Coriolis force

For the psychophysical perception, see Coriolis effect.

(PD) Image: John R. Brews
Wind driven by a pressure gradient is deflected by the Coriolis force, most at the poles and changing sign when crossing the equator.

The Coriolis force is experienced by an object traversing a path in a rotating framework, a force that is proportional to its speed, to the speed of rotation, and also to the sine of the angle between its direction of movement and the axis of rotation.

‘	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

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.

Although sometimes referred to as an apparent force, it can have very real effects, among them the counterclockwise movement of winds around a low pressure area in the northern hemisphere (clockwise in the southern hemisphere).

History

The mathematical expression for the Coriolis force appeared in an 1835 paper Mémoire sur les équations du mouvement relatif des systèmes de corps by the 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.

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that is, in any frame of reference that is in straight-line motion at constant speed relative to the "fixed stars", an historical reference taken today to refer to the entire universe.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 introduction of inertial forces may sound like an accounting trick used for mathematical convenience, but the appearance of these inertial forces in Newton's laws has the very real meaning that these forces are experienced in the rotating frames just as certainly as forces arising from sources like masses, charges, or atomic nuclei.

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.

Rotating sphere

See also the discussion at Foucault pendulum.

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

This example considers the Coriolis force seen by observers on the surface of a rotating sphere. This analysis underlies many observations in meteorology (discussed next) and is useful in understanding how Foucault's pendulum can be used to determine the rate of rotation of the Earth. .

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. This coordinate system sometimes is called the tangent-plane coordinate system. 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:

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

Consequently, the Coriolis acceleration a_Cor becomes:

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

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:

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.

Meteorology

(PD) Image: John R. Brews
In the northern hemisphere, Coriolis force deflects winds flowing into a low pressure region to form a counterclockwise flow.

A pressure gradient in the atmosphere is equalized by driving a wind along the pressure gradient. However, from the viewpoint of meteorology, the motion is observed most naturally within the reference frame of a rotating Earth. As pointed out just above and in the figure in the Introduction, in the Earth's reference frame any motion is affected by a Coriolis force, as given by:

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

where f = 2ω sinφ is the Coriolis parameter. Consequently, the wind takes an apparent direction deflected to the right of the pressure gradient in the northern latitudes, and to the left in the southern latitudes. The Coriolis parameter is a maximum at the poles, zero at the equator and flips sign as one crosses from the northern to southern hemispheres, as indicated in the figure.

One consequence is that a low pressure region that sucks air into it, as shown in the figure on the right, results in a counterclockwise flow of winds in the northern hemisphere (called a cyclone) because of the Coriolis force. The direction is opposite in the southern hemisphere.

It is sometimes suggested that the draining of a bathtub is another example of cyclone motion, but it appears that in this example there are conflicting forces (introduced by the shape of tub near the drain, and the initial conditions of flow) that can overpower the Coriolis force and dictate the direction the vortex spins.

Coriolis flowmeter

(PD) Image: John R. Brews
A fluid forced through a rocking tube experiences a Coriolis acceleration.

The Coriolis force is used to measure mass flows in fluid systems. As shown in the figure, the fluid flows into a U-shaped tube with velocity v and exits with equal but oppositely directed velocity. The tube is rocked, and at a particular moment appears to be rotating about a line joining its openings, as shown. From the viewpoint of the tube, the fluid experiences a Coriolis acceleration normal to the plane of the tube, which has opposite signs on the two legs of the tube. An element of fluid with mass m experiences a Coriolis force:

$\mathbf {F_{Cor}} =2m\mathbf {v\times \Omega } \ ,$

where Ω is the vector angular rate of rotation as found using the right-hand rule and v is the fluid velocity. Idealizing the tube as having right-angle corners at its base, the Coriolis force is the same along the entire length of each leg of the tube, but opposite in sign on each leg, and is zero at the base where the velocity is parallel to the rotation Ω.

As a result of the Coriolis force, the tube is subject to a force couple twisting it about its vertical axis of symmetry. By detecting this torque, the speed of mass flow in the tube can be measured. For example, the torque can be measured by observing the angle of twist of the tube which, like a spring, resists turning due to its mechanical properties.

From the stationary observer's view, an element of fluid moving along the outward leg is increasing its angular velocity at a rate proportional to its speed, because it is moving farther from the axis of rotation. This changing angular momentum requires an acceleration perpendicular to the fluid's path. On the return leg, the element of fluid is losing angular velocity, and so requires an acceleration in the opposite direction to the outward leg. Thus, the torque upon the tube is a consequence of the changing angular momentum of the fluid flipping sign between the two legs.

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. Arrows show the same distance is seen at the moment of closest approach of the ball to the center.

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

(PD) Image: John R. Brews
Path of ball for four rates of rotation. Catcher positioned so the catch is made at 12 o'clock in all cases according to the stationary observer.

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 (orange smiley face) must anticipate just where the catcher will be and tosses the ball toward that future position (green smiley face), 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 (orange smiley face) and the catcher (blue smiley face) are standing at fixed locations, as all three participants turn along with the carousel. Although no-one is moving, 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. The intuitive reaction is that the ball is pushed to the right during flight, and so this force must be countered by throwing the ball somewhat to the left. From a physics standpoint, to the rotating observer the curved path means the Coriolis force must be introduced into Newton's laws, because these laws require a force to cause the ball to curve.

The center figure illustrates that to the rotating observer the ball executes an approximately circular motion about its center of curvature. The slower the rate of rotation, the further the center of curvature of the path is displaced from the center of the carousel, and the greater the radius of curvature. Also, the Coriolis force (causing the rightward deflection of the ball) diminishes. As the rate of rotation slows to zero, the path straightens as the radius of curvature becomes infinite and the center of curvature recedes to infinity, the forces drop to zero, and the path becomes a simple straight-line toss directly toward the catcher.

The lower figure shows more complex paths are possible. Here, for the stationary observer, the ball is thrown from 6 o'clock and caught at 12 o'clock in all cases, but the carousel turns faster and faster, forcing the catcher to start from a greater and greater angle from 12 o'clock in order to arrive in time for the catch. Four cases are shown, where the catcher starts 1/4 turn, then 1/3 turn, 1/2 turn and 3/4 turn from 12 o'clock. The path in the stationary frame always is the same straight line, but the curved path seen by the rotating observer becomes quite complicated.

From the spacing of the points marked on the figures, 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, a 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.

A more detailed discussion follows.

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. The figure shows the calculated inertial forces upon the ball at a particular moment. The motion is nearly circular about a displaced center, but the speed of travel around the circular path varies. 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.

Below are the mathematical details.

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 path from 6 o'clock at angle θ. The position of the ball in the stationary frame at time t is:

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

with i_A, j_A 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 frame.

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

$\mathbf {i_{A}} =\mathbf {i_{B}} \ \cos \Omega t-\mathbf {j_{B}} \ \sin \Omega t\$

$\mathbf {j_{A}} =\mathbf {i_{B}} \ \sin \Omega t+\mathbf {j_{B}} \ \cos \Omega t\ ,$

which, substituted for the unit vectors, lead to the trajectory seen on the carousel as (components are in the i_B, j_B 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 tosser stands on the ground instead of upon the carousel, the original point of toss is at 6 o'clock. 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 the tosser. The distance from the tosser to the ball at any particular time is the same in both frames, and is independent of the rate of rotation, as shown in the figure by the green arrows emanating from the tosser:

$\left|\mathbf {r_{B}} (t)+\mathbf {R} (t)\right|=vt\ .$

The "centripetal" component of acceleration seen in the rotating frame 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.