Lably
Skip to content

Coriolis field

Apparent gravitational field

  • 3 min read
  • 2 chapters

Related on Lably

Through Rotation, Acceleration and Coriolis force

Mathematical expression

Let ω→ be the angular velocity vector of the rotating frame, v→ be the speed of a test particle used to measure the field. Hence, using the expression of the acceleration in a rotating reference frame, it is known that the acceleration of the particle in the rotating frame is:

ar=ai−2ω×v−ω×(ω×r)−dωdt×r

the Coriolis force is assumed to be the fictitious force that compensates the second term:

FCoriolis=−2m(ω×v)=−2(ω×p)

Where p→ denotes the linear momentum. It can be seen that for any object, the coriolis force over it is proportional to its momentum vector. As a vector product, it can be expressed in a tensorial way using the Hodge dual of ω:

FCoriolis=−2(ω×p)=−2(ω×)p=[0−2ω32ω22ω30−2ω1−2ω22ω10][p1p2p3]

This matrix can be seen as a constant tensor field, defined in the whole space, that will yield coriolis forces when multiplied by momentum vectors.

Sign in to keep reading

One more chapter is free with a Lably account.

More in Physics