This preview shows page 1. Sign up to view the full content.
Unformatted text preview: IV1 Chapter IV: Vector Analysis
In 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. Definitions: f / f(x,y,z) / f(r) 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 Cartesian systems. Likewise the f(r) is a "scalar field" and is a zero rank tensor under rotations between Cartesian coordinate systems. The word "field" implies that f and F depend on r and carry information about a physical quantity (say electric field or force). 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^ = [dxi/ds]x i ^ ^ z ^ = cosα1 x + cosα2y + cosα3^ ^ ^ z = v = unit vector along dr ^ Absolute differential, df
df = = = =
Mf/Mx dx + Mf/My dy + Mf/Mz dz dxi Mf/Mxi dxi Li f dr·L f = dqj L'j (in a general coordinate system, qk) Intrinsic (absolute) derivative, df/dt
df/dt = = = [Mf/Mx] dx/dt + [Mf/My] dy/dt + [Mf/Mz] dz/dt = [Mf/Mqk] dqk/dt (in general system) dxi/dt [Mf/Mxi] [dr/dt·L] f Note: x, y and z each functions of one variable, t, ==> use dx/dt (not Mx/Mt). Directional derivative, df/ds
df/ds]v = = [dr/ds·L] f since s replaces the parameter, t. ^·L f v note: ^ =dr/ds must be given. v IV2 One can apply all of the above formulas to F(r): dF = = = dr·L F ^i dr·L Fi x = dr·L Fi^i x since the ^i are constant. x In the general system, the formula is different! dF/dt ^i [dr/dt ·L] Fi x ^i [^ ·L] Fi xv dF/ds]v = Note the advantage of being in a Cartesian system: the ^i are constant. x df ' dr·L f differential df dr ' ·L f derivative dt dt df ' v ·L f directional derivative along v ds /0 v nth order derivatives
d²f/ds² dnf/dsn Similarly, dnF/dsn = ^i [(dr/ds)·L]n Fi x = = = d/ds [df/ds] [(dr/ds)·L]2 f [(dr/ds)·L]n f = d/ds [dr·L] f Example:
Find the directional derivative of f(r) = e²/ar along ^ = [^  ^]//2. v xy 1. First note: ar 2. df/ds]v = = [(ar)·(ar)]1/2 =[a²  2a·r +r²]1/2 [^  ^]//2 · L e²/ar xy 3. evaluate L e²/ar (there is an easy way to do this). First we look at some shortcuts. IV3 Learn to recognize and write down immediately:
1. xixi 2. x i Li ^ 3. LiFi 4. Mxk/MxR 5. = r·r =L = L·F = δkR =3 Mf/Mxi
i j k = r2 = Mqk/MqR (general system) δii 6. Mf/Mqj Mqj/Mxi = 7. C x D 8. W·(C x D) 9. Mr/Mxi = = = = = = = = εijkx C D ^
= (Cartesian system!)
i j k εijk W C D (Cartesian system!) Li [r ·r]1/2 = Li [xjxj]1/2 j 1/2 ½ [xjx ] [xjLixj + xjLixj] ½ [1/r] [xjδij + xjLi δjkxk] ½ [1/r] [xi + xj δjk δik] ½ [1/r] [xi + xj δji ] ½ [1/r] [xi + xi ] xi/r =3 10. L·r 11. dxidxi = dr·dr = ds2 Exercise: M/Mxi r2 = _________________________________ (write down answer). Exercise: M/Mxi ln(r·r) = _____________________________. Exercise: Find M²r/MxiMxk . IV4 Replace M/Mxi with Li:
1. Lif(r) = Lif(r) ^ 2. LiF(r) = x k LiFk 3. Lixk = 4. Li r 5. Li [1/r] 6. Li rn = 7. Li lnr 8. Lir 9. Li (r·a) 10. Li (ra) = [df/dr] Li r = f'xi/r, where f'/ df/dr. δik ****
= = nrn1 Lir = = = = Li r [1/r²]xi /r = [1/r]Li r ^ x k Li xk Li (xkak) ^ x k Li (xkak) Li (xk ak)bk = = xi/r xi/r3 = n rn2 xi n rn1 xi /r = = = = =
k [1/r] xi /r = xi /r² ^ x k δik = ak δi
k k ^ xi ai ^ xi bi = ^ x k δi = bk δi =
k 11. Li (ra)·b = 12. Li sin[(ra)·r] = cos[(ra)·r] [xk δi + (xkak)δkj δij] = cos[(ra)·r] [xi + (xiai)] Li [(ra)·(ra)]1/2 = ½[(ra)·(ra)]1/2 Li (xkak)(xkak) ½[1/ra][(xkak)Li(xkak) + (xkak)Li δkj (xjaj) ] ½[1/ra][(xkak)δik + (xkak)δkj Li(xjaj) ] ½[1/ra][(xkak)δik + (xkak)δkj δij] ½[1/ra][(xiai) + (xkak)δki] ½[1/ra][(xiai) + (xiai)] (xiai)/ra [1/ra²] Li ra = (xiai)/ra3 = cos[(ra)·r] Li (xkak)xk = cos[(ra)·r] [xk Li(xkak) + (xkak)Liδkjxj] 13. Lira = = = = = = = 14. Li [1/ra] = 15. Li lnra 16. Liran 17. Li(rxa)·b = [1/ra] (xiai)/ra = (xiai)/ra² = = nran1 (xiai)/ra = n(xiai)ran2 LiεjkRxjakbR = εjkRδij akbR = εikRakbR = (a x b)i IV5 Finally, manipulate with L:
Some general expressions for L acting on functions: 1. Lf(r) = ^ x iLi f(r) = ^ x i[df/dr]Lir = df/dr Lr 2. Lf(r) = ^ x iLi f(r) i k 3. LF(r) = ^^ x Li x Fk(r) = ^ x k L Fk(r) These occur often and you should learn to do them quickly: 4. Lr = 5. Lrn = 6. Lln(r)= 7. Lr = 8.L r = 9. L· r = 10. Lra= ^ x i Li r = nrn1Lr [1/r]Lr ^ r ^^ x i Li x k xk Li xi ^ x i Li ra = ^ x i xi /r = r/r n1 = nr ^ r = ^/r r = = = ^^ x ix k δik δii = = ^ r = 3 ^^ x ix i note sum, product ^ x i (xiai)/ra= [1/ra²](ra)/ra = (ra)/ra (ra)/ra3 11.L[1/ra]= [1/ra²]Lra L[1/r] = r/r3 12.Lran = nran1Lra = 13.L(r·a) = ^ x iLi xkak nran2(ra) = = ^ x iδikak a/(r·a)² ia exp(ir·a) = ^ x iai = a 14.L[1/(r·a)] = [1/(r·a)²]L(r·a) 15.L exp[i(r·a)]= exp[i(r·a)]Li(r·a) = 16.L sin(r·a)= cos(r·a)L(r·a) = a cos(r·a) These are more complicated: but they are important results. 17. L²exp[i(r·a)]= L·L exp[i(r·a)]= L· ia exp[i(r·a)] = ia· ia exp[i(r·a)] = a²exp[i(r·a)] 18. L²r = L·Lr = L· r/r = [1/r]L· r + r·L[1/r] = = = = 3/r + r·[1/r²]Lr 3/r + r·[1/r²]r/r 3/r + r²[1/r²]/r 3/r + [1/r] = 2/r if r
0 19. L²[1/r] = L· L[1/r] = L·[r/r3] = Note that L²[1/r]
[1/r²]L²r ! [1/r3]L·r  r·L[1/r3] = [1/r3] 3  r·[3/r4] Lr =  3/r3 + r·[3/r4] r/r = 0 if r
0. IV6 Back to the Example on page IV2: Find the directional derivative of f(r) = e²/ar along v = [x  y ]//2. ^ ^^ df/ds]v = [x  y ]//2 · L e²/ar ^^ = [x  y ]//2 · [e²(ra)/ar3] ^^ = [e²//2](x  y )·(ra)/ar3 ^^ = [e²//2](x  y )·(ra)/ar3 ^^ 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] ^ = IV7 Example: Find the directional derivative of f(r) = e² cos[r·a/a²]/ra along v = [x + y ]//2. ^ ^^ df/ds)v = ^ = = = [x + y ]//2·L e² cos[r·a/a²]/ra ^^ e²[x + y ]//2· { [sin[r·a/a²]/ra]L r·a/a² + cos[r·a/a²]L[1/ra] } ^^ e²[x + y ]//2· { [sin[r·a/a²]/ra] a/a² + cos[r·a/a²][(ra)/ra3] } ^^ [e²/(/2ra)] { [x + y ]·[a/a²]sin[r·a/a²] + [x + y ]·[(ra)/ra2] cos[r·a/a²]} ^^ ^^ check: are the units correct? a. assume r and a have length units. b. L 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[ra/r]; leave answer in terms of v . ^ ^ 2. d/ds)v rln(r/a) = ^ IV8 3. d/ds)v L[1/r] = ^ 4. d²/ds²)v L (r·r) = ^ 5. Find df if dr is given as 1012 a/a and f(r) =  105r+ a3. Estimate df at r·a = πa². IV9 FORMULAS FOR OPERATIONS INVOLVING L
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 "operator expressions" intended to operate on whatever function follows the expression. 1. L(f + g) = Lf + Lg 2. L·(F + G) = L·F + L·G 3. L x (F + G) = L x F + L x G 4. L(fg) = [Lf]g + fLg 5. L·(fF) = f(L·F) + F·Lf (when possible, maintain order of functions) (assuming that F and f commute) 6. L · (F x G) = [L x F]·G  F· (L x G) 7. G·L(fF) = F(G·Lf) + f(G·L)F 8. L x (fF) = f(LxF) + (Lf)xF 9. L x (F x G) = (G·L)F  (F·L)G + F(L·G)  G(L·F) 10. L(F·G) = (F·L)G + (G·L)F + Fx(LxG) + Gx(LxF) 11. (G·L)F = ½{ Lx(FxG) + L(F·G)  F(L·G) + G(L·F)  Fx(LxG)  Gx(LxF)} 12. L·L f = L²f 13. (L·L) F = x k L²Fi ^ 14. L(L·F) = L²F + L x (L x F) 15. L x (L x F) = L(L·F)  L²F 16. L x (Lf) = 0 17. L · (L x F) = 0 (unless followed by a function) (unless followed by a function) IV10 When f or F are functions of (rr'), denoted by f(rr') and F(rr') then: L f(rr') = L'f(rr') L F(rr') = L'F(rr') where L' / x k (M/Mx'k). ^ Note that if f is a function of (rr') 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 IV9 5. L·(fF) = = = = = ^ ^ x i · [Li f x kFk] i ^^ x ·x k [Li f Fk] δik [fLi Fk + FkLif] [fLi Fi + FiLif] (assuming that Fi and f commute) f(L·F) + F·Lf 16. L x (Lf) =0 (unless followed by a function) = εijk x iLjLkf ^ (Cartesian system!) ^ = εikj x iLkLjf since the partial derivatives commute =0 Exercise: Derive the final result for 6.: L · (F x G) = IV11 TAYLOR SERIES EXPANSION
f( r %Δr ) ' f( r ) % d 2 f ( Δr ) 2 d 3 f ( Δr ) 3 /0 Δr % / / % %... 2 00 3 00 00 r 2! 3! ds 0 r ds 0 r 2 ( Δr · L ) % ...  f(r) '[ 1 % ( Δr · L ) % 2! ' e Δ r ·L f( r ) df ds similarly, F(r%Δr) ' e Δr·L F(r) if all derivatives of f and Fk exist and if the series converges to the function indicated Δr / Δr
Note that [d²f/ds²]Δr² = (v ·L)²f Δr² = (Δrv ·L)²f = (Δr·L)²f where v is by definition a unit vector ^ ^ ^ along Δr. In using this formalism you must carefully determine at which point, r, you want to evaluate the function and what will be chosen for Δr. All derivatives must be taken before the evaluation dnf/dsnr is made. Often it is less confusing to define the point at which the function is to be evaluated as ro and the shift as Δro so that one finds f(ro+Δro) and the general formula in the box doesn'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 Δr3 you can add the expression: +O(Δr3) to your answer. Warning: Usually when you expand a function about a point you intend to use the expansion only in region where Δr is "small". In this case you should use just enough terms to guarantee that the remainder is negligible (or cancels out in the limit as Δr > 0). IV12 Operators which generate translations in functions
The three dimensional Taylor series expansion provides a compact expression for "space translations", "rotations" and "time translations" operators which act on functions. Recall that in dealing with rotations, we considered only the R(ψ,θ,φ) matrix and its effect on the components of the position vector, r. Now we can talk about the operator which acts on functions of x,y,and z. f(x + ∆x) = = e∆x M/Mx f(x) [1 + ∆x(M/Mx) + ∆x²(M/Mx)²/2! + ...] f(x) This works for any function: h(w + ∆w) = e∆w M/Mw h(w) Example: time translations in quantum mechanics. H ψ(r,t)
= ih M/Mt)ψ(r,t), where H / h L²/(2m) + V(r) /( /² Using iH/h = M/Mt as the "generator" of time translations: / ψ(r,t + ∆t) = = e∆tM/Mt ψ(r,t) / ei∆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·L ψ(r,t) / ei∆r·P/h ψ(r,t) where P / [h L //i] The momentum operator, P / [h L, "generates" the space translations. //i] IV13 Example: rotations about the ^ axis by φ. z
ψ(r,θ,φ+∆φ) = = e∆φ M/Mφ ψ(r,θ,φ) / ei∆φ Lz/h ψ(r,θ,φ) where Lz / [h M/Mφ //i] The n projection of the angular momentum operator, n· ^ ^ generator of rotations about the n axis. When n ^ ^ operator is the "generator" of rotations about the ^ axis. z L / n·(r x P) = n· (r x [h L) is the ^ ^ //i] = ^, this operator is L / [h M/Mφ. The latter z //i]
z 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·L)] 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·L)] r ln(r·a)ro + O(ε²) roln(ro·a) + ε·x kLk [rln(r·a)]ro + ^
k O(ε²)
k ro ^^ roln(ro·a) + ε·x k[ x k ln(r·a) + r[1/r·a]ak]ro +
o o o o k O(ε²) r ln(r ·a) + ε [x ln(r·a) + r[1/(r·a)]a ] + O(ε²) ^ r ln(r ·a) + ε ln(r ·a) + r ε·a [1/(r ·a)] + O(ε²) (r +ε)ln(r ·a) + r [ε·a /(r ·a)]+ O(ε²)
o o o o o o o Exercise: Estimate (rxa)exp[sin(π(k·r)/k)] at r = 10.03 k/k. Note: k / k. IV14 Source position vector, r', and observation point, r
Usually in physics as well as in engineering problems, one needs to identify two position vectors, r'and r in the same coordinate system with basis vectors, ^, ^ and ^. The r vector is generally identified as the observation point and xy z the r' is assigned to the field source point; the source points are often the source points are integrated over to determine the effect of an extended source. The distance between the source point and the observation point is r  r'. L r = ^i M/Mxi x and L r' / ^j M/Mx'j x Note that the basis vectors are the same. If the function on which r  r', then one can convert from Lr to Lr' or vice versa. That is, L r (or Lr') acts is only a function of L r f(r  r') = ^i M/Mxi f(r  r') x = ^i M/Mxi f( ^n[xn x'n] ) x x = ^i M/M[xix'i] f( ^n[xn x'n] ) since x'n always follows xn, x x =  ^i M/M[x'ixi] f( ^n[xn x'n] ) x x =  ^i M/Mx'i f( ^n[xn x'n] ) x x =  Lr' f(r  r') Similarly, L r F(r  r') =  Lr' F(r  r'). Note that the minus sign comes from the conversion of the derivative and the "inner most" functional form, rr', so that, L L L L L L r r f(r  r') F(r  r') =  Lr' f(r  r') and =  Lr' F(r  r'), also. Other variations on this "trick" for converting from one gradient to another are:
r r f(r + r') F(r  r') f(r + ar') F(r + ar') = = L L r' r' f(r + r') F(r + r') and r r = [1/a] Lr' F(r + ar'). = [1/a] Lr' f(r + ar') IV15 Soleniodal and Irrotational vectors
A solenoidal vector field, F(r), is one for which L·F = 0. examples: L·(rxa) = 0 L·B = 0 L·(rxp) = 0 where a is a constant vector. where B is a magnetic field. 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 L x F = 0. examples: Lxr L x rf(r) L x L f(r) = 0 L x d/dt[mr] =ε Lixj^ x ijk = ε Lixjf^ x
ijk
k k δij^ x = = ε [fLixj^ + xjLif^ ] x x
k =ε ijk 0 = 0 + [df/dr]rxr/r = 0 ijk k k = md/dt[Lxr] = 0 where m = mass and mdr/dt = p and p is not an operator. Irrotational fields can have sources and sinks. IV16 Expressions involving the momentum, p
1. L x (r x p) = m ε ^iLj (r x dr/dt)·xk x ^ = m ε ^iLj εknmxn[dxm/dt] x
ijk
i j n m = m L x (r x dr/dt)
ijk ijk ijk = m ε ^iLj (εRnm^Rxn[dxm/dt] )·xk x x ^ εknm ^ L x [dx /dt] x ij i j = m [δ nδ m  δ mδ n] ^ L x [dx x
i j n i i j j =mε LxL = m^i[ [dxj/dt]δj + xi[d/dt(L·r)]  [dxi/dt]δj xj[d/dt(δj )] x = m[dr/dt] + mr[d/dt(3)]  m[dr/dt]3 m·0 = 2m[dr/dt] = 2p
0 = 2p /dt] = m [^iLj x [dx /dt]  ^iLj x [dx /dt] ] x x
i m j i 2. Now suppose that p / [S/i]L , an operator: L x L
0! LxL = = = = = = = = = = = = = = = = εijk^ L (rxp) x ijk ε ^ L εkRmx p x ijk ε εkRm ^ L x p x ij i j [δ Rδ m  δ mδ R ]^ L x p x
i i j j k Rm i j Rm i j Rm ^iLj [ x p  x p ] x ij ji x (S/i)²^iεjst xs[ Lt xiLj  Lt xjLi]; recall Ltxi = δ x (S/i)²^iεjst xs[ δ Lj + xiLtLj δ Li  xjLt Li]
tj ti tj (S/i)²^iεjst xsLt [ xiLj  xjLi] x ^iεjstxspt [ xipj  xjpi] x now replace pj with [S/i]Lj
tn Lnxi = δ tn δni = δti LxL x (S/i)²[^tεsjt xsLj + 0 (S/i)²[ (r x L)] i S/(1)(r x [S/i]L)] i S(r x [S/i]L)] i S(r x p)] iSL x (S/i)²[ ^tεjst xsLj + rεjstxsLtLj  xsεjstδ L  εjstxsxjLt L] 0  0] So, when L is an operator, L x L
0! Another way to write this is ^i[ ε x
ijk εijk ^ L L x
i j k = i S ^iLi x and, LjLk iSL ]
i =0 = = = = iSLi ; iSL εinm
i ε LL εijkεinmL L
j k j ijk and since ^i
0, x multiply by εinm and sum over i note: n and m are not summed over. k LnLm  LmLn
/ [Ln,Lm] iSL εinm
i iSLiεinm LxL = αL ==> [Lj,Lk] = αεRjkLR IV17 Integrations Using Vector Notation I. Defining Surfaces, Curves and Points:
Surface: a surface is defined by f(x,y,z) = 0. examples: r4 =0 (sphere) x²+y²+z²  16 = 0 (sphere) θ  π/4 = 0 (cone with angle π/4, vertex at origin) xy=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 halfplane 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) = 0 f2(x,y,z) = 0 examples: r4 = 0 and z = 2 (a circle of radius 2 %3 in z = 2 plane) φ  π/4 = 0 r 2 = 0 (a halfcircle of radius 2 with x = y > 0 ) (note that tanφ = tan(π/4) = x/y = 1, so xy = 0) Point: a point is specified by the intersection of three surfaces: f1(x,y,z) = 0 f2(x,y,z) = 0 f3(x,y,z) = 0 examples: x² + y² + z² 4 = 0 z 2 = 0 xy = 0 ( x = 0; y = 0, z = 2) φ  π/4 = 0 r 2 = 0 θ  π/2 = 0 exercise: note: θ = π/4 ==> cos θ = z/r = 1//2 or 2z²r² = 0 II. Line Integrals: C is a curve in 3dimensional space ^ x k mC f(r) dxk mC Wk dxk a) mC f(r) dr =x mC f(x,y,z)dx + y mC f(x,y,z)dy + z mC f(x,y,z)dz = ^ ^ ^ b) mC W(r)·dr = mC W1 dx + mC W2 dy + mC W3 dz = c) mC W x dr = x mC [W2dz  W3dy] + y mC [W3dx  W1dz] + z mC [W1dy  W2dx] = ^ ^ ^ (cartesian systems) εijk x ^ i mC Wj dxk 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 must also lie on the curve. IV18 Examples of SIMPLE line integrals: 1. mC 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.) This is done by noting that r is perpendicular to dr. 2. mCr·dr = 0 (for same curve as in 1) = ^mC [xdy ydx] z 3. mCr x dr (for same curve as in 1 and 2) = ^m02π [a sinθcosφ d(a sinθsinφ)  a sinθsinφ d(a sinθcosφ) z = ^ m02π[a²(cos²φ + sin² φ)dφ] z exercise: 4. mC L[1/r]·dr = (where C is the same as in the above examples) = ^ 2πa² z Claim: Lf(r) is perpendicular to the surface f(r) = 0 (An interesting way to "show" this is to use the Taylor series expansion.) 1. Assume Lf(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 > ^r as ∆r  > 0: δ f(r + ∆r) 0 = = f(r) + ∆r·L f(r) + (∆r·L)² f(r)/2! + ... 0 + ∆r·L f(r) + (∆r·L)² f(r)/2! + ... divide by ∆r: and take limit as ∆r > 0: limit [∆r·L f(r)/∆r]∆r>0 =  (∆r·L)² f(r)/(2!∆r)  (∆r·L)3 f(r)/(3!∆r )  ...> 0 and ^r·L f δ =0 δ δ 3. Lf
0 and ^r is an arbitrary tangent vector to surface at r, so Lf must be perpendicular to ^r. ^ ns = Lf / Lf is normal to the surface at r. IV19 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) = constant dq2 / dr·u2u2 = dq2u2 = vector tangent to q1(x,y,z) = constant; (both are tangent to the surface) dA = d q1 x d q2 = dq1u1 x dq2u2 See page III8 for volume element:
i1 j2 = /g εijk dqidqjuk δ = /g dq1dq2 u3 δ dV = /gdq1dq2dq3 dA = /g dq1dq2u3 dV = /g dq1dq2dq3 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/^ij·^s nn IV20 dA = nsdxidxj/ns·nij ^ ^^ Note that the ^s / ^ = Lf/Lf is the normal to the surface at the point r and ^ij is normal to the xixj plane. nn n The surface area of the entire surface is approximated by summing over N small elements ∆Am: nn = limitN >4,∆xi>0 [3m=1 to m = N ∆xi∆xj/^ij·^s
/ mS dA nn = mSij dxidxj/^ij·^ Sij / the projection of S onto the xi xj plane That is, one integrates over the PROJECTION OF S ONTO the xixj plane! Other integrals which can be done this way (providing f(x,y,z) and S are well behaved):
a) mS dA = mSij Lf dxidxj/Lf·nij ^ (Note that here dA has a direction!  each dA is along ^) n b) mS g(r)dA = mSij g(r) Lf dxidxj/Lf·nij ^ ^ c) mS W(r)·dA = mSij W(r)·Lf dxidxj/Lf·nij d) mS W(r) x dA = mSij W(r) x Lf dxidxj/Lf·nij ^ T O 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 = 9 z=2 x² + y² + 4 = 9 ===> x² + y² = 5 is the projection on z = 0 IV21 Example: Evaluate the integral mS r·dA if S is the surface x² + 4y²  z = 0 for which z # 1. 1. Lf = 2x^ + 8y^  ^ x yz 2. mS r·dA 3. = = = =
mSxy [x^ + y^ + z^]·Lf dx dy /Lf·z  note: we let ^ij = ^ x y z ^ nz mSxy [2x² + 8y² z] dx dy /[2x^ + 8y^  ^]·z  x y z^ mSxy [2x² + 8y² z] dx dy /  ^·z  z^ mSx mSy [2x² + 8y² z] dx dy 4. Now put in the value of z (the variable not integrated over) from f(x,y,z) = 0.
mS r·dA = mSx mSy [2x² + 8y² (x² + 4y²)] dx dy = mSx mSy [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 the y integral first. projection: solve for the intersection of x² + 4y²  z = 0 and z = 1 x² + 4y² = 1 ==> 6. mS r·dA = = = = 1 # x # 1 and y1 #y # y1 where y1 / [(1x²)/4]1/2 m1 to +1 dx my1 to y1 [x² + 4y²] dy m1 to +1 dx [x²y + 4y3/3] y1 m1 to +1 [x²/(1x²) + (4·2/3) (1x²)3/2/23] dx m1 to +1 /(1x²)[ 2x²+ 1]/3 dx
y1 = {(2/3)[ x(1x²)3/2/4
/(1x²) + (1/6)sin x }1 to +1
1 + x/(1x²)/8 + (1/8)sin1x] + (1/6)x = (2/3)·(1/8)·[π/2  (π/2)] + (1/6)·[π/2 (π/2)] = π/4 IV22 Special Problem: intersecting cylinders
Find the surface area of x² + z² = a² which is enclosed by x² + y² = a². f(x,y,z) = x² + z²  a² = 0
Lf = 2x^ + 2z^ x z make the projection on the xy plane intersection: x² + y² = a² since this is the cross section of the vertical cylinder! Area z = 2· mmSxy dxdy/^·Lf/Lf  where z $ 0; z x z = 2· mmSxy Lfdxdy/^·(2x^+2z^) Lf = 2[x²+z²]1/2 = 2a Area = 2· mmSxy 2a dxdy/2z note z must be on f(x,y,z)! = 2a· ma to a dx m/(a²x²) to /(a²x) [a² x²]1/2 dy = 2a· 2m0 to a [a² x²]1/2 dx 2m0 to /(a²x) dy = 8a· m0 to a [a² x²]1/2 dx /(a²x) = 8a· m0 to a dx = 8a² IV23 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 the sphere. If S is inside the unit sphere, use picture on above right. Ω = mmprojection of S onto unit sphere sinθ dθ dφ / mm dΩ In order to derive the expression for Ω, we first consider the following with F / r/r3 and the volume, V, enclosed by S, its projection onto the unit sphere, and the radius vectors tracing out the boundary of S (see figures above and note r direction of outer normal, K^, to that part of the enclosing surface on the unit sphere ):
mmmV L·F dV = mmS+Ω+sides F·dA mmmV [(3/r4)r·r +3/r3] dV = mmS r·ns/r3 dA + mmΩ(r/r3)·(K^)r²dΩ + mmsides(r/r3)·(^sides ) dA ^ ^ r n 0 Ω ^ = mmS r·ns/r3 dA + KΩ + 0 ^ = ± mmS r·ns/r3 dA ( take sign which makes Ω $ 0) Solid Angle, Ω Ω = ± mmS [r·ns]/r3 dA ^ 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 = ± mmS [(rro)/rro3]·dA where, r = point on the surface S (defined by f(x,y,z)=0) and dA / ^s dA. n IV24 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.)
Lf = 2xi^i = 2r; x Lf = 2r; ^s = Lf/Lf. n Ω = = = = = = = mmSxy r · dA /r3 mmSxy[(r · ^s)/r3] dxdy/ ^·^s  n zn mmSxy[(r · Lf)/r3] dxdy/ ^·Lf  z mmSxy[(r · 2r)/r3] dxdy/ ^·2r  z mmSxy[(r · 2r)/r3] dxdy/ 2z  mmSxy[1/(rz)] dxdy note that r = a on the surface. keep z on the surface f = 0. mmSxy 1/[a(a²  x²  y²)1/2] dxdy To do this integral, we can change the (xy only!) to cylindrical coordinates, ρ and φ': = = = = =
mmSρφ' 1/[a(a²  x²  y²)1/2] ρdρdφ' let t / (a²  ρ²)1/2/a; m0 to 2π dφ' m0 to ρmax 1/[a(a²  ρ²)1/2] ½d[ρ²]; 2π m1 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 of sight from ro to the "edge" of the surface. Think about this as "something like" a view of the moon. IV25 IV. Volume Integrals
Recall that in general: dV = /gdq1dq2dq3 mmmV F(r) dV = x i mmmV Fi dV ^ mmmV F·G dV = mmmV Fi(r)Gi(r) dV mmmV L x F dV = ε
ijk ^ x i mmmV Lj FkdV whenever possible (and especially if the integration is complicated) try using Cartesian coordinates. IV26 Divergence Theorem
If F(r) is differentiable with continuous LiF in the volume, V enclosed by surface, S, and if the integrals on S exist,
mmmV L·F dV = mmS F·dA where dA = ^sdA n Note that ^s is the outer normal to S. n Stokes Theorem
If F(r) V S C is single valued, differentiable and has continuous partial derivatives in V; is a finite region containing S; is simply connected (no "holes"); 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 not all = 0) (that is, C is piecewise smooth). then Stokes theorem
mmS [L x F]·dA = mC F·dr where the "direction" of the curve, C, is determined from the right hand rule, with the thumb along the outer normal to V on S. 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 with continuous partial derivatives on S:
mmmV[ fL²g  gL²f] dV = mmS [fLg  gLf]·ns dA ^ Derivation: 1. L· [fLg] = Lf·Lg + fL²g L· [gLf] = Lf·Lg + gL²f; subtract and integrate over the closed volume, V: IV27 2. 3. mmmV [ L· [fLg]  L· [gLf] ]dV = mmmV [fL²g  gL²f] dV; mmS [fLg  gLf]·ns dA ^ use divergence th. on left side; = mmmV [fL²g  gL²f] dV. ... B. With the same conditions as in A:
mmmV L·[fLg] dV mmmV L²f dV mmmV[ Lf·Lf + fL²f]dV = mmS fLg·^s dA n = mmS Lf·ns dA ^ = mmS fLf·ns dA ^ Theorems related to Stokes' Theorem:
mS[^ x L] x F dA = mC F x dr n mS[^ x Lf] dA = n mC f(r) dr . Continuity Equation 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 the quantity within an infinitesimal dV is related to the flux of F across the surface dS enclosing dV: Mρ/Mt dV =  F(r)·dA or,
Using the divergence theorem: Mρ/Mt dV + F(r)·dA = 0
Mρ/Mt dV + L·F(r) dV = 0. If dV is constant, one obtains the continuity equation: Mρ/Mt + L·F(r) = 0. Mρ/Mt + L·(ρv) = 0. Note that so that another form of the continuity equation is: dρ/dt = Mρ/Mt + [dr/dt]·L ρ, and L·[vρ] = v·Lρ + ρL·v dρ/dt + ρL·v = 0 IV28 Theorem: If (1) L x W(r) = 0 inside V; (2) W(r) satisfies the conditions of Stokes theorem inside V; (3) V is finite and simplyconnected (no holes); then W(r) = Lf(r) inside V. proof: 1. Consider an arbitrary, simply connected surface, S, inside V bounded by a closed, piecewisesmooth curve, C. Then for all such S and C:
mmS [L x W] · dA = mC W·dr = 0, since L 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 = mC1 W·dr + mC2 W·dr = ma to b (along C1) W·dr + mb to a (along C2) W·dr and
ma to b (along C1) W·dr =  mb to a (along C2) W·dr = ma to b (along C2) W·dr Thus for any a and b in V and for any C1 and C2 between a and b ma to b (along C1) W·dr = ma 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,
ma to b (along any C) W·dr = ma to b df = f(b)  f(a). 4. df = = (dr·L)f (see page IV2)
Lf· dr where L acts only on f. 5. Thus, W·dr = Lf· dr and W(r) = L f(r). ........ IV29 Theorem I: If (a) (b) (c) (d) then L·F(r) = ρ(r) L x F(r) = J(r);
F·ns ^ = α ( rs ) in volume, V, enclosed by S; is specified on S; V is simplyconnected and S is closed; 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) / 2. Then W(r) satisfies: (a') L·W(r) (b') L x W(r) (c') W(r)·ns ^ = ρ(r)  ρ(r) = J(r)  J(r) = α(r)  α(r) = 0; = 0; = 0; F(r)  G(r). 3. Using the theorem on page IV28:
L x W(r) L·W(r) =0 =0 ===> W(r) = L Φ(r); and ===> L²Φ(r) = 0. 4. Now we consider the following integral:
mmS Φ(r) W(r)·ns dA ^ = = = = = = = mmmV L·[Φ(r)W(r)] dV mmmV L·[Φ(r)LΦ(r)] dV mmmV [LΦ(r)·LΦ(r) + Φ(r) L²Φ(r)] dV mmmV LΦ(r)² dV + mmmV [Φ(r) L²Φ(r)] dV mmmV LΦ(r)² dV + mmmV [Φ(r) · 0 ] dV mmmV LΦ(r)² dV mmmV W(r)² dV. 0 Thus, W(r)² = 0 for all r in V and W(r) = 0. Hence F(r) = G(r) for all r. (See page I2 under the properties of a unitary vector space: x² = (x,x) = 0 iff x = 0. ) ... IV30 The Dirac Delta Function, δ(xxo)
Dirac Delta Function In one dimension, δ(xxo) is defined to be such that: ma to b f(x) δ(xxo)dx /
+ *0 if xo is not in [a,b]. *½f(xo) if xo = a or b; *f(xo) if xo ε (a,b). . Properties of δ(xxo): (you should know those marked with *)
*1. δ(xxo) = 0 *2. m4 to +4 δ(x)dx if x
xo =1 3. δ(ax) = δ(x)/a *4. δ(x) = δ(x) 5. δ(x²a²) = [δ(xa) + δ(x+a)]/(2a); a $ 0
+**7. δ(g(x)) = 3i δ(xxoi)/dg/dxx=xoi where g(xoi) = 0 and dg/dx exists at and in a region around xoi. . 6. m4 to +4 δ(xa)δ(xb)dx = δ(ab) *8. f(x)δ(xa) = f(a)δ(xa) 9. δ(x) is a "symbolic" function which provides convenient notation for many mathematical expressions. Often one "uses" δ(x) in expressions which are not integrated over. However, it is understood that eventually these expressions will be integrated over so that the definition of δ (box above) applies. 10. No ordinary function having exactly the properties of δ(x) exists. However, one can approximate δ(x) by the limit of a sequence of (nonunique) functions, δn(x). Some examples of δn(x) which work are given below. In all these cases, m4 to +4 δn(x)dx = 1 n and limitn > 4 m4 to +4 δn(xxo)f(x)dx = f(xo). n.
+ *0 *n *0 . (a) δn(x) / for x < 1/(2n) for 1/(2n) # x # 1/(2n) for x > 1/(2n) (b) δn(x) / n//π exp[n²x²] (c) δn(x) / (n//π)· 1/(1 + n²x²) = (d) δn(x) / sin(nx)/πx [1/(2π)]mn to n exp(ixt)dt IV31 11. ma to b f(x) d /dx δ(xxo) dx =
r r *½[(1) d f/dxrxo if xo = a or b .0 otherwise rr + (1)r drf/dxrxo if xo ε (a,b) f(x) is arbitrary, continuous function at x = xo 12. ma to b xr dr/dxr δ(xxo)dx = ma to b (1)rr! δ(xxo) dx where xo ε (a,b). important expressions involving δ(xxo) 13. δ(xxo) = [1/2π] m4 to +4 eik(xxo)dk 14. δ(rro) / δ(xxo)δ(yyo)δ(zzo)= [1/2π]3mmmall k space exp[ik·(rro)]dkxdkydkz 15. δ(g(x)) = 3n δ(xβn)/dg/dxx = βn where g(βn) = 0. 16. δ(rro) = δ(q1q1o)δ(q2q2o)δ(q3q3o)//g in general system. Dirac Delta Function in 3 Dimensions: δ (r  ro) / δ(xx )δ(yy )δ(zz )
o o o 17. 18. δ(k  ko) = [1/2π]3 m4 to +4m4 to +4m4 to +4 exp[ir·(k  ko)] dxdydz δ(r  ro) = δ(q1q1o)δ(q2q2o)δ(q3q3o)//g derivation: mmmδ(r  ro)f(r) d3x = f(ro) = mmmδ(q1q1o)δ(q2q2o)δ(q3q3o)//g· f(r(qi)) /gdq1dq2dq3 19. 20. δ(r  ro)spherical coordinates = δ(rro)δ(φφo)δ(θθo)/(r²sinθ) ½[δ(xa)+δ(x+a)] = [1/π] m0 to 4 cos(ka) cos(kx) dk Exercises: Evaluate the following integrals. a) m1 to 5 δ(2xπ) exp[ sin3(xπ) ] dx b) mmmall spaceδ(r·ra²)δ(cosθ1//2)δ(sinφ½)exp[ik·r]d3x IV32 Helmholtz Theorem
If (a) L·F(r) = ρ(r) everywhere for finite r; (b) L x F(r) = J(r) " (c) limitr>4ρ(r) = 0; (d) limitr>4J(r) = 0; then F(r) = LΦ(r) + L x A(r) where Φ and A are determined from ρ and J as shown below. proof: 1. Define Φ and A as follows: Φ(r) A(r) = = [1/4π]mmmV= all space ρ(r')/r  r' d3x' + Φo(r), [1/4π]mmmV= all space J(r')/r  r' d3x' + Ao(r), where L²Φo(r) = 0; where Lx(LxAo(r)) = 0; 2. Let F(r) = LΦ(r) + LxA(r); we shall show that this F satisfies the conditions (a) and (b) if (c) and (d) hold:
L·F LxF = = L·[LΦ(r) + LxA(r)] Lx[LΦ(r) + LxA(r)] = = L²Φ(r) Lx[LxA(r)]. 3. L·F = L²Φ(r) = [1/4π]mmmV= all space ρ(r')L²[1/r  r'] d3x' + 0 = L·L[1/r  r'] = L·[(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' 4. But L²[1/r  r'] 5 What happens if r = r'? We shall see that the expression > 4, but with a crucial additional property! CLAIM: L² 1 ' &4π δ(r&r') r&r' Derivation: (a) Consider the following integral, mmmV L·L[1/r] d3x = mmmV L²[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 = 4. 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. IV33 Use the divergence theorem:
mmmV'L·L[1/r] d3x 0 = mmS1L[1/r]·r dA1 + mmcone sides L[1/r]·ns dA2 + mmsmall sphere less circle L[1/r]·^ dA3 ^ ^ r = mmS1 [r/r3]·r dA1 + mmcone sides [r/r3]·ns dA2 + mmsmall sphere less circle [r/r3]·^ dA3 ^ ^ r ^ = mmS1 [r/r3]·r dA1 + 0 + mmsmall sphere less circle [r/r3]·^ dA3 r 0 (b) Now take limit as the angle of the cone, ε, > 0: S1 > complete sphere at r >4 / S4; S3 > complete sphere of radius δ, centered at the origin. 0 0
mmS4 L[1/r]·r dA1 ^ mmmVL·L[1/r] d3x ^ r ^ = mmS4 [r/r3]·r dA1 +  mmsmall sphere with radius δ [^/δ²]·r δ²sinθdθdφ ^ r ^ = mmS4L[1/r]·r dA +  mmsmall sphere with radius δ [^/δ²]·r δ²dΩ = = =  4π independent of δ!  4π if r = 0 is in V. 0 otherwise. This is just the property of the Dirac delta function! All this applies to L²[1/r  r'] also, so
mmmvolume containing r'=r L² [1/r  r'] d3x = 4π mmm δ(r  r') d3x. 6. Thus, L·F = [1/4π]mmmV= all space ρ(r')L²[1/r  r'] d3x' = ρ(r) and F satisfies (a). = Lx[LxA(r)]. = L(L·A)  L²A = L(L·A) + J(r) (note that L² acts on the A integral in the same way it operated on the Φ integral. 7. Next, we examine L x F IV34 8. We now show that L(L·A) = 0:
L(L·A) = [1/4π]mmmV= all space L[L·J(r')/r  r'] d3x' [1/4π]mmmV= all space L[ [J(r')·L[1/r  r'] + [1/r  r']L·J(r') ] d3x'; but L·J(r') =0. [1/4π]mmmV= all space L[J(r')·L[1/r  r'] d3x'; let G / L[1/r  r']. = (J·L)G + (G·L)J(r') + Jx(LxG) + Gx[LxJ(r')] (see page IV9) = (J·L)G + 0 + Jx(LxG) + 0 since J is not a function of r. = (J·L)G + 0 + Jx( L x L[1/r  r'] ) = (J·L)G + = (J·L)G + 0 0 + Jx( ε ^iLjLk [1/r  r'] ) x + Jx( 0 ).
ijk = =
L[J(r')·G(r,r')] Then,
L(L·A) = L(L·A) = [1/4π]mmmV= all space[J(r')·L]L[1/r  r'] d3x'; [1/4π]mmmV= all space[J(r')·L']L'[1/r  r'] d3x'. NOW we can change L to L'. = = = ^k [1/4π]mmm x ^k [1/4π]mmm x
V= all space V= all space [J(r')·L'] G'k] d3x', where G'/ L'[1/r  r']
3 V= all space { L'·[ J(r') G'k]  G'k L'·J(r') } d x'. ^k [1/4π]mmm x ^k [1/4π]mmm x ^k [1/4π]mmm x ^k [1/4π]mm x { L'·[ J(r') G'k ]  G'k L'·[L'xF'(r')] }d3x' from (b). [J = equal to the curl of some vector, F'!] = = = = V= all space { L'·[ J(r') G'k ]  G'k ·
L'·[ J(r') L'k [1/r  r']] d3x' 0 }d3x' V= all space use divergence theorem: S with r > 4 ^s·J(r') L'k [1/r  r'] r'²dΩ' n limitr'>4
# [1/4π]mmSr' ^s·J(r') L' [1/r  r'] r'²dΩ' n L(L·A) limitr'>4 limitr'>4 limitr'>4 limitr'>4 limitr'>4 [1/4π]mmSr' J(r')  (r  r')/r  r'3  ] r'²dΩ' [1/4π]J(r')max mmSr' [1/r  r'²] r'²dΩ' [1/4π]J(r')max mmSr' [1/r'²] r'²dΩ' [1/4π]J(r')max 4π J(r')max = 0 note that J(r')max is the maximum of J at fixed r'. # # # # by condition (d), since even the max of J >0. Thus L(L·A) = 0 and
LxF = L(L·A)  L²A = 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' >4. So one must have ρ(r') 0 as r' >4. IV35 Use of Helmholtz theorem and δ(rr')
Examples: 1. Set up the charge density for a point charge, Q, at r = ro. ρ(r) Q ρ(r) = C δ(r  ro) = mmmV ρ(r')d3x' = mmmV C δ(r'  ro)d3x' = Q δ(r  ro). =C 2. Find the charge density of a uniform shell of charge, with radius a and total charge, Q. ρ(r) Q Q ρ(r) = C δ(r  a) note change to primed variables!
= 4πa²C = mmmV ρ(r')d3x' = mmmV C δ(r'  a) r'²dr'dΩ' = mm C a²dΩ' = [Q/(4πa²)] δ(r  a) 3. Find the charge density of a uniform ring of charge in the xy plane, with radius a and total charge, Q. ρ(r) Q Q ρ(r) = C δ(r  a)δ(θ  π/2) = mmmV ρ(r')d3x' = mmmV C δ(r'  a)δ(θ'π/2) r'²dr'sinθ'dθ'dφ' = m C a²dφ' = 2πa²C = [Q/(2πa²)] δ(r  a)δ(θ  π/2) 4. Find the charge density of a uniform line of charge along ^ from L to +L with total charge, Q. z ρ(r) = C δ(x)δ(y) if L # z # L = Q Q ρ(r) 0 otherwise. = mmmV ρ(r')d3x' = mz' = L to z' = Lmm C δ(x')δ(y') dx'dy'dz' = mz' = 1 to z' = 1 C dz' = C2L = [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 Q ρ(r) = mmmV ρ(r')d3x' = mmmr'=0 to r' = a C δ(θ'π/2) r'²dr'sinθ'dθ'dφ' = C [a3/3]2π = Q[3/(2πa3)] δ(θ  π/2) if r # a =0 for r > a. IV36 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 total charge, Q. The outer radius of the charge distribution is b. IV37 Examples: Assuming that L·E = ρ(r)/ε
page. 1. E(r) =  LΦ(r) o and L x E = 0, find E(r) for the charge distributions on the previous =  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  ro²] 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!
 ,
* +^ = [r² 2rar ·r ' + a²]1/2 is an intractable function of θ' and φ'! ^^ *note: [1/r  ar '] *  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, we can 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 = [r² 2ra{ sinθcosφsinθ'cosφ' + sinθsinφsinθ'sinφ' + cosθcosθ'}+ a²] * . 1/2 [[Q/(4πε )][1 /[r² 2arz ·r '+ a²] ]] d(cosθ')dφ' ^^
o 1/2 = [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)]m1 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)] [ (ra)  (r+a) ] = [Q/(4πεo)][1/r] E(r) =  LΦ(r) = [Q/(4πεo)]r/r3 (Gauss' theorem verifies this result.)  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 IV35 have similar difficulties. IV38 An easier exercise:
Using the Helmholtz theorem find F for ρ(r) = 0 and J(r) = µo mmmall 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. ...
View
Full
Document
This note was uploaded on 12/20/2010 for the course PHYS 402 taught by Professor J. during the Fall '09 term at Missouri S&T.
 Fall '09
 J.
 Work

Click to edit the document details