Consider the equation $\vec{f}=f\hat{z}$. Here, $f$ denotes the Coriolis parameter $f=2\Omega \sin(\phi)$ and $\hat{z}$ denotes the vertical unit vector.

How can we think of $\vec{f}$ physically? I don't understand this concept.

  • $\begingroup$ Btw: Coriolis force is a bit of a misnomer. It should actually be called "Coriolis acceleration". Because the force that is felt depends on the mass of the object in question, whereas the acceleration of the object due to the Coriolis force is independent of its mass. $\endgroup$ Mar 1, 2020 at 10:13
  • $\begingroup$ @cmaster - reinstate monica I don’t think that’s a very good reason to not call something a force. After all, gravity has the same property. $\endgroup$ Mar 20, 2020 at 19:21
  • $\begingroup$ @BobKnighton I've been taught to call $g$ the gravitational acceleration (the term in my mother tongue is a bit different, but it also distinctly calls it an acceleration). You go from the gravitational constant $G$ to the gravitational field, which defines a gravitational acceleration $g$ for every point in space, by taking the field generating mass(es) into account. And you go from the gravitational acceleration to the gravitational force only by also taking the mass of the accelerated object into account. $\endgroup$ Mar 20, 2020 at 19:56

6 Answers 6


enter image description here

1) Find one of these at your local playground

2) Ignore the sign saying "Children only" and get on it, near the edge

3) Have some friends get it going, as fast as they can

4) Swing your arm to and fro in the radial direction, towards/away from the red post

5) The force your arm feels, pushing it sideways as you try to swing radially, is the coriolis force.

  • $\begingroup$ In a more global sense, does $\vec{f}$ point locally up in the Northern Hemisphere and locally down in the Southern Hemisphere? $\endgroup$
    – Steven
    Feb 28, 2020 at 8:54
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    $\begingroup$ In addition you can bring a plumb bob, to observe how the plumb bob responds to motion with respect to the rotating platform. Have your helpers maintain a constant angular velocity. Note the angle of the plumb bob wire at your current angular velocity. Move inward: while you are moving inward the bob is pulling ahead. Move outward, the bob is lagging behind. Move in such a way that the bob is circumnavigating faster than the platform itself, the bob swings wide. Move in such a way that the bob is circumnavigating slower than the platform itself; the bob slumps down. All Coriolis effect. $\endgroup$
    – Cleonis
    Feb 28, 2020 at 10:07
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    $\begingroup$ Another trick is to take a ball with you. When the ride is spinning (not too fast) focus your attention on the platform until you are unaware of the external world and "forget" you are spinning. Now try to roll the ball across the platform in what would be a straight line in your reference frame. To your amazement, a mysterious force will pull the ball to the side! $\endgroup$ Feb 28, 2020 at 10:47
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    $\begingroup$ Pendulums and rolling balls are good... but actually feeling the force on your swinging arm will teach you more about coriolis force than any textbook $\endgroup$ Feb 28, 2020 at 11:23
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    $\begingroup$ Fabri says that acceleration forces are not 'real'. Einstein says that acceleration and gravity are indistinguishable (the equivalence principle). Hopefully we agree that gravity is real. $\endgroup$ Mar 1, 2020 at 9:10

A great resource for this question is provided by MIT's Open Courseware. The notation used in the following is obtained by combining the good ways joined from the books by Zipfel and Etkin.

The Coriolis theorem relates time derivatives of a vector $\vec{s}$ which varies with time as computed using two distinct reference frames, $R$ and $M$, as $\frac{d \vec{s}}{dt} \big|_R = \frac{d \vec{s}}{dt} \big|_M + \vec{\omega}^{MR} \times \vec{s}$. The first term in the this equation is the time derivative or rate of change of the vector as measured using the reference frame $R$, while the second is the time rate of change of the vector as measured using the reference frame $M$ and $\vec{\omega}^{MR}$ is the angular velocity of the reference frame $M$ with respect to the reference frame $R$.

Consider a specific case of kinematics to which we apply this formula. If $\vec{s}$ is the displacement of a particle, $R$ is the inertial frame and $M$ is a frame rotating (but not translating) at a uniform angular velocity with respect to it, then we have obtained a relationship between the velocities of the particle as measured by using the two reference frames.

Now, if we differentiate the left hand once more with respect to time, using the reference frame $R$ and apply the Newton's second law of motion to the material particle of mass $m$, we obtain $\vec{F} = m \frac{d^2 \vec{s}}{dt^2} \big|_R$ where the left hand side is the total external force acting on the particle. Similarly differentiating the right hand side and multiplying by the mass, after a few calculations we obtain (recalling that the angular velocity is assumed constant) the terms $m \frac{d^2 \vec{s}}{dt^2} \big|_M + 2 m \vec{\omega}^{MR} \times \vec{s} + m \vec{\omega}^{MR} \times (\vec{\omega}^{MR} \times \vec{s})$. Therefore, \begin{equation} \vec{F} = m \frac{d^2 \vec{s}}{dt^2} \big|_R = m \frac{d^2 \vec{s}}{dt^2} \big|_M + 2 m \vec{\omega}^{MR} \times \vec{s} + m \vec{\omega}^{MR} \times (\vec{\omega}^{MR} \times \vec{s}). \tag{1} \label{eqn} \end{equation} While the first term on the right hand side is the acceleration of the particle as measured using the reference frame $M$, the last two terms are called the Coriolis and centrifugal terms repectively, by convention. The term named after Coriolis has a form identical to the corresponding term in the Coriolis theorem.

Note that we have derived the last two terms as a consequence of kinematics (book keeping for accounting for measurements using different reference frames) and are fictitious forces, referred to as pseudo, non-Newtonian or inertial forces in the literature. The latter names are used in order to indicate that we have used the non-inertial or non-Newtonian reference frame $M$ to measure the acceleration and consequently, applying Newton's second law is going to be trickier, and will involve subtracting the pseudo centrifugal and Coriolis forces from the real forces (these are distinguishable from the non-inertial ones since they arise out of the interaction of the mass particle with other objects in space).

In other words, non-Newtonian or inertial forces (such as Coriolis) are not physical effects caused on the dynamics of the material particle due to it's interaction with other material particles. These are artificially considered to be forces, in order to be able to write Newton's second law for the particle, \begin{equation} \vec{F} - m \left( 2 \vec{\omega}^{MR} \times \vec{s} + \vec{\omega}^{MR} \times (\vec{\omega}^{MR} \times \vec{s}) \right) = m \frac{d^2 \vec{s}}{dt^2} \big|_M, \tag{2} \end{equation} despite using a non-Newtonian reference frame.

Consider, for instance the case when the left hand side in the equation $\eqref{eqn}$ vanishes, which is reasonable when we place the point mass in an isolated place away from any other material particles. Even in this case, the Coriolis and centrifugal forces are not, in general, negligible.

We present an example application to explain the result. Let us treat the calculation of the acceleration of a particle $P$ using the reference frame $R$ in terms of time derivatives calculated using the reference frame $M$ which moves with respect to the reference frame $R$. Let $M$ have an angular velocity of $\omega^{MR}$ with respect to $R$ and let $\vec{s}_{MR}$ denote the displacement of a reference point on $M$ (typically the origin of the associated coordinate system) with respect to that on $R$. Let us change our notation $\frac{d^R}{dt} := \frac{d}{dt}|_R$ for convenience and denote the displacement of a point $P$ from any point in a reference frame $R$ by $\vec{s}_{PR}$ and so that the velocity of the particle $P$ calculated using a reference frame $R$ is denoted $\vec{v}_P^R := \frac{d^R}{dt}\vec{s}_{PR}$ (the general rule for notation is to denote points in subscripts and rigid bodies and reference frames in superscripts).

Then, on applying the Coriolis theorem we obtain $$\vec{v}_P^R = \frac{d^R}{dt}\vec{s}_{PR} = \frac{d^R}{dt} (\vec{s}_{MR} + \vec{s}_{PM}) \\ = \vec{v}_M^R + \frac{d^M}{dt} \vec{s}_{PM} + \vec{\omega}^{MR} \times \vec{s}_{PM} \\ = \vec{v}_M^R + \vec{v}_P^M + \vec{\omega}^{MR} \times \vec{s}_{PM}.$$ In order to obtain the acceleration we apply the Coriolis theorem again to get $$\vec{a}_P^R := \frac{d^R}{dt} \vec{v}_P^R = \frac{d^R}{dt} (\frac{d^R}{dt} \vec{s}_{PR}) \\ = \frac{d^R}{dt} (\vec{v}_M^R + \frac{d^M}{dt} \vec{s}_{PM} + \vec{\omega}^{MR} \times \vec{s}_{PM}) \\ = \vec{a}_M^R + \vec{a}_P^M + \frac{d^R}{dt}\vec{\omega}^{MR} \times \vec{s}_{PM} + 2 \vec{\omega}^{MR} \times \vec{v}_P^M + \vec{\omega}^{MR} \times (\vec{\omega}^{MR} \times \vec{s}_{PM})$$ where we denote $\vec{a}_M^R := \frac{d^R}{dt}\vec{v}_M^R$ and $\vec{a}_P^M := \frac{d^M}{dt}\vec{v}_P^M$. Denoting the angular acceleration of the reference frame $M$ with respect to the reference frame $R$ by $\vec{\alpha}^{MR} := \frac{d^R}{dt}\vec{\omega}^{MR}$, we then have the directly applicable result $$\vec{a}_P^R = \vec{a}_M^R + \vec{a}_P^M + \vec{\alpha}^{MR} \times \vec{s}_{PM} + 2 \vec{\omega}^{MR} \times \vec{v}_P^M + \vec{\omega}^{MR} \times (\vec{\omega}^{MR} \times \vec{s}_{PM}).$$ The Coriolis and centripetal acceleration terms on the right hand sign are easily identified.

Further, assuming that $R$ is an inertial reference frame, the translational dynamical equations of motion for the material particle $P$ are obtained as $$\vec{F} = m (\vec{a}_M^R + \vec{a}_P^M + \vec{\alpha}^{MR} \times \vec{s}_{PM} + 2 \vec{\omega}^{MR} \times \vec{v}_P^M + \vec{\omega}^{MR} \times (\vec{\omega}^{MR} \times \vec{s}_{PM})),$$ where $\vec{F}$ denotes the net external force acting on the point-mass which has mass $m$. The Coriolis and centripetal acceleration terms (multiplied by $m$) on being moved to the left hand side along with all others except $m \vec{a}_M^R$ (which is the quantity we are tasked to calculate), with the appropriate sign changes ($-2 m \vec{\omega}^{MR} \times \vec{v}_P^M$ and $- m \vec{\omega}^{MR} \times (\vec{\omega}^{MR} \times \vec{s}_{PM})$) are then referred to as the Coriolis force and centrifugal force, which are the (two of the four) non-Newtonian, pseudo or inertial forces acting on the particle.


Imagine that you are standing on a rotating platform near edge and you throw a ball directed towards a platform's center. Throwing ball de-couples it from a rotating reference frame and thus you will see that it moves not in a straight line towards center but instead in a curved line. Schematics : enter image description here

Coriolis force is defined as : $$ \boldsymbol{F} = -2m\,{\boldsymbol {\Omega }}\times {\boldsymbol {v'}} $$

Here $v'$ is ball tangential speed as is measured from the perspective of human in rotating reference frame. And $\Omega$ is a rotational vector, in this case is directed up-wards. Because Coriolis force is a cross product of rotational vector and tangential speed - it is some sort of centripetal force, directed perpendicularly to tangential ball speed vector. Btw, keep in mind that this centripetal force rotational axis will not be the same as platform's rotation axis, they are different. Minus in formula is because Coriolis force is counter-acting against human force pushing ball in rotating reference frame. So for Newton second law to be valid in a rotational reference frame - you need to include this Coriolis fictitious force in a net force calculations.


First of all, it must be noted that the Coriolis force is an inertial force (also known as pseudo force or fictious force) and is used only in a frame which rotates with respect to an inertial frame of reference. We're so used to Newton's laws of motion which are applicable only for inertial frames. In order to adapt Newton's laws of motion to non-inertial frames, we use inertial forces like centrifugal force, Coriolis force, etc. So when you're attached to an inertial frame of reference you don't need to worry about inertial forces.

Let us consider a ball rolling on a frictionless circular disk as depicted in the following animation:


Source: Coriolis force - Wikipedia

Initially, the ball is given some velocity in a radially outward direction. The top part of the animation analyses the motion from an inertial frame of reference whereas the bottom part analyses the motion from a rotating frame of reference (as if you were sitting on the disk and watching the ball roll).

In a disk rotating, the points near the periphery move with larger speeds compared to the point closer to the centre in accordance with one of the most familiar formulas:


where $v$ is the speed, $r$ is the distance of the point from the centre and $\omega$ is the angular speed. As the ball in the first animation moves down, the speed with which the floor it moves on increases progressively. Since the floor of the disk is frictionless, this will not impact the balls motion. However, this has a big effect on what is being observed by the red dot (assuming it's you).

Initially, the ball is projected towards you. But, since you're near the periphery, you move a lot faster than the ball along the axis. So the ball misses you. The motion of the ball as observed by an observer in a non-rotating frame is a simple straight line, but it's a curve as analysed by the observer in the rotating frame. This curve is due to the result of both centrifugal force and Coriolis force.

You may also verify this as suggested by RogerJBarlow, but in a slightly different way. Let's assume you're in the middle of this device:

Rotating reference frame

And let's assume your friends rotate you along with the above device in the counter-clockwise direction (same as the animation above). As you start moving towards the periphery, you'll feel a pushing force towards your right side (assuming you face the outward direction) and this is the Coriolis force. However, for your friends it will look like your head is trying to move in a straight line however it couldn't as it is attached to your legs which are (fixed due to friction) in contact with the floor of the device.

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    $\begingroup$ I made some slow-motion recordings on the playground of throwing a ball with my daughter and her friend. youtube.com/… $\endgroup$
    – user137289
    Mar 13, 2020 at 9:48

I made this animation a while ago to show the physical intuition behind the Coriolis force.

enter image description here

When you travel on a planet and we look outside from space (left of image, red path), the red dot simply slides over the planet.

However, if you measure the path in frame of reference of the rotating planet(right of image, green path) it looks like some fake force is pushing the red dot around. But it's just because the planet is rotating.

Another example

This satellite is orbiting around the earth while earth rotates around another axis independently.

enter image description here

If we map the position of the satellite on earth it looks crazy complex! But reality is very simple.

enter image description here

Coriolis force will happen on whatever rotating object, not just spheres. Guru Vishnu answer brilliantly covers a spinning disc.


One way of understanding fictitious forces that may help is to imagine that you exert the real forces required to balance the fictitious force. For example, if you swing in a circle a rock at the end of a string at constant angular velocity, the tension that you exert on the string so that the rock does in fact go in a circle (that is remains static in a reference frame centered on the circle that rotates at the constant angular velocity relative to an inertial frame of reference) must compensate exactly the centrifugal force. If the string is not present, the rock seems to fly away from an observer static in the rotating frame of reference with an acceleration given by the centrifugal term. This way of understanding centrifugal forces/acceleration is often used.

We can understand the Coriolis force in a similar manner. We can ask: what is the force that needs to be exerted on a particle that moves purely radially in a rotating frame of reference so that an observer static in the rotating frame of reference sees the particle moving at a constant speed along a radius of the rotating frame of reference? Think about a bead moving at constant speed along the seconds hand of an analog clock, or a ring that moves along the radial hand rails in the playground equipment shown in two of the other answers.

Note that we don't care here about radial forces. We exert sufficient forces radially to make sure that the speed is constant in this direction. What we ask for is what is the tangential force, in the circumferential direction? This force is needed to balance the Coriolis acceleration that we would observe if it were not exerted.

We can calculate the Coriolis force in this 2D case by observing that when the particle advances radially in the rotating frame of reference, we need to do two things (when seen from an inertial frame of reference):

1 - we need to deflect the direction of the velocity of the particle, as the radius on which the particle moves is deflected.

2 - we need to change the speed to the particle in the circumferential direction as moving along a radius will change the distance from the center of rotation. If the angular position in the rotating frame of reference must stay constant, the angular speed in the inertial frame of reference will change.

The first term is conveniently given by what we know about centrifugal acceleration. Indeed, we know that to deflect the direction of motion of a particle moving at constant speed v in a rotating frame of reference rotating with an angular speed of $\omega$, we must accelerate the particle with an acceleration $a_1 = \omega v$. Indeed, the rate of change of direction of the velocity vector in the centrifugal acceleration case is the same as in our case. The vectors are just perpendicular to each other when the two cases are compared.

For the second term, when the particle is at a distance $r$ from the center of the rotating frame of reference, its circumferential speed (as seen from the inertial frame of reference) $v_c$ is $\omega r$. If $r$ is increased by $dr$, the increase in $v_c$ is $dv_c = \omega dr$, so that the acceleration $a_2$ due to the second effect in a time $dt$ is $a_2 = dv_c/dt = \omega dr/dt = \omega v$.

Now, to get the total acceleration, we add $a_1$ and $a_2$, which we can do with a simple scalar addition as the two accelerations are in the same (circumferential) orientation, and we get the Coriolis term $2 \omega v$.

From all this, we can see that the Coriolis acceleration is the apparent acceleration of a particle moving at constant velocity in an inertial frame of reference when observed by an observer located in a rotating frame of reference rotating a speed $\omega$. The Coriolis acceleration has two components. The first one is due to the apparent change in movement direction of the particle, and the second one is due to movement away from the center of rotation, which increases the tangential speed of the particle relative to the rotating frame of reference.


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