Does the Coriolis Effect Determine the Direction of Curl in Hemispheres?

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SUMMARY

The Coriolis effect does not dictate that the curl in the northern hemisphere is always negative and in the southern hemisphere always positive. The curl, represented as ∇×, is influenced by the velocity vector (∇v) and cannot be generalized based solely on hemisphere. In the northern hemisphere, the Coriolis force causes trajectories to deflect clockwise, while in the southern hemisphere, they deflect counter-clockwise. The curl at the center of cyclones is indeed zero.

PREREQUISITES
  • Understanding of vector calculus, specifically curl (∇×)
  • Familiarity with the Coriolis effect and its implications in fluid dynamics
  • Basic knowledge of Earth's rotation and its axis
  • Concept of trajectory deflection due to forces
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  • Study the mathematical formulation of the Coriolis effect in fluid dynamics
  • Learn about vector calculus applications in meteorology
  • Research the behavior of cyclones and their characteristics
  • Explore the implications of the Coriolis effect on projectile motion
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Keru
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Two quick questions. Does the Coriolis effect mean that in the northern hemisphere the curl will always have a negative value and in the southern hemisphere a positive one? Is the curl in the eye of the cyclones equal to zero?
Thanks.
 
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The Coriolis effect is proportional to ##-2(\vec{\omega}\times\vec{v})## where ##\vec{\omega}## is pointing up out of the North pole, along the axis of Earth's rotation. Whether this curl is positive or negative depends on the vector ##\vec{v}## and can not be generalized to "northern vs southern hemisphere" effects. However, if something (say a bullet) is moving initially along a straight line in the Northern hemisphere, its trajectory will be deflected (by the Coriolis force) in a clockwise direction, while in the Southern hemisphere the trajectory would be deflected in a counter-clock-wise direction.
 
By "curl" do you mean \nabla\times or something else?
 
Haha! I had a brain fart and associated a cross product ##\vec{\omega}\times\vec{v}## as a curl. Now I'm not sure what the OP is asking.
 
Meir Achuz said:
By "curl" do you mean ∇×\nabla\times or something else?

Yes i meant ∇×. Sorry I am not english native speaker, searched for the translation and that's the best i had.
 
I do not have a good working knowledge of physics yet. I tried to piece this together but after researching this, I couldn’t figure out the correct laws of physics to combine to develop a formula to answer this question. Ex. 1 - A moving object impacts a static object at a constant velocity. Ex. 2 - A moving object impacts a static object at the same velocity but is accelerating at the moment of impact. Assuming the mass of the objects is the same and the velocity at the moment of impact...

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