Lecture 2 : Curvilinear Coordinates - NTNU

Lecture 2 : Curvilinear Coordinates

Fu-Jiun Jiang
October 11, 2010

I. INTRODUCTION

A. Definition and Notations

In 3-dimension Euclidean space, a vector V can be written as V = exVx + eyVy + ezVz
with ex = (1, 0, 0), ey = (0, 1, 0), ez = (0, 0, 1). We call these ex,y,z a basis for the 3-dim
Euclidean space. Notice we have

ei · ei = 1 (1)
ei · ej = 0 for i = j
ex · (ey × ez) = 1 > 0.

We call such a basis an orthonormal basis and x, y, z the orthonormal coordinates of the
3-dim Euclidean space. Notice if the first condition is eq. 1 is removed, then such a basis
is called a orthogonal basis. Let’s assume we have another set of orthogonal coordinates
q1(x, y, z), q2(x, y, z), q3(x, y, z) which are functions of the standard coordinates x, y, z. We
can also consider the standard coordinates as functions of q1, q2, q3. Let r(x, y, z) be the
position vector with respect to the standard basis ex, ey, ez in 3-dim Euclidean space, namely
r(x, y, z) = (x, y, z). Notice if one holds the second and third component fixed and takes
the derivative of r with respect to the coordinate x, one reaches

∂ r(x, y, z) |y=y0 = (1, 0, 0) = ex. (2)
∂x
,z=z0

Similarly, one has

∂ r(x, y , z ) |x=x0,z=z0 = (0, 1, 0) = ey,
∂y

∂r(x, y, z) (3)
∂z |x=x0,y=y0 = (0, 0, 1) = ez.

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In other word, ∂r , ∂r , ∂r are the standard orthonormal basis of 3-dim Euclidean space.
∂x ∂y ∂z

Applying similar arguments to other sets of orthogonal coordinates qi, one arrives at similar

conclusions, namely

∂r /| ∂r | ≡ eqi , i = 1, 2, 3 (4)
∂qi ∂qi

is the orthonormal basis corresponding to the orthogonal coordinates qi. Notice if {q1, q2, q3}

is not an orthogonal coordinate system, then eqi defined above are still a basis (corresponding
to {q1, q2, q3}), but they are not an orthogonal basis.

Now from eq. 4, one immediately has

∂r (5)
∂qi = hieqi,

here hi is called scale factor for each i ∈ 1, 2, 3. To simplify our presentation, we will use ei
instead of eqi if there is no any confusion.

In terms of the curvilinear coordinates qi, the position vector is denoted by r . Further,
the differential distant vector dr takes the form

∂r (6)
dr = i=1,2,3 ∂qi dqi = e1h1dq1 + e2h2dq2 + e3h3dq3.

Next, since the idea of inner product between 2 vectors should be independent of the chosen

basis, one should have

dr · dr = dr · dr . (7)

In other word, we arrive at

dx2 + dy2 + dz2 = h12dq12 + h22dq22 + h23dq32. (8)

As a result, intutively one can just replace dx, dy and dz by h1dq1, h2dq2 and h3dq3, re-

spectively in a expression such as a line integral. Remember for 2 different vectors A, B,

the oriented area exapnded by A, B is given by A × B. Further, for 3 vectors A, B, C not

lying on the same 2-dim plane, the vlolume determined by A, B, C is given by C · (A × B).

Applying these results to {q1, q2, q3} (i.e. ( ∂r , ∂r , ∂r )), one sees that the unit element of
∂q1 ∂q2 ∂q3

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area and volume are given by

dσ = dr|q1,q3fixed × dr|q1,q2fixed + dr|q1,q2fixed × dr|q2,q3fixed (9)
+ dr|q1,q3fixed × dr|q1,q2fixed (10)
= h2h3dq2dq3e1 + h3h1dq3dq1e2 + h1h2dq1dq2e3,

dτ = h1h2h3dq1dq2dq3,

respectively. As a result, the line, surface and volume integrals now become

V · dr = Vihidqi

i

V · dσ = V1h2h3dq2dq3 + V2h3h1dq3dq1 + V3h1h2dq1dq2

φdτ = φh1h2h3dq1dq2dq3.

Example : Let x = 1 (u21 − u22), y = u1u2, z = u3. Then one has
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∂r ∂r = (−u2, u1, 0), ∂r (11)
∂u1 = (u1, u2, 0), ∂u2 = (0, 0, 1).

∂u3

Hence {u1, u2, u3} is an orthogonal coordinate system.

Example : Let x = 3u1 + u2 − u3, y = u1 + 2u2 + 2u3, z = 2u1 − u2 − u3. Then one

finds that

∂r ∂r = (1, 2, −1), ∂r = (−1, 2, −1). (12)
= (3, 1, 2),
∂u1 ∂u2 ∂u3

Hence {u1, u2, u3} is not an orthogonal coordinate system.

II. DIFFERENTIAL VECTOR OPERATORS AGAIN

In general for an orthogonal coordinate system q1, q2, q3, one has similar definitions for
gradient, divengence and curl operators as those of we have learned earlier in our vector
analysis section.

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A. Gradient

For a given scalar function Φ(q1, q2, q3) in a orthogonal coordinates qi, let the gradient of

Φ be

Φ = f1e1 + f2e2 + f3e3, (13)

here e1, e2, e3 is the corresponding orthonormal basis for {q1, q2, q3} and fi are some unknown
functions of {qi} Now using eq. 6, we have

dΦ = Φ · dr = f1h1dq1 + f2h2dq2 + f3h3dq3. (14)

On ther other hand, one can show

∂Φ ∂Φ ∂Φ (15)
dΦ = ∂q1 dq1 + ∂q2 dq2 + ∂q3 dq3. (16)
From eqs. 14 and 15, one immediately finds (17)
(18)
f1 = 1 ∂Φ 1 ∂Φ = 1 ∂Φ (19)
, f2 = , f3 , (20)
h1 ∂q1 h2 ∂q2 h3 ∂q3

namely we have

1 ∂Φ 1 ∂Φ 1 ∂Φ
Φ = h1 ∂q1 e1 + h2 ∂q2 e2 + h3 ∂q3 e3.

Since eq. 17 holds for any scalar function Φ, one arrives at

1∂ 1∂ 1∂
= h1 ∂q1 e1 + h2 ∂q2 e2 + h3 ∂q3 e3.

Notice if we choose Φ = qi for i ∈ {1, 2, 3}, then

q1 = e1 , q2 = e2 , q3 = e3
h1 h2 h3

From eq. 19, one further finds

q2 × q3 = e2 × e3 = e1 ,
h2h3 h2h3

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hence e1 = h2h3( q2 × q3). (21)
Similarly, one has (22)

e2 = h3h1( q3 × q1), e3 = h1h2( q1 × q2).

B. Divergence

For Divengence operator in curilinear coordinate, it is most convenient to obtain the
relevant expression through its integration definition

· V (q1, q2, q3) = lim SV · dσ (23)
.
dτ →0 dτ

Consider the difference of area integrasl for two faces q1 = constant (keeping q2, q3 fixed)

∂ (24)
V1h2h3 + ∂q1 (V1h2h3)dq1 dq2dq3 − V1h2h3dq2dq3

∂
= ∂q1 (V1h2h3)dq1dq2dq3.

By adding other 2 similar formulae (surfaces for constant q2 and q3), we arrive at

V (q1, q2, q3) · dσ = ∂∂∂ (25)
∂q1 (V1h2h3) + ∂q2 (V2h3h1) + ∂q3 (V3h1h2) dq1dq2dq3.

Dividing by dτ = h1h2h3dq1dq2dq3, we finally obtain

· V (q1, q2, q3) = 1 ∂∂∂ (26)
h1h2h3 ∂q1 (V1h2h3) + ∂q2 (V2h3h1) + ∂q3 (V3h1h2) .

Similarly, the Laplacian in curvilinear coordinate qi is given by

· φ(q1, q2, q3) =

· 1 ∂φ + 1 ∂φ + 1 ∂φ
( ∂q1 e1 h2 ∂q2 e2 h3 ∂q3 e3)
h1

1 ∂ ( h2h3 ∂φ ) + ∂ ( h3h1 ∂φ ) + ∂ ( h1h2 ∂φ ) (27)
=
q1q2q3 ∂q1 h1 ∂q1 ∂q2 h2 ∂q2 ∂q3 h3 ∂q3

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FIG. 1: rectangular parallelepiped.

C. Curl

For × V , consider a differential surface element in the curvilinear surface q1 = constant.

The from

× V · dσ = e1 · ( × V )h2h3dq2dq3 (28)

S

and Stokes theorem, we arrive at

e1 · ( × V )h2h3dq2dq3 = V · dr (29)

c

where the line integral (over c) is along the boundary of the constant surface q1 = constant.

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FIG. 2: Curl in orthogonal coordinates.

Notice by the same consideration as we did when deriving the Stoke’s theorem (consid-
ering the line integrals over the boundaries of S), one can show

V (q1, q2, q3) · dr = V2h2dq2 + ∂ dq3
V3h3 + ∂q2 (V3h3)dq2
c

∂
− V3h3dq3 − V2h2 + ∂q3 (V2h2)dq3 dq2

= ∂ (h3 V3 ) − ∂ dq2dq3. (30)
∂q2 ∂q3 (h2V2)

Hence we finally arrive at

×V 1 ∂ (h3V3) − ∂ (h2V2)
= e1 h2h3 ∂q2 ∂q3

1∂ ∂
+ e2 h3h1 ∂q3 (h1V1) − ∂q1 (h3V3)

1 ∂ (h2V2) − ∂ (h1V1)
+ e3 h1h2 ∂q1 ∂q2

1 h1e1 h2e2 h3e3 (31)
=
∂∂∂
h1h2h3 ∂q1 ∂q2 ∂q3

h1V1 h2V2 h3V3

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FIG. 3: Spherical polar coordinate. Source : http://en.citizendium.org

III. SPHERICAL POLAR COORDINATE (r, θ, ϕ)

The relation between cartesian coordinates (x, y, z) and spherical polar coordinates
(r, θ, ϕ) is

x = r sin θ cos ϕ, (32)
y = r sin θ sin ϕ,
z = r cos θ,

or

r = (x2 + y2 + x2)1/2,

θ = arc cos z ,

(x2 + y2 + z2)1/2

y (33)
ϕ = arc tan .

x

The range for spherical polar coordinates are

0 ≤ r < ∞, (34)
0 ≤ θ ≤ π,
0 ≤ ϕ ≤ 2π.

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Now using hj = | ∂r |, one sees
∂qj

h1 = hr = 1, (35)
h2 = hθ = r,
h3 = hϕ = r sin θ.

This gives line and surface (for r = constant) elements

dr = erdr + eθrdθ + eϕr sin θdϕ, (36)
dσθϕ = err2 sin θdθdϕ.

as well as the volume element dτ = r2dr sin θdθdϕ.
One can also express er, eθ and eϕ in terms of ex, ey, ez

er = ex sin θ cos ϕ + ey sin θ sin ϕ + ez cos θ, (37)
eθ = ex cos θ cos ϕ + ey cos θ sin ϕ − ez sin θ,
eϕ = −ex sin ϕ + ey cos ϕ.

With above expressions of er, eθ and eϕ in terms of ex, ey, ez, one can easily verify that er,
eθ and eϕ indeed form a orthogonal basis.

FIG. 4: Orthonormal basis in spherical polar coordinate. Source : http://en.citizendium.org

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FIG. 5: Spherical polar coordinate. Source : http://www.heliotekinc.com/photometry.htm

In spherical polar coordinate, , ·, × and 2 take the form

∂φ 1 ∂φ 1 ∂φ
φ = ∂r er + r ∂θ eθ + r sin θ ∂ϕ eϕ,

·V = 1 ∂ (Vr r2 sin θ) + ∂ ∂ ,
r2 sin θ ∂r ∂θ (Vθr sin θ) + ∂ϕ (Vϕr)

×V = = 1 ∂ (r sin θVϕ) − ∂
er r2 sin θ ∂θ ∂ϕ (rVθ)

1∂ − ∂ 1∂ ∂
+ eθ r sin θ ∂ϕ (Vr) (r sin θVϕ) + eϕ r ∂r (rVθ) − ∂θ (Vr) ,
∂r

· φ= 1 ∂ (r2 ∂φ + ∂ ∂φ ∂ ( 1 ∂φ (38)
sin θ ) (sin θ ) + ).
r2 sin θ ∂r ∂r ∂θ ∂θ ∂ϕ sin θ ∂ϕ

Example : Using spherical polar coordinate, we can rederive some formulae we had
calculated before.

df (r) rn = ernrn−1,
f (r) = er dr ,

· erf (r) = 2 + df · errn = (n + 2)rn−1,
f (r) ,

r dr

2f (r) = 2 df + d2f , 2rn = n(n + 1)rn−2,
r dr dr2

× erf (r) = 0. (39)

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Example : In term of {er, eθ, eϕ}, one has (40)

ex = er sin θ cos ϕ + eθ cos θ cos ϕ − eϕ sin ϕ,
ey = er sin θ sin ϕ + eθ cos θ sin ϕ + eϕ cos ϕ,
ez = er cos θ − eθ sin θ.

Example : In term of { ∂ , ∂ , ∂ }, one can show
∂r ∂θ ∂ϕ

∂ = ∂ 1 ∂ − sin ϕ ∂
sin θ cos ϕ + cos θ cos ϕ ,
∂x ∂r r ∂θ r sin θ ∂ϕ

∂ ∂ 1 ∂ cos ϕ ∂ ,
= sin θ sin ϕ + cos θ sin ϕ +
∂y ∂r r ∂θ r sin θ ∂ϕ

∂ = ∂ − sin θ 1 ∂ . (41)
cos θ
∂z ∂r r ∂ϕ

Example : Using above results of expressing { ∂ , ∂ , ∂ } in term of { ∂ , ∂ , ∂ }, one
∂x ∂y ∂z ∂r ∂θ ∂ϕ

arrives at

−i ∂ −y ∂ = −i ∂ . (42)
x ∂ϕ
∂y ∂x

This is the quantum mechanical operator which corresponds to the z-component of orbital
angular momentum.

IV. CIRCULAR CYLINDRICAL COORDINATES ρ, ϕ, z

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FIG. 6: Circular Cylindrical coordinate. Source : http://www.vias.org/

The relation between cartesian coordinate x, y, z and circular cylindrical coordinates is
given by

ρ = (x2 + y2)1/2, (43)
ϕ = tan−1 y , (44)
(45)
x (46)
z=z

with the limits on ρ, ϕ, z being given by

0 ≤ ρ < ∞, 0 ≤ ϕ ≤ 2π, −∞ < z < ∞.

Or the othter way around

x = ρ cos ϕ, y = ρ sin ϕ, z = z.

As a result, one sees

h1 = hρ = 1, h2 = hϕ = ρ, h3 = hz = 1.

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Also the line surface (for ρ = constant) and volume elements are given by

dr = eρdρ + eϕdϕ + ezdz, (47)
dσϕz = ρdϕdz,

dτ = ρdρdϕdz.

Similarly, the circular cylindrical basis eρ, eϕ, ez can be expressed in terms of ex, ey, ez as
follows

eρ = ex cos ϕ + ey sin ϕ, (48)
eϕ = −ex sin ϕ + ey cos ϕ,

ez = ez.

with which one can easily verify that eρ, eϕ, ez form a orthorgonal basis.
Similarly, in circularl cylindrical coordinate, , ·, × and 2 take the form

∂φ 1 ∂φ ∂φ
φ = ∂ρ eρ + ρ ∂ϕ eϕ + ∂z ez,

1∂ ∂ ∂
· V = ρ ∂ρ (Vρρ) + ∂ϕ (Vϕ) + ∂z (Vzρ) ,

×V = = 1 ∂ − ∂
eρ ρ ∂ϕ (Vz) ∂z (ρVϕ)

∂∂ 1∂ ∂
+ eϕ ∂z (Vρ) − ∂ρ (Vz) + ez ρ ∂ρ (ρVϕ) − ∂ϕ (Vρ) ,

· 1 ∂ ∂φ ∂ 1 ∂φ ∂ ∂φ (49)
φ= (ρ ) + ( ) + (ρ ) .
ρ ∂ρ ∂ρ ∂ϕ ρ ∂ϕ ∂z ∂z

Example : Let A = zex − 2xey + yez. Then from (50)

eρ = ex cos ϕ + ey sin ϕ,
eϕ = −ex sin ϕ + ey cos ϕ,

ez = ez,

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we arrive at (51)
ex = cos ϕeρ − sin ϕeϕ, ey = sin ϕeρ + cos ϕeϕ, ez = ez. (52)

As a result, one finds
A = (z cos ϕ − 2ρ cos ϕ sin ϕ)eρ − (z sin ϕ + 2ρ cos2 ϕ)eϕ + ρ sin ϕez.

Example : From eρ = cos ϕex + sin ϕey and eϕ = − sin ϕex + cos ϕey, one reaches

dd (53)
dt eρ = ϕ˙ eϕ, dt eϕ = −ϕ˙ eρ.

Example : A rigid body is rotating about the z-axis with a constant angular velocity ω.
Then one sees that v = ω × r = ωρeϕ. Also one finds × v = 2ω.

Example : The magnetic vector potential for a conducting wire along the z-axis which

carries a current I is given by

µI 1 (54)
A = ez ln( ).
2π ρ

Then the corresponding magnetic field B is given by

µI (55)
B = eϕ 2πρ

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