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When isobaric surfaces surround the Northern Hemisphere high in the south and low
in the north, the motion follows the trajectory in Fig6. 4.
Now I would like to calculate how much kinetic energy an air parcel obtains when evolving
from its motionless state �� to state �� where the speed reaches maximum. The force a unit volume of air parcel undergoes and the distance it covers constitute work. The energy of this work is converted into kinetic energy. Of the two forces the Coriolis force is perpendicular to the direction of the motion and does not contribute to the work. ( refer to Fig6. 5 )
Therefore, kinetic energy obtained by a unit volume of air parcel during this motion
is provided only by the pressure gradient force. The pressure gradient force is always perpendicular to contours of height. So the total amount of the work during this motion is obtained by integrating from �� to �� ( in Fig.4 ) the inner product of the pressure gradient force and the line segment along the path of the air parcel.
The amount of work given is
∫ −ρ・g・∂h／∂n・ds =ρ・g( height � �height ��) where ρ is the density of air, h is the height of isobaric surface and n and s denote the unit vector directed to the steepest slope of isobaric surface and the unit vector directed to the path of the parcel, respectively.
This equation means that the energy produced during the parcel’s transferring from ��
to �� is equal to the lost amount of potential energy. This amount of work becomes kinetic energy, which is 1／2・ρ・v2
That means an air parcel in ageostrophic motion obtains kinetic energy by being compressed
by the pressure gradient force, and this kinetic energy is equal to the potential energy which is lost while moving down the isobaric surface. In other words, an air parcel obtains
the velocity by moving on isobaric surface and lowering its height.
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