')·][1/|r - r'|] d3x'; NOW we can change to -'.

(·A) = [1/4π]V= all space[J(r')·']'[1/|r - r'|] d3x'.

= xk [1/4π]V= all space[J(r')·'] G'k] d3x', where G' '[1/|r - r'|]

= xk [1/4π]V= all space '·[ J(r') G'k] - G'k '·J(r') d3x'.

= xk [1/4π]V= all space '·[ J(r') G'k ] - G'k '·['xF'(r')] d3x' from (b). [J = equal to the curl of some vector, F'!]

= xk [1/4π]V= all space '·[ J(r') G'k ] - G'k · 0 ] d3x'

= xk [1/4π]V= all space '·[ J(r') 'k [1/|r - r'|]] d3x' use divergence theorem:

= xk [1/4π]S with r ---> ns·J(r') 'k [1/|r - r'|] r'²dΩ'

= limitr'---> [1/4π]Sr' ns·J(r') ' [1/|r - r'|] r'²dΩ'

|(·A)| limitr'---> [1/4π]Sr' |J(r')| | (r - r')/|r - r'|3 | ] r'²dΩ'

limitr'---> [1/4π]|J(r')|max Sr' [1/|r - r'|²] r'²dΩ'

limitr'---> [1/4π]|J(r')|max Sr' [1/|r'|²] r'²dΩ'

limitr'---> [1/4π]|J(r')|max 4π note that |J(r')|max is the maximum of J at fixed r'.

limitr'---> |J(r')|max = 0 by condition (d), since even the max of |J| -->0.

Thus (·A) = 0 and

x F = (·A) - ²A = J(r)

The condition (c) is needed to ensure that the integral form for Φ(r) is finite. Note that the integrand ---> ρ(r')/r'; this is not enough to cancel the divergence of d3x' =r'²dΩ'as r' --->. So one must have ρ(r') --0 as r' -->.

IV-35

Use of Helmholtz theorem and δ(r-r')Examples:

1. Set up the charge density for a point charge, Q, at r = ro.

ρ(r) = C δ(r - ro)

Q = V ρ(r')d3x' = V C δ(r' - ro)d3x' = C

ρ(r) = Q δ(r - ro).

2. Find the charge density of a uniform shell of charge, with radius a and total charge, Q.

ρ(r) = C δ(r - a) note change to primed variables!Q = V ρ(r')d3x' = V C δ(r' - a) r'²dr'dΩ'

Q = C a²dΩ' = 4πa²C

ρ(r) = [Q/(4πa²)] δ(r - a)

3. Find the charge density of a uniform ring of charge in the xy plane, with radius aand total charge, Q.

ρ(r) = C δ(r - a)δ(θ - π/2)

Q = V ρ(r')d3x' = V C δ(r' - a)δ(θ'-π/2) r'²dr'sinθ'dθ'dφ'

Q = C a²dφ' = 2πa²C

ρ(r) = [Q/(2πa²)] δ(r - a)δ(θ - π/2)

4. Find the charge density of a uniform line of charge along z from -L to +L with total charge, Q.

ρ(r) = C δ(x)δ(y) if -L z L

= 0 otherwise.

Q = V ρ(r')d3x' = z' = -L to z' = L C δ(x')δ(y') dx'dy'dz'

Q = z' = -1 to z' = 1 C dz' = C2L

ρ(r) = [Q/(2L)] δ(x)δ(y) if -L z L= 0 otherwise.

5. Find the charge density of a uniform disc of charge (in the xy plane) with radius a, and with total charge, Q.ρ(r) = C δ(θ - π/2) if r a

= 0 for r > a.

Q = V ρ(r')d3x' = r'=0 to r' = a C δ(θ'-π/2) r'²dr'sinθ'dθ'dφ'Q = C [a3/3]2πρ(r) = Q[3/(2πa3)] δ(θ - π/2) if r a

= 0 for r > a.

IV-36

Exercise:

Find the charge density, ρ(r), of a uniform disc of charge (in the xy plane) with a hole of radius a and with a totalcharge, Q. The outer radius of the charge distribution is b.

IV-37

Examples: Assuming that L·E = ρ(r)/εo and L x E = 0, find E(r) for the charge distributions on the previous page.

1. E(r) = - LΦ(r) = - L [1/4π]mmmV= all space ρ(r')/|r - r'| d3x'

= - L [1/4π]mmmV= all space Qδ(r' - ro)/|r - r'| d3x' = - L [1/4πεo] Q/|r - ro| = [Q/4πεo] [(r - ro)/|r - ro|²]

2. Assume r is outside the sphere.

E(r) = - LΦ(r) = - L [1/4π]mmmV= all space[[Q/(4πa²εo)]δ(r' - a) /|r - r'|] d3x'

= - L [1/4π]m0 to πm0 to 2π [[Q/(4πa²εo)]/|r - ar '|] a²sinθ'dθ'dφ' This is hard to do!+----------------------------------------------------------------------------------------------------- --------- ,

*note: [1/|r - ar '|] = [r² -2rar ·r ' + a²]-1/2 is an intractable function of θ' and φ'!* *

* = [r² -2ra sinθcosφsinθ'cosφ' + sinθsinφsinθ'sinφ' + cosθcosθ'+ a²]-1/2

.----------------------------------------------------------------------------------------------------- --------- -

One needs a neat trick (which we will learn in the next chapter) or to find some simplifying symmetry to exploit.For now, we use symmetry. We calculate the integral for the case where r is on the z axis. By symmetry, then, wecan apply the result for all observation points, r.

Φ(rz ) = [1/4π]m0 to π m0 to 2π [[Q/(4πa²εo)][1/|r - ar '|] a²sinθ'dθ'dφ'

= [1/4π]m0 to π m0 to 2π [[Q/(4πa²εo)][1 /|rz -ar '|]] a²sinθ'dθ'dφ'

= [1/4π]m0 to π m0 to 2π [[Q/(4πεo)][1 /[r² -2arz ·r '+ a²]1/2]] d(-cosθ')dφ'

= [1/4π]m1 to -1 m0 to 2π [[Q/(4πεo)][1/[r² -2arcosθ'+ a²]1/2]] d(-cosθ')dφ'

= [1/4π] m1 to -1 2π [[Q/(4πεo)][1 /[r² -2arcosθ'+ a²]1/2]] d(-cosθ')

Let ω / cosθ' Φ(rz ) = [Q/(8πεo)]m-1 to +1 [1 /[r² -2arω+ a²]1/2] dω

= [Q/(8πεo)]mω = -1 to +1 d [r² -2arω+ a²]1/2/(-ar)

= [Q/(8πεo)][-1/(ar)] [ (r-a) - (r+a) ]

= [Q/(4πεo)][1/r] (Gauss' theorem verifies this result.)

E(r) = - LΦ(r) = [Q/(4πεo)]r/r3 -- for r > a.

This only gives the right answer for any r because the charge distribution has spherical symmetry! In the next chapter we shall learn how to do this for more complicated charge distributions.

The remaining cases on page IV-35 have similar difficulties.

IV-38

An easier exercise:Using the Helmholtz theorem find F for ρ(r) = 0 and

J(r) = µo all k space exp[i(k + a)·r -ik·b] [1/[µo x (r x a)]²]d3k.

Hint: Examine all the symbols carefully. µo, a, and b are constant vectors.