Are there equations or even a concept of circular motion/tangential acceleration/centripetal acceleration in three dimensions? Maybe something called "spherical acceleration"? or am I just getting something wrong?
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2 Answers
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We can consider circular motion at constant angular velocity in an arbitrary number of dimensions $D$.
In 2 dimensions, we have formulas like $v=\omega r$, $F=v^2/r=\omega^2 r$. We're dealing with rotations in 2 dimensions without reflection, a group called $SO(2)$. A rotation matrix here would look like follows. $$R(t)=\begin{bmatrix} \cos(\omega t) & -\sin(\omega t) \\ \sin(\omega t) & \cos(\omega t) \end{bmatrix}$$
In 3 dimensions, we can ask what freedom the extra dimension gives us. In fact, rotation in 3 dimensions always leaves one axis fixed. For example, a rotation matrix would look like follows:
$$R(t)=\begin{bmatrix} \cos(\omega t) & -\sin(\omega t) & 0 \\ \sin(\omega t) & \cos(\omega t) & 0 \\ 0 & 0 & 1\end{bmatrix}$$
There is always some vector left fixed, for any rotation in 3 dimensions (or any odd number of dimensions). This means that in 3D, we can take our problem of rotation at constant angular velocity and reduce it to a 2D problem: we have a plane on which things rotate, and a vector perpendicular to that plane. Because we can always do this, we can get by using only 2D circular motion formulas. So there is no need for "spherical acceleration". That said, the fact that we're in 3D does afford us a special tool: the cross product. If we use vector notation and use primes to denote coordinates in a rotating coordinate system, we have: $${\boldsymbol {F}}=m{\bf a}'+2m{\boldsymbol {\omega }}\times {\boldsymbol {v'}}+m{\boldsymbol {\omega }}\times ({\boldsymbol {\omega }}\times {\boldsymbol {r'}})$$
where the first term involving omega is the coriolis force and the second term involving omega is the centrifugal force. For circular motion, in a rotating reference frame, ${\bf v}'=(0,0,0)$, ${\boldsymbol {\omega}}=(0,0,\omega)$, and ${\bf r}'=(r,0,0)$. If we say that the acceleration in the rotating reference frame is zero, we get ${\bf F}=(-m\omega^2 r,0,0)$, which is the circular force of $mv^2/r=m\omega^2 r$ towards the center.
In 4 dimensions, we can now construct rotation matrices like follows: $$R(t)=\begin{bmatrix} \cos(\omega_1 t) & -\sin(\omega_1 t) & 0 & 0 \\ \sin(\omega_1 t) & \cos(\omega_1 t) & 0 & 0 \\ 0 & 0 & \cos(\omega_2 t) & -\sin(\omega_2 t) \\ 0 & 0 & \sin(\omega_2 t) & \cos(\omega_2 t) \end{bmatrix}$$ so now in 4 dimensions your intuition that something might go wrong is correct. We have too much freedom and can no longer reduce the problem to 2-dimensional circular motion, and it is no longer true that there must be a fixed axis of rotation. We could write formulas like $\omega_1^2 r+\omega_2^2 r$, but it might not be clear what we're talking about or how this would be applied in an actual physics situation. So if we lived in 4 dimensions, our introductory physics courses might have a section on "uniform circular motion" and then a second section on "uniform hyperspherical motion". Since we live in 3 dimensions, we don't need to be concerned with uniform hyperspherical motion :)
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Any pattern can have cyclic motion. (Consider tracing a unicursal picture of Snoopy over and over; this is Cyclic Snoopy Motion.) The reason why uniform circular motion is special that it is cyclic motion which involves an acceleration that has constant magnitude and, in the reference frame of the accelerated object, constant direction. We could describe an equation for cyclic spherical motion, although we would need infinite path length for a trajectory to trace out all points on a surface, as opposed to a circle or Snoopy, which has finite path length. However, spherical motion wouldn't be any more special than Snoopy motion, as it would require constantly changing acceleration.
However: uniform circular motion does have higher dimensional analogs. Uniform circular motion is uniform acceleration towards a point at fixed distance from the point. The higher-dimensional analog of this is uniform acceleration normal towards an axis at fixed distance from the axis.
For example, suppose you have a windmill mounted on the top of your car and you affix a light bulb to one of the fins. As your car translates down the road, the light bulb moves in uniform circular motion in the reference frame of the car as it is uniformly accelerated towards the point at the center of the windmill. In the reference frame of a pedestrian watching from the curb, it is uniformly accelerated at right angles towards the line along which the center of the windmill is moving, and its trajectory describes the shape of a coiled spring.
