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Coriolus effect, precession in 4+1D?

  1. Dec 8, 2015 #1
    Here is a derivation of the Coriolus effect on an imaginary 4+1D planet.


    It seems wrong to me. I think that the planet is rigid, so the latitude of a point on the surface, ϑ, is a constant. It's derivative will always be zero. The first two entries to that acceleration vector vanish. The third entry is the acceleration in the ϑ direction, so that has to be zero too. So the whole vector should always be zero.

    Proof: Each point satisfies a^2+b^2+c^2+d^2 = 1, with the two planes of rotation being defined by [a,b] and [c,d]. (Clifford proved this in the 19th century.) So both a^2+b^2=x^2 and c^2+d^2=y^2 are constants. The latitude ϑ=arctan(x/y) is then also a constant.

    Another argument is this. Any rigid 4D object rotates in two planes. If the periods of the planes are equal then the planes aren't unique : any two orthogonal planes will do. In other words, every point on the surface rotates in the same way: it's path has the same shape as the path of any other point. So the Coriolus force has to be the same everywhere. In this case, this person's equation shows the force as nonzero and varying as a function of ϑ. That can't be right.

    You may ask, why don't I ask them? They've been running this forum for a decade, and I don't want to annoy or anger the natives by challenging them. I may be making a wrong assumption.

    Since I couldn't do the math myself right off, so I hoped to work up to it by going to a simpler problem. Suppose I have a 4D gyroscope in a gravitational field. The axle takes away one degree of freedom. Now tilt the axis. Will the gyroscope precess, and if so, how? The familiar 3D math doesn't work. When I try to do it in 4D I get an infinite number of solutions. I'm making a wrong assumption, but I don't know what. The Wikipedia angular velocity page shows how it is done with tensors, but I don't know tensors, Hodge duals, wedge products, and so forth. Hmmm, maybe this area is too hard for me and I should do something else.
  2. jcsd
  3. Dec 10, 2015 #2
    The answer is: such a gyroscope wouldn't precess. I'm not sure what it would do. I guess it would simply fall over

    To get a 4D gyroscope that precesses it has to have a planar axle. One might suppose it would be shaped like an X in order to save on materials, with two perpendicular points touching the "ground."

    In general an ND gyroscope would have a (N-2)D axle.
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