Curvilinear coordinates - University of Rochester

Physics 217, Fall 2002 13 September 2002
1
Today in Physics 217: more vector calculus

Vector derivatives in z φˆ
curvilinear coordinate rˆ θˆ
systems: spherical and
cylindrical coordinates. θ
The Dirac delta function. r

φ y
x

z zˆ
s φˆ

z sˆ

φ y
x

13 September 2002 Physics 217, Fall 2002 1

Curvilinear coordinates

Coordinate systems:
Cartesian coordinates: used to describe systems without
any apparent symmetry.
Curvilinear coordinates: used to describe systems with
symmetry. We will often find spherical symmetry or axial
symmetry in the problems we will do this semester, and
will thus use
• Spherical coordinates
• Cylindrical coordinates
There are other curvilinear coordinate systems (e.g.
ellipsoidal) that have special virtues, but we won’t get to
use them this semester.

13 September 2002 Physics 217, Fall 2002 2

Spherical coordinates

The location of a point P can be z y
defined by specifying the
following three parameters: θ 3
r
Radius r: distance of P from
the origin. φ
Polar angle θ: angle between
the position vector of P and
the z axis. (Like 90o –
latitude.)
Azimuthal angle φ: angle
between the projection of x
the position vector P and the
x axis. (Like longitude.)

13 September 2002 Physics 217, Fall 2002

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Physics 217, Fall 2002 13 September 2002
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Spherical coordinates (continued)

The Cartesian coordinates of z
P are related to the spherical
coordinates as follows: θ
rz
x = r sinθ cosφ r = x2 + y2 + z2 .
y = r sinθ sinφ
( )θ = arctan x2 + y2 z
z = r cosθ φ = arctan (y x)

The unit vectors of spherical φ x y
coordinate systems are not y
constant: their direction x 4
changes when the position of
point P changes.

13 September 2002 Physics 217, Fall 2002

Spherical coordinates (continued)

In Cartesian coordinates, an z
infinitesimal displacement from
point P is equal to
rˆ φˆ
dl = xˆdx + yˆdy + zˆdz
In spherical coordinates, an θ θˆ
r

infinitesimal displacement φy
from point P is equal to
x
dl = rˆdr + θˆrdθ + φˆ r sinθ dφ
where rˆ is parallel to r, θˆ is perpendicular
r-z plane, φˆ is perpendicular to this plane, to rˆ aθˆnadnldiesφˆ in the
and
point in the direction of increasing θ and φ.

13 September 2002 Physics 217, Fall 2002 5

Example: transformation of unit vectors

Griffiths problem 1.37: Express z
the spherical-coordinate unit
vectors in terms of the Cartesian ones. rˆ φˆ
(Worked out on blackboard.) θ θˆ
We get
rˆ = sinθ cosφ xˆ + sinθ sinφ yˆ + cosθ zˆ r
θˆ = cosθ cosφ xˆ + cosθ sinφ yˆ − sinθ zˆ
φˆ = − sinφ xˆ + cosφyˆ φ y

x

13 September 2002 Physics 217, Fall 2002 6

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Physics 217, Fall 2002 13 September 2002
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Spherical coordinates (continued)

In a Cartesian coordinates, an z
infinitesimal volume element
around point P is equal to θ dτ
r
dτ = dx dy dz y
φ
In a spherical coordinates, an
infinitesimal volume element x
around point P is equal to

dτ = dlr dlθ dlφ
= r2 sinθ dr dθ dφ

13 September 2002 Physics 217, Fall 2002 7

Spherical coordinates (continued)

In Cartesian coordinates, an z y
infinitesimal area element on a
plane containing point P is da 8
θ
da = zˆdxdy, or
= xˆdydz, or r
= yˆdzdx. φ

In spherical coordinates, the x
infinitesimal area element on a
sphere through point P is

da = rˆdlθ dlφ = rˆr2 sinθ dθ dφ

13 September 2002 Physics 217, Fall 2002

Vector derivatives in spherical coordinates

What if you want to express a vector derivative in spherical coordinates?
(Or someone asks you to, on a test…)

Start from the Cartesian-coordinate version, and first use the chain
rule to transform the derivatives, e.g.

(∇T )x = ∂T = ∂T ∂r + ∂T ∂θ + ∂T ∂φ
∂x ∂r ∂x ∂θ ∂x ∂φ ∂x

Use the coordinate definitions to reduce the remaining derivatives
and eliminate all Cartesian coordinates, e.g.

∂r = ∂ x2 + y2 + z2 = x = x = sinθ cosφ
∂x ∂x x2 + y2 + z2 r

Transform the unit vectors, as we did earlier today.

Then multiply the whole mess out and simplify.

This is tedious, and takes hours, but is instructive and highly
recommended, even though it’s not on the homework explicitly. See also
Appendix A in Griffiths.

13 September 2002 Physics 217, Fall 2002 9

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Physics 217, Fall 2002 13 September 2002
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Vector derivatives in spherical coordinates
(continued)

The following operations will be encountered frequently enough that
they’re even written on the inside front cover of Griffiths:

∇T = ∂T rˆ + 1 ∂T θˆ + 1 ∂T φˆ
∂r r ∂θ r sinθ ∂φ

1 ∂ 1 ∂ 1 ∂
r2 ∂r r sinθ ∂θ r sinθ ∂φ
( ) ( )∇ ⋅v r2 vr
= + (sinθ vθ )+ vφ

1  ∂ −∂  1 1 ∂ ∂  θˆ
( ) ( )∇ sinθ  ∂θ ∂φ ( vθ )  rˆ ∂φ ( ) ∂r 
×v = r  sinθ vφ + r  sin θ vr − rvφ



+ 1  ∂ (rvθ ) − ∂ ( vr )  φˆ
r  ∂r ∂θ 

∇2T = 1 ∂  r 2 ∂T  + r2 1 ∂  sinθ ∂T  + 2 1 ∂2T
r2 ∂r  ∂r  sinθ ∂θ  ∂θ  sin2 ∂φ 2
r θ

13 September 2002 Physics 217, Fall 2002 10

Cylindrical coordinates

Spherical coordinates are useful z y
mostly for spherically symmetric
situations. In problems involving s 11
symmetry about just one axis, P
cylindrical coordinates are used:
z
The radius s: distance of P from
the z axis. φ
The azimuthal angle φ: angle
between the projection of the x
position vector P and the x axis.
(Same as the spherical coordinate
of the same name.)
The z coordinate: component of
the position vector P along the z
axis. (Same as the Cartesian z).

13 September 2002 Physics 217, Fall 2002

Cylindrical coordinates (continued)

The Cartesian coordinates of z
P are related to the
cylindrical coordinates by

x = s cosφ s = x2 + y2 s
y = s sinφ P
z=z φ = arctan (y x)
z
z=z

Again, the unit vectors of z y
cylindrical coordinate
systems are not constant; φ x 12
their direction changes when y
the position of point P x
changes.

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Physics 217, Fall 2002 13 September 2002
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Cylindrical coordinates (continued)

In cylindrical coordinates: z zˆ φˆ
s sˆ
unit vectors
sˆ = xˆ cosφ + yˆ sinφ z

φˆ = −xˆ sinφ + yˆ cosφ φ y

zˆ = zˆ
infinitesimal displacement

dl = sˆds + φˆ rdφ + zˆdz
infinitesimal volume element x

dτ = rdr dφ dz

infinitesimal area element

da = zˆsdsdφ (top of cylinder), = sˆsdφdz (cylinder wall).

13 September 2002 Physics 217, Fall 2002 13

Cylindrical coordinates (continued)

The more common vector derivatives, in cylindrical
coordinates:

∇T = ∂T sˆ + 1 ∂T φˆ + ∂T zˆ
∂s s ∂φ ∂z

( )∇ 1 ∂ 1 ∂ ∂
⋅v = s ∂s (svs ) + s ∂φ vφ + ∂z ( vz )

×( )∇v=  1 ∂ ( vz ) − ∂ vφ  sˆ +  ∂ ( vs ) − ∂ ( vz )  φˆ
 s ∂φ ∂z   ∂z ∂s 
 

( )+1  ∂ svφ − ∂ ( vs )  zˆ
s  ∂s ∂φ 
 

∇2T = 1 ∂  s ∂T  + 1 ∂2T + ∂2T
s ∂s  ∂s  s2 ∂φ 2 ∂z2

13 September 2002 Physics 217, Fall 2002 14

The Dirac delta function

In problem 1.16, in this week’s homework, you will show that the
following vector function:
1
v(r,φ ,θ ) = r2 rˆ

has divergence that is zero except at the origin, where it’s infinite.
However, you’ll show in problem 1.38 that the integral, over any sphere
centered at the origin, of the divergence of this function, is neither zero
nor infinity, but instead is

∫ ∇ ⋅vdτ = v∫ v ⋅ da = 4π
sphere sphere's
surface
This turns out to be an extremely useful set of characteristics, so we
frequently use the divergence of this function, and give it a special
name…

δ 3 (r) = 1  rˆ  Delta function
4  
∇ ⋅ 2

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Physics 217, Fall 2002 13 September 2002
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The Dirac delta function (continued)

With this definition, the delta function δ 3 (r ) is zero everywhere except r

= 0, at which it is infinite, and the integral over any volume V that
contains the origin is unity:

∫ δ 3 (r ) dτ = 1

V

Because the integral is dimensionless, δ 3 (r ) itself has dimensions of

(1/length)3.

The one-dimensional analogue of this function is also called the delta
function:
a

∫ δ (x) dx = 1 if a ≠ 0.
−a

It has dimensions 1/length, and is related to the other one by

δ 3 (r) = δ 3 (x,y,z) = δ (x)δ (y)δ (z)

13 September 2002 Physics 217, Fall 2002 16

The Dirac delta function (continued)

Why is the delta function useful?
It’s a nice way mathematically to express compact entities such as
point charges, when we have to describe them with differential
equations. For instance, the charge density – electric charge per unit
volum – of a point charge q can be written as (Problem 1.46a):

ρ (r ) = qδ 3 (r )

It’s easy to integrate expressions containing delta functions, because
the integrand is zero everywhere except on the delta function’s
∫“spike:”
f (r )δ 3 (r ) dτ =  f (0) if V contains the origin,

V 0 if not.

∫a f ( x)δ (x) dx =  f (0) if a ≠ 0,
0 if not.
−a

13 September 2002 Physics 217, Fall 2002 17

The Dirac delta function (continued)

Examples, worked out on the blackboard: most from Griffiths problem
1.43.
∞

∫ f (x)δ (x − a) dx = f (a)
−∞

6

( )∫ 3x2 − 2x − 1 δ (x − 3) dx = 20

2
5

∫ cos x δ (x −π ) dx = cos(π ) = −1

0
3

∫ x3δ (x + 1) dx = 0

0

∞

∫ ln (x + 3)δ (x + 2) = ln 1 = 0
−∞

13 September 2002 Physics 217, Fall 2002 18

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The Dirac delta function (continued)

Another example: part of Griffiths problem 1.48:

e−r∫ ∫ ( )V  rˆ  e−r
 ∇ ⋅ r2  dτ = 4πδ 3 (r ) dτ = 4π e0 = 4π .

V

13 September 2002 Physics 217, Fall 2002 19

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