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 system 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 obey 0 ≤ r < ∞, 0 ≤ θ < 2π. Department of ECE, Fall 2014
ECE 206: Advanced Calculus 2
2/27
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= and
p x2 + y 2
x > 0, y > 0, arctan(y/x), θ = 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 Cartesian 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 dγ1 (r) dr r=r0 dγ2 (θ) (1) γ2 (θ0 ) = dθ θ=θ0 (1)
γ1 (r0 ) =
Thus, after normalization, we obtain the two standard vectors (1)
er (r0 , θ0 ) :=
γ1 (r0 ) (1)
kγ1 (r0 )k
= cos θ0 i + sin θ0 j
(1)
eθ (r0 , θ0 ) :=
γ2 (θ0 ) (1)
kγ2 (θ0 )k
= − sin θ0 i + cos θ0 j
The pair of orthogonal basis vectors {er (r0 , θ0 ), eθ (r0 , θ0 )} is called the coordinate frame at the point A. Department of ECE, Fall 2014
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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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6/27
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 coordinates, 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. Department of ECE, Fall 2014
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7/27
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. Department of ECE, Fall 2014
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Cylindrical coordinates (cont.) If A is given by the cylindrical coordinates (r, θ, z) then, in terms of 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 represented 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, (1)
∀z ∈ R.
(1)
Next, we compute the tangent vectors γ1 (r0 ), γ2 (θ0 ), and (1) γ3 (z0 ), followed by their normalization to obtain er (r0 , θ0 , z0 ), eθ (r0 , θ0 , z0 ), and ez (r0 , θ0 , z0 ), respectively.
Department of ECE, Fall 2014
ECE 206: Advanced Calculus 2
9/27
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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ECE 206: Advanced Calculus 2
10/27
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. Department of ECE, Fall 2014
ECE 206: Advanced Calculus 2
11/27
Spherical coordinates
The p radial length of the vector from the origin O to point A is ρ = x2 + y 2 + z 2 .
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ECE 206: Advanced Calculus 2
12/27
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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ECE 206: Advanced Calculus 2
13/27
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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14/27
Spherical coordinates (cont.) The corresponding tangent vectors at (ρ0 , θ0 , φ0 ) are given by (1)
γ1 (ρ0 ) =
dγ1 (ρ) dγ2 (θ) dγ3 (φ) (1) (1) , γ2 (θ0 ) = , γ3 (φ0 ) = . dρ ρ0 dθ θ0 dφ φ0
Normalizing the tangent vectors results in the spherical coordinate 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 other.
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15/27
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 = = kv(ρ + dρ, θ + dθ, φ + dφ) − v(ρ, θ, φ)k2
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16/27
Spherical coordinates (cont.) First we note that v(ρ + dρ, θ + dθ, φ + dφ) − v(ρ, θ, φ) = ∂v ∂v ∂v = (ρ, θ, φ)dρ + (ρ, θ, φ)dθ + (ρ, θ, φ)dφ = ∂ρ ∂θ ∂φ = (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 , or, 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,
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hφ (ρ, θ, φ) := ρ.
ECE 206: Advanced Calculus 2
17/27
Differential operators Let F : R3 → R3 be a C 1 -vector field. The divergence of F is defined as div F(x, y, z) =
∂F1 ∂F2 ∂F3 (x, y, z) + (x, y, z) + (x, y, 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 defined by it.
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18/27
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 ∇=
∂ ∂ ∂ i+ j+ k. ∂x ∂y ∂z
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ECE 206: Advanced Calculus 2
19/27
Gradient in cylindrical coordinates (cont.)
The substitution yields ∂ ∂r ∂ ∂θ ∂ ∂z ∇= + + [cos θ er − sin θ eθ ] + ∂r ∂x ∂θ ∂x ∂z ∂x ∂ ∂r ∂ ∂θ ∂ ∂z + + + [sin θ er + cos θ eθ ] + ∂r ∂y ∂θ ∂y ∂z ∂y ∂ ∂r ∂ ∂θ ∂ ∂z + + + [ez ] . ∂r ∂z ∂θ ∂z ∂z ∂z
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ECE 206: Advanced Calculus 2
20/27
Gradient in cylindrical coordinates (cont.) The substitution yields
∂ ∂r ∂ ∂θ ∂ ∂z ∇ = ∂r ∂x + ∂θ ∂x + ∂z ∂x [cos θ er − sin θ eθ ] + |{z} |{z} |{z} cos θ
0
− sin θ/r
∂ ∂r ∂ ∂z ∂ ∂θ + + + [sin θ er + cos θ eθ ] + ∂z ∂y ∂r ∂y ∂θ ∂y |{z} |{z} |{z} sin θ
0
cos θ/r
∂ ∂r ∂ ∂θ ∂ ∂z + ∂r ∂z + ∂θ ∂z + ∂z ∂z [ez ] . |{z} |{z} |{z} 0
0
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Gradient in cylindrical coordinates (cont.)
After a few trivial steps, we obtain ∇=
∂ 1 ∂ ∂ er + eθ + ez . ∂r r ∂θ ∂z
Let f (ρ, θ, φ) be a C 1 -scalar field. Then ∇f (ρ, θ, φ) =
∂f 1 ∂f ∂f (r, θ, z) er + (r, θ, z) eθ + (r, θ, z) ez . ∂r r ∂θ ∂z
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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ECE 206: Advanced Calculus 2
22/27
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 ∂ 1 ∂ ∂ ∇·F= er + eθ + ez · (F1 er + F2 eθ + F3 ez ) . ∂r r ∂θ ∂z 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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ECE 206: Advanced Calculus 2
23/27
Divergence in cylindrical coordinates (cont.)
∂ er ∂r
∂ ∂F1 ∂er · (F1 er ) = er · (F1 er ) = er · er + F1 = ∂r ∂r ∂r ∂F1 ∂F1 = er · er + F1 er · 0 = . ∂r ∂r
1 ∂ eθ r ∂θ
· (F1 er ) = eθ
1 ∂ 1 · (F1 er ) = eθ r ∂θ r
∂F1 ∂er = F1 . · e + F r 1 ∂θ ∂θ r |{z} eθ
1 ∂ eθ r ∂θ
· (F2 eθ ) = eθ
1 ∂ 1 ∂F2 ∂eθ = 1 ∂F2 . · (F2 eθ ) = eθ · eθ + F2 r ∂θ r ∂θ ∂θ r ∂θ |{z} −er
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ECE 206: Advanced Calculus 2
24/27
Divergence, curl and, Laplacian Proceeding in a similar manner, we finally obtain ∇·F=
∂F1 1 1 ∂F2 ∂F3 + F1 + + = ∂r r r ∂θ ∂z 1 ∂(r F1 ) 1 ∂F2 ∂F3 = + + . r ∂r r ∂θ ∂z
Moreover, one can show that 1 ∂F3 ∂F2 ∂F1 ∂F3 er + eθ + ∇×F= − − r ∂θ ∂z ∂z ∂r 1 ∂(rF2 ) ∂F1 + − ez r ∂r ∂θ and ∆f =
∂2f 1 ∂f 1 ∂2f ∂2f + + + . ∂r2 r ∂r r2 ∂θ2 ∂z 2
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ECE 206: Advanced Calculus 2
25/27
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 ∇=
1 1 ∂ ∂ ∂ eρ + eθ + eφ , ∂ρ ρ sin φ ∂θ ρ ∂φ
so that for a C 1 function f (ρ, θ, φ) we have ∇f (ρ, θ, φ) =
1 ∂f 1 ∂f ∂f (ρ, θ, φ)eρ + (ρ, θ, φ)eθ + (ρ, θ, φ)eφ . ∂ρ ρ sin φ ∂θ ρ ∂φ
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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 , F2 , F3 )) |{z} |{z} |{z} ρ
θ
φ
1 ∂F2 1 ∂(sin φ F3 ) 1 ∂(ρ2 F1 ) + + , ∇·F= 2 ρ ∂ρ ρ sin φ ∂θ ρ sin φ ∂φ
∇×F=
1 ρ sin φ
∂(sin φ F3 ) ∂F2 − ∂φ ∂θ
1 ∂(ρF3 ) ∂F1 eρ + − eθ + ρ ∂ρ ∂φ 1 ∂F1 ∂(ρF2 ) 1 + − eφ , ρ sin φ ∂θ ∂ρ
and 1 ∂ 1 ∂2f 1 ∂ ∂f 2 ∂f ∆f = 2 ρ + + sin φ . ρ ∂ρ ∂ρ sin φ ∂φ ∂φ sin2 φ ∂θ2
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