Editing Rotating reference frame (section)

==Relating rotating frames to stationary frames==
The following is a derivation of the formulas for accelerations as well as fictitious forces in a rotating frame. It begins with the relation between a particle's coordinates in a rotating frame and its coordinates in an inertial (stationary) frame. Then, by taking time derivatives, formulas are derived that relate the velocity of the particle as seen in the two frames, and the acceleration relative to each frame. Using these accelerations, the fictitious forces are identified by comparing Newton's second law as formulated in the two different frames.

=== 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>

=== Time derivatives in the two frames ===
Introduce unit vectors <math>\hat{\boldsymbol{\imath}},\ \hat{\boldsymbol{\jmath}},\ \hat{\boldsymbol{k}}</math>, now representing standard unit basis vectors in the general rotating frame. As they rotate they will remain normalized and perpendicular to each other. If they rotate at the speed of <math>\Omega(t)</math> about an axis along the rotation vector <math>\boldsymbol {\Omega}(t)</math> then each unit vector <math>\hat{\boldsymbol{u}}</math> of the rotating coordinate system (such as <math>\hat{\boldsymbol{\imath}},\ \hat{\boldsymbol{\jmath}},</math> or <math>\hat{\boldsymbol{k}}</math>) abides by the following equation:
<math display=block>\frac{\mathrm{d}}{\mathrm{d}t}\hat{\boldsymbol{u}} = \boldsymbol{\Omega} \times \boldsymbol{\hat{u}} \ .</math>
So if <math>R(t)</math> denotes the transformation taking basis vectors of the inertial- to the rotating frame, with matrix columns equal to the basis vectors of the rotating frame, then the cross product multiplication by the rotation vector is given by <math>\boldsymbol{\Omega}\times = R'(t)\cdot R(t)^T</math>.

If <math>\boldsymbol{f}</math> is a vector function that is written as<ref group=note>So <math>f_1, f_2, f_3</math> are <math>\boldsymbol{f}</math>'s coordinates with respect to the rotating basis vector <math>\hat{\boldsymbol{\imath}},\ \hat{\boldsymbol{\jmath}},\ \hat{\boldsymbol{k}}</math> (<math>\boldsymbol{f}</math>'s coordinates with respect to the inertial frame are not used). Consequently, at any given instant, the rate of change of <math>\boldsymbol{f}</math> with respect to these rotating coordinates is <math>\frac{\mathrm{d}f_1}{\mathrm{d}t}\hat{\boldsymbol{\imath}} + \frac{\mathrm{d}f_2}{\mathrm{d}t}\hat{\boldsymbol{\jmath}} + \frac{\mathrm{d}f_3}{\mathrm{d}t}\hat{\boldsymbol{k}}.</math> So for example, if <math>f_1 \equiv 1</math> and <math>f_2 = f_3 \equiv 0</math> are constants, then <math>\boldsymbol{f} \equiv \hat{\boldsymbol{\imath}}</math> is just one of the rotating basis vectors and (as expected) its time rate of change with respect to these rotating coordinates is identically <math>\boldsymbol{0}</math> (so the formula for <math>\frac{\mathrm{d}}{\mathrm{d}t} \boldsymbol{f}</math> given below implies that the derivative at time <math>t</math> of this rotating basis vector <math>\boldsymbol{f} \equiv \hat{\boldsymbol{\imath}}</math> is <math>\frac{\mathrm{d}}{\mathrm{d}t} \boldsymbol{i} = \boldsymbol{\Omega}(t) \times \boldsymbol{i}(t)</math>); however, its rate of change with respect to the non-rotating inertial frame will not be constantly <math>\boldsymbol{0}</math> except (of course) in the case where <math>\hat{\boldsymbol{\imath}}</math> is not moving in the inertial frame (this happens, for instance, when the axis of rotation is fixed as the <math>z</math>-axis (assuming standard coordinates) in the inertial frame and also <math>\hat{\boldsymbol{\imath}} \equiv (0, 0, 1)</math> or <math>\hat{\boldsymbol{\imath}} \equiv (0, 0, -1)</math>).</ref>
<math display=block>\boldsymbol{f}(t)=f_1(t) \hat{\boldsymbol{\imath}}+f_2(t) \hat{\boldsymbol{\jmath}}+f_3(t) \hat{\boldsymbol{k}}\ ,</math>
and we want to examine its first derivative then (using the [[product rule]] of differentiation):<ref name=Lanczos>{{cite book |url=https://books.google.com/books?num=10&btnG=Google+Search |title=The Variational Principles of Mechanics |author=Cornelius Lanczos |date=1986 |isbn=0-486-65067-7 |publisher=[[Dover Publications]] |edition=Reprint of Fourth Edition of 1970 |no-pp=true |pages=Chapter 4, §5}}</ref><ref name=Taylor>{{cite book |title=Classical Mechanics |author=John R Taylor |page= 342 |publisher=University Science Books |isbn=1-891389-22-X |date=2005 |url=https://books.google.com/books?id=P1kCtNr-pJsC&pg=PP1}}</ref>
<math display=block>\begin{align}
  \frac{\mathrm{d}}{\mathrm{d}t}\boldsymbol{f}
    &= \frac{\mathrm{d}f_1}{\mathrm{d}t}\hat{\boldsymbol{\imath}} + \frac{\mathrm{d}\hat{\boldsymbol{\imath}}}{\mathrm{d}t}f_1 + \frac{\mathrm{d}f_2}{\mathrm{d}t}\hat{\boldsymbol{\jmath}} + \frac{\mathrm{d}\hat{\boldsymbol{\jmath}}}{\mathrm{d}t}f_2 + \frac{\mathrm{d}f_3}{\mathrm{d}t}\hat{\boldsymbol{k}} + \frac{\mathrm{d}\hat{\boldsymbol{k}}}{\mathrm{d}t}f_3 \\
    &= \frac{\mathrm{d}f_1}{\mathrm{d}t}\hat{\boldsymbol{\imath}} + \frac{\mathrm{d}f_2}{\mathrm{d}t}\hat{\boldsymbol{\jmath}} + \frac{\mathrm{d}f_3}{\mathrm{d}t}\hat{\boldsymbol{k}} + \left[\boldsymbol{\Omega} \times \left(f_1 \hat{\boldsymbol{\imath}} + f_2 \hat{\boldsymbol{\jmath}} + f_3 \hat{\boldsymbol{k}}\right)\right] \\
    &= \left( \frac{\mathrm{d}\boldsymbol{f}}{\mathrm{d}t}\right)_{\mathrm{r}} + \boldsymbol{\Omega} \times \boldsymbol{f}
\end{align}</math>
where <math>\left( \frac{\mathrm{d}\boldsymbol{f}}{\mathrm{d}t}\right)_{\mathrm{r}}</math> denotes the rate of change of <math>\boldsymbol{f}</math> as observed in the rotating coordinate system. As a shorthand the differentiation is expressed as:
<math display=block>\frac{\mathrm{d}}{\mathrm{d}t}\boldsymbol{f} = \left[ \left(\frac{\mathrm{d}}{\mathrm{d}t}\right)_{\mathrm{r}} + \boldsymbol{\Omega} \times \right] \boldsymbol{f} \ .</math>

This result is also known as the [[transport theorem]] in analytical dynamics and is also sometimes referred to as the ''basic kinematic equation''.<ref>{{cite web|last=Corless|first=Martin|title=Kinematics|url=https://engineering.purdue.edu/AAE/Academics/Courses/aae203/2003/fall/aae203F03supp.pdf|archive-url=https://web.archive.org/web/20121024121222/https://engineering.purdue.edu/AAE/Academics/Courses/aae203/2003/fall/aae203F03supp.pdf|url-status=dead|archive-date=24 October 2012|work=Aeromechanics I Course Notes|publisher=[[Purdue University]]|access-date=18 July 2011|page=213}}</ref>

=== Relation between velocities in the two frames ===
A velocity of an object is the time-derivative of the object's position, so

:<math>\mathbf{v} \ \stackrel{\mathrm{def}}{=}\   \frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t} \ .</math>

The time derivative of a position <math>\boldsymbol{r}(t)</math> in a rotating reference frame has two components, one from the explicit time dependence due to motion of the object itself in the rotating reference frame, and another from the frame's own rotation.  Applying the result of the previous subsection to the displacement <math>\boldsymbol{r}(t),</math> the [[Velocity|velocities]] in the two reference frames are related by the equation

:<math> 
\mathbf{v_i} \ \stackrel{\mathrm{def}}{=}\   
\left({\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}}\right)_{\mathrm{i}} \ \stackrel{\mathrm{def}}{=}\   
\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t} = 
\left[ \left(\frac{\mathrm{d}}{\mathrm{d}t}\right)_{\mathrm{r}} + \boldsymbol{\Omega} \times \right] \boldsymbol{r} = 
\left(\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}\right)_{\mathrm{r}} + \boldsymbol\Omega \times \mathbf{r} = 
\mathbf{v}_{\mathrm{r}} + \boldsymbol\Omega \times \mathbf{r} \ ,
</math>
where subscript <math>\mathrm{i}</math> means the inertial frame of reference, and <math>\mathrm{r}</math> means the rotating frame of reference.

=== Relation between accelerations in the two frames ===
Acceleration is the second time derivative of position, or the first time derivative of velocity

:<math> 
\mathbf{a}_{\mathrm{i}} \ \stackrel{\mathrm{def}}{=}\   
\left( \frac{\mathrm{d}^{2}\mathbf{r}}{\mathrm{d}t^{2}}\right)_{\mathrm{i}} = 
\left( \frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t} \right)_{\mathrm{i}} = 
\left[  \left( \frac{\mathrm{d}}{\mathrm{d}t} \right)_{\mathrm{r}} + \boldsymbol\Omega \times \right]
\left[\left( \frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t} \right)_{\mathrm{r}} + \boldsymbol\Omega \times \mathbf{r} \right] \ ,
</math>
where subscript <math>\mathrm{i}</math> means the inertial frame of reference, <math>\mathrm{r}</math> the rotating frame of reference, and where the expression, again, <math>\boldsymbol\Omega \times</math> in the bracketed expression on the left is to be interpreted as an [[Operator (mathematics)|operator]] working onto the bracketed expression on the right.

As <math>\boldsymbol\Omega\times\boldsymbol\Omega=\boldsymbol 0</math>, the first time derivatives of <math>\boldsymbol\Omega</math> inside either frame, when expressed with respect to the basis of e.g. the inertial frame, coincide.
Carrying out the [[Derivative|differentiation]]s and re-arranging some terms yields the acceleration ''relative to the rotating'' reference frame, <math>\mathbf{a}_{\mathrm{r}}</math>

:<math> 
\mathbf{a}_{\mathrm{r}} = 
\mathbf{a}_{\mathrm{i}} -
2 \boldsymbol\Omega \times \mathbf{v}_{\mathrm{r}} -
\boldsymbol\Omega \times (\boldsymbol\Omega \times \mathbf{r}) -
\frac{\mathrm{d}\boldsymbol\Omega}{\mathrm{d}t} \times \mathbf{r}
</math>

where <math>\mathbf{a}_{\mathrm{r}} \ \stackrel{\mathrm{def}}{=}\   \left( \tfrac{\mathrm{d}^{2}\mathbf{r}}{\mathrm{d}t^{2}} \right)_{\mathrm{r}}</math> is the apparent acceleration in the rotating reference frame, the term <math>-\boldsymbol\Omega \times (\boldsymbol\Omega \times \mathbf{r})</math> represents [[centrifugal acceleration]], and the term <math>-2 \boldsymbol\Omega \times \mathbf{v}_{\mathrm{r}}</math> is the [[Coriolis acceleration]]. The last term, <math>-\tfrac{\mathrm{d}\boldsymbol\Omega}{\mathrm{d}t} \times \mathbf{r}</math>, is the [[Euler acceleration]] and is zero in uniformly rotating frames.

=== Newton's second law in the two frames ===
When the expression for acceleration is multiplied by the mass of the particle, the three extra terms on the right-hand side result in [[fictitious force]]s in the rotating reference frame, that is, apparent forces that result from being in a [[non-inertial reference frame]], rather than from any physical interaction between bodies.

Using [[Newton's laws of motion|Newton's second law of motion]] <math>\mathbf{F}=m\mathbf{a},</math> we obtain:<ref name=Arnold/><ref name=Lanczos/><ref name=Taylor/><ref name=Landau>{{cite book |title=Mechanics |author=LD Landau |author2=LM Lifshitz |name-list-style=amp |page= 128 |url=https://books.google.com/books?id=e-xASAehg1sC&pg=PA40 |edition=Third |date=1976 |publisher=Butterworth-Heinemann |isbn=978-0-7506-2896-9}}</ref><ref name=Hand/>

* the [[Coriolis force]] <math display="block">
\mathbf{F}_{\mathrm{Coriolis}} = 
-2m \boldsymbol\Omega \times \mathbf{v}_{\mathrm{r}}
</math>
* the [[centrifugal force (fictitious)|centrifugal force]] <math display="block">
\mathbf{F}_{\mathrm{centrifugal}} = 
-m\boldsymbol\Omega \times (\boldsymbol\Omega \times \mathbf{r})
</math>
* and the [[Euler force]] <math display="block">
\mathbf{F}_{\mathrm{Euler}} = 
-m\frac{\mathrm{d}\boldsymbol\Omega}{\mathrm{d}t} \times \mathbf{r}
</math>
where <math>m</math> is the mass of the object being acted upon by these [[fictitious force]]s. Notice that all three forces vanish when the frame is not rotating, that is, when <math>\boldsymbol{\Omega} = 0 \ . </math>

For completeness, the inertial acceleration <math>\mathbf{a}_{\mathrm{i}}</math> due to impressed external forces <math>\mathbf{F}_{\mathrm{imp}}</math> can be determined from the total physical force in the inertial (non-rotating) frame (for example, force from physical interactions such as [[Electromagnetism|electromagnetic forces]]) using [[Newton's laws of motion|Newton's second law]] in the inertial frame:
<math display="block">
\mathbf{F}_{\mathrm{imp}} = m \mathbf{a}_{\mathrm{i}}
</math>
Newton's law in the rotating frame then becomes
::<math>\mathbf{F_{\mathrm{r}}} = \mathbf{F}_{\mathrm{imp}} + \mathbf{F}_{\mathrm{centrifugal}} +\mathbf{F}_{\mathrm{Coriolis}} + \mathbf{F}_{\mathrm{Euler}} = m\mathbf{a_{\mathrm{r}}} \ . </math>
In other words, to handle the laws of motion in a rotating reference frame:<ref name=Hand>{{cite book |title=Analytical Mechanics |author =Louis N. Hand |author2 =Janet D. Finch |page=267 |url=https://books.google.com/books?id=1J2hzvX2Xh8C&q=Hand+inauthor:Finch&pg=PA267
|isbn=0-521-57572-9 |publisher=[[Cambridge University Press]] |date=1998 }}</ref><ref name=Pui>{{cite book |title=Mechanics |author=HS Hans |author2=SP Pui |name-list-style=amp |page=341 |url=https://books.google.com/books?id=mgVW00YV3zAC&q=inertial+force+%22rotating+frame%22&pg=PA341 |isbn=0-07-047360-9 |publisher=Tata McGraw-Hill |date=2003  }}</ref><ref name=Taylor2>{{cite book |title=Classical Mechanics |author=John R Taylor |page= 328 |publisher=University Science Books |isbn=1-891389-22-X |date=2005 |url=https://books.google.com/books?id=P1kCtNr-pJsC&pg=PP1}}</ref>
{{Quotation|Treat the fictitious forces like real forces, and pretend you are in an inertial frame.|Louis N. Hand, Janet D. Finch ''Analytical Mechanics'', p. 267}}
{{Quotation|Obviously, a rotating frame of reference is a case of a non-inertial frame. Thus the particle in addition to the real force is acted upon by a fictitious force...The particle will move according to Newton's second law of motion if the total force acting on it is taken as the sum of the real and fictitious forces.|HS Hans & SP Pui: ''Mechanics''; p. 341}}
{{Quotation|This equation has exactly the form of Newton's second law, ''except'' that in addition to '''F''', the sum of all forces identified in the inertial frame, there is an extra term on the right...This means we can continue to use Newton's second law in the noninertial frame ''provided'' we agree that in the noninertial frame we must add an extra force-like term, often called the '''inertial force'''. |John R. Taylor: ''Classical Mechanics''; p. 328}}