Rotating reference frame
A rotating reference frame is a non-inertial frame of reference that rotates at a constant angular velocity relative to an inertial frame, where the standard form of Newton's laws of motion does not apply directly due to the frame's acceleration.[1] In such frames, the observed motion of objects appears altered, requiring the inclusion of fictitious forces to reconcile descriptions with physical reality.[1] These frames are fundamental in physics for analyzing systems involving rotation, such as planetary motion, engineering designs, and geophysical phenomena.[2] The transformation between an inertial frame and a rotating frame involves relating position, velocity, and acceleration vectors across the frames. The velocity in the inertial frame is given by \mathbf{v} = \mathbf{v}' + \boldsymbol{\Omega} \times \mathbf{r}, where \mathbf{v}' is the velocity relative to the rotating frame, \boldsymbol{\Omega} is the angular velocity vector, and \mathbf{r} is the position vector.[1] Acceleration transforms as \mathbf{a} = \mathbf{a}' + \boldsymbol{\Omega} \times (\boldsymbol{\Omega} \times \mathbf{r}) + 2 \boldsymbol{\Omega} \times \mathbf{v}', introducing terms that correspond to the centrifugal and Coriolis effects.[1] In the rotating frame, Newton's second law becomes m \mathbf{a}' = \mathbf{F} - m \boldsymbol{\Omega} \times (\boldsymbol{\Omega} \times \mathbf{r}) - 2 m \boldsymbol{\Omega} \times \mathbf{v}', where the additional terms act as fictitious forces.[1] The centrifugal force, -m \boldsymbol{\Omega} \times (\boldsymbol{\Omega} \times \mathbf{r}), points radially outward from the axis of rotation and depends on the distance from that axis, contributing to effects like the equatorial bulge of rotating planets.[2] The Coriolis force, -2m \boldsymbol{\Omega} \times \mathbf{v}', is perpendicular to both the angular velocity and the object's velocity in the rotating frame, causing deflections such as the rotation of hurricanes (counterclockwise in the Northern Hemisphere).[2] These forces, though not real interactions, enable accurate predictions of motion within the rotating frame and are crucial in fields like atmospheric science, oceanography, and celestial mechanics.[2]Fundamentals of Reference Frames
Inertial versus Non-Inertial Frames
In classical mechanics, an inertial reference frame is defined as one in which the motion of a body not subject to external forces is rectilinear and uniform, allowing Newton's laws of motion to hold without modification.[3] This frame moves at constant velocity relative to the fixed stars, providing a standard against which other motions can be measured, as the distant stars serve as an approximate inertial backdrop due to their vast separation and minimal relative acceleration.[4] In such frames, the first law of motion states that an object at rest remains at rest, and an object in motion continues in a straight line at constant speed unless acted upon by a net force.[5] Non-inertial reference frames, by contrast, undergo acceleration relative to an inertial frame, either linearly or rotationally, causing Newton's laws to appear invalid without additional corrections.[6] Examples include a car accelerating forward, where objects inside seem to press backward against the seats, or a merry-go-round rotating steadily, where riders experience an outward tendency.[5] In these frames, observers perceive apparent or fictitious forces that account for the observed deviations from inertial motion, effectively restoring the form of Newton's second law by including these pseudo-forces as if they were real interactions.[7] The concept of inertial frames traces back to Isaac Newton's Philosophiæ Naturalis Principia Mathematica (1687), where he introduced absolute space as an unchanging, immovable backdrop independent of external relations, serving as the foundation for true motion and the validity of his laws.[8] Newton argued that relative motions alone could not distinguish absolute rest from uniform motion, but absolute space provided the necessary inertial structure, with the fixed stars offering a practical reference for identifying such frames.[9] This framework resolved earlier debates on motion by positing that only in absolute space do free particles follow straight paths without forces.[10] A qualitative illustration of non-inertial effects occurs in an accelerating elevator: a ball released from rest relative to the elevator appears to accelerate downward faster than gravity alone would predict in an inertial frame, as if an additional backward force acts on it to maintain the form of Newton's laws within the elevator's perspective.[11] Rotating reference frames represent a specific subclass of non-inertial frames, where the acceleration arises from angular velocity, leading to curved paths for free particles as observed from the rotating system.[12]Defining a Rotating Reference Frame
A rotating reference frame is a non-inertial coordinate system that rotates with a constant angular velocity vector \vec{\omega} relative to an inertial reference frame.[13][14] This setup typically employs Cartesian coordinates with the origin located at the axis of rotation, where the z-axis of the rotating frame aligns with the direction of \vec{\omega}.[13] The rotation is assumed to be rigid, meaning all points in the frame maintain fixed relative positions as the entire system rotates uniformly.[1] The angular velocity vector \vec{\omega} specifies both the magnitude \omega (in radians per second) and the direction of the rotation axis, with the direction determined by the right-hand rule: curling the fingers of the right hand in the direction of rotation points the thumb along \vec{\omega}./9:_Rotational_Kinematics_Angular_Momentum_and_Energy/9.7:_Vector_Nature_of_Rotational_Kinematics)[15] A key assumption is that the rotation is steady, with constant \vec{\omega} in both magnitude and direction, which simplifies the kinematic description.[14] Conceptually, consider a point fixed in the rotating frame; in the inertial frame, this point traces a circular path centered on the rotation axis, with the radius equal to its perpendicular distance from the axis and the period of motion given by $2\pi / \omega. This circular trajectory illustrates the relative motion between the frames.[1] A practical example is the Earth, which serves as an approximately rotating reference frame for local observations, with \vec{\omega} directed along its north-south axis and magnitude \omega \approx 7.29 \times 10^{-5} rad/s.[16][14] Such frames, being non-inertial, require the introduction of fictitious forces to describe dynamics accurately.[14]Fictitious Forces in Rotating Frames
Centrifugal Force
In a rotating reference frame, the centrifugal force is a fictitious force acting on an object of mass m, expressed in vector form as \vec{F}_\text{cent} = -m \vec{\omega} \times (\vec{\omega} \times \vec{r}), where \vec{\omega} is the angular velocity vector of the frame and \vec{r} is the position vector of the object relative to the origin on the rotation axis.[14] This expression simplifies using the vector triple product identity, yielding a force directed radially outward from the axis of rotation, perpendicular to \vec{\omega}.[17] The magnitude of the force is F_\text{cent} = m \omega^2 \rho, where \rho = |\vec{r}_\perp| is the perpendicular distance from the rotation axis, highlighting its dependence solely on position and the frame's rotation rate, independent of the object's velocity in the rotating frame.[18] The centrifugal force originates from the inertial tendency of objects to maintain straight-line motion in an inertial frame, which, when observed from the accelerating rotating frame, appears as an outward deflection away from the axis.[14] This apparent force ensures consistency with the conservation of angular momentum: as an object moves relative to the rotating frame, its path curves outward to preserve the angular momentum it holds in the inertial frame, mimicking an expansive push.[19] Unlike real forces from interactions, the centrifugal force has no physical source but emerges purely from the non-inertial nature of the frame. A practical illustration occurs on Earth, where rotation produces a centrifugal force at the equator of approximately $0.034 \, \text{m/s}^2, reducing the effective gravitational acceleration from $9.832 \, \text{m/s}^2 (polar value) to about $9.780 \, \text{m/s}^2, a decrease of roughly 0.3% in apparent weight.[20] This effect diminishes toward the poles, where \rho = 0 and the force vanishes. Although the centrifugal force is non-conservative in general rotating frames where \vec{\omega} may vary, for cases of constant angular velocity, it derives from a scalar potential V_\text{cent} = -\frac{1}{2} m \omega^2 \rho^2, allowing integration along paths independent of velocity and confirming its conservative character under steady rotation./29%3A_Non-Inertial_Frame_and_Coriolis_Effect/29.02%3A_Uniformly_Rotating_Frame) This potential facilitates analysis of equilibrium and motion in systems like rotating fluids or planetary atmospheres.Coriolis Force
The Coriolis force is a velocity-dependent fictitious force that arises in a rotating reference frame with constant angular velocity. It acts on an object of mass m with velocity \vec{v} relative to the rotating frame and is mathematically expressed as -2m \vec{\omega} \times \vec{v}, where \vec{\omega} is the angular velocity vector of the frame.[21] This force is always directed perpendicular to both \vec{v} and \vec{\omega}, resulting in a deflection of the object's path without altering its speed in the rotating frame./12%3A_Non-inertial_Reference_Frames/12.08%3A_Coriolis_Force) The magnitude of the force is given by $2m \omega v \sin\theta, where \theta is the angle between \vec{v} and \vec{\omega}./03%3A_The_Coriolis_Force) In physical terms, the Coriolis force causes a deflection of moving objects that appears to the right of their velocity vector in the Northern Hemisphere due to Earth's rotation and to the left in the Southern Hemisphere.[22] This deflection is a consequence of the frame's rotation and is most pronounced for motions perpendicular to the rotation axis. Because the force is perpendicular to the velocity, it performs no work on the object, thereby conserving kinetic energy while redirecting momentum and altering the trajectory.[21] The effect vanishes at the equator, where \sin\theta = 0 for horizontal motions, and maximizes at the poles./03%3A_The_Coriolis_Force) A classic example is the eastward deflection of objects in free fall on Earth, where the Coriolis force causes the path to deviate from vertical due to the planet's rotation; for a drop from rest at mid-latitudes, this eastward shift can be on the order of centimeters for heights of tens of meters.[23] Another demonstration is the Foucault pendulum, whose plane of oscillation precesses at a rate \Omega = \omega \sin\phi, with \phi denoting the latitude, visibly rotating once per day at the poles and slower elsewhere.[24] This precession rate directly reflects the local vertical component of Earth's angular velocity.[25] The Coriolis force is named after Gaspard-Gustave de Coriolis, who first quantified it in 1835 while analyzing the energy transfer in rotating machinery such as waterwheels, introducing the necessary corrections to Newton's laws for such systems.[26]Euler Force
The Euler force is a fictitious force that arises in a rotating reference frame when the angular velocity \boldsymbol{\omega} is not constant, specifically due to the angular acceleration \dot{\boldsymbol{\omega}}. It is defined as \mathbf{F}_E = -m \dot{\boldsymbol{\omega}} \times \mathbf{r}, where m is the mass of the object, \dot{\boldsymbol{\omega}} is the time derivative of the angular velocity vector, and \mathbf{r} is the position vector from the rotation axis to the object.[27][28] This force accounts for the effects of torque-induced changes in the frame's rotation rate, appearing only when \dot{\boldsymbol{\omega}} \neq 0. In the inertial frame, a real torque \boldsymbol{\tau} produces \dot{\boldsymbol{\omega}} = \boldsymbol{\tau}/I for a rigid body with moment of inertia I; in the rotating frame, this manifests as the Euler force, which adjusts Newton's second law to include this term for apparent equilibrium.[29] The magnitude of the Euler force is m |\dot{\boldsymbol{\omega}}| r \sin \theta, where \theta is the angle between \dot{\boldsymbol{\omega}} and \mathbf{r}, and its direction is perpendicular to both \dot{\boldsymbol{\omega}} and \mathbf{r}, following the right-hand rule for the cross product.[27] A representative example occurs in a spinning top slowing due to friction, where \dot{\boldsymbol{\omega}} points opposite to the spin axis, producing a tangential Euler force that contributes to the top's wobbling or loss of stability as the rotation rate decreases.[29] The Euler force is typically absent in steady-state analyses of constant-\boldsymbol{\omega} rotations but is essential for scenarios involving variable rotation, such as the precession of planetary axes, where the gradual change in \boldsymbol{\omega}'s direction (e.g., Earth's axial precession over 26,000 years) introduces a small but nonzero \dot{\boldsymbol{\omega}}, influencing long-term orbital dynamics.[27] In such cases, the force's scale is often negligible compared to gravitational effects but provides critical insight into non-steady rotational motion.[29]Mathematical Relations Between Frames
Position and Coordinate Transformations
In a rotating reference frame undergoing pure rotation with constant angular velocity \vec{\omega} relative to an inertial frame, the origins of both frames are assumed to coincide, with no relative translation between them. The position vector \vec{r} of any point is identical as a physical entity in both frames, but its representation in coordinates differs due to the orientation of the basis vectors. The components in the inertial frame \vec{r} are related to those in the rotating frame \vec{r}' by the time-dependent rotation matrix R(t), such that \vec{r} = R(t) \vec{r}', where the rotation angle is \theta = \omega t for rotation about a fixed axis.[30] The rotation matrix R(t) is orthogonal, satisfying R^T(t) R(t) = I, which preserves vector lengths and angles under the transformation. Consequently, the inverse relation is \vec{r}' = R^{-1}(t) \vec{r} = R^T(t) \vec{r}, allowing coordinates in the rotating frame to be obtained directly from inertial coordinates.[30] For rotation about the z-axis with angular speed \omega, the explicit coordinate transformations are given by \begin{align*} x' &= x \cos(\omega t) - y \sin(\omega t), \\ y' &= x \sin(\omega t) + y \cos(\omega t), \\ z' &= z. \end{align*} These equations express the rotating-frame coordinates (x', y', z') in terms of the inertial-frame coordinates (x, y, z).[30] A representative example is a point fixed at (x, y, z) = (a, 0, 0) in the inertial frame. In the rotating frame, this point appears to move in a circle of radius a around the z-axis, with coordinates x' = a \cos(\omega t), y' = a \sin(\omega t), z' = 0.[30] This apparent circular motion illustrates how the rotation of the frame induces perceived movement for stationary points. The position mapping forms the basis for deriving velocity transformations via time differentiation.[30]Time Derivatives and Angular Velocity
In a rotating reference frame, the time derivative of a vector \mathbf{A} differs from that in an inertial frame due to the frame's rotation. The fundamental relation is given by \left( \frac{d\mathbf{A}}{dt} \right)_{\text{inertial}} = \left( \frac{d\mathbf{A}}{dt} \right)_{\text{rot}} + \boldsymbol{\omega} \times \mathbf{A}, where \left( \frac{d\mathbf{A}}{dt} \right)_{\text{rot}} is the derivative as measured in the rotating frame, and \boldsymbol{\omega} is the angular velocity vector of the rotating frame relative to the inertial frame.[18][31] This equation, known as the transport theorem, applies to any vector quantity and accounts for the additional rotational contribution.[31] The angular velocity \boldsymbol{\omega} is a vector directed along the axis of rotation, with magnitude equal to the instantaneous rate of rotation d\theta/dt. In three dimensions, it has components \omega_x, \omega_y, and \omega_z along the respective axes of the inertial frame. For a frame rotating with constant angular speed about a fixed axis, \boldsymbol{\omega} remains constant in both magnitude and direction.[18][1] To derive the transport theorem, consider the position vector \mathbf{r} of a point, which transforms between frames via a time-dependent rotation matrix \mathbf{R}(t), such that \mathbf{r}_{\text{inertial}} = \mathbf{R}(t) \mathbf{r}_{\text{rot}}. Differentiating with respect to time using the chain rule yields \left( \frac{d\mathbf{r}}{dt} \right)_{\text{inertial}} = \dot{\mathbf{R}} \mathbf{r}_{\text{rot}} + \mathbf{R} \left( \frac{d\mathbf{r}}{dt} \right)_{\text{rot}}. The term \dot{\mathbf{R}} \mathbf{R}^{-1} is the skew-symmetric matrix representation of \boldsymbol{\omega}, leading to the cross-product form \dot{\mathbf{R}} \mathbf{r}_{\text{rot}} = \boldsymbol{\omega} \times \mathbf{r}_{\text{rot}}, and thus the general relation for any vector.[1][31] For scalar quantities, which lack directional dependence, the time derivative is independent of the frame: \frac{d\phi}{dt}_{\text{inertial}} = \frac{d\phi}{dt}_{\text{rot}}. This follows directly from the absence of a vector cross-product term.[18] The relation also applies to the unit basis vectors \hat{\mathbf{e}}' of the rotating frame, which evolve as \frac{d\hat{\mathbf{e}}'}{dt}_{\text{inertial}} = \boldsymbol{\omega} \times \hat{\mathbf{e}}'. In the rotating frame itself, these basis vectors appear fixed, so \left( \frac{d\hat{\mathbf{e}}'}{dt} \right)_{\text{rot}} = 0, highlighting the rotational contribution.[1][31]Velocity Transformations
In a rotating reference frame with angular velocity \vec{\omega} relative to an inertial frame, the velocity of a particle transforms according to the relation \vec{v} = \vec{v}' + \vec{\omega} \times \vec{r}, where \vec{v} is the velocity in the inertial frame, \vec{v}' is the velocity as measured in the rotating frame, and \vec{r} is the position vector from the common origin of the frames (assuming the origins coincide and are fixed).[14][32] This transformation arises from the general rule for time derivatives in rotating frames: the inertial derivative of a vector \vec{A} is \left( \frac{d\vec{A}}{dt} \right)_{\text{inertial}} = \left( \frac{d\vec{A}}{dt} \right)_{\text{rotating}} + \vec{\omega} \times \vec{A}.[14] Applying this to the position vector \vec{r} yields \vec{v} = \frac{d\vec{r}}{dt}_{\text{inertial}} = \frac{d\vec{r}}{dt}_{\text{rotating}} + \vec{\omega} \times \vec{r} = \vec{v}' + \vec{\omega} \times \vec{r}, since \vec{v}' = \frac{d\vec{r}}{dt}_{\text{rotating}}.[14][32] The term \vec{\omega} \times \vec{r} represents the additional velocity imparted by the rotation of the frame itself, which is perpendicular to both \vec{\omega} and \vec{r} due to the cross-product operation, and has magnitude \omega r \sin\theta, where \theta is the angle between \vec{\omega} and \vec{r}.[14] This perpendicularity ensures that the rotational contribution does not alter the radial component of velocity but adds a tangential component. For steady rotation about a fixed axis, the transformation can be expressed in components aligned with the rotating frame's coordinates, though the vector form suffices for most analyses.[32] Consider a particle at rest in the rotating frame, so \vec{v}' = 0; in the inertial frame, its velocity is then purely \vec{\omega} \times \vec{r}, describing uniform circular motion around the rotation axis with speed \omega r \sin\theta.[14] This example illustrates how observers in the rotating frame might perceive the particle as stationary, while inertial observers see the motion induced by the frame's rotation. The velocities \vec{v} and \vec{v}' are vectors with units of length per time, and the cross product maintains dimensional consistency.[32]Acceleration Transformations
The acceleration of a particle in an inertial frame \mathbf{a}_I relates to its acceleration in a rotating frame \mathbf{a}_R through the transformation \mathbf{a}_I = \mathbf{a}_R + 2 \boldsymbol{\omega} \times \mathbf{v}_R + \boldsymbol{\omega} \times (\boldsymbol{\omega} \times \mathbf{r}) + \dot{\boldsymbol{\omega}} \times \mathbf{r}, where \mathbf{v}_R is the velocity relative to the rotating frame, \mathbf{r} is the position vector from the rotation origin, \boldsymbol{\omega} is the angular velocity of the rotating frame relative to the inertial frame, and \dot{\boldsymbol{\omega}} is the angular acceleration of the rotating frame.[33] This equation arises in classical mechanics for analyzing motion across frames rotating with respect to each other.[14] In this relation, \mathbf{a}_R represents the acceleration as measured by an observer in the rotating frame, while the additional terms account for effects induced by the frame's rotation and any change in its angular velocity.[33] These frame-induced terms modify the apparent dynamics without altering the underlying physics in the inertial frame. To derive this transformation, begin with the velocity relation between the frames: \mathbf{v}_I = \mathbf{v}_R + \boldsymbol{\omega} \times \mathbf{r}.[33] The time derivative of any vector \mathbf{C} in the inertial frame follows the rule \left( \frac{d\mathbf{C}}{dt} \right)_I = \left( \frac{d\mathbf{C}}{dt} \right)_R + \boldsymbol{\omega} \times \mathbf{C}, known as the transport theorem or Coriolis theorem.[14] Apply this rule to differentiate \mathbf{v}_I: \mathbf{a}_I = \left( \frac{d\mathbf{v}_I}{dt} \right)_I = \left( \frac{d\mathbf{v}_I}{dt} \right)_R + \boldsymbol{\omega} \times \mathbf{v}_I. Substitute \mathbf{v}_I = \mathbf{v}_R + \boldsymbol{\omega} \times \mathbf{r} into the rotating-frame derivative: \left( \frac{d\mathbf{v}_I}{dt} \right)_R = \left( \frac{d}{dt} \right)_R (\mathbf{v}_R + \boldsymbol{\omega} \times \mathbf{r}) = \mathbf{a}_R + \dot{\boldsymbol{\omega}} \times \mathbf{r} + \boldsymbol{\omega} \times \mathbf{v}_R, assuming differentiation acts component-wise in the rotating frame.[33] Now substitute back and expand \boldsymbol{\omega} \times \mathbf{v}_I = \boldsymbol{\omega} \times \mathbf{v}_R + \boldsymbol{\omega} \times (\boldsymbol{\omega} \times \mathbf{r}), yielding the full acceleration relation after collecting terms.[14] This second application of the derivative rule introduces the cross-product terms absent in the first derivative (velocity transformation). The key terms in the transformation are the Coriolis-like acceleration $2 \boldsymbol{\omega} \times \mathbf{v}_R, which depends on the velocity in the rotating frame; the centrifugal-like term \boldsymbol{\omega} \times (\boldsymbol{\omega} \times \mathbf{r}), directed radially outward from the axis of rotation; and the Euler-like term \dot{\boldsymbol{\omega}} \times \mathbf{r}, arising from changes in the angular velocity.[33] These terms highlight how rotation distorts acceleration measurements between frames. For example, consider a particle undergoing linear motion with constant velocity \mathbf{v}_R in the rotating frame, so \mathbf{a}_R = 0. In the inertial frame, the path curves due to the nonzero frame-induced accelerations $2 \boldsymbol{\omega} \times \mathbf{v}_R + \boldsymbol{\omega} \times (\boldsymbol{\omega} \times \mathbf{r}) + \dot{\boldsymbol{\omega}} \times \mathbf{r}, assuming \dot{\boldsymbol{\omega}} = 0 for steady rotation.[14] This illustrates the kinematic coupling between frames.Dynamics in Rotating Frames
Newton's Second Law Adaptation
In an inertial reference frame, Newton's second law states that the net real force acting on a particle of mass m is equal to the mass times the acceleration observed in that frame:\mathbf{F}_\text{real} = m \mathbf{a}_\text{inertial}. [14][34] When transitioning to a non-inertial reference frame rotating with angular velocity \boldsymbol{\omega} relative to the inertial frame, the observed acceleration \mathbf{a}_\text{rot} in the rotating frame requires modification to the law to account for the frame's motion. The adapted form incorporates additional terms representing fictitious forces:
\mathbf{F}_\text{real} - m \boldsymbol{\omega} \times (\boldsymbol{\omega} \times \mathbf{r}) - 2 m \boldsymbol{\omega} \times \mathbf{v}_\text{rot} - m \frac{d \boldsymbol{\omega}}{dt} \times \mathbf{r} = m \mathbf{a}_\text{rot},
where \mathbf{r} is the position vector from the rotation origin, \mathbf{v}_\text{rot} is the velocity relative to the rotating frame, and d \boldsymbol{\omega}/dt is the angular acceleration of the frame.[14][28][34] This equation interprets the dynamics such that the real forces, augmented by the fictitious centrifugal, Coriolis, and Euler forces, balance the mass times the acceleration measured in the rotating frame. The fictitious terms arise solely from the kinematics of the frame transformation and vanish in inertial frames, allowing Newton's second law to be applied as if the rotating frame were inertial.[14][28][34] The adaptation assumes a classical mechanical framework with no relativistic effects, treating the rotation as rigid and the masses as point particles without quantum considerations. It holds for moderate angular velocities where special relativity is negligible.[14][34] A representative example is the analysis of planetary motion in a frame rotating with the angular velocity of the planet's orbit around the Sun, centered on the Sun. For a circular orbit, the planet appears stationary in this frame (\mathbf{a}_\text{rot} = 0, \mathbf{v}_\text{rot} = 0), so the gravitational force balances the centrifugal term: \mathbf{F}_\text{grav} = m \boldsymbol{\omega} \times (\boldsymbol{\omega} \times \mathbf{r}), yielding GMm / r^2 = m \omega^2 r and Kepler's third law relation \omega^2 = GM / r^3.[34]