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bottom, no slip at the top and free slip at the bottom, and perfectly conducting or
insulating for each of the former eight boundary conditions.
2.9.2 Convection in a resting atmosphere with rotation
Just for illustrative purposes, we add rotation to the problem and consider only
how it qualitatively changes the results. To do this, we simply add the acceleration
due to the Coriolis force in (2.203) and (2.204). The new set of horizontal
equations of motion are:
2
ð@=@
t r
Þ u ¼ 1
= @
p
=@
x þ 2
O
ð 2
:
249 Þ
2
ð@=@
t r
Þv ¼ 1
= @
p
=@
y 2
O
ð 2
:
250 Þ
where
is the local rotation rate (the Coriolis force due to the rotation of the
spherical Earth is not considered here). To get an equation in terms of w alone,
we first form a vertical vorticity equation from (2.249) and (2.250) and make use
of the continuity equation (2.207) to eliminate
O
@
u
=@
x þ@v=@
y; we also express the
vertical vorticity as
. In addition, we form a horizontal divergence equation from
(2.203) and (2.204) and again use (2.207) to eliminate
@
u
=@
x þ@v=@
y; once more,
we express the vertical vorticity as
. We then eliminate p from the vorticity and
divergence equations to obtain the following:
2
2 w ¼ 2
2
x 2
2
y 2
ð@=@
t r
Þr
O @=@
z þð@
=@
þ@
=@
Þ B
ð 2
:
251 Þ
We now eliminate B from (2.251) and the thermodynamic equation (2.213) to get
2
2
2
2 w ¼ ð@=@
2
2
x 2
2
y 2
ð@=@
t r
Þð@=@
t r
Þ
r
t r
Þð@
=@
þ@
=@
Þ w
2
2
2
Oð@=@
t r
Þð@=@
t r
Þ@=@
z ð 2
:
252 Þ
Finally, we eliminate vorticity from (2.252) by differentiating the vorticity equation
with respect to z (not shown):
2
2 w
z 2
ð@=@
t r
Þ@=@
z ¼ 2
O @
=@
ð 2
:
253 Þ
The final form of the equation for w is as follows:
2
2
2
2 w ¼ ð@=@
2
2
x 2
2
y 2
ð@=@
t r
Þð@=@
t r
Þ
r
t r
Þð@
=@
þ@
=@
Þ w
2
2
2 w
z 2
ð 2
ð@=@
t r
Þ@
=@
ð 2
:
254 Þ
This equation is of order eight, two up from what it had been before rotation was
added (2.220). Although the equation has a very high order, at least it is still
linear. As earlier, we scale the independent variables x, y, and z,byH and time
by H 2
=
to arrive at the following:
2
2
2
2 w ¼ Ra ð@=@
2
2
x 2
2
y 2
ð@=@
t r
Þð@=@
t r
Þ
r
t r
Þð@
=@
þ@
=@
Þ w
T 0 ð@=@
2
2 w
z 2
t r
Þ@
=@
ð 2
:
255 Þ
where T 0 , the Taylor number, is
2 H 4
2
T 0 ¼ð 2
=
ð 2
:
256 Þ
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