# What's the relationship between velocity components in the Coriolis effect

I still haven't understood why Coriolis parameter in $$(x,y,z)$$ plane has its velocity components described as follows: $$\frac{du}{dt} = f\cdot v$$

and $$\frac{dv}{dt} = -f\cdot u$$

Why does the rate of change of the $$u$$ velocity component depends on $$v$$ and vice-versa?

I am seriously confused about this relationship and I would appreciate if someone could clarify this for me.

• Can you define all of your variables? Oct 4, 2019 at 19:03
• Hello Aaron, In most physical oceanography introductory books I have seem these two equations demonstrating a relationship: du/dt= f v dv/dt=- f u. Where f as being the coriolis parameter and u,v being velocity vectors in x y respectively. Sorry i can provide a clear breakdown of all variables. Oct 4, 2019 at 19:15

For the equation of motion for motion relative to a rotating coordinate system the coriolis force in the equation of motion is at right angles to the current velocity.

The coriolis force in the equation of motion represents acceleration with respect to the rotating coordinate system. Since the coriolis force is at right angles to the current velocity the acceleration due to the coriolis force is at right angles to the current velocity.

The two velocity components $$u$$ and $$v$$ are defined as perpendicular to each other.

First the two simplest cases:

• velocity in the $$u$$ direction. The acceleration due to the coriolis force is at right angles to that, hence acceleration in the $$v$$ direction.

• velocity in the $$v$$ direction. The acceleration due to the coriolis force is at right angles to that, hence acceleration in the $$u$$ direction.

But of course in general the current velocity will be at some angle in between the $$u$$ direction and the $$v$$ direction. So: you decompose the current velocity in a $$u$$ component and a $$v$$ component. That gives you two acceleration components, and when you recombine those acceleration components you get an acceleration that is at right angles to the current velocity.

It may be that you are actually wondering why the coriolis force is at right angles to the current velocity. The reason for that is mathematical.

When you use a rotating coordinate system it is convenient to decompose the acceleration relative to the rotating coordinate system in the following two components:

• an acceleration that is a function of your distance to the axis of rotation (referred to as 'centrifugal force')

• an acceleation that depends on your velocity relative to the rotating coordinate system (referred to as 'coriolis force').

In the case of constant angular velocity of the rotating coordinate system those two combined are exhaustive, so you have everything covered.

When you do the above decomposition in 'centrifugal' and 'coriolis' it follows mathematically that the coriolis force comes out at right angles to the current velocity. That is not a physics thing, it's a mathematical thing.

The other 3 answers cover things pretty well, but it seems to me that one potential source of confusion is glossed over in all of them: the angular velocity vector of the earth points toward the North Star, which is not straight up unless you happen to be at the North Pole; so why does everyone seem to pretend that only the $$z$$ component of $$\Omega$$ is nonzero?

The reason is that for geophysical flows (unlike, say, ballistic trajectories) motion is constrained to the horizontal plane. What this ends up meaning is that in the analysis of these flows, you can ignore the horizontal components of the angular velocity vector of the earth—you can just use its projection onto the local vertical. Instead of treating the earth as a sphere, you pretend it’s flat, and rotating like a record player at a rate that’s proportional to the sine of the latitude—at 30 degrees N, the effective rate is once every two days.

You could get the same answers of course by using the actual angular velocity vector of the earth and including the locally vertical components of the Coriolis force, and only apply the constraint that the motion is horizontal at the end. But the math is simpler if you don’t.

I think the terms are the components of the derivative on a rotating frame.

A fixed vector $$\vec{A}$$ attached to a rotating frame with rotational velocity $$\vec{\Omega}$$ has changing components according to the law

$$\frac{{\rm d}}{{\rm d}t} \vec{A} = \vec{\Omega} \times \vec{A}$$

If the vector is changing with time, then the above is

$$\frac{{\rm d}}{{\rm d}t} \vec{A} = \frac{\partial}{\partial t} \vec{A} + \vec{\Omega} \times \vec{A}$$

When expand out the vector cross product $$\times$$ by component you will find the relationships mentioned.

$$\vec{\Omega} \times \vec{A} = \begin{vmatrix} 0 & -\Omega_z & \Omega_y \\ \Omega_z & 0 & -\Omega_x \\ -\Omega_y & \Omega_x & 0 \end{vmatrix} \pmatrix{A_x \\ A_y \\ A_z} = \pmatrix{\Omega_y A_z - \Omega_z A_y \\ \Omega_z A_x - \Omega_x A_z \\ \Omega_x A_y - \Omega_y A_x}$$

Now consider the rotating frame in the plane where $$\Omega_x =0$$ and $$\Omega_y=0$$ and $$A_z = 0$$

$$\vec{\Omega} \times \vec{A} = \pmatrix{- \Omega_z A_y \\ \Omega_z A_x \\ 0}$$

The Coriolis acceleration is:

$$\vec{a}_c=-2\,\vec{\omega}\times \vec{v}$$

$$\vec{\omega}$$ angular velocity

$$\vec{v}$$ translation velocity

with:

$$\vec{\omega}=\begin{bmatrix} 0 \\ 0 \\ \Omega \\ \end{bmatrix}$$

and

$$\vec{v}=\begin{bmatrix} U \\ V \\ 0 \\ \end{bmatrix}$$

you get

$$\vec{a}_c=\begin{bmatrix} \frac{dU}{dt} \\ \frac{dV}{dt} \\ \end{bmatrix}=\vec{\omega}\times \vec{v}=-2\begin{bmatrix} -\Omega\,V \\ \Omega\,U \\ \end{bmatrix}$$

$$\Rightarrow$$

$$\frac{dU}{dt}=2\Omega\,V\quad \surd$$ $$\frac{dV}{dt}=-2\Omega\,U\quad \surd$$