Cylindrical and Spherical Coordinates The Cartesian coordinate - - PowerPoint PPT Presentation

Cylindrical and Spherical Coordinates

The Cartesian coordinate system is by far the simplest, the most universal and the most important. There are some situations for which the Cartesian coordinate sys- tem is not entirely ideal. These typically involve scalar or vector fields which exhibit some kind of inherent symmetry. The cylindrical and spherical coordinate systems are designed for just this purpose.

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Polar coordinates

The point A is represented by (r, θ), which has a very different interpretation from the Cartesian pair (x, y). While −∞ < x < ∞ and −∞ < y < ∞, the polar coordinates

• bey

0 ≤ r < ∞, 0 ≤ θ < 2π.

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Polar coordinates (cont.)

The Cartesian and polar coordinates are related as x = r cos θ, y = r sin θ. Thus, in terms of the Cartesian basis, any point in R2 can be represented as v(r, θ) = r cos θ i + r sin θ j. Conversely, r =

• x2 + y2

and θ =      arctan(y/x), x > 0, y > 0, π + arctan(y/x), x < 0, y ∈ R, 2π + arctan(y/x), x > 0, y < 0.

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Standard vectors

Any v ∈ R2 can be represented in terms of the standard Carte- sian unit vectors i and j as v = x i + y j. What are the “standard” vectors which are better suited to polar coordinates? To answer this question, for A with coordinates (r0, θ0), we define the following two paths: γ1(r) := v(r, θ0) = r cos θ0 i + r sin θ0 j and γ2(θ) := v(r0, θ) = r0 cos θ i + r0 sin θ j.

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Standard vectors (cont.)

The tangent vectors at (r0, θ0) are given by γ(1)

1 (r0) = dγ1(r)

dr

• r=r0

γ(1)

2 (θ0) = dγ2(θ)

dθ

• θ=θ0

Thus, after normalization, we obtain the two standard vectors er(r0, θ0) := γ(1)

1 (r0)

γ(1)

1 (r0)

= cos θ0 i + sin θ0 j eθ(r0, θ0) := γ(1)

2 (θ0)

γ(1)

2 (θ0)

= − sin θ0 i + cos θ0 j The pair of orthogonal basis vectors {er(r0, θ0), eθ(r0, θ0)} is cal- led the coordinate frame at the point A.

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Coordinate frame

Notice that {er(r, θ), eθ(r, θ)} change direction as the point (r, θ) moves, so that we have a moving coordinate frame. This is in direct contrast to the Cartesian unit basis vectors {i, j} which are constant throughout R2.

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Riemannian scale functions

How does length changes when we change the polar coordinates? When “moving” from A to B, the Euclidean distance is defined by (ds)2 = (dx)2 + (dy)2. At the same time, in the polar co-

• rdinates, ds is equal to

(ds)2 = (dr)2 + r2(dθ)2 = = [hr(r, θ)dr]2 + [hθ(r, θ)dθ]2, where hr(r, θ) := 1, hθ(r, θ) = r are the Riemann scale functions.

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Cylindrical coordinates

Let (r, θ) be the polar coordinates of the point B. Then the triplet of real numbers (r, θ, z) (with z ∈ R) completely specifies the point A.

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Cylindrical coordinates (cont.)

If A is given by the cylindrical coordinates (r, θ, z) then, in terms

• f the Cartesian basis {i, j, k}, it must be given by the vector

v(r, θ, z) = r cos(θ) i + r sin(θ) j + z k. To find the standard coordinate frame, we define three curves re- presented by the following paths: γ1(r) := v(r, θ0, z0) = r cos(θ0) i + r sin(θ0) j + z0 k, ∀r ∈ (−∞, 0], γ2(θ) := v(r0, θ, z0) = r0 cos(θ) i + r0 sin(θ) j + z0 k, ∀θ ∈ [0, 2π), γ3(z) := v(r0, θ0, z) = r0 cos(θ0) i + r0 sin(θ0) j + z k, ∀z ∈ R. Next, we compute the tangent vectors γ(1)

1 (r0), γ(1) 2 (θ0), and

γ(1)

3 (z0), followed by their normalization to obtain er(r0, θ0, z0),

eθ(r0, θ0, z0), and ez(r0, θ0, z0), respectively.

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Cylindrical coordinates (cont.)

The cylindrical coordinate frame is defined by er(r, θ, z) = cos θ i + sin θ j + 0 k, eθ(r, θ, z) = − sin θ i + cos θ j + 0 k, ez(r, θ, z) = 0 i + 0 j + 1 k. One can see that the unit vectors are orthogonal with respect to each other. Also, the triplet of vectors constitutes a moving coordinate frame.

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Cylindrical coordinates (cont.)

Due to a small “permutation” the length ds can be defined as follows. (ds)2 = (length AB)2 = (length AD)2+ +(length AC)2 + (length AE)2 = = (dr)2 + r2(dθ)2 + (dz)2. Alternatively, we can write (ds)2 = [hr(r, θ, z)dr]2 + [hθ(r, θ, z)dθ]2 + [hz(r, θ, z)dz]2, where hr(r, θ, z) := 1, hθ(r, θ, z) := r, hz(r, θ, z) := 1 are the Riemannian scale functions.

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Spherical coordinates

The radial length of the vector from the origin O to point A is ρ =

• x2 + y2 + z2.

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Spherical coordinates (cont.)

The Cartesian coordinates (x, y, z) are given in terms of the spherical coordinates (ρ, θ, φ) by x = ρ sin φ cos θ, y = ρ sin φ sin θ, z = ρ cos φ, with ρ ∈ [0, ∞), θ ∈ [0, 2π), and φ ∈ [0, π]. Then, in terms of the Cartesian basis {i, j, k}, any point A with coordinates (ρ, θ, φ) can be expressed as v(ρ, θ, φ) = ρ sin φ cos θ i + ρ sin φ sin θ j + ρ cos φ k. This shows how the Cartesian representation of a point changes when we change its spherical coordinates.

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Spherical coordinates (cont.)

To define a moving (orthonormal) frame, we start with γ1(ρ) = v(ρ, θ0, φ0) = ρ sin φ0 cos θ0 i + ρ sin φ0 sin θ0 j + ρ cos φ0 k, γ2(θ) = v(ρ0, θ, φ0) = ρ0 sin φ0 cos θ i + ρ0 sin φ0 sin θ j + ρ0 cos φ0 k, γ3(φ) = v(ρ, θ0, φ) = ρ0 sin φ cos θ0 i + ρ0 sin φ sin θ0 j + ρ0 cos φ k.

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Spherical coordinates (cont.)

The corresponding tangent vectors at (ρ0, θ0, φ0) are given by γ(1)

1 (ρ0) = dγ1(ρ)

dρ

• ρ0, γ(1)

2 (θ0) = dγ2(θ)

dθ

• θ0, γ(1)

3 (φ0) = dγ3(φ)

dφ

• φ0.

Normalizing the tangent vectors results in the spherical coordi- nate frame {eρ, eθ, eφ} which are eρ(ρ, θ, φ) = sin φ cos θ i + sin φ sin θ j + cos φ k, eθ(ρ, θ, φ) = − sin θ i + cos θ j + 0 k, eφ(ρ, θ, φ) = cos φ cos θ i + cos φ sin θ j − sin φ k, under the replacement of the generic (i.e., “naughted”) spherical coordinates (ρ0, θ0, φ0) with (ρ, θ, φ) (to simplify the notations). One can check that the unit vectors indeed orthogonal w.r.t. each

• ther.

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Spherical coordinates (cont.)

Suppose a point A is given by v(ρ, θ, φ), while a close-by point B is given by (ρ + dρ, θ + dθ, φ + dφ). Let’s find an expression of ds in terms of (ρ, θ, φ). Formally we have (ds)2 = (length AB)2 = = v(ρ + dρ, θ + dθ, φ + dφ) − v(ρ, θ, φ)2

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Spherical coordinates (cont.)

First we note that v(ρ + dρ, θ + dθ, φ + dφ) − v(ρ, θ, φ) = = ∂v ∂ρ (ρ, θ, φ)dρ + ∂v ∂θ (ρ, θ, φ)dθ + ∂v ∂φ(ρ, θ, φ)dφ = = (dρ) eρ(ρ, θ, φ) + (ρ sin φ dθ) eθ(ρ, θ, φ) + (ρ dφ) eφ(ρ, θ, φ). Thus, we have (ds)2 = (dρ)2 + (ρ sin φ)2(dθ)2 + (ρ)2(dφ)2,

• r, alternatively (in the Riemannian form),

(ds)2 = (hρ(ρ, θ, φ)dρ)2 + (hθ(ρ, θ, φ)dθ)2 + (hφ(ρ, θ, φ)dφ)2, with the Riemannian scaling functions defined as hρ(ρ, θ, φ) := 1, hθ(ρ, θ, φ) := ρ sin φ, hφ(ρ, θ, φ) := ρ.

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Differential operators

Let F : R3 → R3 be a C1-vector field. The divergence of F is defined as div F(x, y, z) = ∂F1 ∂x (x, y, z) + ∂F2 ∂y (x, y, z) + ∂F3 ∂z (x, y, z), for all (x, y, z) in R3. Alternatively, we can write div F = ∇ · F, where ∇ := (∂/∂x) i + (∂/∂y) j + (∂/∂z) k. The gradient operator ∇ is of fundamental importance, since all the other operators (viz., divergence, curl, and Laplacian) are de- fined by it.

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Gradient in cylindrical coordinates

Recall that the cylindrical coordinate frame is defined as er(r, θ, z) = cos θ i + sin θ j + 0 k (= cos θ i + sin θ j), eθ(r, θ, z) = − sin θ i + cos θ j + 0 k (= − sin θ i + cos θ j), ez(r, θ, z) = 0 i + 0 j + 1 k (= k). By simple manipulations, we obtain i = cos θ er − sin θ eθ, j = sin θ er + cos θ eθ, k = ez. Next, we are going to substitute the above into ∇ = ∂ ∂x i + ∂ ∂y j + ∂ ∂z k.

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Gradient in cylindrical coordinates (cont.)

The substitution yields ∇ = ∂ ∂r ∂r ∂x + ∂ ∂θ ∂θ ∂x + ∂ ∂z ∂z ∂x

• [cos θ er − sin θ eθ] +

+ ∂ ∂r ∂r ∂y + ∂ ∂θ ∂θ ∂y + ∂ ∂z ∂z ∂y

• [sin θ er + cos θ eθ] +

+ ∂ ∂r ∂r ∂z + ∂ ∂θ ∂θ ∂z + ∂ ∂z ∂z ∂z

• [ez] .

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Gradient in cylindrical coordinates (cont.)

The substitution yields ∇ =     ∂ ∂r ∂r ∂x

• cos θ

+ ∂ ∂θ ∂θ ∂x

• − sin θ/r

+ ∂ ∂z ∂z ∂x

• 

   [cos θ er − sin θ eθ] + +      ∂ ∂r ∂r ∂y

• sin θ

+ ∂ ∂θ ∂θ ∂y

• cos θ/r

+ ∂ ∂z ∂z ∂y

• 

    [sin θ er + cos θ eθ] + +     ∂ ∂r ∂r ∂z

• + ∂

∂θ ∂θ ∂z

• + ∂

∂z ∂z ∂z

• 1

    [ez] .

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Gradient in cylindrical coordinates (cont.)

After a few trivial steps, we obtain ∇ = ∂ ∂r er + 1 r ∂ ∂θ eθ + ∂ ∂z ez. Let f(ρ, θ, φ) be a C1-scalar field. Then ∇f(ρ, θ, φ) = ∂f ∂r (r, θ, z) er + 1 r ∂f ∂θ (r, θ, z) eθ + ∂f ∂z (r, θ, z) ez. Using the above definition, we can now work out ∇·, ∇×, and ∆ for any given field. In doing so, it will be important to remember that the cylindrical coordinate frame is not constant.

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Divergence in cylindrical coordinates

Given a vector field F(r, θ, z) = F1(r, θ, z) er + F2(r, θ, z) eθ + F3(r, θ, z) ez, consider its divergence ∇ · F = ∂ ∂r er + 1 r ∂ ∂θ eθ + ∂ ∂z ez

• · (F1 er + F2 eθ + F3 ez) .

Note that “opening the brackets” yields a total of nine terms. To compute these terms, we do the differentiation first and then the dot (or cross) product. Let’s look at some specific cases.

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Divergence in cylindrical coordinates (cont.)

∂ ∂r er

• · (F1 er) = er · ∂

∂r (F1 er) = er · ∂F1 ∂r er + F1 ∂er ∂r

• =

= ∂F1 ∂r er · er + F1er · 0 = ∂F1 ∂r . 1 r ∂ ∂θ eθ

• · (F1 er) = eθ

1 r · ∂ ∂θ (F1 er) = eθ 1 r ·     ∂F1 ∂θ er + F1 ∂er ∂θ

• eθ

    = F1 r . 1 r ∂ ∂θ eθ

• · (F2 eθ) = eθ

1 r · ∂ ∂θ(F2eθ) = eθ 1 r ·     ∂F2 ∂θ eθ + F2 ∂eθ ∂θ

• −er

    = 1 r ∂F2 ∂θ .

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Divergence, curl and, Laplacian

Proceeding in a similar manner, we finally obtain ∇ · F = ∂F1 ∂r + 1 r F1 + 1 r ∂F2 ∂θ + ∂F3 ∂z = = 1 r ∂(r F1) ∂r + 1 r ∂F2 ∂θ + ∂F3 ∂z . Moreover, one can show that ∇ × F = 1 r ∂F3 ∂θ − ∂F2 ∂z

• er +

∂F1 ∂z − ∂F3 ∂r

• eθ+

+1 r ∂(rF2) ∂r − ∂F1 ∂θ

• ez

and ∆f = ∂2f ∂r2 + 1 r ∂f ∂r + 1 r2 ∂2f ∂θ2 + ∂2f ∂z2 .

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Gradient in spherical coordinates

Consider a scalar field f(ρ, θ, φ) and a vector field F(ρ, θ, φ) = F1(ρ, θ, φ)eρ + F2(ρ, θ, φ)eθ + F3(ρ, θ, φ)eφ defined in the spherical coordinate systems. In this case, the derivation of ∇, ∇·, ∇×, and ∆ are analogous to the previous case. Particularly, one can show that ∇ = ∂ ∂ρeρ + 1 ρ sin φ ∂ ∂θeθ + 1 ρ ∂ ∂φeφ, so that for a C1 function f(ρ, θ, φ) we have ∇f(ρ, θ, φ) = ∂f ∂ρ (ρ, θ, φ)eρ+ 1 ρ sin φ ∂f ∂θ (ρ, θ, φ)eθ+1 ρ ∂f ∂φ(ρ, θ, φ)eφ.

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Other differential operators

Proceeding in a manner analogous to the cylindrical case, we also get (for a scalar f field and a vector field F = ( F1

• ρ

, F2

• θ

, F3

• φ

)) ∇ · F = 1 ρ2 ∂(ρ2F1) ∂ρ + 1 ρ sin φ ∂F2 ∂θ + 1 ρ sin φ ∂(sin φ F3) ∂φ , ∇ × F = 1 ρ sin φ ∂(sin φ F3) ∂φ − ∂F2 ∂θ

• eρ + 1

ρ ∂(ρF3) ∂ρ − ∂F1 ∂φ

• eθ+

+1 ρ

• 1

sin φ ∂F1 ∂θ − ∂(ρF2) ∂ρ

• eφ,

and ∆f = 1 ρ2 ∂ ∂ρ

• ρ2 ∂f

∂ρ

• +

1 sin2 φ ∂2f ∂θ2 + 1 sin φ ∂ ∂φ

• sin φ∂f

∂φ

• .

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