If we interchange velocity with charge and omega that is rotation of a system with $B$, magnetic field, we get the same thing. Is there any deeper meaning to this same mathematical form?


1 Answer 1


I believe you want to replace mass with charge and angular velocity with the magnetic induction.

The Coriolis effect is an apparent force due to the fact that the observer is measuring with respect to a rotating frame of reference. There is no actual force acting on the body, so this can be made to disappear by changing the frame of reference. Classical electrodynamics is best described in 4-vector notation. Using Lorentz transformations (that is, changing frame of reference) it is possible to put an observer in a situation where only an electric field or just a magnetic induction is present.

From a mathematical point of view the two formulae $2m\mathbf v\times\boldsymbol\omega$ and $q\mathbf v\times\mathbf B$ have deep analogies. Both $\boldsymbol\omega$ and $\mathbf B$ are not vectors, but rather pseudo-vector (this notion stems from the fact that these two objects don't change under parity, contrary to a genuine vector). More precisely I'd say that they are 2-forms that can be represented as "vectors" because of the Hodge duality between $\bigwedge^2 V$ and $V$, where $V$ is a Euclidean space of dimension 3 (here I'm also identifying $V$ with its dual $V^*$).

  • $\begingroup$ Yes, mass and charge, sorry about that. I am aware of the fact about pseudo-vectors, I was thinking more in the direction of physical deeper meaning..I was too thinking in a direction of Lorentz transformations but is there anything else? I guess nope? $\endgroup$ Dec 26, 2014 at 14:40
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    $\begingroup$ Well I'd say that the physical connection is that they're both "apparent" phenomena in the appropriate sense. One big difference is that the standard Coriolis is derived within the framework of classical (non-relativistic) mechanics, while electrodynamics requires relativity. $\endgroup$
    – Phoenix87
    Dec 26, 2014 at 14:49
  • $\begingroup$ Would agree with that, good verbalization Phoenix87. Its funny how the trajectory of a particle in a rotating frame looks like one of charged particle in a magnetic field. Of course, if we are talking about a flat, rotating disc.Well, you could say its just geometrical coincidence... $\endgroup$ Dec 26, 2014 at 21:20

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