6 Vector Operators. 6.1 The Gradient Operator

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1 6 Vecto Opeatos 6. The Gadient Opeato In the B2 couse ou wee intoduced to the gadient opeato in Catesian coodinates. Fo an diffeentiable scala function f(x,, z), we can define a vecto function though ( ) ( ) ( ) f f f gad f = f = (f x ê x + f ê + f z ê z ) = ê x x + ê + ê z z. (6.) Thee is a slight diffeence with the notation of the fist ea couse since î, ĵ and ˆk have been eplaced b ê x, ê and ê z to allow a moe staightfowad genealisation to pola coodinates. ometimes we expess the esult in tems of an opeato equation = ê x x + ê + ê z z, (6.2) whee it is undestood that the opeato, ponounced Del, is to act upon the function f. If the point (x,, z) is changed b an infinitesimal amount d = (dx, d, dz), the function changes b = (f x ê x + f ê + f z ê z ) (dx ê x + d ê + dz, ê z ) = f x dx + f d + f z dz, (6.3) which is just the expession fo the total deivative in tems of the slopes in the thee diections. Upon the suface f(x,, z) = constant, =, so that = f d. (6.4) In ode that the point stas along the contou, small changes d ae pependicula to the gadient. Examples. Find the gadient vectos φ and ψ of the functions φ = x z 2 + 3, ψ = x z + zx, at the point (3, 2, 4), and the acute angle between these two diections coect to.. The two gadients ae At the point (3, 2, 4), If θ is the angle between these two vectos, then Hence the acute angle θ φ = (2x, 2, 2z), ψ = ( + z, x z, x ). φ = (6, 4, 8), ψ = (6,, ). cos θ = (6, 4, 8) (6,, ) = In the case whee f is onl a function of = x z 2, then ( ) ( f = ) x z 2 = x x d x whee we have used a chain ule fo diffeentiating. Thus gad f = (x,, z) whee ê is a unit vecto pointing in the diection of. 68 d = ê d, d,

2 6.2 The Divegence Opeato In the pevious section, we used the (Del) opeato to poduce a vecto field gad f fom a scala field f. The divegence opeato does the opposite it ceates a scala field fom a vecto using the scala poduct. Fo a diffeentiable vecto function v(x,, z), ( ) div v = v = ê x x + ê + ê z (v ) x ê z x + v ê + v z ê z Examples. In the case whee v =, = v x x + v + v z z. (6.5) div = = x x + + z z = If f is a function of the magnitude of, div [ f()] = [x f()] x + [ f()] + [z f()] z = 3f() + d. = f + x2 d + f + 2 d + f + z2 3. uppose that f(x,, z) and v(x,, z) ae espectivel scala and vecto functions of the coodinates (x,, z). Then (f v) = x (f v x) + (f v ) + z (f v z) f = v x x + v f + v f z z + f v x x + f v + f v z z = v f + f v. The vecto pat of the opeato obes the ules fo vectos, wheeas the diffeentiation pat woks obes the nomal ules fo diffeentiation, including that fo the deivative of a poduct. d 6.3 The Cul Opeato We saw in the B2 couse that the diffeential is pefect if and onl if dw = v x dx + v d + v z dz, (6.6) v x v x = v z v z = v z x v z =. (6.7) Let us intoduce the moe compact notation using the cul of a vecto function v = (v x, v, v z ), defined as cul v = v. (6.8) Using the expession fo the coss poduct given in the B2 couse, togethe with the expession fo Del given b Eq. (6.2), it is seen that this has components ( v) x = v z v z, ( v) = v x z v z x, ( v) z = v x v x (6.9) 69

3 Thus, if cul v =, then the integal W (x,, z) = (x,,z) (x,,z ) v d (6.) defines a unique function W (x,, z), whose value does not depend upon the path of integation. Examples. We saw, in the discussion of the divegence of a poduct, that obes simultaneousl the ules of diffeentiation and vecto algeba. This is also the case fo cul. Thus (f v) = f ( v) + ( f) v, whee the convention is that the diffeentiation on the ight hand side onl takes place inside the backet. The simplest poof is in tems of components. Taking just the x-component of the LH, = and similal fo the othe components. (LH) x = (f v z) z (f v ) ( f v z f ) ( z v vz + f v ) = [( f) v] z x + f [ v] x, 2. If now v = and f = f(), what is ( f())? = x z x z ê x ê ê z =, since the x patial diffeentiation acts hee on and z, but not x. We have alead shown that ince this means that 6.4 Opeatos quadatic in f() = ê d = ( ) =, d ( f()) =. The gadient opeato takes a scala into a vecto. Acting on the esult with the divegence opeato gives a scala again. div gad f = ( f) = 2 f = 2 f x f f z 2 (6.) The esulting opeato 2 is called the Laplacian opeato and it has alead been used when discussing the Legende polnomials. Most of Phsics seems to be govened b second ode diffeential equations involving the Laplacian opeato. In the Quantum Mechanics couse ou have been looking at the chödinge equation descibing the motion of a paticle of mass m with eneg E in a potential V (); d. h2 2m 2 Ψ + V () Ψ = E Ψ. (6.2) In electostatics, ou leaned that the potential Φ() due to a chage densit ρ() satisfies the equation 2 Φ = 4πε ρ. (6.3) 7

4 Thee ae, howeve, man moe examples. Thee ae othe opeatos which ae quadatic in, e.g. A = cul gad φ = ( φ) = x φ x φ z φ z ê x ê ê z (6.4) This has an x-component of fo an easonable function φ(x,, z). Thus A x = z φ z φ = (6.5) ( φ) =. (6.6) You should all ecognise this esult in the case of an electostatic field with E = φ. The electostatic field is iotational E =. (6.7) B witing things out in component fom, one can easil show that ( A) = ( Az x A ) + ( Ax z z A z x ) + z ( A x A ) x =. (6.8) This is anothe useful esult in electomagnetism. The magnetic induction field B is solenoidal, i.e. B =. Hence, using Eq. (6.8), we can wite B = A. (6.9) In Electomagnetism, A is called the magnetic vecto potential. Though this is not cuentl used in the second ea E&M couse, it will be needed late in the quantum desciption of the inteaction of adiation with matte. The magnetic potential seems just to be some atificial constuct intoduced to make B automaticall solenoidal and so it needn t coespond diectl to a measuable phsical quantit. Nevetheless, the Ahaanov-Bohm effect in Quantum Mechanics shows that cetain featues of the magnetic potential have expeimental consequences! Anothe quadatic elation is ( A) = ( A) 2 A. (6.2) In wods this is cul (cul A) = gad (div A) del squaed A. This can be poved b witing evething explicitl in tems of components, but thee ae othe methods. In the B2 couse, it was shown that fo odina vectos d = a (b c) = (a c) b (a b) c. (6.2) The touble about using this elation is that, when b is also a diffeential opeato, we ae not allowed to change the ode at will. Hence wite it in the smbolic component fom d i = j (a j b i c j a j b j c i ), (6.22) whee we have NOT alteed the ode of the vectos. Now put a = b = and c = A. ince a and b ae now the same opeatos, it doesn t matte in which ode we wite them. Hence d i = j ( i j A j j j A i ), (6.23) which is just the component epesentation of Eq. (6.2). 7

5 6.5 The Divegence Theoem Two-dimensional integals wee discussed in the B2 couse. These have to be genealised a bit in ode to intoduce the Flux of a Vecto Field. An infinitesimal integation element d = dx d is a small bit of the x plane of aea dx d. Planes, howeve, have diections given b thei nomals. In this case, the nomal is along the z-axis, which means that the integation element is eall the vecto d = dx d ê z. (6.24) This is just one paticula example, but in geneal d = d ˆn, (6.25) whee ˆn is the nomal to this small bit of plane. To undestand the concept of vecto flux, conside a simple example fom fluid flow whee a liquid of densit ρ is moving with velocit v though a suface. The amount of mass that passes though an element d pe unit time depends upon the component of v pependicula to d; dm = ρ v cos θ d = ρ v d, (6.26) whee θ is the angle between v and ˆn. If we define a vecto field b F = ρ v, then the ate of flow (flux) of mass pe unit time is and the total flux though the suface is m = dm = F d, (6.27) F d = F ˆn d. (6.28) You will have seen integals of this tpe in Faada s law of Magnetic Induction. Conside now the flux of a vecto though a closed suface, such as the infinitesimal cube in the pictue. 2 dz d (x,, z) dx 4 We ae going to build up the total flux though the six sides b evaluating each face sepaatel, stating with face-. The nomal to this suface points in the positive x-diection, so that d = d dz ê x. Note howeve that the value of the x-coodinate is x + dx, since the lowe ight-hand cone has been taken to be (x,, z). Theefoe the flux is I = F (x + dx,, z) ê x d dz = F x (x + dx,, z) d dz. (6.29) On face-2 the nomal to the suface points in the negative x-diection, so that d = d dz ê x and I 2 = F x (x,, z) d dz. The sum of the flux though these two faces is I + I 2 = (F x (x + dx,, z) F x (x,, z)) d dz. (6.3) 72

6 ince dx is ve small, we can expand F x in a Talo seies in dx, keeping just the fist two tems: F x (x + dx,, z) F x (x,, z) + dx x F x(x,, z). (6.3) I + I 2 = x F x(x,, z) dx d dz. (6.32) Adding the contibutions fom the othe two pais of sides, F d = ( F ) dx d dz = ( F ) dv. (6.33) This is the infinitesimal fom of the divegence theoem. The integal ove a closed suface of the flux of a vecto field is equal to the volume integal of the divegence of the vecto. ( F ) dv = F d. (6.34) V If we put two such cubes togethe, the flux tems cancel on the common suface because the nomals ae in opposite diections. The volume tems just add. Hence the integal fom of Eq. (6.34) is valid fo the two cubes togethe. We can build up an closed shape if we take enough infinitesimal cubes. T building a model of Big Ben out of Lego bicks! The divegence theoem is theefoe tue geneall. The divegence theoem is much used in electostatics, whee the divegence of the displacement vecto D is popotional to the chage densit; D = ε ρ(). (6.35) Integating this ove a volume V and using the divegence theoem, we have Q = ρ dv = ( D) dv = D d. (6.36) ε ε V V The electic flux though a closed suface is equal to the amount of chage contained theein. This is known as Gauss s theoem. As a simple application of the law, conside a spheicall smmetic chage distibution. This gives ise to a spheicall smmetic electic field D = D ê. If we take the suface to be that of a sphee of adius, then the nomal to the suface lies along the adius vecto. Hence D d = 4π 2 D = Q (6.37) ε This gives the standad esult that D = Q/4πε 2. Example Find the divegence A of the field A = xê x + ê + z 3 ê z. Fo a ight clinde defined b x and z +, deive the volume integal of the divegence of A. Veif the divegence theoem in this case. Integating ove the clinde, I = V A dv = A = (x) x d 2π dθ + + () + (z3 ) = + + 3z 2. z dz ( + + 3z 2 ) = 73 d 2π dθ [ ( + )z + z 3] +.

7 To poceed futhe, change to pola coodinates = sin θ. ince the integal of sin θ fom to 2π then vanishes, we ae left with I = d(8π) = 4π. To veif the divegence theoem, split the suface integal into ones ove the caps of the clinde and one ove the cuved pat. On the top cap, ˆn = ê z and A ˆn = z 3 =. I = d 2π dθ = π. On the bottom cap, the outwad nomal changes sign, but so does z 3, which means that this suface integal is also I 2 = π. On the cuved suface ˆn = ˆ = cos θ ê x + sin θ ê, so that A ˆn = x cos θ + sin θ = cos 2 θ sin θ + sin 2 θ. I 3 = 2π dθ + dz ( cos 2 θ sin θ + sin 2 θ ) = 2 2π The fist integal on the RH vanishes because dθ ( cos 2 θ sin θ + sin 2 θ ). On the othe hand, 2π dθ cos 2 θ sin θ = θ=2π θ= cos 2 θ d(cos θ) = cos 3 θ θ=2π θ= =. 2π dθ sin 2 θ = 2 2π dθ ( cos 2θ) = 2 [θ sin 2θ/2]2π = π. Hence I 3 = 2π and I + I 2 + I 3 = 4π, which veifies the divegence theoem in this case. 6.6 tokes Theoem tokes theoem states that the suface integal of the cul of a vecto is equal to the line integal of the vecto along the contou suounding this aea ( F ) d = F dl. (6.38) We have to be slightl caeful hee. The sign of the nomal to the suface is given b the cokscew (ighthand) ule when going ound the contou L. The method of poof is ve simila to that of the divegence theoem. We stat b taking an infinitesimal ectangle in the x plane, as in the pictue. L 74

8 2 3 d (x, ) 4 dx tat at the bottom-ight cone and go ound anti-clockwise,, 2, 3, and 4. paallel to ê, I = +d ince, along path-, dl is F (x + dx, ) d F (x + dx, ) d (6.39) fo small d. On the othe hand, along path-3, dl points in the negative -diection, so that I 3 F (x, ) d. The sum of these two contibutions is I + I 3 = (F (x + dx, ) F (x, )) d x F (x, ) dx d. (6.4) Adding the contibutions fom the othe two paths, we see that ( F dl = x F (x, ) ) F x(x, ) dx d = ( F ) d, (6.4) L since d lies in the positive z-diection (using the ight-hand ule convention). This poves tokes theoem fo the ve small ectangle. An suface can be constucted as a sum of infinitesimal plane ectangles so that we can sew them togethe and pove the theoem fo an aea of abita shape in thee dimensions. Note that along the common line between two such ectangles, the line integals cancel, so that onl the path aound the peiphe of the combined aea is left. Example Veif tokes theoem fo the vecto field A = 2 ê x +3x ê z 2 ê z and a hemispheical suface x z 2 = 9 fo which z. tat b woking out cul A. ê x ê ê z A = x z = [ ]ê 2 3x z 2 x [ ]ê + [3 2]ê z = ê z. ince the suface is that of a hemisphee cented at the oigin, the nomal to the suface is and ˆn = ˆ = xê x + ê + zê z ( A) ˆn = z = cos θ., 75

9 It is easiest to do the integation ove the suface in pola coodinates, whee d = 9 sin θ dθ dφ; ( F ) d = 9 π sin θ cos θ dθ 2π dφ = 9π π sin 2θ dθ = 9π 2 [cos 2θ]π = 9π. On the othe hand, along the cicle x = 9, we use plane pola coodinates x = 3 cos φ, = 3 sin φ, dl = ( 3 sin φ ê x + 3 cos φ ê )dφ. L A dl = = 2π 2π (6 sin φ ê x + 9 cos φ ê ) ( 3 sin φ ê x + 3 cos φ ê ) dφ (27 cos 2 φ 8 sin 2 φ) dφ = 2π [ ] = 9π. The last integal can be done with the standad tigonometic identities, but I used the esult that, aveaged ove one peiod, < cos 2 φ >=< sin 2 φ >= Coodinate-Independent Definitions The foms of div, gad and cul have been defined in Catesian coodinates, but we must now evaluate them in othe coodinate sstems. The thee fomulae needed ae = ( f) d, (6.42) fo an infinitesimal change in the position vecto. { F = lim V V fo a ve small volume. fo the component of cul in the Ŝ diection. } F d { } ( F ) Ŝ = lim F dl L (6.43) (6.44) 6.8 pheical Pola Coodinates It would be idiculous to wok out the potential due to a chaged sphee in Catesian coodinates athe than spheical polas. imilal, fo a dielectic clinde, use clindical pola coodinates. Thee-dimensional poblems ae quite difficult enough; the smmet of the poblem must be used to simplif the poblems as much as possible. In mathematics books like Afken and Webe, ou will find that thee ae 4 useful coodinate sstems! Howeve, in all m eseach aeas, I have onl eve used fou and in this couse I am onl going to discuss thee, viz Catesian, pheical Pola, and Clindical Pola coodinates. The Catesian components of a point in spheical pola coodinates (, θ, φ) ae given b x = sin θ cos φ, = sin θ sin φ, z = cos θ. (6.45) The basis vectos in this sstem can be found geometicall if ou ae good at dawing in thee dimensions. The ae ê = sin θ cos φ ê x + sin θ sin φ ê + cos θ ê z, ê θ = cos θ cos φ ê x + cos θ sin φ ê sin θ ê z, ê φ = sin φ ê x + cos φ ê, (6.46) 76

10 which satisf ê ê θ = ê φ. (6.47) This shows that the basis vectos in this sstem fom a ight-handed pependicula sstem. z ê φ θ φ x Diffeentiating Eq. (6.45) with espect to, θ, and φ, we see that ê θ ê d = d ê + dθ ê θ + sin θ dφ ê φ. (6.48) This can also be seen geometicall with a nice pictue. The volume element is the poduct of these thee tems (much simple than woking out the Jacobian) and the elements of aea pointing in the diections of the basis vectos dv = 2 sin θ d dθ dφ, (6.49) d = 2 sin θ dθ dφ, d θ = sin θ d dφ, d φ = d dθ. (6.5) Boas, and othe books, evaluate evething fo a geneal coodinate sstem in tems of the so-called scalefactos of the metic h i, defined b 3 d = h i (dx) i ê i. (6.5) i= ince we ae hee onl going to look at two coodinate sstems, I pefe to do things explicitl and not use the h i. We fist wok out the expession fo the gadient in spheical pola coodinates: Fom Eq. (6.42), f = ( f) ê + ( f) θ ê θ + ( f) φ ê φ. (6.52) = ( f) d = (( f) ê + ( f) θ ê θ + ( f) φ ê φ ) (d ê + dθ ê θ + sin θ dφ ê φ ) But, fom the chain ule fo patial diffeentiation, = ( f) d + ( f) θ dθ + ( f) φ sin θ dφ. (6.53) = f f f d + dθ + dφ. (6.54) θ φ 77

11 Compaing the Eqs. (6.53) and (6.54), we can ead off the spheical pola components of the gadient; ( f) = f, ( f) θ = f θ, ( f) φ = sin θ f φ (6.55) To find the expession fo the divegence, evaluate the flux of the vecto F (, θ φ) though the sides of the little hpecube (d, dθ, sin θ dφ). The net flux though the two sides with constant ae I + I 2 = F ( + d, θ, φ) ( + d) 2 sin θ dθ dφ F (, θ, φ) 2 sin θ dθ dφ ( 2 F (, θ, φ) ) d sin θ dθ dφ. (6.56) Note that, not onl does F change when inceases, but so does the aea facto 2. Doing the same thing fo the change in the θ and φ coodinates, I 3 + I 4 = F θ (, θ + dθ, φ) sin(θ + dθ) d dφ F θ (, θ, φ) sin θ d dφ θ (sin θ F θ(, θ, φ)) d dθ dφ. (6.57) I 5 + I 6 = F φ (, θ, φ + dφ) d dθ F φ (, θ, φ) d dθ φ (F φ(, θ, φ)) d dθ dφ. (6.58) Thus, fom Eq. (6.43), F = 2 sin θ d dθ dφ { ( 2 F (, θ, φ) ) d sin θ dθ dφ + θ (sin θ F θ(, θ, φ)) d dθ dφ + } φ (F φ(, θ, φ)) d dθ dφ = ( 2 2 F (, θ, φ) ) + sin θ θ (sin θ F θ(, θ, φ)) + sin θ φ (F φ(, θ, φ)). (6.59) Note that, even if all of the components of F ae constant, the divegence does not vanish because of the changing geomet in spheical pola coodinates. ince we now have expessions fo both the gadient and divegence in spheical pola coodinates, we can also wok out that fo the Laplacian b substituting Eq. (6.55) into Eq. (6.59); 2 V = 2 ( 2 V ) + 2 sin θ θ ( sin θ V θ ) + 2 sin 2 θ ( ) V. (6.6) φ φ This agees with the fomula of Eq. (2.29) that we obtained b manipulating the patial diffeentiatals. Using exactl the same techniques on tokes theoem allows us to obtain the expession fo the cul. To get the -component, look at the line integal aound the elementa contou of the figue. sin(θ) dφ dθ dθ sin(θ + dθ) dφ 78

12 On cuve-, the pola angle is fixed at θ + dθ and the length is sin(θ + dθ) dφ, wheeas on cuve-3 the angle and length ae θ and sin θ espectivel. On cuves-4 and -2, the azimuthal angles ae fixed as φ and φ + dφ espectivel, wheeas both lengths ae equal to dθ. The line integal of F is F dl = L F φ (, θ + dθ, φ) sin(θ + dθ) dφ F θ (, θ, φ + dφ) dθ F φ (, θ, φ) sin θ dφ + F θ (, θ, φ) dθ { θ {sin θ F φ(, θ, φ)} } φ F φ(, θ, φ) dθ dφ. (6.6) Fom Eq. (6.44), { } ( F ) Ŝ = lim F dl, L so that ( F ) = { sin θ θ {sin θ F φ(, θ, φ)} } φ F θ(, θ, φ). (6.62) The othe two components ae evaluated in much the same wa. The esults ae ( F ) θ = { sin θ φ F (, θ, φ) sin θ } { F φ(, θ, φ)}, (6.63) ( F ) φ = { { F θ(, θ, φ)} } θ F (, θ, φ) (6.64) Example But Evaluating the divegence of F = ê θ in spheical pola coodinates gives In Catesian coodinates, F = ( sin θ) = cot θ. sin θ θ F = cos θ cos φ ê x + cos θ sin φ ê sin θ ê z. cos θ = z, sin θ = x2 + 2, x cos φ = x2 +, sin φ = 2 x2 + 2 Thus F = zx z x2 + 2 êx + x2 + 2 ê x ê z. F = z (x ) zx 2 /2 (x ) + z 3/2 (x ) z 2 /2 (x ) 3/2 = z(x2 + 2 ) (x ) 3/2 = z x2 + 2 = cot θ. When woking out the cul of F, onl the φ component suvives and we ae left with In Catesian coodinates, F = F = 2 ê φ. ê x ê ê z x zx x2 + 2 z z x2 + 2 x = 2 x2 + 2 êx + 2 x x2 + 2 ê = 2 sin φ ê x + 2 cos φ ê = 2 ê φ. 79

13 6.9 Clindical pola coodinates The othe coodinate sstem intoduced in B2 is clindical pola coodinates, defined in tems of Catesians b x = cos θ, = sin θ, z = z. (6.65) The basis vectos can be obtained geometicall; the z coodinate woks just like a standad Catesian coodinate and the pola pat can be plot in a plane. ê, ê θ, and ê z ae defined as the diections in which the point P moves when the coodinates and θ and z ae inceased b a ve small amount. ê = cos θ ê x + sin θ ê, ê θ = sin θ ê x + cos θ ê, ê z = ê z (6.66) ê ê θ P θ ê ê x The basis vectos fom an othogonal sstem with ê ê θ = ê z. (6.67) Note that is used fo the adius vecto in the plane and cannot be used simultaneousl fo the position vecto. Fo this, we shall use a capital; R = ê + z ê z, (6.68) When the coodinates change b infinitesimal amounts, the position vecto changes The volume element is the poduct of these thee tems and the elements of aea pointing in the diections of the basis vectos dr = d ê + dθ ê θ + dz ê z. (6.69) dv = d dθ dz, (6.7) d = dθ dz, d θ = d dz, The gadient in clindical pola coodinates is obtained fom d z = d dθ. (6.7) = (( f) ê + ( f) θ ê θ + ( f) z ê z ) (d ê + dθ ê θ + dz z) = ( f) d + ( f) θ dθ + ( f) z dz. (6.72) 8

14 We can now ead off the clindical pola components of the gadient using the chain ule fo patial diffeentiation: ( f) = f, ( f) θ = f θ, ( f) z = f z (6.73) To wok out the divegence, we must fist evaluate the flux though the hpecube. Poceeding as fo the spheical pola coodinates, this is Hence Thus, b Eq. (6.44), F d F = [F ( + d, θ, z)( + d) F (, θ, z)] dθ dz + [F θ (, θ + dθ, z) F θ (, θ, z)] d dz + [F z (, θ, z + dz) F z (, θ, z)] d dθ. (6.74) { ( F (, θ, z)) + θ F θ(, θ, z) + } z F z(, θ, z) d dθ dz. (6.75) { ( F (, θ, z)) + θ F θ(, θ, z) + } z F z(, θ, z). (6.76) It is then staightfowad to use Eqs. (6.73) and (6.76) togethe to get an expession fo the Laplacian opeato in clindical pola coodinates: 2 V = ( V ) + 2 V 2 θ V z 2 (6.77) To find the z-component of cul, look at the line integal aound the elementa contou of the figue. On cuve-, the adial vaiable is fixed at + d and the length is ( + d) dθ, wheeas on cuve-3 the adius and length ae and dθ espectivel. On cuves-4 and -2, the angles ae fixed as θ and θ + dθ espectivel, wheeas both lengths ae equal to d. The line integal of F is F dl = F θ ( + d, θ, z) ( + d) dθ F (, θ + dθ, z) d F θ (, θ, z) dθ + F (, θ, z) d L { { F θ(, θ, φ)} } θ F (, θ, φ) d dθ. (6.78) sin(θ) dφ dθ dθ Hence ( F ) z = sin(θ + dθ) dφ { { F θ(, θ, φ)} } θ F (, θ, φ). (6.79) 8

15 The othe two components ae a little easie to wok out: ( F ) = F z θ F θ z (6.8) ( F ) θ = F z F z (6.8) 82

B da = 0. Q E da = ε. E da = E dv

B da = 0. Q E da = ε. E da = E dv lectomagnetic Theo Pof Ruiz, UNC Asheville, doctophs on YouTube Chapte Notes The Maxwell quations in Diffeential Fom 1 The Maxwell quations in Diffeential Fom We will now tansfom the integal fom of the