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According to Coriolis force an air parcel is deflected in the absence of other forces by the acceleration (Holton, 4th Edition):

enter image description here

These equations lead to circular motion.

This figure is from a Matlab script, provided by Holton:

enter image description here

But how is this consistent with the picture, where an object moves along a straight line in the inertial system and the deflection is only apparently because we are in a rotating frame? If the object moves along a straight line in the inertial system, the curve cannot be circular in the rotating frame.

It's due to the disappearance of vertical motion, I think. But in Holton, the third equation is simply left out without comment, and instead only the above two equations are finally written. It is clear that the above equations come out when the vertical velocity is zero, but this is not justified in the explanations. Is this just an empirical observation?

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    $\begingroup$ After playing around with MATLAB examples it became finally clear. Question is answered. $\endgroup$
    – MichaelW
    Commented Jan 24, 2023 at 23:02

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The parcel in question is not moving in a straight line. It is a particle at rest in the reference frame rotating earth. A "fixed" observer in space would see a particle moving in circles with angular frequency $\Omega$ (earths angular frequency) and angular velocity $u_a$ which is given by $u_a=\Omega R$, where $R$ is the distance of the particle to the axis of rotation (compare to figures 1.6 and 1.8 in Holton).

To unterstand the circular motion we need to take the angular momentum $L$ into account which is given by $L = mRu_a$, $m$ is the particles mass. $L$ is a conserved quantity.

Let's see what this means based on the plot you provided: Suppose the particle is initially at rest at 44° latitude. If we push the parcel towards the north pole while keeping its vertical distance to sea level constant (compare to Holton: "Suppose that an object of unit mass, initially at latitude φ moving zonally at speed u, relative to the surface of the earth, is displaced in latitude or in altitude by an impulsive force"), we reduce the distance $R$ of the particle to the axis of rotation. In order for $L$ to stay the same either $u_a$ needs to increase (or the mass has to change, which by mass conservation does not happen). An observer on earth at 44 degrees north looking towards the North Pole would recognise a deflection of the particle towards the east as the particle is accelerating in order to conserve momentum.

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  • $\begingroup$ It assumes, that vertical distance is kept constant all the time while the particle is moving? I had in mind a scenario like a gun bullet shot in zonal direction on a planet without atmosphere. What happens to the bullet? It must follow a straight line in the inertial system. So it must necessarily change vertical position. From the viewpoint of the rotating system the bullet is subjected to a vertical force. $\endgroup$
    – MichaelW
    Commented Jan 23, 2023 at 21:03
  • $\begingroup$ It makes, of course, no sense for a bullet, because it is attracted by gravitation, therefore the straight line is not possible under gravitation. If I throw a stone wit lets say 10m/s (this is typical speed of wind), it will clearly fall down finally to earth and I can safely neglect air friction. So the stone doesn't circulate like an air parcel, because the air parcel (naturally) doesn't fall down. Somehow the air parcel keeps its height "automatically" while moving around. Then a circular trajectory makes sense. But I cannot see, why this follows from the equations of motion. $\endgroup$
    – MichaelW
    Commented Jan 23, 2023 at 21:36
  • $\begingroup$ I think that Holton in the first chapter just assumes that vertical position doesn't change without giving a proof for it. I think that I have to accept that fact first and only get the detailed reasoning later. $\endgroup$
    – MichaelW
    Commented Jan 23, 2023 at 22:04

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