General Accelerating Frames 61EXERCISE 2.1.14. Explain the formation of the temperate westerlies and
the polar easterlies.
EXERCISE 2.1.15.'^4 The flight time from Los Angeles to Boston is usually
different from the flight time from Boston to Los Angeles. Which flight
takes longer? Which of the following factors contributes the most to this
difference, and how: (a) The Earth's rotation under the airplane; (b) The
Coriolis force acting on the airplane; (c) The atmospheric winds? Hint: If
in doubt, check the schedules of direct flights between the two cities.
The complete mathematical model of atmospheric physics is vastly
more complicated and is outside the scope of this book; possible refer-
ence on the subject is the book An Introduction to Dynamic Meteorology
by J. R. Holton, 2004, and some partial differential equations appearing in
the modelling of flows of gases and liquids are discussed below in Section
6.3.5.
In 1963, while studying the differential equations of fluid convection, the
American mathematician and meteorologist EDWARD NORTON LORENZ (b.
1917) discovered a chaotic behavior of the solution and a strange attrac-
tor. These Lorentz differential equations are a prime example of a chaotic
flow. They also illustrate the intrinsic difficulty of accurate weather predic-
tion. His book The Nature And Theory of The General Circulation of The
Atmosphere 1967, is another standard reference in atmospheric physics.
2.1.4 General Accelerating Frames
The analysis in the previous section essentially relied on the equation
(2.1.18) on page 51, which was derived for uniformly rotating frames. In
what follows, we will use linear algebra to show that, with a proper def-
inition of the vector u>, relation (2.1.18) continues to hold for arbitrary
rotating frames.
Consider two cartesian coordinate systems: (i, j, k) with origin O, and
(£i, jl7 KI) with origin 0. We assume that O = 0\ see Figure 2.1.9.
Consider a point P in R^3. This point has coordinates (x, y, z) in
(z, j, k) and {x\, j/i, z{) in (fj, jlt ki). Then
x i + y j + z k = xi ii + yi j 1 + zi ki. (2.1.33)We now take the dot product of both sides of (2.1.33) with i to get
x = xi(ii -i) + yi(j 1 •i) + zi(ki -i). (2.1.34)