Vector Algebra and Calculus

1. Revision of vector algebra, scalar product, vector product 2. Triple products, multiple products, applications to geometry 3. Diﬀerentiation of vector functions, applications to mechanics 4. Scalar and vector ﬁelds. Line, surface and volume integrals, curvilinear co-ordinates 5. Vector operators — grad, div and curl 6. Vector Identities, curvilinear co-ordinate systems 7. Gauss’ and Stokes’ Theorems and extensions 8. Engineering Applications 6. Vector Operator Identities & Curvi Coords

• In this lecture we look at identities built from vector operators. • These operators behave both as vectors and as diﬀerential operators, so that the usual rules of taking the derivative of, say, a product must be observed. • We are laying the groundwork for the use of these identities in later parts of the Engineering course.

• We then turn to derive expressions for grad, div and curl in curvilinear coordinates. • After deriving general expressions, we will specialize to the Polar family. Identity 1: curl grad U = 0 6.2

• U(x,y,z) is a scalar ﬁeld. Then ˆı ˆ kˆ ∇∇×∇∇U = ∂/∂x ∂/∂y ∂/∂z

∂U/∂x ∂U/∂y ∂U/∂z

∂2U ∂2U = ˆ ı − +ˆ () + kˆ () ∂y∂z ∂z∂y = 0 .

• ∇∇×∇∇ can be thought of as a null operator. Identity 2: div curl a =0 6.3

• For a(x,y,z) a vector ﬁeld:

∂/∂x ∂/∂y ∂/∂z ∇∇· (∇∇× a) = ∂/∂x ∂/∂y ∂/∂z

a a a x y z ∂2a ∂2a = z − y ∂x∂y ∂x∂z ∂2a ∂2a − z + x ∂y∂x ∂y∂z ∂2a ∂2a + y − x ∂z∂x ∂z∂y = 0 Identity 3: divergence of Uv 6.4 • Suppose that • For example – U(r) is a scalar field – U(r) could be fluid density; and – v(r) is a vector field – v(r) its instantaneous velocity and we are interested in the divergence of the product The product would be the mass flux per unit area. Uv. • The product Uv is a vector field, so we can compute its divergence ... ∇∇· (Uv) = U(∇∇· v) + (∇U) · v = Udivv + (gradU) · v

• In steps: ∂ ∂ ∂ ∇∇· (Uv) = (Uv ) + (Uv ) + (Uv ) ∂x x ∂y y ∂z z ∂v ∂v ∂v ∂U ∂U ∂U = U x + U y + U z + v + v + v ∂x ∂y ∂z x ∂x y ∂y z ∂z = Udivv + v · gradU Identity 3: curl of Ua 6.5

• In a similar way, we can take the curl of the product of a scalar and vector field field Uv. • The result should be a vector field. • And you’re probably happy now to write down ∇∇× (Uv) = U(∇∇× v) + (∇U) × v . Identity 4: div of a × b 6.6

• But things get trickier to guess when vector or scalar products are involved! • Eg, not at all obvious that: div(a × b) = curla · b − a · curlb • To show this, use the determinant: ∂/∂x ∂/∂y ∂/∂z a a a x y z b b b x y z ∂ ∂ ∂ = [a b − a b ] + [a b − a b ] + [a b − a b ] ∂x y z z y ∂y z x x z ∂z x y y x

= ... Vector operator identities in HLT 6.7

• We could carry on inventing vector identities for some time, but it is a bit, er, dull. • Why bother at all, as they are in HLT?

1. Since grad, div and curl describe key aspects of vectors ﬁelds, they often arise often in practice. The identities can save you a lot of time and hacking of partial derivatives, as we will see when we consider Maxwell’s equation as an example later. 2. Secondly, they help to identify other practically important vector operators.

• We now look at such an example. Identity 5: curl(a × b) 6.8

ˆı ˆ kˆ curl(a × b) = ∂/∂x ∂/∂y ∂/∂z

a b − a b a b − a b a b − a b y z z y z x x z x y y x ∂ ∂ ⇒ curl(a × b) = (a b − a b ) − (a b − a b ) x ∂y x y y x ∂z z x x z This can be written as the sum of four terms: ∂b ∂b ∂a ∂a a • + y + z − b ∗ + y + z + x ∂y ∂z x ∂y ∂z ∂ ∂ ∂ ∂ ∗ + b + b a − • + a + a b y ∂y z ∂z x y ∂y z ∂z x

∂bx •: ax ∂x add to term1, sub from term4 ∂ax ∗: bx ∂x : sub from term2, add to term3 Hence ∇∇× (a × b) = (∇∇· b)a − (∇∇· a)b +[b ·∇∇]a − [a ·∇∇]b [a ·∇∇] can be regarded as new, and very useful, scalar diﬀerential operator. Deﬁnition of the operator [a · ∇] 6.9

• This is a scalar operator ...

∂ ∂ ∂ [a ·∇∇] ≡ a + a + a . x ∂x y ∂y z ∂z • Notice that the components of a don’t get touched by the differentiation. • Applied to a scalar field, results in a scalar field • Applied to a vector field results in a vector field Identity 6: curl(curla) for you to derive 6.10

• Amuse yourself by deriving the following important identity ... curl(curla) = grad(diva) −∇2a where 2 2 2 2 ∇ a = ∇ axˆı + ∇ ayˆ + ∇ az kˆ • We are about to use it .... ♣ Eg using Identity 6: electromagnetic waves 6.11

• Background: Maxwell established a set of four vector equations which are fundamental to working out how eletro- magnetic waves propagate. The entire telecommunications industry is built on these! divD = ρ divB = 0 ∂ curlE = − B ∂t ∂ curlH = J + D ∂t

• In addition, we can assume the following

B = µr µ0H J = σE

D = ǫr ǫ0E Example ctd 6.12

Question: Show that in a material with no free charge, ρ = 0, and with zero conductivity, σ = 0, the electric ﬁeld E 2 2 2 must be a solution of the wave equation ∇ E = µr µ0ǫr ǫ0(∂ E/∂t ) . Answer: divD = div(ǫr ǫ0E) = ǫr ǫ0divE = ρ = 0; ⇒divE =0

divB = div(µr µ0H) = µr µ0divH =0 ⇒divH =0

curlE = −∂B/∂t = −µr µ0(∂H/∂t)

curlH = J + ∂D/∂t = 0 + ǫr ǫ0(∂E/∂t)

But curlcurlE = ∇(∇∇· E) −∇2E, so Then

2 ∂ ∂E 2 curl [−µr µ0(∂H/∂t)] = −∇ E −µ µ ǫ ǫ = −∇ E r 0 ∂t r 0 ∂t ∂ 2 −µr µ0 [curlH] = −∇ E ∂2E ∂t ⇒µ µ ǫ ǫ = ∇2E r 0 r 0 ∂t2 2 Grad, div, curl and ∇ in curvilinear coords 6.13

• It is possible to obtain general expressions for grad, div and curl in any orthogonal curvilinear co-ordinate system ... • Need the scale factors h ... • We recall that the unit vector in the direction of increasing u, with v and w being kept constant, is 1 ∂r uˆ = hu ∂u where r is the general position vector, and ∂r h = u ∂u

and similar expressions apply for the other co-ordinate direct ions. Then

dr = huduuˆ + hv dvvˆ + hw dwwˆ . Grad in curvilinear coordinates 6.14

• Using the properties of the gradient of a scalar field obtained previously, ∂U ∂U ∂U ∇U · dr = dU and dU = du + dv + dw ∂u ∂v ∂w It follows that ∂U ∂U ∂U ∇U · (h uˆdu + h vˆdv + h wˆdw) = du + dv + dw u v w ∂u ∂v ∂w • The only way this can be satisfied for independent du, dv, dw is when Grad U in curvilinear coords: 1 ∂U 1 ∂U 1 ∂U ∇U = û + ˆv + ˆw hu ∂u hv ∂v hw ∂w Divergence in curvilinear coordinates 6.15

• If the curvilinear coordinates are orthogonal then δvolume is a cuboid (to 1st order in small things) and

dV = hu hv hw dudvdw .

w h (v) dw w h (v+dv) dw w • However, it is not quite a cuboid: the area of two opposite faces will diﬀer as the scale parameters are v h (v) du functions of u, v, w. u h u (v+dv) du

h v dv u The scale params are functions of u,v,w • So the nett eﬄux from the two faces in the ˆv dirn is ∂a ∂h ∂h = a + v dv h + u dv h + w dv dudw − a h h dudw v ∂v u ∂v w ∂v v u w ∂(a h h ) ≈ v u w dudvdw ∂v Divergence in curvilinear coordinates /ctd 6.16

• Repeat: the nett eﬄux from the two faces in the ˆv dirn is ∂a ∂h ∂h = a + v dv h + u dv h + w dv dudw − a h h dudw v ∂v u ∂v w ∂v v u w ∂(a h h ) = v u w dudvdw ∂v • Now div is net eﬄux per unit volume, so sum up other faces: ∂(a h h ) ∂(a h h ) ∂(a h h ) diva dV = u v w + v u w + w u v dudvdw ∂u ∂v ∂w

• Then divide by dV = huhv hw dudvdw ... Conclude: div in curvi coords is: 1 ∂(a h h ) ∂(a h h ) ∂(a h h ) diva = u v w + v u w + w u v huhv hw ∂u ∂v ∂w Curl in curvilinear coordinates 6.17 v au(v+dv) v+dv

• For an orthogonal curvi coord system h u (v+dv) du dS = huhv dudw. • But the opposite sides are not of same length! dv Lengths are h (v) du hu(v)du, and hu(v + dv)du. u v u u+du au (v) • Summing this pair contributes to circulation (in ˆw dirn) ∂(h a ) a (v)h (v)du − a (v + dv)h (v + dv)du = − u u dvdu u u u u ∂v • Add in the other pair to ﬁnd circulation per unit area dC 1 ∂(h a ) ∂(h a ) = − u u + v v huhv dudv huhv ∂v ∂u Curl in curvilinear coordinates, ctd 6.18

• To repeat, the part related to ˆw is: dC 1 ∂(h a ) ∂(h a ) = − u u + v v huhv dudv huhv ∂v ∂u • Adding in the other two components gives: 1 ∂(h a ) ∂(h a ) curla(u,v,w) = w w − v v ˆu + hv hw ∂v ∂w 1 ∂(h a ) ∂(h a ) u u − w w ˆv + hw hu ∂w ∂u 1 ∂(h a ) ∂(h a ) v v − u u ˆw huhv ∂u ∂v

• You should show that can be written more compactly as: Curl in curvi coords is: h ˆu h ˆv h ˆw 1 u v w curla(u,v,w) = ∂ ∂ ∂ h h h ∂u ∂v ∂w u v w h a h a h a u u v v w w

The Laplacian in curvilinear coordinates 6.19

• Substitute the components of gradU into the expression for diva ... • Much grinding gives the following expression for the Laplacian in general orthogonal co-ordinates:

Laplacian in curvi coords is: ∇2U = 1 ∂ h h ∂U ∂ h h ∂U ∂ h h ∂U v w + w u + u v . huhv hw ∂u hu ∂u ∂v hv ∂v ∂w hw ∂w Grad, etc, the 3D polar coordinate 6.20

• There is no need slavishly to memorize the above derivations or their results. • More important is to realize why the expressions look suddenly more complicated in curvilinear coordinates • We are now going to specialize our expressions for the polar family • As they are 3D entities, we need consider only cylindrical and spherical polars. Grad, etc, in cylindrical polars 6.21

• We recall that r = r cos θˆı + r sin θˆ + zkˆ, and that hu = |∂r/∂u|, and so

2 2 hr = (cos θ + sin θ)=1,

p 2 2 2 2 hθ = (r sin θ + r cos θ) = r, q hz = 1

• Hence, using these and U(r,θ,z) and a = ar êr + aθêθ + az kˆ ∂U 1 ∂U ∂U gradU = ê + ê + kˆ ∂r r r ∂θ θ ∂z 1 ∂(ra ) ∂a ∂a diva = r + θ + z r ∂r ∂θ ∂z 1 ∂a ∂a ∂a ∂a 1 ∂(ra ) ∂a curla = z − θ ê + r − z ê + θ − r kˆ r ∂θ ∂z r ∂z ∂r θ r ∂r ∂θ

• The derivation of the expression for ∇2U in cylindrical polar co-ordinates is set as a tutorial exercise. Grad, etc, in spherical polars 6.22 • We recall that r = r sin θ cos φˆı + r sin θ sin φˆ + r cos θkˆ so that

2 2 2 2 hr = (sin θ(cos φ + sin φ)+cos θ)=1 q 2 2 2 2 2 2 hθ = (r cos θ(cos φ + sin φ) + r sin θ) = r q 2 2 2 2 hφ = (r sin θ(sin φ + cos φ) = r sin θ q • Hence ∂U 1 ∂U 1 ∂U gradU = ê + ê + ê ∂r r r ∂θ θ r sin θ ∂φ φ 1 ∂(r 2a ) 1 ∂(a sin θ) 1 ∂a diva = r + θ + φ r 2 ∂r r sin θ ∂θ r sin θ ∂φ ê ∂ ∂ ê ∂ ∂ curla = r (a sin θ) − (a ) + θ (a ) − (a r sin θ) r sin θ ∂θ φ ∂φ θ r sin θ ∂φ r ∂r φ ê ∂ ∂ + φ (a r) − (a ) r ∂r θ ∂θ r ♣ Examples 6.23

Question: Find curla in (i) Cartesians and (ii) Spherical polars when a = x(xˆı + yˆ + zkˆ). Answer (i): • In Cartesians, using the pseudo determinant gives ˆı ˆ kˆ curla = ∂/∂x ∂/∂y ∂/∂z = −zˆ + ykˆ

x 2 xy xz

♣ Example /ctd 6.24

Answer (ii): • We were told a = x(xˆı + yˆ + zkˆ). • In spherical polars x = r sin θ cos φ and (xˆı + yˆ + zkˆ) = r

2 • Hence a = r sin θ cos φr = r sin θ cos φ ˆer 2 or in component form: ar = r sin θ cos φ; aθ = 0; aφ =0 . • Expression for curl (earlier, and HLT):

êr ∂ ∂ êθ ∂ ∂ êφ ∂ ∂ curla = (aφ sin θ) − (aθ) + (ar ) − (aφr sin θ) + (aθr) − (ar ) r sin θ ∂θ ∂φ r sin θ ∂φ ∂r r ∂r ∂θ • Hence ê ∂ ê ∂ curla = θ (r 2 sin θ cos φ) + φ − (r 2 sin θ cos φ) r sin θ ∂φ r ∂θ ê ê = θ (−r 2 sin θ sin φ) + φ −r 2 cos θ cos φ) r sin θ r = êθ(−r sin φ) + êφ(−r cos θ cos φ) Example Check: These two results should be the same! 6.25

• To check we need êr , êθ, êφ in terms ofˆı,ˆ, kˆ ... • Use r = xˆı + yˆ + zkˆ and 1 ∂r 1 ∂r 1 ∂r êr = ; êθ = ; êφ = hr ∂r hθ ∂θ hφ ∂φ

• Hence, doing the first of these, as hr =1 ∂ ê = r sin θ cos φˆı + r sin θ sin φˆ + r cos θkˆ = sθcφˆı + sθsφˆ +cθkˆ r ∂r • Which gives the top row of the matrix. Grind to find the rest ...

êr sin θ cos φ sin θ sin φ cos θ ˆı ˆı  êθ  =  cos θ cos φ cos θ sin φ − sin θ   ˆ  = [R]  ˆ  ê − sin φ cos φ 0 kˆ kˆ  φ        • Don’t be shocked to see a rotation matrix [R]! We are rotating one right-handed orthogonal coord system into another. Check /ctd 6.26

• Now we convert the spherical polar expression in Cartesians ...

êr curla = êθ(−r sin φ) + êφ(−r cos θ cos φ) = −r[0, sin φ, cos θ cos φ] êθ  ê  φ  sin θ cos φ sin θ sin φ cos θ ˆı = −r[0, sin φ, cos θ cos φ]  cos θ cos φ cos θ sin φ − sin θ   ˆ  − sin φ cos φ 0 kˆ     = −r sin φ(cos θ cos φˆı + cos θ sin φˆ − sin θkˆ) + (−r cos θ cos φ)(− sin φˆı + cos φˆ) = −r cos θˆ + r sin θ sin φkˆ = −zˆ + ykˆ

• This is exactly what we got before! Summary 6.27

Take home messages ...

• The key thing when combining operators is to remember that each partial derivative operates on everything to its right. • The identities (eg in HLT) are not mysterious. They merely provide useful short cuts. • There is no need slavishly to learn the expressions for grad, div and curl in curvi coords. They are in HLT, but – you need to know how they originate. – you need to be able to hack them out when asked. • Ditto with the specializations to polars. • Just as physical vectors are independent of their coordinate systems, so are diﬀerential operators.