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11. 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 pressure gradient force
1
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 equation of horizontal motion in polar


Du h
1
 f  u h    h p (where the h subscript indicates just the iˆ and ĵ terms,
Dt

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)
coordinates,
Figure 1 – Representation of a radial vector in the Cartesian plane.
From figure 1 we can see that
x  r cos 
1
(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)
j  sin   r  cos 
^
^
^
2
^
^
Figure 2 shows the unit vectors r and  in the x-y plane.
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 rectangular

r r 
counterparts:
,
,
and
.
x y x
y
r
r
To find
and
, simply take the derivative of equation (2) with respect to x and y:
x
y
r  2

x  y2
x x


1
2


1 2
x  y2
2


1
2
2x 
x
 cos 
r
(7)
3
r
 2

x  y2
y y


1
2


1 2
x  y2
2


1
2
y
 sin  
r
2y 
Rather than use the tedious tangent function or its inverse in equation (2) to find

and
x

, we can use the orthogonality of the x and y coordinates and equations (1) as follows:
y
y
r


sin  
 0  sin    r cos 


x
x
x
x
r
(8)
x
r

 cos 
 0  cos   r sin  


y
y
y
y
r
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.
x
from Eq (5)
y
from Eq (5)
4
^
^
^
^
^
^




r  x i  y j  r cos( ) cos  r  sin     r sin(  ) sin   r  cos  




2
2
ˆ
r  (r cos ( )  r sin ( ))rˆ  (r sin( ) cos( )  r cos( ) sin( ))  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  
u
r

u


r




 r


r  
r 
r 
r 
r 

u u r ^ u u u u r ^ u



r 
r 
r
r
2
 u u r u  ^  u u u u r
 r 
  

  r   r
r


r


2 ^
^
r
r
^


Combining both terms we see that
 u r u u r u 2  ^  u u u u u r 
    ur
u  u  r  u 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 (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.
6
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
fr
r p
p
 fr 
1) Normal Cyclonic flow: u      
; u  0 and
0
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 radical are of the same
fr
sign and the sum is greater than
implying that the axial flow is counter-clockwise or
2
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.
fr
r p
p
 fr 
   
; u  0 and
0
r
2
 o r
2
2
2) Normal anti-cyclonic flow : u  
:
The decreasing radial pressure gradient indicates a high pressure system. The root term is
fr
positive but less than
indicating the axial flow is clockwise. There is a force imbalance where
2
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 gradient of pressure is less than 0, there
p  o f 2 r
p
is a limit on how large
can be to ensure the radical is real. The restriction is

r
r
4
fr
r p
 fr 
3): Anomalous anti-cyclonic flow about a high: u      
; u  0 and
2
 o r
2
p
0
r
2
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
p
 fr 
   
; u  0 and
0
r
2
 2  o r
This fourth type of flow can exist theoretically and occasionally is seen in nature (i.e., an anticyclonically 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
(negative u ) .
2
4): Anomalous anti-cyclonic flow about a low: u  
8