Cylindrical and Spherical Coordinatesece206/Lectures/slides/section... · 2014-11-24 ·...

Cylindrical and Spherical Coordinates

The Cartesian coordinate system is by far the simplest, the mostuniversal 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 somekind of inherent symmetry.

The cylindrical and spherical coordinate systems are designed forjust this purpose.

Department of ECE, Fall 2014 ECE 206: Advanced Calculus 2 1/27

Polar coordinates

The point A is represented by (r, θ), which has a very differentinterpretation from the Cartesian pair (x, y).

While −∞ < x <∞ and −∞ < y <∞, the polar coordinatesobey

0 ≤ r <∞, 0 ≤ θ < 2π.


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 berepresented 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.


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 polarcoordinates?

To answer this question, for A with coordinates (r0, θ0), we definethe following two paths:

γ1(r) := v(r, θ0) = r cos θ0 i + r sin θ0 j

andγ2(θ) := v(r0, θ) = r0 cos θ i + r0 sin θ j.


Standard vectors (cont.)

The tangent vectors at (r0, θ0) aregiven 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.


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.


Riemannian scale functions

How does length changes when we change the polar coordinates?

When “moving” from A to B, theEuclidean distance is defined by

(ds)2 = (dx)2 + (dy)2.

At the same time, in the polar co-ordinates, 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.


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) completelyspecifies the point A.


Cylindrical coordinates (cont.)

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



The cylindrical coordinate frame isdefined by

er(r, θ, z) = cos θ i + sin θ j + 0k,

eθ(r, θ, z) = − sin θ i + cos θ j + 0k,

ez(r, θ, z) = 0 i + 0 j + 1k.

One can see that the unit vectors are orthogonal with respect toeach other.

Also, the triplet of vectors constitutes a moving coordinate frame.



Due to a small “permutation” thelength 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,

wherehr(r, θ, z) := 1, hθ(r, θ, z) := r, hz(r, θ, z) := 1

are the Riemannian scale functions.


Spherical coordinates

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

√x2 + y2 + z2.


Spherical coordinates (cont.)

The Cartesian coordinates (x, y, z) are given in terms of thespherical coordinates (ρ, θ, φ) by

x = ρ sinφ cos θ, y = ρ sinφ sin θ, z = ρ cosφ,

with ρ ∈ [0,∞), θ ∈ [0, 2π), and φ ∈ [0, π].

Then, in terms of the Cartesian basis {i, j,k}, any point A withcoordinates (ρ, θ, φ) can be expressed as

v(ρ, θ, φ) = ρ sinφ cos θ i + ρ sinφ sin θ j + ρ cosφk.

This shows how the Cartesian representation of a point changeswhen we change its spherical coordinates.



To define a moving (orthonormal) frame, we start with

γ1(ρ) = v(ρ, θ0, φ0) = ρ sinφ0 cos θ0 i + ρ sinφ0 sin θ0 j + ρ cosφ0 k,

γ2(θ) = v(ρ0, θ, φ0) = ρ0 sinφ0 cos θ i + ρ0 sinφ0 sin θ j + ρ0 cosφ0 k,

γ3(φ) = v(ρ, θ0, φ) = ρ0 sinφ cos θ0 i + ρ0 sinφ sin θ0 j + ρ0 cosφk.



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) = sinφ cos θ i + sinφ sin θ j + cosφk,

eθ(ρ, θ, φ) = − sin θ i + cos θ j + 0k,

eφ(ρ, θ, φ) = cosφ cos θ i + cosφ sin θ j− sinφk,

under the replacement of the generic (i.e., “naughted”) sphericalcoordinates (ρ0, θ0, φ0) with (ρ, θ, φ) (to simplify the notations).

One can check that the unit vectors indeed orthogonal w.r.t. eachother.



Suppose a point A is given by v(ρ, θ, φ), while a close-by point Bis 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



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 sinφdθ) eθ(ρ, θ, φ) + (ρ dφ) eφ(ρ, θ, φ).

Thus, we have

(ds)2 = (dρ)2 + (ρ sinφ)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 sinφ, hφ(ρ, θ, φ) := ρ.


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 allthe other operators (viz., divergence, curl, and Laplacian) are de-fined by it.


Gradient in cylindrical coordinates

Recall that the cylindrical coordinate frame is defined as

er(r, θ, z) = cos θ i + sin θ j + 0k (= cos θ i + sin θ j),

eθ(r, θ, z) = − sin θ i + cos θ j + 0k (= − sin θ i + cos θ j),

ez(r, θ, z) = 0 i + 0 j + 1k (= 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

∇ =∂

∂xi +

∂

∂yj +

∂

∂zk.


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] .



The substitution yields

∇ =

∂

∂r

∂r

∂x︸︷︷︸cos θ

+∂

∂θ

∂θ

∂x︸︷︷︸− sin θ/r

+∂

∂z

∂z

∂x︸︷︷︸0

[cos θ er − sin θ eθ] +

+

∂

∂r

∂r

∂y︸︷︷︸sin θ

+∂

∂θ

∂θ

∂y︸︷︷︸cos θ/r

+∂

∂z

∂z

∂y︸︷︷︸0

[sin θ er + cos θ eθ] +

+

∂

∂r

∂r

∂z︸︷︷︸0

+∂

∂θ

∂θ

∂z︸︷︷︸0

+∂

∂z

∂z

∂z︸︷︷︸1

[ez] .



After a few trivial steps, we obtain

∇ =∂

∂rer +

1

r

∂

∂θeθ +

∂

∂zez.

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 cylindricalcoordinate frame is not constant.


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 =

(∂

∂rer +

1

r

∂

∂θeθ +

∂

∂zez

)· (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 thenthe dot (or cross) product.

Let’s look at some specific cases.


Divergence in cylindrical coordinates (cont.)

(∂

∂rer

)· (F1 er) = er ·

∂

∂r(F1 er) = er ·

(∂F1

∂rer + F1

∂er∂r

)=

=∂F1

∂rer · er + F1er · 0 =

∂F1

∂r.

(1

r

∂

∂θeθ

)· (F1 er) = eθ

1

r· ∂∂θ

(F1 er) = eθ1

r·

∂F1

∂θer + F1

∂er∂θ︸︷︷︸eθ

=F1

r.

(1

r

∂

∂θeθ

)· (F2 eθ) = eθ

1

r· ∂∂θ

(F2eθ) = eθ1

r·

∂F2

∂θeθ + F2

∂eθ∂θ︸︷︷︸−er

=1

r

∂F2

∂θ.


Divergence, curl and, Laplacian

Proceeding in a similar manner, we finally obtain

∇ · F =∂F1

∂r+

1

rF1 +

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.


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 tothe previous case.

Particularly, one can show that

∇ =∂

∂ρeρ +

1

ρ sinφ

∂

∂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 sinφ

∂f

∂θ(ρ, θ, φ)eθ+

1

ρ

∂f

∂φ(ρ, θ, φ)eφ.


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 sinφ

∂F2

∂θ+

1

ρ sinφ

∂(sinφF3)

∂φ,

∇× F =1

ρ sinφ

(∂(sinφF3)

∂φ− ∂F2

∂θ

)eρ +

1

ρ

(∂(ρF3)

∂ρ− ∂F1

∂φ

)eθ+

+1

ρ

(1

sinφ

∂F1

∂θ− ∂(ρF2)

∂ρ

)eφ,

and

∆f =1

ρ2

[∂

∂ρ

(ρ2∂f

∂ρ

)+

1

sin2 φ

∂2f

∂θ2+

1

sinφ

∂

∂φ

(sinφ

∂f

∂φ

)].


Cylindrical and Spherical Coordinatesece206/Lectures/slides/section... · 2014-11-24 ·...

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