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Vector Operators in Curvilinear Coordinate Systems

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Vector Operators in Curvilinear Coordinate Systems 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 ^1h1dx1 +e ^2h2dx2 +e ^3h3dx3 • Back to our example of cylindrical coordiantes,e ^1 =e ^r ,e ^2 =e ^θ, ande ^3 =e ^z , and ~ ds =e ^r dr +e ^θrdθ +e ^z 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 = rφ · ~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 ^r , then ds = dr @φ r~ φ(r; θ; z) · e^ = r @r Vector operators in general curvilinear coordinates • Next ~u =e ^rθ, then ds = rdθ (h2 = r) 1 @φ r~ φ(r; θ; z) · e^ = θ r @θ • It is also easy to show, @φ r~ φ(r; θ; z) · e^ = z @z • Now that we have the projections, we can find r~ φ in cylindrical coordinates, @φ 1 @φ @φ r~ φ = e^ + e^ + e^ @r r r @θ θ @z z Gradient in curvilinear (orthogonal) coordinate system • Most generally, we have 3 X 1 @φ r~ φ = e^i hi @xi i=1 • In Cartesian, obviously h1 = h2 = h3, and x1 = x, x2 = y, and x3 = z, @φ @φ @φ r~ φ = ^i + ^j + k^ @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 r~ u = e^ + e^ + e^ @r r r @θ θ r sin θ @φ φ Divergence in curvilinear coordinates • We recall in Cartesian coordinates, with V~ = Vx^i + Vy^j + Vz k^, ~ @ ^ @ ^ @ ^ and the gradient operator r = @x i + @y j + @z k @V @V @V r~ · V~ = x + y + z @x @y @z • In a general orthogonal system, V~ = V1e^1 + V2e^2 + V3e^3 • 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 r~ · e^3 = 0 (Problem 1) h1h2 • Assumee ^1 × e^2 =e ^3 (orthogonal coordinate system), and then obviously rx = e^1 and rx = e^2 , and r~ x × r~ x = e^3 , and 1 h1 2 h2 1 2 h1h2 next e^3 r~ · = r~ · r~ x1 × r~ x2 h1h2 • The vector relations at the end of Chapter 6 help to work out the right-hand side, r~ · r~ x1 × r~ x2 = r~ x2 · r~ × r~ x1 − r~ x1 · r~ × r~ x2 • But we have r~ × r~ x = r~ × r~ x = 0, so we have shown 1 2 r~ · e^3 = 0 h1h2 Divergence in curvilinear coordinates, continued • We use then that r~ · e^3 = 0, and also r~ · e^1 = 0 and h1h2 h2h3 r~ · e^2 = 0 h1h3 e^1 e^2 e^3 r~ · V~ = r~ · h2h3V1 + h1h3V2 + h1h2V3 h2h3 h1h3 h1h2 • We use then r~ · (φV~ ) = V~ · r~ φ + φr~ · V~ e^1 e^2 e^3 r~ · V~ = · r~ (h2h3V1) + · r~ (h1h3V2) + · r~ (h1h2V3) h2h3 h1h3 h1h2 • Then we see thate ^ · r~ (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 @ @ @ r~ · 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 e^r + Vθe^θ + Vz e^z 1 @ @ @ r~ · V~ = (rV ) + (V ) + (rV ) r @r r @θ θ @z z • Finally we simplify, 1 @ 1 @V @V r~ · V~ = (rV ) + θ + z r @r r r @θ @z Another important example: Divergence in spherical coordinates ~ • In spherical coordinates V = Vr e^r + +Vθe^θ + Vφe^φ • Here we have x1 = r, x2 = θ, and x3 = φ • The scale factors are h1 = 1, h2 = r, and h3 = r sin θ 1 @ @ @ r~ · 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 r~ · V~ = r 2V + θ + φ r 2 @r r r @θ r sin θ @φ Laplacian • We want to have an expression for r2u for a general curvilinear system • We start with r~ u = e^1 @u + e^2 @u + e^3 @u h1 @x1 h2 @x2 h3 @x3 • Then with V~ = V e^ + V e^ + V e^ , 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 @ @ @ r~ · V~ = (h2h3V1) + (h1h3V2) + (h1h2V3) h1h2h3 @x1 @x2 @x3 • The r2u = r~ · r~ u is 1 @ h h @u @ h h @u @ h h @u r2u = 2 3 + 1 3 + 1 2 h1h2h3 @x1 h1 @x1 @x2 h2 @x2 @x3 h3 @x3 Example: r2u in cylindrical coordinates 1 @ h h @u @ h h @u @ h h @u r2u = 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 r2u = r + + r r @r @r @θ r @θ @z @z • This simplifies to 1 @ @u 1 @2u @2u r2u = r + + r @r @r r 2 @θ2 @z2 Example: r2u in spherical coordinates 1 @ h h @u @ h h @u @ h h @u r2u = 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 r2u = 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 r2u = 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~ = V1e^1 + V2e^2 + V3e^3 • Derivation following problem 2, start with r~ x = e^1 , and then 1 h1 r~ × r~ x1 = 0 • Hence we see r~ × e^1 = r~ × e^2 = r~ × e^3 = 0 h1 h2 h3 • Write V~ = e^1 (h V ) + e^2 (h V ) + e^3 (h V ) h1 1 1 h2 2 2 h3 3 3 • Now we use the relation r~ × (φU~ ) = φ(r~ × U~ ) − U~ × (r~ φ) e^1 e^2 e^3 r~ × V~ = − × r~ (h1V1) − × r~ (h2V2) − × r~ (h3V3) h1 h2 h3 • Consider just the first term, to keep it simple, e^1 e^1 e^1 @(h1V1) e^2 @(h1V1) e^3 @(h1V1) − ×r~ (h1V1) = − × + + h1 h1 h1 @x1 h2 @x2 h3 @x3 Curl in curvilinear coordinates, continued • Next we just usee ^1 × e^1 = 0,e ^1 × e^2 =e ^3, ande ^1 × e^3 = −e^2 e^1 1 @(h1V1) @(h1V1) − × r~ (h1V1) = − h3e^3 − h2e^2 h1 h1h2h3 @x2 @x3 • We can do the other terms as well, and the final result is expressed as a determinant h1e^1 h2e^2 h3e^3 ~ ~ 1 @ @ @ r × V = @x @x @x h1h2h3 1 2 3 h1V1 h2V2 h3V3 Example: Curl in cylindrical coordiates h1e^1 h2e^2 h3e^3 ~ ~ 1 @ @ @ r × 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 e^r re^θ e^z 1 r~ × V~ = @ @ @ r @r @θ @z Vr rVθ Vz • We can evaluate the determinant 1 @V @V @V @V 1 @ @V ∇×~ V~ = z − θ e^ + r − z e^ + (rV ) − r e^ r @θ @z r @z @r θ r @r θ @θ z Example: Curl in spherical coordinate h1e^1 h2e^2 h3e^3 ~ ~ 1 @ @ @ r × 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 e^r re^θ r sin θe^φ 1 @ @ @ r~ × V~ = r 2 sin θ @r @θ @φ Vr rVθ r sin θVφ • We can evaluate the determinant to get r~ × V~.
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