# Could we measure fictitious forces arising from rotation about an axis in a fourth spatial dimension?

Two-dimensional creatures confined to a flat sheet that was uniformly rotating around an axis perpendicular to the sheet would have no trouble identifying that their sheet was rotating. They would be able to measure the centrifugal and Coriolis forces. With some elementary calculations, they could then figure out the center of rotation as well as the rate and direction of rotation.

But what about three-dimensional creatures confined to a three-dimensional volume? Could they tell whether they were being subjected to uniform rotation around a fourth axis perpendicular to that volume? Presumably there is a corresponding set of fictitious forces that can be measured? I have zero intuition on this topic, but I assume the mathematics exists to describe this situation and answer the question?

[edit #2: correcting example per @Laff70 comment]

2D sheet in 3D space: Say the 2D sheet, x',y', lies initially in the x,y plane and can be rotated around the x, y, or z axis. We measure the centrifugal force, $$F$$, only, including the direcion vector (e.g. $$\hat x$$)

1. rotate in x,y plane (invariant along z): $$F \propto \omega^2r'_{xy} \hat r'_{xy}$$ where $$r'_{xy} = \sqrt{x'^2+y'^2}$$
2. rotate in y,z plane (invariant along x): $$F \propto \omega^2 x' \hat x'$$
3. rotate in z,x plane (invariant along y): $$F \propto \omega^2 y' \hat y'$$

3D volume in 4D space: Say the 3D volume, x',y',z', lies initially in the x,y,z volume and is subjected to single-parameter rotation with respect to the six possible planes, x-y, y-z, z-w, w-x, x-z, y-w. We measure the centrifugal force, $$F$$, only. Forces along the fourth dimension, w, are invisible.

1. rotate in x,y-plane (invariant along z & w): $$F \propto \omega^2r'_{xy} \hat r'_{xy}$$ where $$r'_{xy} = \sqrt{x'^2+y'^2}$$
2. rotate in y,z-plane (invariant along x & w): $$F \propto \omega^2r'_{yz} \hat r'_{yz}$$ where $$r'_{yz} = \sqrt{y'^2+z'^2}$$
3. rotate in z,w-plane (invariant along x & y): $$F \propto \omega^2 z' \hat z'$$
4. rotate in w,x-plane (invariant along y & z): $$F \propto \omega^2 x' \hat x'$$
5. rotate in x,z-plane (invariant along y & w): $$F \propto \omega^2r'_{xz} \hat r'_{xz}$$ where $$r'_{xz} = \sqrt{x'^2+z'^2}$$
6. rotate in y,w-plane (invariant along x & z): $$F \propto \omega^2 y' \hat y'$$

Of these, 1, 2, and 5 correspond to simple rotations in 3D space and 3, 4, and 6 correspond to a weird rotation that includes the fourth dimension.

It's interesting that these rotations can be grouped in independent pairs: 1&3, 2&4, 5&6. Applying the same rotation rate to each member of a pair, you seem to end up with a spherically symmetric centrifugal force $$F \propto r'_{xyz} \hat r'_{xyz}$$ where $$r'_{xyz} = \sqrt{x'^2+y'^2+z'^2}$$. This is pretty weird.

• Rotations in 4D do not have axes the way you're thinking. 2D rotations have a point as the axis, 3D rotations have a line, and (simple) 4D rotations have a whole plane as the axis. (You can also combine two simple 4D rotations into either a "double" or "isoclinic" rotation that has no axis of fixed points, but (at least) two planes that points stay on.) A simple 4D rotation could have its axis plane intersect the 3D volume at a line (reduces to your 2D case), have the axis plane inside the 3D volume, or have its axis plane be parallel to the volume.
– HTNW
Apr 16, 2022 at 4:25
• @HTNW That's good feedback. I'm unable to visualize this, but in that middle case with the axis-plane inside the volume, is there a simple example? For instance, when the axis-plane is coincident with the x-y plane, is the motion of points easily described? Apr 16, 2022 at 5:33
• 4D rotation: mikelortega.github.io/tesseract Apr 17, 2022 at 18:51
• @PM2Ring absolutely beautiful, but I'm no wiser. I think that's a 4D wire cube with a perspective projection into 3D space and then outlined with pipes. It looks like there's a single parameter rotation happening somehow? Apr 17, 2022 at 19:35
• I had a brief look at the source code, but it's (almost) devoid of comments, and I'm not quite sure how the mouse (or touch) events get turned into rotation parameter(s). But as Laff70 said, it's best to think about rotation as a transformation in a plane, since that applies to any number of dimensions >=2. Apr 17, 2022 at 19:49

Ah, I remember the mathematics behind all this quite well(well, I remember working on it and possess several files of equations, the specifics are somewhat blurry), after it caused me to burn a theory I had been working on for a while. You see, our description of rotation as a vector is inherently erroneous, as this representation is only valid in 3D. It is most accurately described as the plane of rotation, which can be represented as an antisymmetric matrix or a bivector. Let’s look at how we can properly describe rotation with an antisymmetric matrix. Let’s say we have an object rotating about the origin in a space. The inertial frame of reference is designed $${\vec{r}}_{space}$$ whilst the frame of reference where the object is still is designated $${\vec{r}}_{body}$$. We can convert between them using an orientation matrix. $${\vec{r}}_{space}=\widetilde{A}{\vec{r}}_{body}$$ We can find the velocity of a chunk of the object at a given point in the inertial frame with $$\vec{v}=\vec{\omega}\times\vec{r}=\widetilde{\omega}\vec{r}=\left[\begin{matrix}0&-\omega_z&\omega_y\\\omega_z&0&-\omega_x\\-\omega_y&\omega_x&0\\\end{matrix}\right]\vec{r}$$ where $$\widetilde{\omega}=\left[\begin{matrix}0&-\omega_3&\omega_2\\\omega_3&0&-\omega_1\\-\omega_2&\omega_1&0\\\end{matrix}\right]=\left(\frac{d\widetilde{A}}{dt}\right){\widetilde{A}}^T=\dot{\widetilde{A}}{\widetilde{A}}^T$$ You’ll notice that the vector representation of angular velocity only works in 3D, whilst the matrix representation works in any dimension. You may be wondering about angular momentum and moments of inertia. I was too, and IIRC, I couldn't find much information on it, so I did some math and found their hyperdimensional equivalents. I derived a hyperdimensional version of the moment of inertia tensor which I designate with $$\widetilde{K}$$. For a particle with mass m located at $$\vec{r}$$, $$\begin{matrix}\widetilde{K}=m\vec{r}{\vec{r}}^T&\widetilde{L}=\widetilde{\omega}\widetilde{K}+\widetilde{K}\widetilde{\omega}\\\end{matrix}$$ Several different $$\widetilde{K}$$ matrices can be added together to represent several connected masses without issue. Mass distributions can also be integrated over to get a $$\widetilde{K}$$. In 3D only, $$\widetilde{K}$$ is related to the moment of inertia tensor by $$\begin{matrix}\widetilde{K}=m\left|\vec{r}\right|^2{\widetilde{I}}_3-\widetilde{I}=\frac{Tr{\left(\widetilde{I}\right)}}{2}{\widetilde{I}}_3-\widetilde{I}\\\widetilde{I}=Tr{\left(\widetilde{K}\right)}{\widetilde{I}}_3-\widetilde{K}=m\left(\left|\vec{r}\right|^2{\widetilde{I}}_3-\vec{r}{\vec{r}}^T\right)\\\end{matrix}$$ While in 3D, getting $$\widetilde{\omega}$$ from $$\widetilde{L}$$ and $$\widetilde{K}$$ isn’t too hard. However, finding a dimensionally universal version of said operation is difficult. By this I mean some simple matrix equation which could involve inverses and matrix multiplications. Something like $$\vec{\omega}={\widetilde{I}}^{-1}\vec{L}$$ but for matrices rather than pseudovectors. I wasn’t too focused on rotation when I was doing all this and I really only cared about this as an analog for magnetic fields. Essentially, magnetic fields, like rotations, are most accurately described by an antisymmetric tensor field, and I wanted some way to convert between B and H fields in any dimension using a magnetic permeability tensor. This failed cataclysmically and caused me to outright abandon and reject the concept of magnetic permeability tensors. I was eventually able to convert between B and H fields in any dimension, but it was with them being described with the covariant and contravariant Faraday tensor. There could very well be a better representation of rotation using antisymmetric matrices than what I’ve showcased here. I didn’t have the motivation to search for it though. Maybe you will. If so, I wish you luck.