MT11B Weather Systems Analysis
Eleanor Highwood, December 2003
Lecture T5
Thermal wind shear
Reading: FWC section 7.10
T5 1 - Graphical illustration:
We start from two principles:

The hydrostatic relationship of AP3 indicates that thickness is
proportional to temperature.

The geostrophic relationship of D2 indicates that the wind blows
parallel to contours of constant height of a pressure surface, with a
speed proportional to the slope of the pressure surface.
Consider a slab of air (between 2 pressure surfaces) across which there
is a gradient of temperature.
At A, the air is cold, and its thickness is smaller.
At point B, it is warm and so has larger thickness.
As a result, the lower level pressure surface slopes downward, from A
to B. The geostrophic wind, with low pressure on its left, must blow out
of the plane of the cross section.
Similarly, the upper level pressure surface slopes upwards from A to B. The geostrophic
wind on this surface must blow into the plane of the cross section.
Thus the wind must change with height i.e. the wind must be “sheared”. This shear is
called the “thermal wind shear”.
[Note that these considerations do not tell us the actual wind speed. For instance, the
thickness equation would still be satisfied if the p1 surface were flat but the p2 surface
had twice the slope.]
MT11B Weather Systems Analysis
Eleanor Highwood, December 2003
T5 2 - Mathematical derivation:
Let us derive a formula for this shear. The shear is merely the change in wind with
height, and we know from above that it depends on the temperature gradient in the
horizontal. Doing the usual thing of splitting the geostrophic wind into components in the
N-S and E-W direction (remember EC7) we need to find:
a) Change of E-W component of geostrophic wind with pressure,
u g
Z
u g
p
, or with height,
.
And
b) Change of S-N component of geostrophic wind with pressure,
c) And relate these to the horizontal gradients in temperature,
vg
p
T
x
, or with height,
and
T
y
vg
Z
.
.
From the hydrostatic relationship we know that how Z varies with p depends on T,
Z
RT
.

p
pg
Differentiate with respect to y to give us a temperature gradient on one side of the
equation:
 2Z
R T

.
 p y
pg  y
(1)
Secondly, we can define ug in terms of the horizontal gradient of geopotential height by
re-arranging the geostrophic relationship in the form:
Remember that
vg
p
is an
example of partial
differentiation. This means
that although vg is really a
function of may variables, we
are just considering the rate of
change of vg with respect to p
while holding the remaining
variables fixed.
In general for a variable G
being a function of x,y,p and t
G(x,y,p,t)=x2+y3+2p+t
Then partial differentiation
gives
G
G
G
 2x ,
 2,
 3x 2 ,
x
p
y
G
1
t
The rule are the same as in
ordinary differentiation. Those
of you doing maths lectures in
the summer term will cover the
theory.
MT11B Weather Systems Analysis
Eleanor Highwood, December 2003
fu g
Z

y
g
We want the change in ug with pressure so we differentiate both sides with respect to p:
 2Z
f  ug

.
 y p
g p
(2)
But in a mixed second derivative (as on the left hand sides of both equations 1 and 2),
the order of differentiation is irrelevant. So Z can be eliminated between the two
expressions to give:
 ug
R T

.
 p pf  y
Similarly we can obtain
 vg
R T

.
p
pf  x
This is a mathematical expression for the thermal wind shear. Sometimes, it is more
convenient to express the thermal wind shear using height rather than pressure as a
vertical coordinate. The corresponding expressions are:
 ug
 vg
g T
g T

and

.
Z
fT  y
Z
fT  x
T5 3 - The thermal wind:
Meteorologists frequently consider the thickness of the 100-50 kPa layer, and so are
concerned with the wind shear between these two levels.
“The thermal wind” = the vector difference between the geostrophic wind at 100 kPa and
50 kPa.
For revision purposes make
sure you can derive (i.e. follow
the maths from the start of
T5.2) for both these
relationships and get to the
versions using height rather
than pressure (hint – use the
hydrostatic relationship again).
MT11B Weather Systems Analysis
Eleanor Highwood, December 2003
The thermal wind, denoted v T , is a vector parallel to the contours of thickness, with cold
air on the left (on the right in the southern hemisphere).
The magnitude of the “thermal wind” VT is easily calculated using the formulae for the
geostrophic wind. Consider two levels denoted by subscript 1 and 2. Suppose for
simplicity that the height contours are parallel at both levels. Then:
Vg1 
g  Z1
g  Z2
.
, Vg 2 
f y
f y
Subtract these two equations to find:
VT  Vg 2  Vg1 
g 
( Z2  Z1 )
f y
And so the thermal wind is easily related to the horizontal gradient in thickness (and so
could be found from a thickness chart for example).
This relationship between horizontal temperature gradients and vertical wind shears is
what gives rise to many important features of atmospheric circulations, e.g. the jets
above the frontal surfaces in frontal cross sections, and on a larger scale, the mid-latitude
westerly jets are a result of the equator to Pole temperature gradient.
You can use this relationship to calculate the temperature gradient if you know the wind
shear, or vice versa. You’ll see more examples in EC9.
MT11B Weather Systems Analysis
Eleanor Highwood, December 2003
Worked Example:
A radiosonde ascent at 52N records a wind of 10 m s-1, bearing 240 at 90 kPa and
20 m s-1, bearing 270 at 50 kPa. Work out the direction of the isotherms and the
magnitude of the temperature gradient, assuming it to be constant between 90 and
50 kPa.
Step 1: decompose the winds into components parallel to the x (W-E) and y (S-N) axes.
The results are:
1 - At 90 kPa, u = 8.7 m s-1, v = 5.0 m s-1,
2 - At 50 kPa, u = 20.0 m s-1, v = 0.0 m s-1.
Step 2: Use  u /  p  u / p , etc., so that  u /  p is -2.8310-4m s-1Pa-1 and  v /  p is
1.2510-4m s-1Pa-1. (We must take care with signs in these examples remembering that
pressure decreases as you go to higher heights in the atmosphere.)
Step 3: From the thermal wind equations we can find the horizontal temperature
gradients:
T
 T pf  u
pf  v

 350
.  10 6 K m -1 .

 7.93  10 6 K m-1 ,

x
R

p
y
R  p
Step 4: The temperature gradient is a vector with components  T /  x,  T /  y : its
magnitude is
 T /  x
2
  T /  y
2
 , i.e., 8.710-6 K m-1 (8.7 K per 1000 km).
Step 5: The direction of the temperature gradient is calculated from
 , and works out at 66 to the zonal direction (draw it if you aren’t
tan 1   T /  y


T
/

x


sure). Then the direction of the isotherms is perpendicular to the direction of the
temperature gradient and works out at 90-66 = 24 to horizontal.
You will get plenty of practice in this in EC9.