Vector operators in curvilinear coordinate systems • In a Cartesian system, take x1 = x, x2 = y , and x3 = z, then an element of arc length ds 2 is, ds 2 = dx12 + dx22 + dx32 • In a general system of coordinates, we still have x1 , x2 , and x3 • For example, in cylindrical coordinates, we have x1 = r , x2 = θ, and x3 = z • We have already shown how we can write ds 2 in cylindrical coordinates, ds 2 = dr 2 + r 2 dθ + dz 2 = dx12 + x12 dx22 + dx32 • We write this in a general form, with hi being the scale factors ds 2 = h12 dx12 + h22 dx22 + h32 dx32 • We see then for cylindrical coordinates, h1 = 1, h2 = r , and h3 = 1

Curvilinear coordinates ~ • For an vector displacement ds ~ = eˆ1 h1 dx1 + eˆ2 h2 dx2 + eˆ3 h3 dx3 ds • Back to our example of cylindrical coordiantes, eˆ1 = eˆr , eˆ2 = eˆθ , and eˆ3 = eˆz , and ~ = eˆr dr + eˆθ rdθ + eˆz dz ds • These are orthogonal systems, but it would not have to be! ds 2 =

3 X 3 X

gij dxi dxi

i=1 j=1

• The gij is the metric tensor, and for an orthogonal system it is diagonal with gi = hi2

Vector operators in general curvilinear coordinates • Recall the directional derivative vector

dφ ds

along ~u , where ~u was a unit

dφ = ∇φ · ~u ds • Now the ~u becomes the unit vectors in an orthogonal system, for example in cylindrical coordinates • Now we recall that ds 2 = ds 2 = h12 dx12 + h22 dx22 + h32 dx32 • Let’s take a cylindrical system, first consider ~u = eˆr , then ds = dr ∂φ ~ ∇φ(r , θ, z) · eˆr = ∂r

Vector operators in general curvilinear coordinates • Next ~u = eˆr θ , then ds = rdθ (h2 = r ) 1 ∂φ ~ ∇φ(r , θ, z) · eˆθ = r ∂θ • It is also easy to show, ∂φ ~ ∇φ(r , θ, z) · eˆz = ∂z ~ in cylindrical • Now that we have the projections, we can find ∇φ coordinates, ~ = ∂φ eˆr + 1 ∂φ eˆθ + ∂φ eˆz ∇φ ∂r r ∂θ ∂z

Gradient in curvilinear (orthogonal) coordinate system • Most generally, we have ~ = ∇φ

3 X i=1

eˆi

1 ∂φ hi ∂xi

• In Cartesian, obviously h1 = h2 = h3 , and x1 = x, x2 = y , and x3 = z, ~ = ∂φˆi + ∂φ ˆj + ∂φ kˆ ∇φ ∂x ∂y ∂z • In a spherical coordinate system x1 = r , x2 = θ, and x3 = φ, then h1 = 1, h2 = r , and h3 = r sin θ ~ = ∂u eˆr + 1 ∂u eˆθ + 1 ∂u eˆφ ∇u ∂r r ∂θ r sin θ ∂φ

Divergence in curvilinear coordinates

~ = Vx ˆi + Vy ˆj + Vz k, ˆ • We recall in Cartesian coordinates, with V ∂ ∂ ∂ ~ = ˆi + ˆj + kˆ and the gradient operator ∇ ∂x ∂y ∂z ~ = ∂Vx + ∂Vy + ∂Vz ~ ·V ∇ ∂x ∂y ∂z ~ = V1 eˆ1 + V2 eˆ2 + V3 eˆ3 • In a general orthogonal system, V • The difficultly comes because the unit vectors in a general orthogonal coordinate system may not be fixed

Divergence in curvilinear coordinates, continued   ~ · eˆ3 = 0 (Problem 1) • First show that ∇ h1 h2 • Assume eˆ1 × eˆ2 = eˆ3 (orthogonal coordinate system), and then ~ 1 × ∇x ~ 2 = eˆ3 , and obviously ∇x1 = heˆ11 and ∇x2 = heˆ22 , and ∇x h1 h2 next     eˆ3 ~ · ∇x ~ 1 × ∇x ~ 2 ~ =∇ ∇· h1 h2 • The vector relations at the end of Chapter 6 help to work out the right-hand side,       ~ · ∇x ~ 1 × ∇x ~ 2 = ∇x ~ 2· ∇ ~ × ∇x ~ 1 − ∇x ~ 1· ∇ ~ × ∇x ~ 2 ∇ ~ ~ ~ ~ • But  we have ∇ × ∇x1 = ∇ × ∇x2 = 0, so we have shown e ˆ 3 ~ · ∇ h1 h2 = 0

Divergence in curvilinear coordinates, continued ~ · • We use then that ∇   ~ · eˆ2 = 0 ∇ h1 h3 ~ =∇ ~ ·V ~ · ∇





eˆ3 h1 h2



~ · = 0, and also ∇



eˆ1 h2 h3



= 0 and

eˆ2 eˆ3 eˆ1 h2 h3 V1 + h1 h3 V2 + h1 h2 V3 h2 h3 h1 h3 h1 h2



~)=V ~ · ∇φ ~ ~ · (φV ~ + φ∇ ~ ·V • We use then ∇ ~ = eˆ1 · ∇(h ~ ·V ~ 2 h3 V1 ) + eˆ2 · ∇(h ~ 1 h3 V2 ) + eˆ3 · ∇(h ~ 1 h2 V3 ) ∇ h2 h3 h1 h3 h1 h2 ~ 2 h3 V1 ) = • Then we see that eˆ1 · ∇(h

1 ∂ h1 ∂x1 (h2 h3 V1 )

,etc.

Divergence in curvilinear coordinates, final result! • Finally we get, ~ = ~ ·V ∇

  1 ∂ ∂ ∂ (h2 h3 V1 ) + (h1 h3 V2 ) + (h1 h2 V3 ) h1 h2 h3 ∂x1 ∂x2 ∂x3

• Example: Cylindrical coordinates, x1 = r , x2 = θ, and x3 = z, with h1 = 1, h2 = r , and h3 = 1 ~ = Vr eˆr + Vθ eˆθ + Vz eˆz • In cylindrical coordinates, V   1 ∂ ∂ ∂ ~ ~ ∇·V = (rVr ) + (Vθ ) + (rVz ) r ∂r ∂θ ∂z • Finally we simplify, ~ = 1 ∂ (rVr ) + 1 ∂Vθ + ∂Vz ~ ·V ∇ r ∂r r ∂θ ∂z

Another important example: Divergence in spherical coordinates ~ = Vr eˆr + +Vθ eˆθ + Vφ eˆφ • In spherical coordinates V • Here we have x1 = r , x2 = θ, and x3 = φ • The scale factors are h1 = 1, h2 = r , and h3 = r sin θ  
1 ∂ ∂ ∂ 2 ~ = ~ ·V ∇ r sin θVr + (r sin θVθ ) + (rVφ ) + r 2 sin θ ∂r ∂θ ∂φ • This simplifies to
~ = 1 ∂ r 2 Vr + 1 ∂Vθ + 1 ∂Vφ ~ ·V ∇ 2 r ∂r r ∂θ r sin θ ∂φ

Laplacian • We want to have an expression for ∇2 u for a general curvilinear system ~ = eˆ1 ∂u + eˆ2 ∂u + eˆ3 ∂u • We start with ∇u h1 ∂x1 h2 ∂x2 h3 ∂x3 ~ = V1 eˆ1 + V2 eˆ2 + V3 eˆ3 , we have V1 = 1 ∂u and • Then with V ∂u ∂u V2 = h12 ∂x and V3 = h13 ∂x 2 3 • Now we go back to our formula for the divergence,

~ = ~ ·V ∇

h1 ∂x1

  1 ∂ ∂ ∂ (h2 h3 V1 ) + (h1 h3 V2 ) + (h1 h2 V3 ) h1 h2 h3 ∂x1 ∂x2 ∂x3

~ · ∇u ~ is • The ∇2 u = ∇

       1 ∂ h2 h3 ∂u ∂ h1 h3 ∂u ∂ h1 h2 ∂u + + ∇ u= h1 h2 h3 ∂x1 h1 ∂x1 ∂x2 h2 ∂x2 ∂x3 h3 ∂x3 2

Example: ∇2 u in cylindrical coordinates

∇2 u =

       ∂ h2 h3 ∂u ∂ h1 h3 ∂u ∂ h1 h2 ∂u 1 + + h1 h2 h3 ∂x1 h1 ∂x1 ∂x2 h2 ∂x2 ∂x3 h3 ∂x3

• In cylindrical coordinates we have x1 = r , x2 = θ, and x3 = z, with h1 = 1, h2 = r , and h3 = 1        1 ∂ ∂u ∂ 1 ∂u ∂ ∂u 2 ∇ u= r + + r r ∂r ∂r ∂θ r ∂θ ∂z ∂z • This simplifies to 1 ∂ ∇ u= r ∂r 2



∂u r ∂r

 +

1 ∂2u ∂2u + 2 r 2 ∂θ2 ∂z

Example: ∇2 u in spherical coordinates

       1 ∂ h2 h3 ∂u ∂ h1 h3 ∂u ∂ h1 h2 ∂u ∇ u= + + h1 h2 h3 ∂x1 h1 ∂x1 ∂x2 h2 ∂x2 ∂x3 h3 ∂x3 2

• In spherical coordinates we have x1 = r , x2 = θ, and x3 = φ, with h1 = 1, h2 = r , and h3 = r sin θ

∇2 u =

       ∂ 1 ∂u 1 ∂u ∂ ∂u ∂ 2 r sin θ + sin θ + r 2 sin θ ∂r ∂r ∂θ ∂θ ∂φ sin θ ∂φ

• This simplifies to

∇2 u =

1 ∂ r 2 ∂r

    ∂u 1 ∂ ∂u 1 ∂2u r2 + 2 sin θ + 2 2 ∂r r sin θ ∂θ ∂θ r sin θ ∂φ2

Curl in curvilinear coordinates • As before we work with some general system and vector ~ = V1 eˆ1 + V2 eˆ2 + V3 eˆ3 V ~ 1 = eˆ1 , and then • Derivation following problem 2, start with ∇x h1 ~ × ∇x ~ 1=0 ∇ ~ × eˆ1 = ∇ ~ × eˆ2 = ∇ ~ × eˆ3 = 0 • Hence we see ∇ h1 h2 h3 ~ = eˆ1 (h1 V1 ) + eˆ2 (h2 V2 ) + eˆ3 (h3 V3 ) • Write V h1

h2

h3

~ = φ(∇ ~ −U ~ × (∇φ) ~ × (φU) ~ × U) ~ • Now we use the relation ∇ ~ = − eˆ1 × ∇(h ~ 1 V1 ) − eˆ2 × ∇(h ~ 2 V2 ) − eˆ3 × ∇(h ~ 3 V3 ) ~ ×V ∇ h1 h2 h3 • Consider just the first term, to keep it simple,   eˆ1 ~ eˆ1 eˆ1 ∂(h1 V1 ) eˆ2 ∂(h1 V1 ) eˆ3 ∂(h1 V1 ) − ×∇(h1 V1 ) = − × + + h1 h1 h1 ∂x1 h2 ∂x2 h3 ∂x3

Curl in curvilinear coordinates, continued

• Next we just use eˆ1 × eˆ1 = 0, eˆ1 × eˆ2 = eˆ3 , and eˆ1 × eˆ3 = −ˆ e2

−

  ∂(h1 V1 ) ∂(h1 V1 ) eˆ1 ~ 1 V1 ) = − 1 × ∇(h h3 eˆ3 − h2 eˆ2 h1 h1 h2 h3 ∂x2 ∂x3

• We can do the other terms as well, and the final result is expressed as a determinant h1 eˆ1 h2 eˆ2 h3 eˆ3 1 ∂ ∂ ∂ ~ = ~ ×V ∇ ∂x1 ∂x2 ∂x3 h1 h2 h3 h1 V1 h2 V2 h3 V3

Example: Curl in cylindrical coordiates h eˆ h2 eˆ2 h3 eˆ3 1 1∂ 1 ∂ ∂ ~ ~ ∇×V = ∂x2 ∂x3 h1 h2 h3 ∂x1 h1 V1 h2 V2 h3 V3 • In cylindrical coordinates we have x1 = r , x2 with h1 = 1, h2 = r , and h3 = 1 eˆ r eˆθ eˆz 1 ∂r ∂ ∂ ~ ~ ∇ × V = ∂r ∂θ ∂z r Vr rVθ Vz



= θ, and x3 = z,

• We can evaluate the determinant

~ = ~ V ∇×



1 ∂Vz ∂Vθ − r ∂θ ∂z



    ∂Vr ∂Vz 1 ∂ ∂Vr − (rVθ ) − eˆr + eˆθ + eˆz ∂z ∂r r ∂r ∂θ

Example: Curl in spherical coordinate

1 ~ = ~ ×V ∇ h1 h2 h3

h1 eˆ1 h2 eˆ2 h3 eˆ3 ∂ ∂ ∂ ∂x2 ∂x3 ∂x1 h1 V1 h2 V2 h3 V3



• In spherical coordinates we have x1 = r , x2 = φ, and x3 = θ, with h1 = 1, h2 = r sin θ, and h3 = r eˆr r eˆθ r sin θˆ eφ 1 ∂ ∂ ∂ ~ = ~ ×V ∇ ∂φ r 2 sin θ ∂r ∂θ Vr rVθ r sin θVφ ~ ~ ×V • We can evaluate the determinant to get ∇