Coriolis force

For the psychophysical perception, see Coriolis effect.

The Coriolis force is a force experienced by an object traversing a path in a rotating framework that is proportional to its speed and also to the sine of the angle between its direction of movement and the axis of rotation. In vector notation, using the right-hand rule to associate a vector Ω with the rotation, the Coriolis force is F_Cor = 2m v × Ω, where × denotes the vector cross product, v is the body velocity and m is its mass. For a counterclockwise rotation of the observer's frame of reference, the Coriolis force is a push to the right when facing in the direction of the velocity. It is one of three such inertial forces that appear in an accelerating frame of reference due to the acceleration of the frame, the other two being the centrifugal force and the Euler force. The mathematical expression for the Coriolis force appeared in an 1835 paper by a French scientist Gaspard-Gustave Coriolis in connection with the theory of water wheels, and also in the tidal equations of Pierre-Simon Laplace in 1778.

Although sometimes referred to as an apparent force, it can have very real effects.

Reference frames

Newton's laws of motion are expressed for observations made in an inertial frame of reference, that is, in any frame of reference that is in straight-line motion at constant speed relative to the "fixed stars", an historical reference taken today to refer to the entire universe. However, everyday experience does not take place in such a reference frame. For example, we live upon planet Earth, which rotates about its axis (an accelerated motion), orbits the Sun (another accelerated motion), and moves with the Milky Way (still another accelerated motion).

The question then arises as to how to connect experiences in accelerating frames with Newton's laws that are not formulated for such situations. The answer lies in the introduction of inertial forces, which are forces observed in the accelerating reference frame, due to its motion, but are not forces recognized in an inertial frame. These inertial forces are included in Newton's laws of motion, and with their inclusion Newton's laws work just as they would in an inertial frame.

The Coriolis force affects the aiming of artillery pieces and plotting transoceanic air flights, for example. A few examples below illustrate how Coriolis forces work.

Carousel

See also the discussion at Crossing a carousel.

(PD) Image: John R. Brews
Tossed ball on carousel. At the center of the carousel, the path is a straight line for a stationary observer, and is an arc for a rotating observer.

(PD) Image: John R. Brews
The ball follows a nearly circular path about the center of curvature.

The rotating carousel is perhaps the most common example used to illustrate the effects of rotation upon the formulation of Newton's laws of motion and the introduction of inertial forces into these laws. A simple case is playing catch on the carousel. Standing on the rim of the carousel, a ball is tossed to another player standing at a different position on the rim.

As shown in the figure, from the viewpoint of a stationary observer, the tossed ball is a free body (apart from downward gravity) so its horizontal motion is that of a free body: a straight line at constant speed. The catcher of the ball rotates during the flight of the ball, so the tosser must anticipate just where the catcher will be and tosses the ball toward that future position, rather than directly at the catcher.

From the viewpoint of an observer at the center of the carousel but rotating with it, both the person tossing the ball and the catcher are standing at fixed locations, as all three participants turn along with the carousel. However, the tosser cannot toss the ball directly toward the catcher, because the ball veers off to the right, assuming the carousel in counter-clockwise rotation. To the rotating observer, the curved path means that Newton's laws require a force to cause the ball to curve. The intuitive reaction is that the ball is pushed to the right during flight (by the Coriolis force), and so this force must be countered by throwing the ball somewhat to the left.

From the spacing of the points marked on the figure it may be seen that the ball moves more rapidly for the rotating observer. That can be understood as an increased speed because the carousel is rotating, moving toward the ball. To put it differently, the curved path is longer than the straight one seen by the inertial observer, so the ball has to move faster to cover the curved path in the same time.

For both the rotating and the fixed observer, the distance of the ball from the center of the carousel at any moment in time is exactly the same. For example, the arrows in the figure show that the same distance is seen at the moment of closest approach of the ball to the center.

The lower figure illustrates that to the rotating observer the ball executes an approximately circular motion about its center of curvature. The center of curvature of the path is displaced from the center of the carousel by a greater amount the slower the rate of rotation, and the radius of curvature increases. The Coriolis force (causing the rightward deflection of the ball) diminishes. According to the rotating observer, the ball is subject to a centripetal force directed toward the center of curvature, as shown by the arrow. As the rate of rotation slows to zero, the centripetal force drops to zero as the center of curvature recedes to infinity, and the path is the straight-line toss to the catcher.

Inertial frame

(PD) Image: John R. Brews
Some useful notation for the ball toss on a carousel.

In the stationary frame, the path of the ball is given in x,y coordinates as:

${\begin{aligned}x&=vt\ \cos \theta \\y&=-R+vt\ \sin \theta \ ,\\\end{aligned}}$

where R is the radius of the carousel, θ is the angle of launch of the ball, v is the speed of launch, and t is the time. At t=0, the tosser is at 6 o'clock, as viewed in the figure, and θ is the angle from the horizontal to the line of throw. The catcher at this time is an angle φ from 3 o'clock. If the carousel turns at the rate Ω, then the catcher has turned from the starting position by an angle Ωt_c, placing the catcher in position provided:

$2\theta =\Omega t_{c}+\varphi +\pi /2\ ,$

where t_c is the time of the catch and φ is the starting angle of the catcher. This equation results from the lower diagram and the equality of corresponding angles of parallel lines:

$\theta =(\varphi +\Omega t_{c})+(\pi /2-\theta )\ .$

The ball arrives at the rim for the arriving catcher when the path length equals the chord of the circle:

$vt_{c}=2R\ \sin \theta \ .$

These two equations either determine the angle of launch in terms of a known speed by use of a nonlinear equation or, by assuming a time of catch, the first equation determines the angle of launch and the last equation determines the necessary speed of throw:

$v=2{\frac {R}{t_{c}}}\ \sin \left({\frac {\Omega t_{c}+\varphi +\pi /2}{2}}\right)\ .$

Rotating frame

(PD) Image: John R. Brews
The inertial accelerations on the ball combine to provide the net inward acceleration necessary for the curved path seen on the carousel. The two components of the Coriolis acceleration are shown, one radially inward toward the center of rotation and one acting to the right of the path. Notation v refers to the speed of the straight-line motion seen in the stationary frame.

The description in the rotating frame is more complex than in the inertial frame. An approximate approach to the analysis of this case is to recognize that for the rotating observer at the center of the carousel, the ball traces a very nearly circular path from the tosser to the catcher. The path has a radius, say ρ, measured from the center of curvature of the path, which is off-center from the carousel. To travel this circular path of radius ρ at a speed s requires a centripetal acceleration s²/ρ, where the speed s is set by the way the ball is tossed in combination with the rotation. The speed s along the curved path is not the same speed v as that on the straight-line path seen by the stationary observer, and s varies somewhat along the path.

The figure shows the calculated inertial forces upon the ball at a particular moment. The motion is approximately circular about a displaced center, but the radius of the path varies slightly, and the speed of travel around the circular path also varies somewhat. Consequently, the motion is only approximately uniform motion in a circular path, and the net acceleration (the sum of the two inertial accelerations, the outward centrifugal Ω²r and the Coriolis) is only approximately directed toward the center of the circle.

Mathematical details follow.

Mathematical background

Variables in the inertial (non-rotating) frame of reference are identified by subscript A, and those for the rotating frame of the carousel by subscript B.

The trajectory in the inertial frame (denoted A) is a straight line radial path at angle θ. The position of the ball in the stationary frame at time t is:

$\mathbf {r} _{A}(t)=vt\ (\cos \theta \mathbf {i_{0}} +\ \sin \theta \mathbf {j_{0}} )-R\mathbf {j_{0}} \ ,$

with i₀, j₀ the unit vectors in the x- and y-directions in the non-rotating frame, and assuming the ball is tossed from a 6 o'clock position at time t = 0. The speed v of the ball refers to the straight-line path seen in the stationary plane.

In the turntable frame (denoted B), the x- y axes rotate at angular rate Ω , and in terms of the unit vectors i, j taken to be stationary on the carousel,

$\mathbf {i_{0}} =\mathbf {i} \cos \Omega t-\mathbf {j} \sin \Omega t\$

$\mathbf {j_{0}} =\mathbf {i} \sin \Omega t-\mathbf {j} \cos \Omega t\ ,$

which, substituted for the unit vectors, lead to the trajectory seen on the carousel as (components are in the i, j directions):

$\mathbf {r} _{B}(t)$   $=\left(vt\ \cos(\theta -\Omega t)-R\sin \Omega t,\right.$$\left.{\color {white}..}vt\sin(\theta -\Omega t)-R\cos \Omega t\right)\ .$

Introducing α as:

$\alpha =\theta -\Omega t\ ,$

the velocity becomes:

$\mathbf {v_{B}} =\left(v\ \cos \alpha +vt\Omega \ \sin \alpha -\Omega R\ \cos \Omega t,\ \ v\ \sin \alpha -vt\Omega \ \cos \alpha +\Omega R\ \sin \Omega t\right)\ .$

If this motion were exactly uniform circular motion, the speed would be constant. Instead, the speed is given by:

$v_{B}^{2}=v^{2}+(\Omega R)^{2}+(\Omega vt)^{2}-2\Omega ^{2}Rvt\sin \theta -2\Omega Rv\cos \theta \ .$

To determine the components of acceleration, a general expression is used from the article inertial forces:

$\mathbf {a} _{B}=\mathbf {a} _{A}-2{\boldsymbol {\Omega }}\times \mathbf {v} _{B}-{\boldsymbol {\Omega }}\times ({\boldsymbol {\Omega }}\times \mathbf {r} _{B})-{\frac {d{\boldsymbol {\Omega }}}{dt}}\times \mathbf {r} _{B}\ ,$

in which the term in Ω × v_B is the Coriolis acceleration and the term in −Ω × ( Ω × r_B) is the centrifugal acceleration. The results are (let α = θ − Ω t):

${\boldsymbol {\Omega }}\mathbf {\times r_{B}} ={\begin{vmatrix}{\boldsymbol {i}}&{\boldsymbol {j}}&{\boldsymbol {k}}\\0&0&\Omega \\vt\cos \alpha -R\sin \Omega t&vt\sin \alpha -R\cos \Omega t&0\end{vmatrix}}\$

$=\Omega tv\left(-\sin \alpha ,\cos \alpha \right)+\left(R\cos \Omega t,\ -R\sin \Omega t\right)\ ,$

${\boldsymbol {\Omega \ \times }}\left({\boldsymbol {\Omega }}\mathbf {\times r_{B}} \right)={\begin{vmatrix}{\boldsymbol {i}}&{\boldsymbol {j}}&{\boldsymbol {k}}\\0&0&\Omega \\-\Omega tv\sin \alpha +R\cos \Omega t&\Omega tv\cos \alpha -R\sin \Omega t&0\end{vmatrix}}\ \$

producing a centrifugal acceleration:

$\mathbf {a_{\mathrm {Cfgl} }} =\Omega ^{2}\mathbf {r_{B}} (t)\ .$

Proceeding with the calculation of the Coriolis force:

${\boldsymbol {\Omega }}\mathbf {\times v_{B}} ={\begin{vmatrix}\!{\boldsymbol {i}}&\!{\boldsymbol {j}}&\!{\boldsymbol {k}}\\0&0&\Omega \\v\cos \alpha \quad &v\sin \alpha \quad &\quad \\\;+\Omega t\ v\sin \alpha -\Omega R\cos \Omega t&\;-\Omega t\ v\cos \alpha +\Omega R\sin \Omega t&0\end{vmatrix}}\ \ ,$

producing a Coriolis acceleration:

$\mathbf {a_{\mathrm {Cor} }} =-2{\boldsymbol {\Omega }}\times \mathbf {v} _{B}=2\Omega v\left(\sin \alpha ,\ -\cos \alpha \right)-2\Omega ^{2}\mathbf {r_{B}} (t)\ .$

It is seen that the last term in the Coriolis acceleration not only cancels the centrifugal acceleration, but together they provide a net "centripetal" component of acceleration, radially inward (that is, directed toward the center of rotation):

$\mathbf {a_{\mathrm {Cptl} }} =-\Omega ^{2}\mathbf {r_{B}} (t)\ .$

The other component of Coriolis acceleration given by:

$\mathbf {a_{C\perp }} =2\Omega v\left(\sin \alpha ,\ -\cos \alpha \right)\ ,$

is perpendicular to r_B (t) + R (t) with the vector R (t) = R (sinΩt, cosΩt):

$\mathbf {r_{B}} (t)+\mathbf {R} (t)=vt\ \left(\cos \alpha ,\right.$$\left.{\color {white}..}\sin \alpha \right)\ .$

If the original point of toss at 6 o'cock is marked in the inertial frame, then this position appears to rotate clockwise in the rotating frame and is connected by R (t) to the axis of rotation (dotted smiley face). The vector r_B (t) + R (t) locates the ball from this original point of toss.

The "centripetal" component of acceleration resembles that for circular motion at radius r_B, while the perpendicular component is velocity dependent, increasing with the speed v and directed to the right of the velocity. The situation could be described as a circular motion about the center of rotation combined with a sideways "Coriolis acceleration" of 2Ω v. However, this is a rough labeling: a careful designation of the true centripetal force refers to a local reference frame that employs the directions normal and tangential to the path, not coordinates referred to the axis of rotation.

These results also can be obtained directly by two time differentiations of r_B (t).

Observations

As a result of this analysis an important point appears: all the inertial forces must be included to obtain the correct trajectory. In particular, besides the Coriolis force, the centrifugal force plays an essential role. It is easy to get the impression from verbal discussions of the ball problem, which are focused on displaying the Coriolis force particularly, that the Coriolis force is the only factor that must be considered; emphatically, that is not so. A turntable for which the Coriolis force is the only factor is the parabolic turntable, a concave turntable shaped to provide the centrifugal force of a curved path through its curvature in the manner of a banked turn.

A somewhat more complex situation is the idealized example of flight routes over long distances, where the centrifugal force of the path and aeronautical lift are countered by gravitational attraction.

(PD) Image: John R. Brews
Coordinate system at latitude φ with x-axis east, y-axis north and z-axis upward (that is, radially outward from center of sphere).

Rotating sphere

Consider a location with latitude φ on a sphere that is rotating around the north-south axis. A local coordinate system is set up with the x-axis horizontally due east, the y-axis horizontally due north and the z-axis vertically upwards. The rotation vector, velocity of movement and Coriolis acceleration expressed in this local coordinate system (listing components in the order East (e), North (n) and Upward (u)) are:

${\boldsymbol {\Omega }}=\omega {\begin{pmatrix}0\\\cos \varphi \\\sin \varphi \end{pmatrix}}\ ,$     ${\boldsymbol {v}}={\begin{pmatrix}v_{e}\\v_{n}\\v_{u}\end{pmatrix}}\ ,$

${\boldsymbol {a_{Cor}}}=-2{\boldsymbol {\Omega \times v}}=2\,\omega \,{\begin{pmatrix}v_{n}\sin \varphi -v_{u}\cos \varphi \\-v_{e}\sin \varphi \\v_{e}\cos \varphi \end{pmatrix}}\ .$

When considering atmospheric or oceanic dynamics, the vertical velocity is small and the vertical component of the Coriolis acceleration is small compared to gravity. For such cases, only the horizontal (East and North) components matter. The restriction of the above to the horizontal plane is (setting v_u=0):

${\boldsymbol {v}}={\begin{pmatrix}v_{e}\\v_{n}\end{pmatrix}}\ ,$     ${\boldsymbol {a_{Cor}}}={\begin{pmatrix}v_{n}\\-v_{e}\end{pmatrix}}\ f\ ,$

where f = 2ω sinφ is called the Coriolis parameter.

By setting v_n = 0, it can be seen immediately that (for positive φ and ω) a movement due east results in an acceleration due south. Similarly, setting v_e = 0, it is seen that a movement due north results in an acceleration due east. In general, observed horizontally, looking along the direction of the movement causing the acceleration, the acceleration always is turned 90° to the right and of the same size regardless of the horizontal orientation. That is:

‘	On a merry-go-round in the night Coriolis was shaken with fright Despite how he walked 'Twas like he was stalked By some fiend always pushing him right	’
—David Morin, Eric Zaslow, E'beth Haley, John Golden, and Nathan Salwen

As a different case, consider equatorial motion setting φ = 0°. In this case, Ω is parallel to the North or n-axis, and:

${\boldsymbol {\Omega }}=\omega {\begin{pmatrix}0\\1\\0\end{pmatrix}}\ ,$     ${\boldsymbol {v}}={\begin{pmatrix}v_{e}\\v_{n}\\v_{u}\end{pmatrix}}\ ,$  ${\boldsymbol {a_{Cor}}}=-2{\boldsymbol {\Omega \times v}}=2\,\omega \,{\begin{pmatrix}-v_{u}\\0\\v_{e}\end{pmatrix}}\ .$

Accordingly, an eastward motion (that is, in the same direction as the rotation of the sphere) provides an upward acceleration known as the Eötvös effect, and an upward motion produces an acceleration due west.