I'm taking the example of a rotating wheel space station, in which artificial gravity is created through rotation due to the inertia inside.

The idea in the 3D environment is that the wheel (2D) is rotating with respect to a third axis (the one perpendicular to the wheel plane), so generalising this concept for a further dimension, the wheel would be a hollow sphere and everything would be rotating in another axis, that should be perpendicular to the other three.

Could you design this hollow sphere (very big, like 100 km in diameter) that is rotating in that 4th spatial dimension (assuming there exists one) so that humans inside feel artificial gravity? Would that make sense, at least mathematically?

If so, would people inside feel the spherical shape is changing?

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    $\begingroup$ You must of course first describe how gravitation works in a four-dimensional space. Hint: it's not trivial. $\endgroup$
    – AlexP
    Sep 4, 2018 at 10:37
  • $\begingroup$ Well I don't know if I need gravity at all. If that hollow sphere is a spacecraft deep in space where gravity is negligible, and you want to create a sense for it just by rotating, then it's not gravity, it's just inertia, right? $\endgroup$ Sep 4, 2018 at 10:47
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    $\begingroup$ You need gravity because that is what you want to simulate. If we don't know how gravity works in a four-dimensional space we don't know what we are supposed to simulate. $\endgroup$
    – AlexP
    Sep 4, 2018 at 10:59
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    $\begingroup$ Just a comment, it's perfectly fine to assume that things work out, but I don't think you can assume anything as a given in 4 dimensions. Can people even exist? Biomolecules heavily rely on having the right geometry. What is a 4d protein? How would that even work? Well, you need to do 4d quantum mechanics now, congratulations. While this is a cool idea of course, you are running into horribly complicated problems pretty fast. I personally wouldn't touch it without a Ph.D. in math, at least not in a super serious setup. If it's comedic or a homage or something, of course, who cares. $\endgroup$
    – Raditz_35
    Sep 4, 2018 at 11:01
  • $\begingroup$ To back up AlexP's and Raditz_35's comments, take a look at physics.stackexchange.com/q/50142/79374 - the TL;DR is that there aren't stable orbits in 4D. So if you've got a fourth spatial dimension, gravity has to behave differently in it or else there are no solar systems and galaxies. That opens a can of worms, though - how else is the fourth spatial dimension not the same as the other ones? $\endgroup$
    – Rob Watts
    Sep 4, 2018 at 17:25

1 Answer 1


This setup wouldn't work, not even in a mathematical sense, at least as you described it.

In four dimensions, you can't actually rotate with respect to an axis. Instead, a 4D rotation leaves either a plane or a single point invariant.

In more detail, a rotation in a space of arbitrary dimension can be thought of as composed of simple rotations. It can be seen mathematically that a simple rotation needs a two-dimensional subspace to take place in (you can't rotate anything in 1D). This means that:

  • In 2D, all rotations are simple. They leave a $2-2=0$-dimensional subspace invariant, i.e. a point.

  • In 3D, all rotations are simple. They leave a $3-2=1$-dimensional subspace invariant (a rotation axis).

  • In 4D, a rotation can be simple, leaving a $4-2=2$-dimensional subspace invariant (a plane), or it can be a double rotation composed of two simple rotations, leaving in total a $4-2-2=0$-dimensional subspace invariant (a point). These simple rotations can even have different angular speeds (if they have the same speed, it's called an isoclinic rotation).

  • In 5D, rotations are again of two types, which leave invariant either a 3D subspace or an axis.

  • In 6D, rotations can be of three types...

...and so on. Going back to your setup, we can find two cases. In the first case, assuming (in the spirit of your question) that the invariant plane contains the extra four-dimensional axis, the sphere would rotate normally as the Earth does, leaving two "North" and "South" poles invariant. Here the extra dimension is redundant, our ordinary 3D physics already tells us what would happen: there would be artificial gravity near the equator and no gravity near the poles.

In the second case, the sphere will disappear from view almost all the time, only reappearing periodically at certain instants which depend on the rotation's angular speeds. Obviously this wouldn't be feasible as a space station of any kind, since all the air would quickly escape from it.

In conclusion, you can't rotate all points of the sphere at the same time while keeping them inside your 3D environment.

In fact, the setup wouldn't work no matter how many extra dimensions one adds. This is because of the so-called hairy ball theorem. This theorem says that all possible continuous tangent vector fields on a sphere must vanish at some point, and it is sometimes popularly stated as "you can't comb a hairy ball without creating at least one cowlick".

Hairy ball with two cowlicks

An infinitesimal rotation defines a smooth vector field on the sphere (you can think of a small arrow attached to each point, and pointing at where it will move), and the theorem then implies there must be at least one arrow of length zero (the center of the "cowlick"), which means a point which won't move.

  • $\begingroup$ After this, my answer is worth deletion. $\endgroup$
    – L.Dutch
    Sep 4, 2018 at 12:28
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    $\begingroup$ Wow, excellent answer! I suppose I had been secretly wondering about this myself for some time. $\endgroup$
    – user44399
    Sep 4, 2018 at 12:32
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    $\begingroup$ Once a day, I will find some way to tell people ""you can't comb a hairy ball without creating at least one cowlick". $\endgroup$
    – Willk
    Sep 4, 2018 at 15:10
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    $\begingroup$ In other words, you can create artificial gravity, but no matter how many extra dimensions you use there will always be at least one point where there is still no artificial gravity? $\endgroup$
    – Rob Watts
    Sep 4, 2018 at 17:36
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    $\begingroup$ @pregunton: Wouldn't said force just be a 4D analogue of centripetal force? In regular 2D rotational scenarios we don't see such cancellation... $\endgroup$
    – Kevin
    Sep 4, 2018 at 22:20

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