10 - Conversion from Rectangular to polar coordinates
and gradient wind
1. Derivation of horizontal equation of motion in polar coordinates
Many problems in Meteorology and Oceanography have circular symmetries
which make them much easier to deal with in a polar coordinate system. As we learned
in our discussion of vectors, the results of a physical problem are independent of our
choice of coordinate system. The coordinate system choice is a matter of convenience in
the calculation of the problem (in the case of rectangular versus polar coordinates – it is
mainly a matter of how much trigonometry we wish to endure). It also allows us to
examine matters from a different point of view, from which we can gain better insight
into the nature of the physical problem.
For example, an extremely simple but surprisingly accurate representation of
atmospheric flow is the Geostrophic wind where the Coriolis force is balanced by the
1
pressure gradient force in the horizontal plane. f  u    h P . The resulting flow is

axi-symmetric (no dependence on  ) and, by examining this result in polar coordinates,
we can reduce two differential equations to just one. If we examine the more general

Du h
 f u
Dt

h

1
 h p (where the

h subscript indicates just the iˆ and ĵ terms, we can come to a better physical
understanding on what it means to neglect the advection term in the Geostrophic
approximation.
To start the process of relating rectangular and polar coordinates, consider a radial
vector in the Cartesian coordinate plane as in figure 1. (A radial vector or ray is a
specific position vector which always starts at the origin or pole)
equation of horizontal motion in polar coordinates,
^
j
^
x
^
r  x i y j
y

^
i
Figure 1 – Representation of a radial vector in the Cartesian plane.
1
From figure 1 we can see that
x  r cos 
(1)
y  r sin  
Using equations (1) and figure 1 we can also see that
r
x2  y2
(2)
 y
x
  tan 1  
Equations (2) describe the two parameters of the polar CS which can be used to
define any point in the 2-D plane. Since we are interested in polar vectors, we need to
derive the basis vectors in the new polar coordinate system. We wish to express any
vector in terms of a radial component from the origin and its angular distance,  , from
^
the x axis. The radial basis vector, r , is simply a unit vector in the same direction as the
^
radial vector r shown in figure 1. Using the requirement that r  1 along with equations
^
^
(1) we can relate the radial unit vector to our rectangular unit vectors i and j :
^
^


r
cos

i

sin

j





^
r
x i y j
  cos  i^  sin  ^j
r 
 
 
 
r
r
x2  y 2
^
^
(3)
Notice that r is a function of the azimuthal angle, r  r   . This will be important later in
the discussions.
^
^
^
^
^
To find the azimuthal unit vector, with respect to i and j , we simply require it
^
^
to be orthogonal to r . Use of simple trigonometric identities allow us to find  :
  cos  90 i  sin   90 j   sin   i  cos  j
^
^
^
^
^
(4)
Notice that  is also a function of the azimuthal angle,      .
^
^
^
^
^
Equation (3) and (4) are two simultaneous equations with unknowns i and j .
With some algebra we can show that
i  cos  r  sin  
^
^
^
(5)
2
j  sin   r  cos 
^
^
^
^
^
Figure 2 shows the unit vectors r and  in the x-y plane.
^
j
^
^
r  x i y j
^

^
r
r
r

^
i
Figure 2. Graph showing the radial and azimuthal unit vectors, Note, as is the case with
^
^
all unit vectors, r  1 and   1
We now have the tools to relate the polar components and associated basis vectors
to their rectangular counterparts. Since the equations of motion deal with temporal and
spatial variations of the velocity vector, we also need to derive the variation of our polar
coordinates in equation (2) with respect to our rectangular coordinates so that we can then
define the gradient or advection operator in polar coordinates. For starters, we can see by
taking the derivative of equations (3) and (4) with respect to  that
^
^
^
^
dr
  sin   i  cos  j  
d
(6)
^
^
^
^
d
  cos  i  sin   j   r
d
Finally, we need to relate the variation of our polar components with respect to their
r r 

rectangular counterparts:
,
,
and
.
x y x
y
r
r
To find
and
, simply take the derivative of equation (2) with respect to x
x
y
and y:
3
r  2

x  y2
x x

r

 x 2  y
y y


1
2

1
 x2  y2
2


1
2
x
 cos 
r
2x 
(7)
1
2 2

1 2
x  y
2

1
2 2
y
 sin  
r
2y 
Rather than use the tedious tangent function or its inverse in equation (2) to find


and
, we can use the orthogonality of the x and y coordinates and equations (1) as
x
y
follows:
y
r


sin  
 0  sin    r cos 


x
x
x
x
r
x
r

 cos 
 0  cos   r sin  


y
y
y
y
r
(8)
Gradient operator
We can now find the gradient in terms of polar coordinates and their associated
basis vectors by using equations (5), (7) and (8). First use the chain rule to express
variations with respect to x and y as variations with respect to r and  :
^
h  i
 ^  ^ r    
j
 i


x
y  x r x  
 r    

j

 y r y  
^
Now substitute equation (5) to obtain the polar basis vectors
^
^
^
^
 sin     
 cos   



 h   cos  r  sin    cos  
   sin   r  cos   sin   

r
r   
r
r  



^ 
^ 1 
(9)
h  r  
r
r 
Spatial and velocity components in polar coordinates:
Equation (3) already established the relationship of a radial vector in terms of the
^
polar components and unit vectors as r  r r . It is instructive to verify this relationship
by directly converting spatial rectangular coordinates and unit vectors into polar
coordinates as defined above.
4
x
from Eq (5)
y
from Eq (5)
^
^
^
^
^
^




r  x i  y j  r cos( ) cos  r  sin     r sin(  ) sin   r  cos  




2
2
r  (r cos ( )  r sin ( ))rˆ  (r sin( ) cos( )  r cos( ) sin( ))ˆ  r rˆ (cos 2 ( )  sin 2 ( ))
^
r  rr
(10)
Now, to find the associated velocity components in polar form, take the time derivative of
^
equation (10) and use equation (6) for the variation of the unit vector r with respect to 
^
d r d  ^  dr ^
d r d dr ^
d ^
u
 r r 
r r

r r

dt dt   dt
d dt
dt
dt
We now have the velocity vector in the form of polar components and unit vectors.
We will define the associated velocity components of the velocity vector as follows:
dr
dt
d
u  r
dt
so
ur 
^
^
u  u r r  u 
(11)
Advection derivative:
From equations (9) and (11) and the use of the dot product, the conversion of the
advective derivative from rectangular to polar coordinates is straightforward:
^
 u 
 ^
 ^  ^ 1  
u     u r r  u      r  

  u r
r  
r
r 

  r
(12)
Equation (12) is valid for a scalar function,  r,  . For a vector however, we saw from
equations (6) that the polar unit vectors vary with respect to  so we must take this into
consideration in our results. For the advection of a flow field for example
u  u  u r
^
^
  ^
 u   ^

 u r r  u   
 u r r  u  
r 
r





Let us examine each term in the above equation. The first term is simplified to:
5
ur
^
  ^
 ^ u r ^ u
  ur
 u r r  u    r u r
r 
r
r

The second term is a bit more complicated due to the  -dependence of the basis vectors.
It simplifies to:
^
^
2
^
u   ^
 ^ u ur ^ u u u ur  r u  



 ur r  u    r
r  
r 
r 
r 
r 

u
r 
r
u
  
 r
^
u r ^ u u u u r ^ u ^



r

r 
r
r
2
u r u  ^  u u u u r  ^
 r 




r   r 
r 
2
Combining both terms we see that
2
^
u u r u  ^  u u u u u r 
u
    ur
u  u  r  u r r  






r
r


r

r
r


r 



The resulting components of the equation of motion,

Du
 f u
Dt

h

1

h p ,
in polar coordinates are:
u
u r   u  
1 p
  ur

u r    fu  
t  r r  
r
 r
2
(13)
u   u  
u u
1 p
  ur

u   r  fur  
t  r r  
r
r 
We can see equations (13) contain an advective operator just like in equation (12) when
operating on a scalar. The above equations also contain an additional term that accounts
for the curvature of the problem. The additional term in the radial equation above is
simply the centrifugal acceleration.
2. Gradient Wind:
Looking back at Figure 1, we notice that curvature is significant in the flow field.
The continually changing direction of the fluid parcel as it travels around the low
pressure system will lead to a centrifugal acceleration and associated force to the
equations of motion. By converting the original equations of motion, given by equations
6
(1), to cylindrical polar coordinates we can derive a simple relationship for the axial flow
around a low or high pressure system that includes the effects of rotation.
Recall the results of equation (13) and including conservation of mass
u
u r   u  
1 p
  ur

u r    fu  
t  r r  
r
 r
2
(14)
u   u  
u u
1 p
  ur

u   r  fur  
t  r r  
r
r 
u v w


0
x y z
We now make the following assumptions:
1) The flow is steady – therefore

0
t

0

3) Assume the flow has no radial component (tangential): is u r  0
2) The flow is axi-symmetric – therefore
Given the above assumptions, the radial component of the conservation of momentum is
u
1 p
 fu 
r
 o r
2
(15)
All terms of the axial component of conservation of momentum are eliminated and the
vertical component of momentum conservation is the equation of hydrostatic balance.
The axial flow in equation (15) is called the gradient wind and accounts for both the
Coriolis force and pressure gradient force. Unlike a geostrophic flow however, there is
force imbalance in such a way that we obtain a steady state curved flow where the only
acceleration is a continual change in direction caused by the centripetal acceleration.
Since equation (15) is simply a quadratic equation for u , we can obtain a simple
analytic solution for the gradient wind:
fr
r p
 fr 
u      
2
 o r
2
2
(8)
There are four possible physical situations for equation (8) depending on the sign of the
square root and the sign of the pressure gradient. They are:
7
p
fr
r p
 fr 
0
1) Normal Cyclonic flow: u      
; u  0 and
r
2
 o r
2
2
Since the radial gradient of the pressure field is greater than 0, we are talking about a
pressure field that increases as one travels away from the origin. Both terms in the
fr
radical are of the same sign and the sum is greater than
implying that the axial flow is
2
counter-clockwise or cyclonic. From a force relationship perspective, consider first a
geostrophic balance. The straight flow is a balance between an inward directed pressure
gradient and an outward directed Coriolis force. In the case of the gradient wind, there is
a force imbalance where the pressure gradient force is larger than Coriolis force. This
leads to net force directed inward and a resultant centripetal acceleration causing a curved
cyclonic flow.
p
fr
r p
 fr 
0
2) Normal anti-cyclonic flow : u      
; u  0 and
r
2
 o r
2
:
The decreasing radial pressure gradient indicates a high pressure system. The root term
fr
is positive but less than
indicating the axial flow is clockwise. There is a force
2
imbalance where the outward directed pressure gradient force is less than inward directed
Coriolis force. This force imbalance leads to net force directed inward and a resultant
centripetal acceleration causing a curved anti-cyclonic flow. Notice that, since the radial
p
gradient of pressure is less than 0, there is a limit on how large
can be to ensure the
r
p  o f 2 r
radical is real. The restriction is

r
4
2
fr
r p
 fr 
; u  0
   
2
 o r
2
2
3): Anomalous anti-cyclonic flow about a high: u  
and
p
0
r
This third type of flow can exist theoretically but is not really seen empirically. It has the
same force balance as example 2 but utilizes the negative root of the quadratic equation.
Simple calculation shows that the required pressure gradients for the above flow to exist
as a gradient wind are extremely small.
fr
r p
 fr 
4): Anomalous anti-cyclonic flow about a low: u      
; u  0 and
2
 2  o r
p
0
r
2
8
This fourth type of flow can exist theoretically and occasionally is seen in nature (i.e., an
anti-cyclonically rotating tornado). It has the same force balance as example 1 but
requires that the pressure gradient (which is a positive number) be large enough to
support anti-cyclonic flow (positive u ) .
9