0
$\begingroup$

Can Coriolis force and deflection of wind patterns towards their right in northern hemisphere and towards left in southern hemisphere be explained with the help of vectors?I have tried but could not do it.I could not understand why the direction is towards left in southern hemisphere and towards right in northern hemisphere.

$\endgroup$
1
$\begingroup$

The Coriolis force is analogous to $\vec{u}$ x $\vec{Ω}$. This is the cross product between the velocity of the particle and Ω which is the angular momentum of the rotating frame with respect to the moving mass. So, for a particle moving near one of the hemispheres, the angular velocity vector is pointing upwards(normal to the Earth's surface) while for a particle moving near the other hemisphere it points downwards(again normal to the surface). In order to understand this, you just need to visualize the angular velocity vector seen by two observers, each on one hemisphere. So, if the mass in both scenarios is moving say to the center, the Force being the cross product of the velocity and the angular momentum will point to the right for the first hemisphere and left for the second hemisphere(right and left here mean with respect to an observer on that hemisphere). In both cases, if the mass moves with the same velocity in magnitude, then the force will be equal in magnitude. Also, I think that checking out the Wikipedia page for the Coriolis force will help a lot, especially with the visualization part.
(Note: the Coriolis force direction can be found with the right hand rule just like the magnetic force can be found in the same way when a charge is moving in a magnetic field)

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Coriolis force is a virtual force experienced in non inertial frames, and it is a pure cinematic phenomenon and so it can be derived with vectors through frame transformation. Be $O, \pmb i, \pmb j, \pmb k$ our fixed inertial frame and $O', \pmb i', \pmb j', \pmb k'$ the relative frame with $\pmb v_{O'}, \pmb a_{O'}$ the relative velocity and acceleration of O' and $\pmb \omega$ the angular velocity of the moving tern. We have a vector $\pmb x-O=x_1\pmb i+x_2\pmb j+x_3\pmb k$ we want to fine $x'=x-O'= x_1'\pmb i'+x_2'\pmb j'+x_3'\pmb k'$ We can write $\pmb x-O'=(\pmb x-O)-(O'-O)\Rightarrow x_1'\pmb i'+x_2'\pmb j'+x_3'\pmb k'=x_1\pmb i+x_2\pmb j+x_3\pmb k-(O'-O)$ now remembering ${d\pmb e_i \over dt}=\omega\times e_i$ we calculate the derivative of the expression above:

$\pmb v_{x'}+\pmb \omega\times \pmb x'=\pmb v_x -\pmb v_{O'} \Rightarrow \pmb v_{x'}=\pmb v_x -\pmb\omega\times \pmb x'-\pmb v_{O'}$ which is the classical velocity composition, now we derive another time paying attention to ${d\pmb v_{x'}\over dt}=\ddot x_1'\pmb i'+\ddot x_2'\pmb j'+\ddot x_3'\pmb k'+ \dot x_1'(\pmb\omega\times \pmb i')+\dot x_2'(\pmb\omega\times \pmb j')+\dot x_3'(\pmb \omega\times\pmb k')=\pmb a_{x'}+\pmb\omega\times\pmb v_{x'}$

So $\pmb a_{x'}+\pmb\omega\times\pmb v_{x'}=\pmb a_x -\pmb\omega\times\pmb v_{x'}-\pmb x'\times\dot{\pmb\omega}-\pmb\omega\times(\pmb\omega\times\pmb x')-\pmb a_{O'}$

so $\pmb a_{x'}=\pmb a_x - 2\pmb\omega\times\pmb v_{x'}-\pmb x'\times\dot{\pmb\omega}-\pmb\omega\times(\pmb\omega\times\pmb x')-\pmb a_{O'}$

The term $2\pmb\omega\times\pmb v_{x'}$ is the coriolis acceleration, the correspondent force is found multiplying it by the mass of the corp. Since clouds move toward the equator if you calculate that vector product you will obtain the direction of the shifting.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks but I'm having trouble visualizing the thing without a diagram...+1 $\endgroup$ – tatan May 15 '17 at 11:51
  • $\begingroup$ I don't have an image but I try to explain. If you think of earth as normally ( northern hemisphere up) it turns counterclockwise, so angular velocity points up. Now take a point in the northern hemisphere, think of the tangent direction toward the equator which is the speed of a cloud, it points 45 degree down if you take it in the middle . If you apply the rule for the vector product -WxV you will find that it is directed west. You can do the same for the southern hemisphere. I hope it helps. $\endgroup$ – Claudio P May 15 '17 at 21:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.