Rotating reference frame

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 ${\displaystyle \left(x^{\prime },y^{\prime },z^{\prime }\right)}$ of the rotating reference frame and the coordinates ${\displaystyle (x,y,z)}$ of an inertial reference frame with the same origin. If the rotation is about the ${\displaystyle z}$ axis with a constant angular velocity ${\displaystyle \mathrm{\Omega }}$ (so ${\displaystyle z^{\prime }=z}$ and ${\displaystyle \frac{\mathrm{d}\theta }{\mathrm{d}t}\equiv \mathrm{\Omega },}$ which implies ${\displaystyle \theta (t)=\mathrm{\Omega }t+{\theta }_0}$ for some constant ${\displaystyle {\theta }_0}$ where ${\displaystyle \theta (t)}$ denotes the angle in the ${\displaystyle x-y}$-plane formed at time ${\displaystyle t}$ by ${\displaystyle \left(x^{\prime },y^{\prime }\right)}$ and the ${\displaystyle x}$-axis), and if the two reference frames coincide at time ${\displaystyle t=0}$ (meaning ${\displaystyle \left(x^{\prime },y^{\prime },z^{\prime }\right)=(x,y,z)}$ when ${\displaystyle t=0,}$ so take ${\displaystyle {\theta }_0=0}$ or some other integer multiple of ${\displaystyle 2\pi }$), the transformation from rotating coordinates to inertial coordinates can be written $${\displaystyle x=x^{\prime }\cos (\theta (t))-y^{\prime }\sin (\theta (t))}$$ $${\displaystyle y=x^{\prime }\sin (\theta (t))+y^{\prime }\cos (\theta (t))}$$ whereas the reverse transformation is $${\displaystyle x^{\prime }=x\cos (-\theta (t))-y\sin (-\theta (t))}$$ $${\displaystyle y^{\prime }=x\sin (-\theta (t))+y\cos (-\theta (t)).}$$

This result can be obtained from a rotation matrix.

Introduce the unit vectors ${\displaystyle \hat{\mathit{ı}},\hat{\mathit{ȷ}},\hat{\mathit{k}}}$ 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 ${\displaystyle t=0}$ and the ${\displaystyle z}$-axis is the axis of rotation. Then for a counterclockwise rotation through angle ${\displaystyle \mathrm{\Omega }t}$: $${\displaystyle \hat{\mathit{ı}}(t)=(\cos \theta (t),\sin \theta (t))}$$ where the ${\displaystyle (x,y)}$ components are expressed in the stationary frame. Likewise, $${\displaystyle \hat{\mathit{ȷ}}(t)=(-\sin \theta (t),\cos \theta (t)).}$$

Thus the time derivative of these vectors, which rotate without changing magnitude, is $${\displaystyle \frac{\mathrm{d}}{\mathrm{d}t}\hat{\mathit{ı}}(t)=\mathrm{\Omega }(-\sin \theta (t),\cos \theta (t))=\mathrm{\Omega }\hat{\mathit{ȷ}};}$$ $${\displaystyle \frac{\mathrm{d}}{\mathrm{d}t}\hat{\mathit{ȷ}}(t)=\mathrm{\Omega }(-\cos \theta (t),-\sin \theta (t))=-\mathrm{\Omega }\hat{\mathit{ı}},}$$ where ${\displaystyle \mathrm{\Omega }\equiv \frac{\mathrm{d}}{\mathrm{d}t}\theta (t).}$ This result is the same as found using a vector cross product with the rotation vector ${\displaystyle \mathbf{\Omega }}$ pointed along the z-axis of rotation ${\displaystyle \mathbf{\Omega }=(0,0,\mathrm{\Omega }),}$ namely, $${\displaystyle \frac{\mathrm{d}}{\mathrm{d}t}\hat{\mathit{u}}=\mathbf{\Omega }\mathbf{\times }\hat{\mathit{u}},}$$ where ${\displaystyle \hat{\mathit{u}}}$ is either ${\displaystyle \hat{\mathit{ı}}}$ or ${\displaystyle \hat{\mathit{ȷ}}.}$

Time derivatives in the two frames

Introduce unit vectors ${\displaystyle \hat{\mathit{ı}},\hat{\mathit{ȷ}},\hat{\mathit{k}}}$, 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 ${\displaystyle \mathrm{\Omega }(t)}$ about an axis along the rotation vector ${\displaystyle \mathbf{\Omega }(t)}$ then each unit vector ${\displaystyle \hat{\mathit{u}}}$ of the rotating coordinate system (such as ${\displaystyle \hat{\mathit{ı}},\hat{\mathit{ȷ}},}$ or ${\displaystyle \hat{\mathit{k}}}$) abides by the following equation: $${\displaystyle \frac{\mathrm{d}}{\mathrm{d}t}\hat{\mathit{u}}=\mathbf{\Omega }\times \hat{\mathit{u}}.}$$ So if ${\displaystyle R(t)}$ 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 ${\displaystyle \mathbf{\Omega }\times =R^{\prime }(t)\cdot R(t{)}^T}$.

If ${\displaystyle \mathit{f}}$ is a vector function that is written as $${\displaystyle \mathit{f}(t)=f_1(t)\hat{\mathit{ı}}+f_2(t)\hat{\mathit{ȷ}}+f_3(t)\hat{\mathit{k}},}$$ and we want to examine its first derivative then (using the product rule of differentiation): $${\displaystyle \begin{array}{cc}\frac{\mathrm{d}}{\mathrm{d}t}\mathit{f}& =\frac{\mathrm{d}{f}_1}{\mathrm{d}t}\hat{\mathit{ı}}+\frac{\mathrm{d}\hat{\mathit{ı}}}{\mathrm{d}t}{f}_1+\frac{\mathrm{d}{f}_2}{\mathrm{d}t}\hat{\mathit{ȷ}}+\frac{\mathrm{d}\hat{\mathit{ȷ}}}{\mathrm{d}t}{f}_2+\frac{\mathrm{d}{f}_3}{\mathrm{d}t}\hat{\mathit{k}}+\frac{\mathrm{d}\hat{\mathit{k}}}{\mathrm{d}t}{f}_3\\ & =\left[\frac{\mathrm{d}{f}_1}{\mathrm{d}t}\hat{\mathit{ı}}+\frac{\mathrm{d}{f}_2}{\mathrm{d}t}\hat{\mathit{ȷ}}+\frac{\mathrm{d}{f}_3}{\mathrm{d}t}\hat{\mathit{k}}\right]+\left[\frac{\mathrm{d}\hat{\mathit{ı}}}{\mathrm{d}t}{f}_1+\frac{\mathrm{d}\hat{\mathit{ȷ}}}{\mathrm{d}t}{f}_2+\frac{\mathrm{d}\hat{\mathit{k}}}{\mathrm{d}t}{f}_3\right]\\ & =\left[\frac{\mathrm{d}{f}_1}{\mathrm{d}t}\hat{\mathit{ı}}+\frac{\mathrm{d}{f}_2}{\mathrm{d}t}\hat{\mathit{ȷ}}+\frac{\mathrm{d}{f}_3}{\mathrm{d}t}\hat{\mathit{k}}\right]+\left[\mathbf{\Omega }\times \left(f_1\hat{\mathit{ı}}+f_2\hat{\mathit{ȷ}}+f_3\hat{\mathit{k}}\right)\right]\\ & ={\left(\frac{\mathrm{d}\mathit{f}}{\mathrm{d}t}\right)}_{\mathrm{r}}+\mathbf{\Omega }\times \mathit{f}\end{array}}$$ where ${\displaystyle {\left(\frac{\mathrm{d}\mathit{f}}{\mathrm{d}t}\right)}_{\mathrm{r}}}$ denotes the rate of change of ${\displaystyle \mathit{f}}$ as observed in the rotating coordinate system. As a shorthand the differentiation is expressed as: $${\displaystyle \frac{\mathrm{d}}{\mathrm{d}t}\mathit{f}=\left[{\left(\frac{\mathrm{d}}{\mathrm{d}t}\right)}_{\mathrm{r}}+\mathbf{\Omega }\times \right]\mathit{f}.}$$

This result is also known as the transport theorem in analytical dynamics and is also sometimes referred to as the basic kinematic equation.

Relation between velocities in the two frames

A velocity of an object is the time-derivative of the object's position, so

${\displaystyle \mathbf{v}\stackrel{\mathrm{d}\mathrm{e}\mathrm{f}}{=}\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}.}$

The time derivative of a position ${\displaystyle \mathit{r}(t)}$ 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 ${\displaystyle \mathit{r}(t),}$ the velocities in the two reference frames are related by the equation

${\displaystyle {\mathbf{v}}_{\mathbf{i}}\stackrel{\mathrm{d}\mathrm{e}\mathrm{f}}{=}{\left(\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}\right)}_{\mathrm{i}}\stackrel{\mathrm{d}\mathrm{e}\mathrm{f}}{=}\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}=\left[{\left(\frac{\mathrm{d}}{\mathrm{d}t}\right)}_{\mathrm{r}}+\mathbf{\Omega }\times \right]\mathit{r}={\left(\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}\right)}_{\mathrm{r}}+\mathbf{\Omega }\times \mathbf{r}={\mathbf{v}}_{\mathrm{r}}+\mathbf{\Omega }\times \mathbf{r},}$

where subscript ${\displaystyle \mathrm{i}}$ means the inertial frame of reference, and ${\displaystyle \mathrm{r}}$ 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

${\displaystyle {\mathbf{a}}_{\mathrm{i}}\stackrel{\mathrm{d}\mathrm{e}\mathrm{f}}{=}{\left(\frac{{\mathrm{d}}^2\mathbf{r}}{\mathrm{d}{t}^2}\right)}_{\mathrm{i}}={\left(\frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t}\right)}_{\mathrm{i}}=\left[{\left(\frac{\mathrm{d}}{\mathrm{d}t}\right)}_{\mathrm{r}}+\mathbf{\Omega }\times \right]\left[{\left(\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}\right)}_{\mathrm{r}}+\mathbf{\Omega }\times \mathbf{r}\right],}$

where subscript ${\displaystyle \mathrm{i}}$ means the inertial frame of reference, ${\displaystyle \mathrm{r}}$ the rotating frame of reference, and where the expression, again, ${\displaystyle \mathbf{\Omega }\times }$ in the bracketed expression on the left is to be interpreted as an operator working onto the bracketed expression on the right.

As ${\displaystyle \mathbf{\Omega }\times \mathbf{\Omega }=\mathbf{0}}$, the first time derivatives of ${\displaystyle \mathbf{\Omega }}$ inside either frame, when expressed with respect to the basis of e.g. the inertial frame, coincide. Carrying out the differentiations and re-arranging some terms yields the acceleration relative to the rotating reference frame, ${\displaystyle {\mathbf{a}}_{\mathrm{r}}}$

${\displaystyle {\mathbf{a}}_{\mathrm{r}}={\mathbf{a}}_{\mathrm{i}}-2\mathbf{\Omega }\times {\mathbf{v}}_{\mathrm{r}}-\mathbf{\Omega }\times (\mathbf{\Omega }\times \mathbf{r})-\frac{\mathrm{d}\mathbf{\Omega }}{\mathrm{d}t}\times \mathbf{r}}$

where ${\displaystyle {\mathbf{a}}_{\mathrm{r}}\stackrel{\mathrm{d}\mathrm{e}\mathrm{f}}{=}{\left(\frac{{\mathrm{d}}^2\mathbf{r}}{\mathrm{d}{t}^2}\right)}_{\mathrm{r}}}$ is the apparent acceleration in the rotating reference frame, the term ${\displaystyle -\mathbf{\Omega }\times (\mathbf{\Omega }\times \mathbf{r})}$ represents centrifugal acceleration, and the term ${\displaystyle -2\mathbf{\Omega }\times {\mathbf{v}}_{\mathrm{r}}}$ is the Coriolis acceleration. The last term, ${\displaystyle -\frac{\mathrm{d}\mathbf{\Omega }}{\mathrm{d}t}\times \mathbf{r}}$, 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 forces 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 second law of motion ${\displaystyle \mathbf{F}=m\mathbf{a},}$ we obtain:

• the Coriolis force $${\displaystyle {\mathbf{F}}_{\mathrm{C}\mathrm{o}\mathrm{r}\mathrm{i}\mathrm{o}\mathrm{l}\mathrm{i}\mathrm{s}}=-2m\mathbf{\Omega }\times {\mathbf{v}}_{\mathrm{r}}}$$
• the centrifugal force $${\displaystyle {\mathbf{F}}_{\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{f}\mathrm{u}\mathrm{g}\mathrm{a}\mathrm{l}}=-m\mathbf{\Omega }\times (\mathbf{\Omega }\times \mathbf{r})}$$
• and the Euler force $${\displaystyle {\mathbf{F}}_{\mathrm{E}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{r}}=-m\frac{\mathrm{d}\mathbf{\Omega }}{\mathrm{d}t}\times \mathbf{r}}$$

where ${\displaystyle m}$ is the mass of the object being acted upon by these fictitious forces. Notice that all three forces vanish when the frame is not rotating, that is, when ${\displaystyle \mathbf{\Omega }=0.}$

For completeness, the inertial acceleration ${\displaystyle {\mathbf{a}}_{\mathrm{i}}}$ due to impressed external forces ${\displaystyle {\mathbf{F}}_{\mathrm{i}\mathrm{m}\mathrm{p}}}$ can be determined from the total physical force in the inertial (non-rotating) frame (for example, force from physical interactions such as electromagnetic forces) using Newton's second law in the inertial frame: $${\displaystyle {\mathbf{F}}_{\mathrm{i}\mathrm{m}\mathrm{p}}=m{\mathbf{a}}_{\mathrm{i}}}$$ Newton's law in the rotating frame then becomes

${\displaystyle {\mathbf{F}}_{\mathrm{r}}={\mathbf{F}}_{\mathrm{i}\mathrm{m}\mathrm{p}}+{\mathbf{F}}_{\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{f}\mathrm{u}\mathrm{g}\mathrm{a}\mathrm{l}}+{\mathbf{F}}_{\mathrm{C}\mathrm{o}\mathrm{r}\mathrm{i}\mathrm{o}\mathrm{l}\mathrm{i}\mathrm{s}}+{\mathbf{F}}_{\mathrm{E}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{r}}=m{\mathbf{a}}_{\mathrm{r}}.}$

In other words, to handle the laws of motion in a rotating reference frame:

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

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

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

Use in magnetic resonance

It is convenient to consider magnetic resonance in a frame that rotates at the Larmor frequency of the spins. This is illustrated in the animation below. The rotating wave approximation may also be used.