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Mathematical Vector Derivation

From a Fixed to a Spinning View

To understand the Coriolis force mathematically, we must first distinguish between how motion is observed from a fixed (inertial) frame of reference versus a rotating one. Let's call the fixed frame S (think of it as viewing Earth from space) and the rotating frame S' (our perspective on Earth's surface). Any vector, like the position of an object r⃗\vec{r}, will have its rate of change measured differently in each frame.

The relationship between the time derivative of any vector A⃗\vec{A} in the fixed frame and the rotating frame is given by a fundamental transformation rule. This rule accounts for the rotation of the S' frame, which has an angular velocity vector Ω⃗\vec{\Omega}.

(dA⃗dt)S=(dA⃗dt)S′+Ω⃗×A⃗\left( \frac{d\vec{A}}{dt} \right)_S = \left( \frac{d\vec{A}}{dt} \right)_{S'} + \vec{\Omega} \times \vec{A}

This formula is our key. It tells us that the change we see from a fixed point in space is the sum of the change seen by an observer on the rotating platform and the change caused by the platform's own spin. Let's apply this to the position vector r⃗\vec{r} to find the velocity.

v⃗S=(dr⃗dt)S=(dr⃗dt)S′+Ω⃗×r⃗=v⃗S′+Ω⃗×r⃗\vec{v}_S = \left( \frac{d\vec{r}}{dt} \right)_S = \left( \frac{d\vec{r}}{dt} \right)_{S'} + \vec{\Omega} \times \vec{r} = \vec{v}_{S'} + \vec{\Omega} \times \vec{r}

Deriving Fictitious Forces

Newton's second law, F⃗=ma⃗\vec{F} = m\vec{a}, holds true in inertial frames. So, the true force F⃗true\vec{F}_{true} is equal to mass times the acceleration in the fixed frame, a⃗S\vec{a}_S. To find a⃗S\vec{a}_S, we apply our transformation rule again, this time to the velocity vector v⃗S\vec{v}_S we just found.

a⃗S=(dv⃗Sdt)S=(ddt)S′(v⃗S′+Ω⃗×r⃗)+Ω⃗×(v⃗S′+Ω⃗×r⃗)\vec{a}_S = \left( \frac{d\vec{v}_S}{dt} \right)_S = \left( \frac{d}{dt} \right)_{S'} (\vec{v}_{S'} + \vec{\Omega} \times \vec{r}) + \vec{\Omega} \times (\vec{v}_{S'} + \vec{\Omega} \times \vec{r})

Assuming Earth's rotation Ω⃗\vec{\Omega} is constant, its derivative is zero. After expanding and collecting terms, we arrive at the relationship between accelerations in the two frames.

a⃗S=a⃗S′+2(Ω⃗×v⃗S′)+Ω⃗×(Ω⃗×r⃗)\vec{a}_S = \vec{a}_{S'} + 2(\vec{\Omega} \times \vec{v}_{S'}) + \vec{\Omega} \times (\vec{\Omega} \times \vec{r})

Now, we can finally see where the fictitious forces come from. We start with Newton's second law in the inertial frame, F⃗true=ma⃗S\vec{F}_{true} = m\vec{a}_S, and substitute our expression for a⃗S\vec{a}_S.

F⃗true=ma⃗S′+2m(Ω⃗×v⃗S′)+m(Ω⃗×(Ω⃗×r⃗))\vec{F}_{true} = m\vec{a}_{S'} + 2m(\vec{\Omega} \times \vec{v}_{S'}) + m(\vec{\Omega} \times (\vec{\Omega} \times \vec{r}))

To make this equation look like Newton's second law from the perspective of the rotating frame (ma⃗S′=...m\vec{a}_{S'} = ...), we simply rearrange the terms.

ma⃗S′=F⃗true−2m(Ω⃗×v⃗S′)−m(Ω⃗×(Ω⃗×r⃗))m\vec{a}_{S'} = \vec{F}_{true} - 2m(\vec{\Omega} \times \vec{v}_{S'}) - m(\vec{\Omega} \times (\vec{\Omega} \times \vec{r}))

From this, we identify the terms being subtracted from the true force. These are the fictitious forces that must be 'invented' to make the physics work in our rotating world.

The Coriolis force is:

F⃗Coriolis=−2m(Ω⃗×v⃗S′)\vec{F}_{Coriolis} = -2m(\vec{\Omega} \times \vec{v}_{S'})

And the other term is the centrifugal force:

F⃗centrifugal=−m(Ω⃗×(Ω⃗×r⃗))\vec{F}_{centrifugal} = -m(\vec{\Omega} \times (\vec{\Omega} \times \vec{r}))

The Cross Product's Role

The vector cross product in the Coriolis formula, Ω⃗×v⃗S′\vec{\Omega} \times \vec{v}_{S'}, is crucial. It dictates that the Coriolis force is always perpendicular to both the planet's axis of rotation (the direction of Ω⃗\vec{\Omega}) and the object's direction of motion (the direction of v⃗S′\vec{v}_{S'}).

This perpendicular nature is why the Coriolis force is a deflecting force. It doesn't change the speed of an object, only its direction. For an object moving north in the Northern Hemisphere, the cross product results in a force pointing to the east, causing a rightward deflection. The opposite happens in the Southern Hemisphere.

This mathematical framework moves beyond simple analogies. It provides a robust way to calculate the precise effects of Earth's rotation on any moving object, from missile trajectories to large-scale ocean currents.

Quiz Questions 1/5

Why are the Coriolis and centrifugal forces referred to as 'fictitious forces'?

Quiz Questions 2/5

The formula for the Coriolis force is F⃗Coriolis=−2m(Ω⃗×v⃗S′)\vec{F}_{Coriolis} = -2m(\vec{\Omega} \times \vec{v}_{S'}). What does the vector cross product Ω⃗×v⃗S′\vec{\Omega} \times \vec{v}_{S'} imply about the direction of the force?