Consider two frames of reference, one rotating and the other not; the co-ordinates at which an event occurs as described by one are obtained by applying a rotation to the co-ordinates of that event as described by the other; the required angle of rotation varies linearly with time.

Chose a point on the axis of rotation as origin and use cylindrical polar co-ordinates; the two systems will then share an axial co-ordinate z and a radial co-ordinate r, measuring distance from the axis, each of which is a length. Ignore relativistic effects, so the two also share a time co-ordinate, t. Complete the systems of co-ordinates with an angle p in the non-rotating one, P in the rotating one; have P and p coincide at t=0, so that P = p −w.t (modulo whole turns) for some angular velocity w.

Measure angles, p or P, positive if they are in the same direction as the rotation, thereby making w positive. Looking along the z axis in one direction will see such angles as clockwise; looking along it the other way, as anti-clockwise. Let z increase as one moves along the axis in the former direction, decreasing in the latter.

To obtain cartesian co-ordinates, replace

- r and p with: x = r.Cos(p), y = r.Sin(p)
- r and P with: X = r.Cos(P), Y = r.Sin(P)

(I use Cos and
Sin as the names of functions ({scalars}: |{angles}), whose inputs have
units of angle; in contrast, cos and sin are ({scalars}: |{scalars}), equal to
the composites of Cos and Sin after ({angles}: radian.i ←i :{scalars});
i.e. cos and sin measure angles in radians

.)

Now,

- Cos(P) = Cos(p−w.t) = Cos(p).Cos(w.t) +Sin(p).Sin(w.t)
- Sin(P) = Sin(p−w.t) = Sin(p).Cos(w.t) −Cos(p).Sin(w.t)

whence

- X = x.Cos(w.t) +y.Sin(w.t)
- Y = y.Cos(w.t) −x.Sin(w.t)

This tells us how positions are related to one another in the two frames: we shall now look into how the velocity and acceleration of a body, as seen by the two frames, are related. To do this, take x, y and z as the co-ordinates of the body, hence functions of time, and see how the time-derivatives of X and Y depend on those of x and y [that of z will pass seemlessly between the frames and have no impact on the others, so I ignore it]. We obtain

- dX/dt
- = (dx/dt).Cos(w.t) +(dy/dt).Sin(w.t) −x.Sin(w.t).w/radian +y.Cos(w.t).w/radian
- = (dx/dt).Cos(w.t) +(dy/dt).Sin(w.t) +Y.w/radian
- dY/dt
- = (dy/dt).Cos(w.t) −(dx/dt).Sin(w.t) −(y.Sin(w.t) +x.Cos(w.t)).w/radian
- = (dy/dt).Cos(w.t) −(dx/dt).Sin(w.t) −X.w/radian

Applying differentiation once more,

- ddX/dt/dt
- = (ddx/dt/dt).Cos(w.t) +(ddy/dt/dt).Sin(w.t) −(dx/dt).Sin(w.t).w/radian +(dy/dt).Cos(w.t).w/radian +(dY/dt).w/radian
- = (ddx/dt/dt).Cos(w.t) +(ddy/dt/dt).Sin(w.t) +2.(dY/dt).w/radian +X.w.w/radian/radian
- ddY/dt/dt
- = (ddy/dt/dt).Cos(w.t) −(ddx/dt/dt).Sin(w.t) −((dy/dt).Sin(w.t) +(dx/dt).Cos(w.t)).w/radian −(dX/dt).w/radian
- = (ddy/dt/dt).Cos(w.t) −(ddx/dt/dt).Sin(w.t) −2.(dX/dt).w/radian +Y.w.w/radian/radian

This enables us to describe the acceleration of the body, as seen by the rotating frame, as a sum of three terms:

- rotated
actual

acceleration obtained, from the components of the acceleration seen by the non-rotating frame, by the same transformation as we had to apply to the components of position; this will exactly equal the force acting on our body, as described by the rotating frame, divided by the body's mass. We can include the z-component of

actual

accelleration as a natural component of this term; it is unaffected by the rotation.- the
Coriolis

term which is 2.w/radian times the result of rotating the velocity of the body (as seen by the rotating frame) through a right angle in the sense which maps a purely Y-wards vector into a purely X-wards one. (The name comes from someone who famously documented this effect.) The axial component of the velocity is ignored in this term.

- the
centrifugal

term which is the body's position, ignoring axial component, scaled by the square of w/radian. (The name means

flying away from the centre

; it would be better namedaxifugal

, since it actually flys away from the*axis*on which r = 0, not from the cental*point*where z and r are zero.)

If someone in the rotating frame attempts to describe the physics they
see, they thus see force = mass.acceleration

modified by, in effect, the
addition of two forces

which arise as artefacts of their choice of frame
of reference. [Interestingly, modern physics is much more at ease with this than
its pre-Einstein antecedents; general relativity is quite happy to let us use
the rotating frame, and effectively regards the two artificial

forces as
part of gravitation – but I digress.]

outerproduct

In three dimensions (and, let me stress, specifically in three dimensions)
there is a product operator (which actually depends on your metric) which
combines two 3-vectors, antisymmetrically, to produce a third. Specifically, if
the [x,y,z] components of vectors u and v are [a,b,c] and [e,f,g], respectively,
then those of their outer product, u^v, are [b.g−c.f, c.e−a.g,
a.f−b.e] = (: sum(: sign([i,j,k]).[a,b,c](j).[d,e,f](k) ← [i,j,k]
:Σ(3)) ←i :) where Σ(3) = {permutations of [0,1,2]} and
({±1}:sign:) is the signature mapping from each permutation group to the
multiplicative group {±1}. [This relies on x, y, z being
orthonormal

co-ordinates for our metric; along with a presumption that
the rotation which takes the y axis to the x axis appears clockwise when viewed
from a position with positive z co-ordinate.] Crucially: u^v is perpendicular
to both u and v; applying any scaling to either u or v scales u^v to the same
degree; and v^u = −u^v.

Now, both the Coriolis and centrifugal terms in our transformed acceleration have zero component in the z-direction, i.e. parallel to the axis; furthermore, the Coriolis term is perpendicular to the transformed velocity's projection into the plane perpendicular to the axis – it is, consequently, perpendicular to the transformed velocity itself (which is the sum of its given projection and its component parallel to the axis; adding two vectors perpendicular to the Coriolis term yields another). So we should compare the Coriolis term with the outer product of the transformed velocity and some vector in the direction of the axis. Since the Coriolis term is also proportional to w/radian, we may as well use this as the scale of the axial vector.

So let W be a vector parallel to our axis, of magnitude w/radian (which has units 1/time), pointing in the direction of increasing z. In the non-rotating frame, let s be our body's position, v = ds/dt its velocity and a = dv/dt its acceleration, with components

- [x, y, z]
- [dx/dt, dy/dt, dz/dt]
- [ddx/dt/dt, ddy/dt/dt, ddz/dt/dt]

respectively. Let S, V and A, likewise, be the position, velocity and acceleration seen by the rotating frame, with components

- [X, Y, z]
- [dX/dt, dY/dt, dz/dt]
- [ddX/dt/dt, ddY/dt/dt, ddz/dt/dt]

Then S^W has co-ordinates [Y,−X,0].w/radian, giving us V = v +S^W. Likewise, the Coriolis term in A is simply 2.V^W. Furthermore, (S^W)^W has co-ordinates [−X, −Y, 0].w.w/radian/radian; so we are able to write our transformation, above, simply as

- v = V −S^W
- a = A −2.V^W +(S^W)^W

and the force

that our rotating frame sees act on a body with
velocity V at position S is then just that seen by the non-rotating frame with
an added term 2.V^W +W^(S^W) per unit mass.