M402-Chapter4

M402-Chapter4 - IV-1 Chapter IV Vector Analysis In this chapter we shall be working primarily in the Cartesian system Unless stated otherwise

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Unformatted text preview: IV-1 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 IV-2 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/|a-r| along ^ = [^ - ^]//2. v xy 1. First note: |a-r| 2. df/ds]v = = [(a-r)·(a-r)]1/2 =[a² - 2a·r +r²]1/2 [^ - ^]//2 · L e²/|a-r| xy 3. evaluate L 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 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 . IV-4 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 (r-a) = [df/dr] Li r = f'xi/r, where f'/ df/dr. δik **** = = nrn-1 Lir = = = = Li |r| -[1/r²]xi /r = [1/r]Li r ^ x k Li xk Li (xkak) ^ x k Li (xk-ak) Li (xk- ak)bk = = xi/r -xi/r3 = n rn-2 xi n rn-1 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 (r-a)·b = 12. Li sin[(r-a)·r] = cos[(r-a)·r] [xk δi + (xk-ak)δkj δij] = cos[(r-a)·r] [xi + (xi-ai)] Li [(r-a)·(r-a)]1/2 = ½[(r-a)·(r-a)]-1/2 Li (xk-ak)(xk-ak) ½[1/|r-a|][(xk-ak)Li(xk-ak) + (xk-ak)Li δkj (xj-aj) ] ½[1/|r-a|][(xk-ak)δik + (xk-ak)δkj Li(xj-aj) ] ½[1/|r-a|][(xk-ak)δik + (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| [-1/|r-a|²] Li |r-a| = -(xi-ai)/|r-a|3 = cos[(r-a)·r] Li (xk-ak)xk = cos[(r-a)·r] [xk Li(xk-ak) + (xk-ak)Liδkjxj] 13. Li|r-a| = = = = = = = 14. Li [1/|r-a|] = 15. Li ln|r-a| 16. Li|r-a|n 17. Li(rxa)·b = [1/|r-a|] (xi-ai)/|r-a| = (xi-ai)/|r-a|² = = n|r-a|n-1 (xi-ai)/|r-a| = n(xi-ai)|r-a|n-2 LiεjkRxjakbR = εjkRδij akbR = εikRakbR = (a x b)i IV-5 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. L|r| = 8.L r = 9. L· r = 10. L|r-a|= ^ x i Li r = nrn-1Lr [1/r]Lr ^ r ^^ x i Li x k xk Li xi ^ x i Li |r-a| = ^ x i xi /r = r/r n-1 = nr ^ r = ^/r r = = = ^^ x ix k δik δii = = ^ r = 3 ^^ x ix i note sum, product ^ x i (xi-ai)/|r-a|= -[1/|r-a|²](r-a)/|r-a| = (r-a)/|r-a| -(r-a)/|r-a|3 11.L[1/|r-a|]= -[1/|r-a|²]L|r-a| L[1/r] = -r/r3 12.L|r-a|n = n|r-a|n-1L|r-a| = 13.L(r·a) = ^ x iLi xkak n|r-a|n-2(r-a) = = ^ 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. 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 · L 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·L e² cos[r·a/a²]/|r-a| ^^ e²[x + y ]//2· { -[sin[r·a/a²]/|r-a|]L r·a/a² + cos[r·a/a²]L[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. 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[|r-a|/r]; leave answer in terms of v . ^ ^ 2. d/ds)v rln(r/a) = ^ IV-8 3. d/ds)v L[1/r] = ^ 4. d²/ds²)v L (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 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) IV-10 When f or F are functions of (r-r'), denoted by f(r-r') and F(r-r') then: L f(r-r') = -L'f(r-r') L F(r-r') = -L'F(r-r') where L' / x k (M/Mx'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. 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) = IV-11 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]|Δr|² = (v ·L)²f |Δr|² = (|Δr|v ·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/dsn|r 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 |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 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). IV-12 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) / 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·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] IV-13 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|. IV-14 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[xi-x'i] f( ^n[xn -x'n] ) since -x'n always follows xn, x x = - ^i M/M[x'i-xi] 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, r-r', 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') IV-15 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. IV-16 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 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) = 0 f2(x,y,z) = 0 examples: r-4 = 0 and z = 2 (a circle of radius 2 %3 in z = 2 plane) φ - π/4 = 0 r -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) = 0 f2(x,y,z) = 0 f3(x,y,z) = 0 examples: x² + y² + z² -4 = 0 z -2 = 0 x-y = 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 2z²-r² = 0 II. Line Integrals: C is a curve in 3-dimensional 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. IV-18 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. IV-19 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 III-8 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 IV-20 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 IV-21 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 / [(1-x²)/4]1/2 m-1 to +1 dx m-y1 to y1 [x² + 4y²] dy m-1 to +1 dx [x²y + 4y3/3]| -y1 m-1 to +1 [x²/(1-x²) + (4·2/3) (1-x²)3/2/23] dx m-1 to +1 /(1-x²)[ 2x²+ 1]/3 dx y1 = {(2/3)[ -x(1-x²)3/2/4 /(1-x²) + (1/6)sin x }|-1 to +1 -1 + x/(1-x²)/8 + (1/8)sin-1x] + (1/6)x = (2/3)·(1/8)·[π/2 - (-π/2)] + (1/6)·[π/2 -(-π/2)] = π/4 IV-22 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 |Lf|dxdy/|^·(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· m-a to a dx m-/(a²-x²) to /(a²-x) [a² -x²]-1/2 dy = 2a· 2m0 to a [a² -x²]-1/2 dx 2m0 to /(a²-x) dy = 8a· m0 to a [a² -x²]-1/2 dx /(a²-x) = 8a· m0 to a dx = 8a² IV-23 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 [(r-ro)/|r-ro|3]·dA where, r = point on the surface S (defined by f(x,y,z)=0) and dA / ^s dA. n 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.) 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 m-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 of sight from ro to the "edge" of the surface. Think about this as "something like" a view of the moon. IV-25 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. IV-26 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: IV-27 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 IV-28 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 simply-connected (no holes); then W(r) = Lf(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: 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 IV-2) Lf· dr where L acts only on f. 5. Thus, W·dr = Lf· dr and W(r) = L f(r). ........ IV-29 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 simply-connected 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 IV-28: 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 I-2 under the properties of a unitary vector space: |x|² = (x,x) = 0 iff x = 0. ) ... IV-30 The Dirac Delta Function, δ(x-xo) Dirac Delta Function In one dimension, δ(x-xo) is defined to be such that: ma to b f(x) δ(x-xo)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(x-xo): (you should know those marked with *) *1. δ(x-xo) = 0 *2. m-4 to +4 δ(x)dx if x … xo =1 3. δ(ax) = δ(x)/|a| *4. δ(-x) = δ(x) 5. δ(x²-a²) = [δ(x-a) + δ(x+a)]/(2a); a $ 0 +------------------------------------------------------------------------------------------------------------------------------------------**7. δ(g(x)) = 3i δ(x-xoi)/|dg/dx|x=xoi where g(xoi) = 0 and dg/dx exists at and in a region around xoi. .------------------------------------------------------------------------------------------------------------------------------------------ 6. m-4 to +4 δ(x-a)δ(x-b)dx = δ(a-b) *8. f(x)δ(x-a) = f(a)δ(x-a) 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 (non-unique) functions, δn(x). Some examples of δn(x) which work are given below. In all these cases, m-4 to +4 δn(x)dx = 1 œ n and limitn --> 4 m-4 to +4 δn(x-xo)f(x)dx = f(xo). œ 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)]m-n to n exp(ixt)dt IV-31 11. ma to b f(x) d /dx δ(x-xo) dx = r r *½[(-1) d f/dxr|xo if xo = a or b .0 otherwise rr + (-1)r drf/dxr|xo if xo ε (a,b) f(x) is arbitrary, continuous function at x = xo 12. ma to b xr dr/dxr δ(x-xo)dx = ma to b (-1)rr! δ(x-xo) dx where xo ε (a,b). important expressions involving δ(x-xo) 13. δ(x-xo) = [1/2π] m-4 to +4 eik(x-xo)dk 14. δ(r-ro) / δ(x-xo)δ(y-yo)δ(z-zo)= [1/2π]3mmmall k space exp[ik·(r-ro)]dkxdkydkz 15. δ(g(x)) = 3n δ(x-βn)/|dg/dx|x = βn where g(βn) = 0. 16. δ(r-ro) = δ(q1-q1o)δ(q2-q2o)δ(q3-q3o)//g in general system. Dirac Delta Function in 3 Dimensions: δ (r - ro) / δ(x-x )δ(y-y )δ(z-z ) o o o 17. 18. δ(k - ko) = [1/2π]3 m-4 to +4m-4 to +4m-4 to +4 exp[ir·(k - ko)] dxdydz δ(r - ro) = δ(q1-q1o)δ(q2-q2o)δ(q3-q3o)//g derivation: mmmδ(r - ro)f(r) d3x = f(ro) = mmmδ(q1-q1o)δ(q2-q2o)δ(q3-q3o)//g· f(r(qi)) /gdq1dq2dq3 19. 20. δ(r - ro)spherical coordinates = δ(r-ro)δ(φ-φo)δ(θ-θo)/(r²sinθ) ½[δ(x-a)+δ(x+a)] = [1/π] m0 to 4 cos(ka) cos(kx) dk Exercises: Evaluate the following integrals. a) m-1 to 5 δ(2x-π) exp[ sin3(x-π) ] dx b) mmmall spaceδ(r·r-a²)δ(cosθ-1//2)δ(sinφ-½)exp[ik·r]d3x IV-32 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-->4|J(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. IV-33 Use the divergence theorem: mmmV'-L·L[1/r] d3x 0 = mmS1-L[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 ^ = mmS4-L[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 IV-34 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 IV-9) = (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. 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) 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. 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 total charge, Q. The outer radius of the charge distribution is b. IV-37 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 - 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! --------- , * +----------------------------------------------------------------------------------------------------^ = [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)]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] 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 IV-35 have similar difficulties. IV-38 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. ...
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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.

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