Vector operators in curvilinear coordinate systems

• In a Cartesian system, take x1 = x, x2 = y, and x3 = z, then an element of arc length ds2 is,

2 2 2 2 ds = dx1 + dx2 + dx3

• 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 ds2 in cylindrical coordinates,

2 2 2 2 2 2 2 2 ds = dr + r dθ + dz = dx1 + x1 dx2 + dx3

• We write this in a general form, with hi being the scale factors

2 2 2 2 2 2 2 ds = h1dx1 + h2dx2 + h3dx3

• We see then for cylindrical coordinates, h1 = 1, h2 = r, and h3 = 1 Curvilinear coordinates

• For an vector displacement ds~

ds~ =e ˆ1h1dx1 +e ˆ2h2dx2 +e ˆ3h3dx3

• Back to our example of cylindrical coordiantes,e ˆ1 =e ˆr ,e ˆ2 =e ˆθ, ande ˆ3 =e ˆz , and

~ ds =e ˆr dr +e ˆθrdθ +e ˆz dz • These are orthogonal systems, but it would not have to be!

3 3 2 X X ds = gij dxi dxi i=1 j=1

• The gij is the metric tensor, and for an orthogonal system it is 2 diagonal with gi = hi Vector operators in general curvilinear coordinates

dφ • Recall the directional derivative ds along ~u, where ~u was a unit vector

dφ = ∇φ · ~u ds • Now the ~u becomes the unit vectors in an orthogonal system, for example in cylindrical coordinates 2 2 2 2 2 2 2 2 • Now we recall that ds = ds = h1dx1 + h2dx2 + h3dx3 • Let’s take a cylindrical system, first consider ~u =e ˆr , then ds = dr

∂φ ∇~ φ(r, θ, z) · eˆ = r ∂r Vector operators in general curvilinear coordinates

• Next ~u =e ˆrθ, then ds = rdθ (h2 = r) 1 ∂φ ∇~ φ(r, θ, z) · eˆ = θ r ∂θ • It is also easy to show,

∂φ ∇~ φ(r, θ, z) · eˆ = z ∂z • Now that we have the projections, we can find ∇~ φ in cylindrical coordinates,

∂φ 1 ∂φ ∂φ ∇~ φ = eˆ + eˆ + eˆ ∂r r r ∂θ θ ∂z z Gradient in curvilinear (orthogonal) coordinate system

• Most generally, we have

3 X 1 ∂φ ∇~ φ = eˆi hi ∂xi i=1

• In Cartesian, obviously h1 = h2 = h3, and x1 = x, x2 = y, and x3 = z, ∂φ ∂φ ∂φ ∇~ φ = ˆi + ˆj + kˆ ∂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 1 ∂u 1 ∂u ∇~ u = eˆ + eˆ + eˆ ∂r r r ∂θ θ r sin θ ∂φ φ Divergence in curvilinear coordinates

• We recall in Cartesian coordinates, with V~ = Vxˆi + Vyˆj + Vz kˆ, ~ ∂ ˆ ∂ ˆ ∂ ˆ and the gradient operator ∇ = ∂x i + ∂y j + ∂z k ∂V ∂V ∂V ∇~ · V~ = x + y + z ∂x ∂y ∂z

• In a general orthogonal system, V~ = V1eˆ1 + V2eˆ2 + V3eˆ3 • The difficultly comes because the unit vectors in a general orthogonal coordinate system may not be fixed Divergence in curvilinear coordinates, continued   • First show that ∇~ · eˆ3 = 0 (Problem 1) h1h2 • Assumee ˆ1 × eˆ2 =e ˆ3 (orthogonal coordinate system), and then obviously ∇x = eˆ1 and ∇x = eˆ2 , and ∇~ x × ∇~ x = eˆ3 , and 1 h1 2 h2 1 2 h1h2 next   eˆ3   ∇~ · = ∇~ · ∇~ x1 × ∇~ x2 h1h2 • The vector relations at the end of Chapter 6 help to work out the right-hand side,

      ∇~ · ∇~ x1 × ∇~ x2 = ∇~ x2 · ∇~ × ∇~ x1 − ∇~ x1 · ∇~ × ∇~ x2

• But we have ∇~ × ∇~ x = ∇~ × ∇~ x = 0, so we have shown   1 2 ∇~ · eˆ3 = 0 h1h2 Divergence in curvilinear coordinates, continued

    • We use then that ∇~ · eˆ3 = 0, and also ∇~ · eˆ1 = 0 and h1h2 h2h3   ∇~ · eˆ2 = 0 h1h3

  eˆ1 eˆ2 eˆ3 ∇~ · V~ = ∇~ · h2h3V1 + h1h3V2 + h1h2V3 h2h3 h1h3 h1h2

• We use then ∇~ · (φV~ ) = V~ · ∇~ φ + φ∇~ · V~

eˆ1 eˆ2 eˆ3 ∇~ · V~ = · ∇~ (h2h3V1) + · ∇~ (h1h3V2) + · ∇~ (h1h2V3) h2h3 h1h3 h1h2

• Then we see thate ˆ · ∇~ (h h V ) = 1 ∂ (h h V ) ,etc. 1 2 3 1 h1 ∂x1 2 3 1 Divergence in curvilinear coordinates, final result!

• Finally we get,

1  ∂ ∂ ∂  ∇~ · V~ = (h2h3V1) + (h1h3V2) + (h1h2V3) h1h2h3 ∂x1 ∂x2 ∂x3

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

1 ∂ 1 ∂V ∂V ∇~ · V~ = (rV ) + θ + z r ∂r r r ∂θ ∂z Another important example: Divergence in spherical coordinates

~ • In spherical coordinates V = Vr eˆr + +Vθeˆθ + Vφeˆφ • Here we have x1 = r, x2 = θ, and x3 = φ • The scale factors are h1 = 1, h2 = r, and h3 = r sin θ

1  ∂ ∂ ∂  ∇~ · V~ = r 2 sin θV
+ (r sin θV ) + (rV ) + r 2 sin θ ∂r r ∂θ θ ∂φ φ

• This simplifies to

1 ∂ 1 ∂V 1 ∂V ∇~ · V~ = r 2V
+ θ + φ r 2 ∂r r r ∂θ r sin θ ∂φ Laplacian

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

1  ∂ ∂ ∂  ∇~ · V~ = (h2h3V1) + (h1h3V2) + (h1h2V3) h1h2h3 ∂x1 ∂x2 ∂x3

• The ∇2u = ∇~ · ∇~ u is

1  ∂ h h ∂u  ∂ h h ∂u  ∂ h h ∂u  ∇2u = 2 3 + 1 3 + 1 2 h1h2h3 ∂x1 h1 ∂x1 ∂x2 h2 ∂x2 ∂x3 h3 ∂x3 Example: ∇2u in cylindrical coordinates

1  ∂ h h ∂u  ∂ h h ∂u  ∂ h h ∂u  ∇2u = 2 3 + 1 3 + 1 2 h1h2h3 ∂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  ∇2u = r + + r r ∂r ∂r ∂θ r ∂θ ∂z ∂z • This simplifies to

1 ∂  ∂u  1 ∂2u ∂2u ∇2u = r + + r ∂r ∂r r 2 ∂θ2 ∂z2 Example: ∇2u in spherical coordinates

1  ∂ h h ∂u  ∂ h h ∂u  ∂ h h ∂u  ∇2u = 2 3 + 1 3 + 1 2 h1h2h3 ∂x1 h1 ∂x1 ∂x2 h2 ∂x2 ∂x3 h3 ∂x3

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

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

• This simplifies to

1 ∂  ∂u  1 ∂  ∂u  1 ∂2u ∇2u = r 2 + sin θ + r 2 ∂r ∂r r 2 sin θ ∂θ ∂θ r 2 sin2 θ ∂φ2 Curl in curvilinear coordinates

• As before we work with some general system and vector V~ = V1eˆ1 + V2eˆ2 + V3eˆ3 • Derivation following problem 2, start with ∇~ x = eˆ1 , and then 1 h1 ∇~ × ∇~ x1 = 0 • Hence we see ∇~ × eˆ1 = ∇~ × eˆ2 = ∇~ × eˆ3 = 0 h1 h2 h3 • Write V~ = eˆ1 (h V ) + eˆ2 (h V ) + eˆ3 (h V ) h1 1 1 h2 2 2 h3 3 3 • Now we use the relation ∇~ × (φU~ ) = φ(∇~ × U~ ) − U~ × (∇~ φ)

eˆ1 eˆ2 eˆ3 ∇~ × V~ = − × ∇~ (h1V1) − × ∇~ (h2V2) − × ∇~ (h3V3) h1 h2 h3 • Consider just the first term, to keep it simple,

  eˆ1 eˆ1 eˆ1 ∂(h1V1) eˆ2 ∂(h1V1) eˆ3 ∂(h1V1) − ×∇~ (h1V1) = − × + + h1 h1 h1 ∂x1 h2 ∂x2 h3 ∂x3 Curl in curvilinear coordinates, continued

• Next we just usee ˆ1 × eˆ1 = 0,e ˆ1 × eˆ2 =e ˆ3, ande ˆ1 × eˆ3 = −eˆ2

  eˆ1 1 ∂(h1V1) ∂(h1V1) − × ∇~ (h1V1) = − h3eˆ3 − h2eˆ2 h1 h1h2h3 ∂x2 ∂x3 • We can do the other terms as well, and the final result is expressed as a determinant

h1eˆ1 h2eˆ2 h3eˆ3 ~ ~ 1 ∂ ∂ ∂ ∇ × V = ∂x ∂x ∂x h1h2h3 1 2 3 h1V1 h2V2 h3V3 Example: Curl in cylindrical coordiates

h1eˆ1 h2eˆ2 h3eˆ3 ~ ~ 1 ∂ ∂ ∂ ∇ × V = ∂x ∂x ∂x h1h2h3 1 2 3 h1V1 h2V2 h3V3

• In cylindrical coordinates we have x1 = r, x2 = θ, and x3 = z, with h1 = 1, h2 = r, and h3 = 1

eˆr reˆθ eˆz 1 ∇~ × V~ = ∂ ∂ ∂ r ∂r ∂θ ∂z Vr rVθ Vz • We can evaluate the determinant

1 ∂V ∂V  ∂V ∂V  1  ∂ ∂V  ∇×~ V~ = z − θ eˆ + r − z eˆ + (rV ) − r eˆ r ∂θ ∂z r ∂z ∂r θ r ∂r θ ∂θ z Example: Curl in spherical coordinate

h1eˆ1 h2eˆ2 h3eˆ3 ~ ~ 1 ∂ ∂ ∂ ∇ × V = ∂x ∂x ∂x h1h2h3 1 2 3 h1V1 h2V2 h3V3

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

eˆr reˆθ r sin θeˆφ 1 ∂ ∂ ∂ ∇~ × V~ = r 2 sin θ ∂r ∂θ ∂φ Vr rVθ r sin θVφ • We can evaluate the determinant to get ∇~ × V~