Classical Mechanics

Dynamics in Rotating Reference Frames

Velocity and acceleration transformations, inertial forces, and Newtonian dynamics in non-inertial reference frames.

Building upon the kinematics of rotating reference frames, this essay derives the equations of motion observed by a rotating observer. Beginning with the transformation laws for velocity and acceleration, we obtain the inertial forces that arise in non-inertial frames, including the Euler, Coriolis, and centrifugal forces, and discuss their physical interpretation.

Prerequisites

The Kinematics of Rigid Body Motion
Newton's laws of motion

Time derivatives in rotating frames
Velocity transformations
Acceleration transformations
Euler force
Coriolis force
Centrifugal force
Equations of motion in rotating frames

I. Introduction
1. Kinematical Preliminaries
II. Equations of Motion in a Rotating Frame
1. The Centrifugal Force
2. The Coriolis Force
3. Lagrange Equations of Motion in a Rotating Frame
III. Life in a Rotating Reference Frame
1. A Final Comment

I. Introduction

This essay develops the dynamics of particles as observed in a rotating reference frame. We begin by recalling the kinematical relations required to compare time derivatives in inertial and rotating frames, then derive the transformation of the acceleration and the inertial forces that arise from it.

1. Kinematical Preliminaries

Let

A (t)

be the orthogonal matrix that maps the components of a vector in the inertial frame to its components in the rotating frame:

Rotating Frames:Frame Transformation

g^{'} (t) = A (t) g (t), g (t) = A^{⊤} (t) g^{'} (t) .

The angular velocity vector is defined as the time derivative of the infinitesimal rotation vector,

Rotating Frames:Angular Velocity Definition

ω = Δ t \to 0 lim \frac{d Ω}{Δ t} .

The corresponding transformation law for time derivatives is

Rotating Frames:Time Derivative Rule

\frac{d g}{d t} = A^{⊤} (t) (ω \times g^{'} + \frac{d g ^{'}}{d t}) .

These equations summarize all the results from rigid-body kinematics required in the remainder of this essay.

II. Equations of Motion in a Rotating Frame

In this section we derive the equations of motion of a particle as observed in a rotating frame of reference with respect to an inertial one.We restrict our discussion to the case of constant angular velocity. The resulting equations introduce the Coriolis and centrifugal forces that arise in a rotating frame.

We begin with the transformation law for time derivatives, Eq. time derivative rule . Differentiating once more gives the second derivative of the position vector in the inertial frame,

\frac{d ^{2} r ( t )}{d t ^{2}} = \frac{d A ^{⊤}}{d t} (ω \times r^{'} + \frac{d r ^{'}}{d t}) + A^{⊤} (ω \times \frac{d r ^{'}}{d t} + \frac{d ^{2} r ^{'}}{d t ^{2}})

From the orthogonality of the rotation matrix,

A^{⊤} A = I

\frac{d A ^{⊤}}{d t} A + A \frac{d A ^{⊤}}{d t} = 0

we obtain:

\frac{d A ^{⊤}}{d t} = - A^{⊤} \frac{d A}{d t} A^{⊤}

Inserting this expression in the equation above, and using the invariance of the cross product under proper orthogonal transformations,,

\frac{d ^{2} r ( t )}{d t ^{2}} = - A^{⊤} \frac{d A}{d t} A^{⊤} (ω \times r^{'} + \frac{d r ^{'}}{d t}) + A^{⊤} (ω \times \frac{d r ^{'}}{d t} + \frac{d ^{2} r ^{'}}{d t ^{2}}) = - A^{⊤} \frac{d A}{d t} (A^{⊤} ω \times A^{⊤} r^{'} + A^{⊤} \frac{d r ^{'}}{d t}) + A^{⊤} (ω \times \frac{d r ^{'}}{d t} + \frac{d ^{2} r ^{'}}{d t ^{2}}) = - A^{⊤} (A (A^{⊤} ω \times A^{⊤} r^{'} + A^{⊤} \frac{d r ^{'}}{d t})) \times ω + A^{⊤} (ω \times \frac{d r ^{'}}{d t} + \frac{d ^{2} r ^{'}}{d t ^{2}}) = - A^{⊤} (ω \times r^{'} + \frac{d r ^{'}}{d t}) \times ω + A^{⊤} (ω \times \frac{d r ^{'}}{d t} + \frac{d ^{2} r ^{'}}{d t ^{2}})

we arrive at the equation:

A \frac{d ^{2} r ( t )}{d t ^{2}} = ω \times (ω \times r^{'}) + 2 ω \times \frac{d r ^{'}}{d t} + \frac{d ^{2} r ^{'}}{d t ^{2}} .

Introducing the inertial acceleration

a = \frac{d ^{2} r ( t )}{d t ^{2}}

, and the acceleration measured in the rotating frame as

a^{'} = \frac{d ^{2} r ^{'} ( t )}{d t ^{2}}

, we obtain the acceleration transformation:

Rotating Frames:Acceleration Transformation

a^{'} = - ω \times (ω \times r^{'}) - 2 ω \times \frac{d r ^{'}}{d t} + A a

Equation acceleration transformation is purely kinematical. No dynamical assumptions have been introduced yet. By applying Newton's second law to the rotating frame, we identify

F^{'} = m a^{'},

and Eq. acceleration transformation becomes

Rotating Frames:Forces Transformation

F^{'} = - m ω \times (ω \times r^{'}) - 2 m ω \times \frac{d r ^{'}}{d t} + m A a

In this equation there are terms that originate entirely from the kinematics of the rotating frame. They are interpreted as apparent forces and are responsible for the distinctive dynamics observed in non-inertial reference frames.

1. The Centrifugal Force

The centrifugal force is the first term of equation forces transformation ,

Rotating Frames:Centrifugal Force

F_{cf} = - m ω \times (ω \times r^{'}) = m (ω^{2} r^{'} - ω ⟨ ω ∣ r^{'} ⟩)

which has a magnitude proportional to the square of the angular velocity, and is directed perpendicular to the rotation axis and away from it. This force depends only on the angular velocity of the rotating frame and on the position of the body. It is independent of the body's velocity in the rotating frame.

The centrifugal force admits the potential

V_{cf} = \frac{m}{2} (- ⟨ r^{'} ∣ r^{'} ⟩ ⟨ ω ∣ ω ⟩ + ⟨ ω ∣ r^{'} ⟩^{2}),

and it is therefore a conservative force, since

F_{cf} = - \nabla_{r^{'}} V_{cf} .

2. The Coriolis Force

The Coriolis force is the term proportional to the velocity of the body in the rotating frame appearing in equation forces transformation ,

Rotating Frames:Coriolis Force

F_{cor} = - 2 m ω \times \frac{d r ^{'}}{d t}

This force arises only from the motion of the body with respect to the rotating frame. Since it is always perpendicular to the velocity, it performs no work on the body.

The Coriolis force has the same mathematical structure as the Lorentz force in electromagnetism. Although it is not conservative, since it depends explicitly on the velocity, it can nevertheless be derived from the generalized potential

V_{cor}

determined as,

V_{cor} = - m ⟨ ω \times r^{'} ∣ \dot{r^{'}} ⟩

so that the Coriolis force derives from a generalized potential:

F_{cor} = \frac{d}{d t} \frac{\partial V _{cor}}{\partial r ^{'} ˙} - \frac{\partial V _{cor}}{\partial r ^{'}}

To verify that the generalized potential indeed reproduces the Coriolis force, we use the cyclic permutation identity of the scalar triple product,

⟨ a \times b ∣ c ⟩ = ⟨ b \times c ∣ a ⟩ = ⟨ c \times a ∣ b ⟩,

to rewrite the scalar product so that the vector with respect to which the gradient is taken appears directly in the scalar product rather than in the cross product. This allows the required derivatives with respect to

r

and

\dot{r}

to be computed immediately.

Unlike an ordinary potential, a generalized potential may depend explicitly on both the position and the velocity of the body. Such potentials naturally arise in systems whose forces are linear in the velocity, such as the Coriolis force and the Lorentz force in electromagnetism.

3. Lagrange Equations of Motion in a Rotating Frame

In the remainder of this section, we will denote the time derivative

d a / d t

\overset{a}{˙}

to simplify the equations. Unless otherwise stated, all vectors will be expressed in the rotating frame, and

r

will denote the position vector in that frame. In this section we would like to briefly touch on the functional form of the Lagrangian in the rotating frame of a body subject to zero net force in the inertial frame

The Lagrangian is obtained by adding to the kinetic energy all potential contributions, including generalized potentials when present.

L = T - V

As the centrifugal and Coriolis forces admit potentials, the Lagrangian of the system reads:

L = \frac{1}{2} m ⟨ \dot{r} ∣ \dot{r} ⟩ + \frac{m}{2} (⟨ r ∣ r ⟩ ⟨ ω ∣ ω ⟩ - ⟨ ω ∣ r ⟩^{2}) + m ⟨ ω \times r ∣ \dot{r} ⟩ .

The Lagrange equations of motion are given by:

\frac{d}{d t} \frac{\partial L}{\partial r ˙} - \frac{\partial L}{\partial r} = 0

The first term of the Lagrangian gives:

\frac{d}{d t} \frac{\partial L}{\partial r ˙} = m \ddot{r} + m ω \times \dot{r}

The second term of the Lagrangian gives:

\frac{\partial L}{\partial r} = m (ω^{2} r - ⟨ ω ∣ r ⟩ ω) + m \dot{r} \times ω

The Lagrange equations of motion in the rotating frame are therefore

m \ddot{r} + 2 m ω \times \dot{r} + m (⟨ ω ∣ r ⟩ ω - ω^{2} r) = 0.

Dividing by

m

, we obtain

\ddot{r} = - 2 ω \times \dot{r} + ω^{2} r - ⟨ ω ∣ r ⟩ ω,

which is identical to equation acceleration transformation .

It has to be noted that a conservative force in the inertial frame is not necessarily conservative in the rotating frame. A simple example can be easily constructed by considering a constant force in the plane

z = 0

of the inertial frame such as:

F = F_{x} i + F_{y} j

where

F_{x}

and

F_{y}

are constants and

i

j

are the unit vectors of the inertial frame. The force in a frame rotating with constant angular velocity

ω

around the axis

z

is given by:

F^{'} = cos ω t - sin ω t 0 sin ω t cos ω t 0 001 F = (F_{x} cos (ω t) + F_{y} sin (ω t)) i^{'} + (- F_{x} sin (ω t) + F_{y} cos (ω t)) j^{'}

where

i^{'}

j^{'}

are the unit vectors of the rotating frame. The force in the rotating frame is time-dependent, and does not have a generalized potential as it does not explicitly depend on the velocity. Each case will have to be considered individually to determine whether the dynamics can be described in a Lagrangian formalism, and in general the equations of motion acceleration transformation will have to be used. The existence of a Lagrangian for the inertial forces provides an elegant alternative formulation of the dynamics in rotating reference frames. Nevertheless, when additional external forces are present, each case must be analyzed individually to determine whether a Lagrangian description exists or whether the equations of motion must instead be used directly.

III. Life in a Rotating Reference Frame

The dynamics of rotating reference frames is often counterintuitive, and even situations with which we are very familiar become surprisingly subtle when examined in detail. One such example is the carousel.

If you are an observer at rest in the inertial frame fixed to the Earth, life is quite simple for you. All people standing in front of the carousel are not moving and you see the people on the carousel moving in circles around the centre of rotation of the carousel, which has an angular velocity

ω

parallel to the axis perpendicular to the Earth surface, which we call

e_{z}

If you are on the carousel, reality is more complicated First, you have to hold on to something in order not to fly away. You clearly perceive and measure a force that pushes you away from the center of rotation, whose magnitude increases with your distance from the center. Furthermore, you see the people standing on the Earth moving around you in circles with constant angular velocity. This is quite a strange observation, because you cannot explain why they are not pushed away by the same force that you yourself must resist in order not to fly away. Even more interestingly, if something anchored to the carousel for some reason slips away, it also starts to go in circle although following a more complicated dynamics as the people on the Earth.

The resolution of these apparent paradoxes lies in the combined action of the centrifugal and Coriolis forces in the rotating frame.

Consider a person standing on the Earth. Since the Earth is at rest in the inertial frame, its position vector

g (t)

is constant in time, and therefore

\frac{d g}{d t} = 0.

Using the transformation law relating the components of a vector in the fixed and rotating frames,

\frac{d g}{d t} = A^{- 1} (t) (ω \times g^{'} + \frac{d g ^{'}}{d t}),

we conclude that

\frac{d g ^{'}}{d t} = - ω \times g^{'} .

Thus, an observer on the carousel sees every point fixed on the Earth moving with a velocity equal to the opposite of the local rotational velocity of the carousel. The differential equation

\frac{d g ^{'}}{d t} = - ω \times g^{'}

can be solved explicitly. We seek a solution of the form

g^{'} (t) = k e^{- i α t},

where

k

is a complex constant vector. Substituting into the differential equation gives

i α k = - ω \times k .

Writing the complex vector as

k = a + i b,

with

a

and

b

real vectors, we obtain

i α (a + i b) = - ω \times (a + i b),

which, upon separating the real and imaginary parts, yields

α a α b = - ω \times b, = - ω \times a .

The second equation implies

b = \frac{1}{α} ω \times a .

Substituting the expression for

b

into the first equation yields

α a = - \frac{1}{α} ω \times (ω \times a),

or equivalently,

α^{2} a = - ω \times (ω \times a) .

Using the vector identity

u \times (v \times w) = ⟨ u ∣ w ⟩ v - ⟨ u ∣ v ⟩ w,

we obtain

α^{2} a = - (⟨ ω ∣ a ⟩ ω - ω^{2} a),

which may be rewritten as

(α^{2} - ω^{2}) a + ⟨ ω ∣ a ⟩ ω = 0.

Since the two vectors

a

and

ω

are linearly independent, the above equation implies

α^{2} = ω^{2}, ⟨ ω ∣ a ⟩ = 0.

Hence,

α = \pm ω,

The general solution is therefore given by

g^{'} (t) = (a + \frac{i}{α} ω \times a) e^{- i α t} + g_{0}^{'}, α = \pm ω,

where

g_{0}^{'}

is any constant vector parallel to

ω

. To obtain a more explicit expression, we define the orthonormal basis

a = a e_{x}, ω = ω e_{z},

where

e_{x}

is chosen along the direction of

a

and

e_{z}

is parallel to the axis of rotation. Since

e_{y} = e_{z} \times e_{x},

we obtain

ω \times a = ω a e_{y},

and the solution becomes

g^{'} (t) = a (e_{x} + i \frac{ω}{α} e_{y}) e^{- i α t} + g_{0}^{'} .

The physical solutions are obtained by taking the real part of the complex solutions. Using

e^{- i α t} = cos (α t) - i sin (α t),

we find

g^{'} (t) = ℜ [a (e_{x} - i \frac{ω}{α} e_{y}) e^{- i α t}] + g_{0}^{'} = a [cos (α t) e_{x} - \frac{ω}{α} sin (α t) e_{y}] + g_{0}^{'},

where

g_{0}^{'}

is constant and parallel to the axis of rotation. For

α = ω

, this gives

g^{'} (t) = a [cos (ω t) e_{x} - sin (ω t) e_{y}] + g_{0}^{'} .

For

α = - ω

, we obtain

g^{'} (t) = a [cos (- ω t) e_{x} - \frac{ω}{- ω} sin (- ω t) e_{y}] + g_{0}^{'} = a [cos (ω t) e_{x} - sin (ω t) e_{y}] + g_{0}^{'} .

Thus, both values of

α

produce the same real solution. The same result can be obtained directly from the rotation matrix. Consider, for simplicity, a point whose position in the fixed frame is

g = a e_{x} + g_{0},

where

g_{0}

is parallel to the axis of rotation. Its components in the rotating frame are obtained by applying the rotation matrix

A (t)

g^{'} (t) = A (t) g = a cos (ω t) - sin (ω t) 0 sin (ω t) cos (ω t) 0 001 100 + g_{0} = a [cos (ω t) e_{x} - sin (ω t) e_{y}] + g_{0} .

This is precisely the same solution obtained from the differential equation

\frac{d g ^{'}}{d t} = - ω \times g^{'} .

Now that we have determined the trajectory of a person standing on the Earth as observed from the rotating frame, we can compute the centrifugal force acting on this person,

F_{cf} = m ω^{2} g^{'} (t),

where we have used the fact that the solution satisfies

⟨ ω ∣ g^{'} ⟩ = 0.

The Coriolis force is,

F_{cor} = - 2 m ω \times \frac{d g ^{'}}{d t} = 2 m ω \times (ω \times g^{'}),

where we have used

\frac{d g ^{'}}{d t} = - ω \times g^{'} .

Since

g^{'}

is orthogonal to

ω

, we obtain

F_{cor} = - 2 m ω^{2} g^{'} (t) .

The total apparent force acting on the person in the rotating frame is therefore

F_{tot} = F_{cf} + F_{cor} = m ω^{2} g^{'} (t) - 2 m ω^{2} g^{'} (t) = - m ω^{2} g^{'} (t) .

This is precisely the centripetal force required to keep the person on the Earth moving on the circular trajectory observed from the rotating frame.

1. A Final Comment

At this point it is worth pausing to discuss a few important conceptual points that, although implicit throughout our derivation, play a fundamental role in the dynamics of rotating reference frames.

So far, we have assumed that Newton's first principle of dynamics holds in the rotating frame. An observer standing on the carousel could verify this experimentally by measuring the force required to keep a mass

m

at rest at different positions on the carousel. The measured force would be precisely the centrifugal force. Likewise, the observer could measure the force required to keep a body moving with constant velocity. Repeating this experiment for different velocities would allow the observer to determine the combined action of the centrifugal and Coriolis forces. The fact that such experiments can be performed and lead to well-defined force fields is a remarkable property of rotating frames and is by no means obvious.

Once these force fields have been determined - assuming no other forces are present - the observer can predict the motion of any body from its initial conditions by solving Newton's second law or, equivalently, the Lagrange equations. Thus, in the rotating frame, Newton's second law, Hamilton's principle, the Lagrange equations, and the Hamilton equations remain perfectly valid descriptions of the dynamics.

What would nevertheless surprise the observer is that the apparent forces do not satisfy Newton's third principle of dynamics. In particular, there is no body in the universe on which acts a force equal and opposite to the centrifugal force. This lack of an action-reaction pair is one of the reasons why the centrifugal and Coriolis forces are referred to as apparent forces.

In classical mechanics, inertial reference frames are precisely those in which Newton's three principles of dynamics hold. The rotating frame considered in this chapter is therefore not an inertial frame. Remarkably, by introducing the centrifugal and Coriolis forces, Newton's first and second principles can be restored, allowing the dynamics to be described exactly as in an inertial frame. Newton's third principle, however, cannot be recovered: there is no body on which acts a force equal and opposite to the centrifugal or Coriolis force. This is one of the defining characteristics of apparent forces.

One should not conclude that every force that fails to satisfy the simple action-reaction principle is therefore an apparent force. The Lorentz force on a charge moving in a magnetic field provides an important counterexample. Unlike the centrifugal and Coriolis forces, the Lorentz force cannot be eliminated by changing the reference frame. It represents a genuine physical interaction and therefore belongs to a completely different category.

There is, however, an even deeper reason why these forces are regarded as apparent. The Lorentz force is fundamentally different from the centrifugal and Coriolis forces. No change of coordinates can eliminate the Lorentz force, since it originates from the electromagnetic interaction and, in special relativity, arises from the transformation of the electric field of the charge in its rest frame. The centrifugal and Coriolis forces, on the other hand, can be made to disappear completely simply by observing the same motion from an inertial frame, such as the frame of an observer standing on the Earth. It is this dependence on the choice of reference frame that ultimately distinguishes apparent forces from genuine physical interactions.

Dynamics in Rotating Reference Frames

Velocity and acceleration transformations, inertial forces, and Newtonian dynamics in non-inertial reference frames.

Prerequisites

Contents

Contents

I. Introduction

1. Kinematical Preliminaries

II. Equations of Motion in a Rotating Frame

1. The Centrifugal Force

2. The Coriolis Force

3. Lagrange Equations of Motion in a Rotating Frame

III. Life in a Rotating Reference Frame

1. A Final Comment