PHYS 3323 Lesson 3
Coordinate Systems, Delta Functions and Integration
A.
Why So Many Coordinate Systems?
Solving a physics problem usually involves solving a differential
equation under some constraint (i.e. The electric potential is constant
along a sphere of radius 0.2 m.). These constraints along with the
symmetry of the problem usually dictate the choice of coordinate
systems in order to simplify mathematical difficulties. For instance, a
sphere centered at the origin can be described as r = constant in
spherical coordinates while its equation is x 2  y 2  z 2  constant in
Cartesian coordinates. The trade-off for simplifying constraint
equations is the need to use more complicated forms of the divergence
and other vector operators.
B.
Cylindrical Coordinate System
1.
Coordinates
The three coordinates used in cylindrical coordinates are , , and z as
shown below.
z
k̂
z
y

̂

x
ρ̂
2.
Transforming Between Cartesian Coordinates
Using the figure above, we can obtain the following relationships
1.
ρ  x 2  y2
2.
x  ρ cos 
3.
x  ρ sin  
4.
z=z
In order to develop transformation equations, we need to use our
knowledge of the scalar product to develop relationships between the
Cartesian and Cylindrical coordinate unit vectors. These relationships
can be found using the following diagram of the unit vectors in the xy plane.
y
ˆ  ˆj
̂
 ˆ  î

ρ̂

ρ̂  ĵ
ρ̂  î
x
From the diagram, we evaluate the scalar products as
î  ρ̂  ρ̂  î  (1)(1)cos   cos 
ĵ  ρ̂  ρ̂  ĵ  (1)(1)sin   sin
î  ˆ ˆ  î   (1)(1)sin    sin
ĵ  ˆ ˆ  ĵ  (1)(1)cos  cos
We can use these relationships to build our substitution equations. For
instance, we find the x and y components of the ρ̂ unit vector using the
scalar product.
ρ̂   ρ̂  î î   ρ̂  ĵ ĵ   ρ̂  k̂ k̂  cos î  sin ĵ






Using the same technique, we find the other four relationships as
ˆ  ˆ  î î  ˆ  ĵ ĵ  ˆ  k̂ k̂   sin î  cos ĵ






î   î  ρ̂ ρ̂   î  ˆ ˆ   î  k̂ k̂  cos ρ̂  sin ˆ






ĵ   ĵ  ρ̂ ρ̂   ĵ  ˆ ˆ   ĵ  k̂ k̂  sin ρ̂  cos ˆ






k̂  k̂
It is now easy to convert any vector between the two coordinate
systems.

EXAMPLE: Convert the vector E  E x î  E y ĵ  E z k̂ to cylindrical
coordinates.
SOLUTION:
We use our knowledge of the scalar product to write the vector in
cylindrical coordinates. The radial component is found by taking the
scalar product of the vector and the radial unit vector.

E ρ  E  ρ̂   E î  E ĵ  E k̂    cos î  sin ĵ  E x cos  E ysin
y
z  

 x
Following the same procedure, we find the remaining two components
of the vector as

E  E  ˆ   E î  E ĵ  E k̂    - sin î  sin ĵ   E x sin  E y cos
y
z  

 x
Ez  Ez
If you have really mastered the vector material in lesson 1, you should have
no problem converting vectors between different coordinate systems. Thus, I
will leave the remaining transformations for you to work out.
Learn the technique and don’t just try and memorize the results!
3.
Volume and Surface Elements
In order to successfully write down up surface and volume integrals,
you must be able to able to determine differential surface and volume
elements for various coordinate systems. This is usually not difficult
provided you understand the coordinate system that you are using and
you draw a good picture.
EXAMPLE: Write the differential volume element for cylindrical
coordinates.
z
dS3 = dz
y
d

x
dS1 = d
dS2 = d
SOLUTION:
In the figure above, I have drawn an expanded view of a differential
volume element. To make the picture easier to read, I have drawn the
changes in the coordinates so large that they are no longer
infinitesimal and the volume element has some curvature. However,
for infinitesimal changes in the coordinates, the volume element
would be a small cube of sides dS1, dS2, and dS3. The volume of any
cube is found by the formula
Volume = Base x Width x Height
Using our drawing, we see that the infinitesimal volume element in
cylindrical coordinates is
dV  ρ dρ d dz
You can also use the relationship Area = Base x Height and the figure
to find the infinitesimal area of any of the sides.
4.
Vector Operators in Cylindrical Coordinate Representation
The vector operators (Gradient, Divergence, and Curl) are more
complicated in non-Cartesian coordinate systems.
f 
f
1 f
f
ρ̂ 
ˆ  k̂
ρ ρ 
z
 1 
ρEρ  1  E   E z 
  E
ρ ρ
ρ 
z
  1 E z E    E ρ E z 
E ρ 
1 
ˆ


 E

ρ̂




ρE

k̂

z   z
ρ 
ρ  ρ
 
 ρ 
1   f  1  2 f  2 f
  
 f

      2  2 z 2
2


EXAMPLE 1: Evaluate the volume integral  E dv for E  3 r 2 r̂
v
and a sphere of radius R.
SOLUTION:
First, you must find the differential volume element dv for the sphere
dv =
So our integral in spherical coordinates is given as

E
 dv 
v
Important Point: The unit vectors in spherical coordinates are not
constant (i.e. They change direction) so you can pull them out of the
integral.
Trick: Replace the unit vectors with their representations in Cartesian
coordinates. The Cartesian unit vectors ( î , ˆj , and k̂ ) are constants and can
be removed from the integral.
EXAMPLE 1 (CONTINUED)
We start by writing r̂ using the scalar product as
     
r̂  r̂  î î  r̂  ĵ ĵ  r̂  k̂ k̂
We now use the following diagrams to evaluate the scalar products
z
rxy
r̂
y


y
rxy
x
x
The projection of r̂ on the z-axis is given by
The projection of r̂ on the xy plane is given by
Thus, the projection of r̂ on the x-axis and y-axis are given by
EXAMPLE 1 (CONTINUED)
Therefore, r̂ in Cartesian coordinates is given by
We substitute our results into our integral and get
R
2π
π
2π
π
2π

 π 2

4
2
v E dv  3 0 r dr î 0 sin θ dθ 0 cos d  ĵ 0 sin θ dθ 0 sin d  k̂ 0 cosθsinθ dθ 0 d 
Thus, our final result is

 E dv  0
v

ANALOGY: If E was a force, we would have an equal force all the way
around the sphere. Thus, the net external force would be zero!! You should
notice that you don’t get the correct result if you pull r̂ out of the integral!!
EXAMPLE 2: A cylinder of radius a and length 1whose bottom is centered

at the origin is placed in an electric field, E  x î  y ĵ  z 2 1 k̂ E o .



a
1
A.
Evaluate the electric flux across the closed surface of the cylinder by
direct integration.
SOLUTION:
The total flux integral can be broken into three surface integrals


E

d
A


s






E

d
A

E

d
A

E

d
A



bottom
top
side

For the bottom integral, we have z=0 and the surface element dA given by

dA 
Therefore, the flux through the bottom is given by


E

d
A


bottom
EXAMPLE 2 (CONTINUED)
For the top integral, we have z = 1. Thus, the electric field has no zcomponent at the top of the cylinder. Since the surface element dA points in
the  k̂ direction on the top of the cylinder, electric flux through the top of
the cylinder is zero!!
Thus, we are left with just the side integral where  = a. The surface element

dA is given by

dA 
We now convert ρ̂ into its Cartesian representation using our previous work.

Thus, the surface element dA is given by

dA 
Therefore, the electric flux through the side of the cylinder is given by


E

d
A


side


E

d
A


side
Thus, the total electric flux through the cylinder is


E

d
A


s
EXAMPLE 2 (CONTINUED)
B.
Evaluate the electric flux across the closed surface of the cylinder by
applying Gauss’s Theorem.
SOLUTION:
Since the electric field is given in Cartesian coordinates, we use the
Cartesian representation for the divergence. Thus, the divergence of the
electric field is given by
 E E y E z
  E x 


x
y
z
The volume element in cylindrical coordinates is given by
dv 
Using Gauss’ Theorem, we find the electric flux by



E
dv 

v



E
dv 

v
C.
According to Gauss’ Law of Electrostatics, what is the interpretation
of your results from parts A&B.
z
. Compute
x  y2
the gradient of the electric potential (negative of the electrostatic field).
EXAMPLE 3: The electric potential V is given by V 
SOLUTION:
2
EXAMPLE 4: Starting with the Cartesian representation for the Laplacian
in two dimensions derive the Laplacian’s Polar representation.
SOLUTION:
The Cartesian representation of the Laplacian is
2f 
By the chain rule of Calculus, we have
f

x
f

y
We need to express the Cartesian coordinates x and y and their differentials
dx and dy in their polar representations
x=
y=
dx =
dy =
We now perform some algebra to obtain the polar coordinate differentials d
and d.
cos dρ 
sin dρ 
EXAMPLE 4 (CONTINUED)
Multiplying the first equation by cos and the second one by sin, we get
cos2 dρ 
sin 2 dρ 
We now add the equations together to obtain d.
dρ 
We now do a similar algebraic set of tricks to get d.
sin d 
cos d 
sin 2 d 
cos2 d 
d 
From the chain rule and our equations for d and d, we find
ρ

x
ρ

y


x


y
EXAMPLE 4 (CONTINUED)
Substituting our results back into our equations for the partials with respect
To x and y, we get
f

x
f

y
We now find the second derivatives by repeating our process
 2 f   f  
  
x 2 x  x  x
 2f

x 2
 2f

x 2
 2 f   f  
  
y 2 y  y  y
EXAMPLE 4 (CONTINUED)
 2f

y 2
 2f

y 2
Thus, our final result for the Laplacian in polar coordinates is
2f 
2f 
C.
Generalized Orthogonal Coordinates
You need to become familiar with the mathematics of generalized
orthogonal coordinate systems. This material is covered in Chapter 2
of Mathematical Methods of Physics by Arfkin and Weber and my
class notes for Mathematical Methods. I strongly suggest that you
become familiar with the material available on vector analysis and
coordinate systems in your Math Handbook. If you are using the
Schaum’s Outline Mathematical Handbook by Spiegel, the material is
on pages 116-130.
D.
Dirac Delta Function
The Dirac delta function is not actually a function since it isn’t finite.
It is defined under the theory of distributions (see Mathematical
Physics by Butkov for more information).
The Dirac delta function is defined by its assigned properties:
δx   0 when x  0
f 0 when x  0 is within a, b




f
x
δ
x
dx


a
 0 otherwise
b
It is important to remember that the Dirac Delta function is only
defined under an integral!! Thus, you usually derive its properties by
showing that two integrals are equal.
Another way of deriving properties of the Dirac delta function are to
represent it in terms of well behaved functions that approach the delta
function under a suitable limit (See my Math Methods notes and
Chapter 1 of Mathematical Methods for Physicists by Arfkin and
Weber).
The Dirac Delta function’s practical purpose is to represent point
sources. For example, we know from PHYS2424 that we can obtain
the net charge in an interval by integrating the linear charge density
over the interval.
Q enclosed   λ dx
line
We now consider a problem in which three point charges each of
charge q exist along the x-axis as shown below:
x = -3
x=2
x=4
How are we going to write the linear charge density for this charge
distribution?
ANSWERE:
λ  q δx  3  q δx  2  qδx  4
EXAMPLE PROBLEM:
For the charge distribution shown above, find the charge in the
interval between x=-5 and x=3.
SOLUTION:
3
3
3
5
5
5
Q enclosed  q  δx  3dx  q  δx  2dx  q  δx  4dx
The first two integrals are equal to one as the Dirac delta function’s
argument goes to zero over the interval. For the same reason, the last
integral is zero!! Thus, we have
Q enclosed  2 q
We can also write three dimensional Dirac delta functions by taking
the product of three 1-dimensional Dirac delta functions.
3-Dimensional Dirac Delta Function in Cartesian Coordinates
δx, y, z   δx δy δz 
EXAMPLE: Evaluate the following integral
x 5 y 4 z 8
   3x
x 0 y 0 z  5
SOLUTION:
2
y z 3δx  1δy  2δz  3dxdydz