Lesson 1: Vectors and Coordinate Systems
I.
A.
Position Vector
Definition:
Consider an object (a baseball for instance) which is located at some
arbitrary point in space (point P). We need to develop a way of
describing the location of the ball in terms of mathematics. The
method that physicists have developed is called the position vector!!
The position vector for an object is the vector drawn from the origin
of a coordinate system to the location of the object.

In the drawing below, the position vector r (shown in green) is drawn
from the origin to the location of our baseball at point P.
P

r
B.
Is the position vector unique?
Since the position vector is defined mathematically in terms of a
coordinate system, it is not unique since the coordinate system is not
unique. For instance, John might specify the position of the library
(green arrow) with a coordinate system’s whose origin is the science
building while Susan specifies the position of the library (blue arrow)
with respect to the math building.

r

R

r
P
Mathematically, these are two different ways of specifying the same
physical point in space (the location of the library). Thus, these two
mathematical descriptions must be related to each other in some way.
Such relationships are known in physics as transformation equations
and are very important. In this case, the connection is simply the
relationship for adding vectors. The library’s position vector as given
by Susan is the sum of John’s position vector according to Susan
(shown in red) and the library’s position vector according to John.
  
r   R  r
You might be inclined to remark that all of this is trivial and even that
it is a waist of time. All of this material should be review from your
introductory physics course. I encourage you to reject such quick
judgments and to reflect deeply upon each concept that we study in
this course. The reason that I have chosen to include this material is to
remind you of some very important physics and to highlight the
course goals. First, you should remember from your introductory
physics course that many other physics quantities are derived from the
position vector including displacement, velocity, acceleration, linear
momentum, angular momentum, etc. Thus, the question of reference
frames and transformation equations is extremely important in
physics. Secondly, a major reason for the course is not to solve
problems involving balls rolling down inclined planes, but to develop
an understanding of the way in which physical systems can be
represented in mathematics. By learning to apply mathematics to
mechanical systems where you have some physical intuition, you
build the skills needed for more advanced physics courses (Relativity,
Quantum Electrodynamics, etc) where intuition based upon everyday
experience will be of little help.
II.
Velocity Vector
A.
Definition:
The velocity of an object with respect to a coordinate system is
defined as the time rate of change of the object’s position vector in
that coordinate system.

 dr
v
dt
B.
Is the velocity vector unique?
No, the velocity vector is not unique. Observers using two different
coordinate systems might disagree on the velocity of an object.
You might be inclined to believe that this is a mathematical artifact
and that you could do an experiment to determine if the object is
moving or not. However, one of the fundamental principles of physics
states that it is impossible to devise an experiment that can actually
detect absolute motion. All motion is dependent on the observer.
III.
Acceleration Vector
A.
Definition:
The acceleration of an object with respect to a coordinate system is
defined as the time rate of change of the object’s velocity vector in
that coordinate system.

 dv
a
dt
B.
Is the acceleration vector unique?
No, the acceleration vector is not unique. Observers using two
different coordinate systems might disagree on the acceleration of an
object.
IV.
Cartesian Coordinate System
You should have been introduced to the Cartesian Coordinate system
in your introductory physics course. In this system, an arbitrary point
in space such as point P show below is defined by three coordinates
(x,y,z).
P (x,y,z)

r
z
x
y
A.
Position Vector
We have already discussed that a position vector can be written
mathematically in more than one way. In this case, we choose
 the

write
the
position
vector
as
the
sum
of
three
other
vectors
(
,
, and
X
Y

Z ) as shown below.

r

X

Y
   
r XY Z

Z
You may remember that a unit vector is a vector of magnitude
(length) 1 with no units that is used solely to provide direction.To
 

specify the direction of these three vectors ( X , Y , and Z ), we define
three unit vectors (one for each of our coordinates). These unit vector
are
î which points in the +x direction
ˆj which points in the +y direction
k̂ which points in the +z direction.
By multiplying each unit vector by its associated coordinate, we can
produce our three required vectors.

X  x î

Y  y ĵ

Z  z k̂
This is just a case of multiplication of a vector by a scalar. If the scalar
(coordinate) is positive then the resultant vector points in the same
direction as the unit vector. If the scalar (coordinate) is negative then
the resultant vector is in the opposite direction (rotated 180 degrees)
from the direction of the unit vectors. Putting this into our relationship
for the unit vector, we get

r  x î  y ĵ  z k̂
Again, you should have already seen this result in your introductory
physics course. Did you understand the process by which it came
about or did you just memorize it? Often in introductory physics
classes, we can take advantage of our students’ every day experiences
to teach certain physics principles. Your instructor may have
discussed using a combination of North/South and East/West to drive
between your house and the grocery store. Because the Earth is
relatively flat based upon your everyday experience, most students
have an intuitive feel for Cartesian Coordinates. However, we will be
working with other coordinate systems so you need to understand
actual mathematical process.
B.
Velocity Vector
We now use the definition of velocity to develop the relationship for
Cartesian coordinates.

 dr d
v   ( x î  y ĵ  z k̂ )
dt dt
 d
d
d
v  ( x î )  ( y ĵ )  ( z k̂ )
dt
dt
dt
 d
d
d
d
d
d
v  ( x ) î  x ( î )  ( y ) ĵ  y ( ĵ )  ( z ) k̂  z ( k̂)
dt
dt
dt
dt
dt
dt
The three unit vectors ( î , ˆj , and k̂ ) are constant (remember that for a
vector this means constant in direction as well as constant in
magnitude) so their derivatives are zero!!!
 d
d
d
v  ( x ) î  ( y ) ĵ  ( z ) k̂
dt
dt
dt

v  x î  y ĵ  z k̂
There are several important things that you should take away from
this exercise. First, one of the advantages of working with Cartesian
coordinates is that the unit vectors don’t enter into the Calculus. In
other coordinate systems, you will have to take derivatives of unit
vectors!! Secondly, the result shows that the velocity component in
each direction depends only on the time rate of change of that
coordinate!! Therefore, motion along one axis is independent of
motion along either of the other axis. This is why a ball thrown
horizontally will hit the Earth at the same time as a ball dropped from
the same height provided we neglect the minute curvature of the
Earth. It is also why we were able to break rectilinear motion
problems into parts in your introductory physics course.
C.
Acceleration Vector
We now use the definition of acceleration to develop the relationship
for Cartesian coordinates.

 dv d
a   ( x î  y ĵ  z k̂ )
dt dt
 d
d
d
a  ( x î )  ( y ĵ )  ( z k̂ )
dt
dt
dt
 d
d
d
d
d
d
a  ( x ) î  x ( î )  ( y ) ĵ  y ( ĵ )  ( z ) k̂  z ( k̂)
dt
dt
dt
dt
dt
dt
Again, we use the fact that the derivative of our three unit vectors ( î ,
ˆj , and k̂ ) are zero!!!
 d
d
d
a  ( x ) î  ( y ) ĵ  ( z ) k̂
dt
dt
dt

a  x î  y ĵ  z k̂
Here we see that the acceleration components in each direction are
independent.
V.
Cylindrical Coordinate System
In this system, an arbitrary point in space such as point P show below
is defined by three coordinates (, , and z) as shown below.
z
P (,,z)

r
z

y

x
In the drawing, point P is on the surface of an imaginary cylinder of
radius , height h, and with the projection of its position vector onto
the x-y plane at an angle  from the x-axis. You also used this
coordinate system in introductory physics to handle rotation problems.
However, you probably just memorized the results of the math like the
formula for centripetal acceleration instead of actually performing the
math operations!!
A.
Unit Vectors
Following the procedure for our past work, we are going to define a
unit vector for each of the three coordinates. The first unit vector is
found by only allowing  to vary while keeping  and z constant. This
amounts to moving along the radius of the cylinder in the direction of
an increasing radius. The second unit vector is found by only allowing
 to vary while keeping  and z constant. This amounts to walking
counter clockwise around a circle. The third unit vector is found by
only allowing z to vary while keeping  and  constant. This amounts
to walking in the +z direction.
ρ̂ points in the + radial direction
̂ points in the direction of increasing angle 
k̂ points in the +z direction.
z
P (,,z)

r
̂
k̂
y

ρ̂
x
Since the unit vectors ρ̂ and ̂ lie in the x-y plane and do not depend
on z, it is often useful to have the following additional diagram when
working out the Calculus:
y
̂

ρ̂

x
Using our knowledge of the dot product and components, we see that
ρ̂  Cos( ) î  Sin(  ) ĵ
ˆ   Sin( ) î  Cos( ) ĵ
These are useful relations for doing the Calculus!!
B.
Position Vector
In cylindrical coordinates, we can write the position vector in as the
sum of only two vectors:
  
r ρ  Z
z
P (,,z)

r

Z z k̂
y


ρ  ρ ρ̂
x
Thus, we have the position vector in cylindrical coordinates as

r  ρ ρ̂  z k̂
You might be surprised that there is no  component. This is just one
of many places where your intuition from Cartesian coordinates can
get you into trouble. Also, you should consider the fact that  is an
angle and not a distance. Thus, a component in the  direction would
also have the wrong units unless the term also contained either z or .
C.
Velocity Vector
We now use the definition of velocity to develop the relationship for
cylindrical coordinates.

 dr d
v   ( ρρ̂  z k̂ )
dt dt
 d
d
v  ( ρρ̂ )  ( z k̂ )
dt
dt
 d
d
d
d
v  ( ρ ) ρ̂  ρ ( ρ̂ )  ( z ) k̂  z ( k̂)
dt
dt
dt
dt
The unit vectors ( ρ̂ and ̂ ) are not constant since their direction
depends on the location of point P. Therefore, we will have to take
their derivatives!! We do this by replacing each unit vector with its
representation in Cartesian coordinates and then do the Calculus.
Starting with ρ̂ , we have
d
d
( ρ̂ )  [ Cos( ) î  Sin ( ) ĵ ]
dt
dt
d
d
d
( ρ̂ )  [ Cos( )] î  [ Sin ( ) ] ĵ
dt
dt
dt
d
d
d
( ρ̂ )   Sin( ) [  ] î  Cos( ) [  ] ĵ
dt
dt
dt
d
d
( ρ̂ )  [Sin( ) î  Cos( ) ĵ] [  ]
dt
dt
d
( ρ̂ )   ˆ  ωˆ
dt
The time rate of change of the angular position  was called the
angular velocity, , for rotation problems in your introductory physics
course!!
We now calculate the time derivative of ̂ by the same procedure.
d
d
(ˆ )  [ - Sin( ) î  Cos( ) ĵ ]
dt
dt
d
d
d
(ˆ )   [ Sin( )] î  [ Cos( ) ] ĵ
dt
dt
dt
d
d
d
(ˆ )   Cos( ) [  ] î  Sin ( ) [  ] ĵ
dt
dt
dt
d
d
(ˆ )  [Cos( ) î  Sin ( ) ĵ] [  ]
dt
dt
d
(ˆ )   ρ̂   ω ρ̂
dt
A graph of the unit vectors and their derivatives may help you
visualize the results. It also reminds you that the derivative of a vector
produces a vector tangent to the original vector!! The length of the
derivative vectors depend on the angular velocity. If there is no
rotation then the motion is in a plane and the derivatives of the unit
vectors are zero!! If the angular velocity is high then the magnitudes
of the derivatives of the unit vectors will be large and these terms will
dominate the velocity results.
y
dρ̂
dt
̂
dˆ
dt


ρ̂
x

We are now ready to complete our relationship for the velocity of an
object in Cylindrical coordinates by substituting in our derivative
results and remembering that k̂ is constant.
 d
d
v  ( ρ ) ρ̂  ρ ˆ  ( z ) k̂
dt
dt

v  ρ ρ̂  ρ ˆ  z k̂  ρ ρ̂  ρ ωˆ  z k̂
Let us now consider our results. The first term is the radial velocity
component. The last term is the velocity component in the z-direction.
If we consider the special case of a particle traveling in a circle in the
x-y plane then both the first and third terms in the velocity
relationship are zero. Thus, an object moving in a circular path as in
your introductory physics course would have a velocity with a
component only in the ̂ direction with a magnitude of ρ ω . This is
just the tangential velocity from your introductory physics course!!
In other words, we have

v  v radial ρ̂  v tangential ˆ  v z k̂
D.
Acceleration Vector
We now use the definition of acceleration to develop the relationship
for cylindrical coordinates.

 dv d
a   ( ρ ρ̂  ρˆ  z k̂ )
dt dt
 d
d
d
a  ( ρ ρ̂)  (ρˆ )  (z k̂ )
dt
dt
dt
 d
d
d
d
d
d
d
a  ( ρ )ρ̂  ρ ( ρ̂)  (ρ)ˆ  ρ ( )ˆ  ρ (ˆ )  (z )k̂  z ( k̂ )
dt
dt
dt
dt
dt
dt
dt
Using our relationships for the time derivatives of the unit vectors, we
have

a  ρρ̂  ρ ˆ  ρ ˆ  ρˆ  ρ (- ρ̂)  z k̂
Rearranging and combining terms, we have that

a  ( ρ - ρ 2 )ρ̂  (2 ρ   ρ)ˆ  z k̂
The z-component of the acceleration is just due to the change in the zcomponent of the objects velocity as in the Cartesian coordinate
system. However, we see that there are two terms contributing to the
acceleration in both the radial and angular directions. The first term in
the radial direction occurs if the radial velocity of the object is
changing. The first term in the angular acceleration occurs if the
object has both angular and radial velocity.
To understand the two remaining terms, we again consider an object
moving in a circle in the x-y plane. In this case,  and z are constant
so ρ  ρ  z  0 . Thus, the acceleration becomes

a  - ρ 2ρ̂  ρˆ  - ρω2ρ̂  ραˆ
The radial term is the centripetal acceleration and the angular term is
the tangential acceleration of the object from introductory physics
class where α  is the angular acceleration.
VI.
Transformations Between Coordinate Systems
A.
An important use of the scalar product is to find the component of a
vector in the direction of anarbitrary unit vector. To illustrate this
property consider a vector A . In Cartesian coordinates, the vector


A is represented as A  A x î  A y j  A z k̂ . If we take the scalar

product of vector A and the unit vector î , we get the x-component of

vector A as shown below:

î  A  A x ( î  î )  A y ( î  ĵ )  A z (î  k̂)

î  A  A x (1 )  A y ( 0 )  A z (0)

î  A  A x
In a similar manner, we can show that

ĵ  A  A y

k̂  A  A z
This procedure is not limited to just the Cartesian coordinate
representation. It will work for any coordinate system whose unit
vectors are orthogonal (i.e. the scalar product of two different
 unit
vectors is zero). For instance if we represented the vector A using

cylindrical coordinates where A  Aρ ρ̂  A ˆ  A z k̂ then we have

ρ̂  A  Aρ ( ρ̂  ρ̂)  A ( ρ̂  ˆ )  A z ( ρ̂  k̂)

ρ̂  A  Aρ (1)  A ( 0)  A z (0)

ρ̂  A  Aρ
By a similar procedure, we can also show that

ˆ  A  A
The key thing is not to remember these results, but to understand the
process so that you are able to work with any orthogonal coordinate
system that you might encounter!!
B. Let us now use the material in the previous section to develop
relationships for transforming a vector’s representation in one
coordinate
system to another. Let us suppose that we know vector

A ’s representation in Cartesian coordinates and want to know its
representation in cylindrical coordinates.

Aρ  ρ̂  A
Aρ  ρ̂  ( A x î  A y ĵ  A z k̂)
Aρ  A x (ρ̂  î)  A y (ρ̂  ĵ)  A z (ρ̂  k̂)

A  ˆ  A
A ˆ  ( A x î  A y ĵ  A z k̂)
A  A x (ˆ  î)  A y (ˆ  ĵ)  A z (ˆ  k̂)

A z  k̂  A
A z  k̂  ( A x î  A y j  A z k̂)
A z  A x (k̂  î)  A y (k̂  j )  A z (k̂  k̂)
I could simplify these expressions, but I have left them in their most
general form so that I can rewrite them in terms of matrices.
  ρ̂  î 
 Aρ   


 

 A    ˆ  î 


 
A
 z    k̂  î 


 ρ̂  ĵ


ˆ  ĵ


 k̂  ĵ


 ρ̂  k̂  

  A x 


ˆ  k̂   A 
y

 

 k̂  k̂   A z 


Our result indicates that we can translate between any two orthogonal
coordinate systems by multiplying our vector by a transformation
matrix whose elements are the scalar product of the unit vectors in
these two coordinate systems!!! This process is well suited for
computer analysis since computers are extremely efficient at handling
matrices (called arrays in CS jargon). In fact, this is the way that
rotations and other motion calculations are often handled in video
games!!!
1)
2)
3)
4)
Summary:
Find the scalar products between each of the unit vectors
Write the transformation matrix whose elements were found in
step 1.
Write your original vector as a column matrix whose elements
are the components of the vector in the original coordinate
system
Multiply the transformation matrix by the original vector’s
column matrix to produce a new column matrix whose elements
are the components of the vector in the new coordinate system.
As for our example, we know that
ρ̂  î  Cos( )
ρ̂  ĵ  Sin( )
ρ̂  k̂  0
ˆ  î   Sin( )
ˆ  ĵ  Cos( )
ˆ  k̂  0
k̂  î  0
k̂  ĵ  0
k̂  k̂ 1
This gives us a transformation matrix of
 Cos( ) Sin( ) 0 


T    Sin( ) Cos( ) 0 

0
0
1 

This happens to be the rotation matrix about the z-axis developed in
the textbook!!
VII. Other Sources of Information
1. Schaum’s Outline Series: Math Handbook – Section on Coordinate
Systems
2. Mathematical Methods For Physicists: 5th Ed. By Arfken and
Weber : Chapter 1 Sections 1-5, Chapter 2 Sections 1-5, and
Chapter 3 Sections 1-3.
VIII. Problems & Questions To Consider
I have left spherical coordinates for you to work out. Some of the
question that you might consider if you think that you understand this
material are:
1. What is the determinant of the transformation matrix? What is the
physical significance of this value?
2. What is the transformation matrix for converting from cylindrical
to spherical coordinates?
3. What is the transformation matrix for converting from Cartesian
to spherical coordinates?
4. How do you write the position vector
for a particle in spherical coordinates?
5. What is the time-derivative of the unit vectors in spherical
coordinates?
6. How do you write the velocity, and acceleration vectors
for a particle in spherical coordinates?
7. The scalar product of two vector in Cartesian coordinates is
 
A  B  A x Bx  A y B y  A z Bz . What are the formulas for the
scalar product in cylindrical and spherical coordinates?