Editing Rotating reference frame (section)

=== Relation between positions in the two frames ===

To derive these fictitious forces, it's helpful to be able to convert between the coordinates <math>\left(x', y', z'\right)</math> of the rotating reference frame and the coordinates <math>(x, y, z)</math> of an [[inertial reference frame]] with the same origin.<ref group=note>So <math>x', y', z'</math> are functions of <math>x, y, z,</math> and time <math>t.</math> Similarly <math>x, y, z</math> are functions of <math>x', y', z',</math> and <math>t.</math> That these reference frames have the same origin means that for all <math>t,</math> <math>\left(x', y', z'\right) = (0, 0, 0)</math> if and only if <math>(x, y, z) = (0, 0, 0).</math></ref> 
If the rotation is about the <math>z</math> axis with a constant [[angular velocity]] <math>\Omega</math> (so <math>z' = z</math> and <math>\frac{\mathrm{d} \theta}{\mathrm{d} t} \equiv \Omega,</math> which implies <math>\theta(t) = \Omega t + \theta_0</math> for some constant <math>\theta_0</math> where <math>\theta(t)</math> denotes the angle in the <math>x-y</math>-plane formed at time <math>t</math> by <math>\left(x', y'\right)</math> and the <math>x</math>-axis), 
and if the two reference frames coincide at time <math>t = 0</math> (meaning <math>\left(x', y', z'\right) = (x, y, z)</math> when <math>t = 0,</math> so take <math>\theta_0 = 0</math> or some other integer multiple of <math>2\pi</math>), the transformation from rotating coordinates to inertial coordinates can be written
<math display=block>x = x'\cos(\theta(t)) - y'\sin(\theta(t))</math>
<math display=block>y = x'\sin(\theta(t)) + y'\cos(\theta(t))</math>
whereas the reverse transformation is
<math display=block>x' = x\cos(-\theta(t)) - y\sin(-\theta(t))</math>
<math display=block>y' = x\sin( -\theta(t)) + y\cos(-\theta(t)) \ .</math>

This result can be obtained from a [[rotation matrix]].

Introduce the unit vectors <math>\hat{\boldsymbol{\imath}},\ \hat{\boldsymbol{\jmath}},\ \hat{\boldsymbol{k}}</math> representing standard unit basis vectors in the rotating frame. The time-derivatives of these unit vectors are found next. Suppose the frames are aligned at <math>t = 0</math> and the <math>z</math>-axis is the axis of rotation. Then for a counterclockwise rotation through angle <math>\Omega t</math>:
<math display=block>\hat{\boldsymbol{\imath}}(t) = (\cos\theta(t),\ \sin \theta(t))</math>
where the <math>(x, y)</math> components are expressed in the stationary frame. Likewise,
<math display=block>\hat{\boldsymbol{\jmath}}(t) = (-\sin \theta(t),\ \cos \theta(t)) \ .</math>

Thus the time derivative of these vectors, which rotate without changing magnitude, is
<math display=block>\frac{\mathrm{d}}{\mathrm{d}t}\hat{\boldsymbol{\imath}}(t) = \Omega (-\sin \theta(t), \ \cos \theta(t))=  \Omega \hat{\boldsymbol{\jmath}} \ ; </math>
<math display=block>\frac{\mathrm{d}}{\mathrm{d}t}\hat{\boldsymbol{\jmath}}(t) = \Omega (-\cos \theta(t), \ -\sin \theta(t))= - \Omega \hat{\boldsymbol{\imath}} \ ,</math>
where <math>\Omega \equiv \frac{\mathrm{d}}{\mathrm{d}t}\theta(t).</math> 
This result is the same as found using a [[vector cross product]] with the rotation vector <math>\boldsymbol{\Omega}</math> pointed along the z-axis of rotation <math>\boldsymbol{\Omega} = (0,\ 0,\ \Omega),</math> namely,
<math display=block>\frac{\mathrm{d}}{\mathrm{d}t}\hat{\boldsymbol{u}} = \boldsymbol{\Omega \times}\hat{\boldsymbol{u}} \ , </math>
where <math>\hat{\boldsymbol{u}}</math> is either <math>\hat{\boldsymbol{\imath}}</math> or <math>\hat{\boldsymbol{\jmath}}.</math>