# Derivation of the centrifugal and coriolis force

I was wondering how easily these two pseudo-forces can be derived mathematically in order to exhibit a clear physical meaning.

How would you proceed?

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Good question. I'll try to have a go at this shortly. – Noldorin Nov 8 '10 at 22:50
I also just gave a slightly mathy-er version here physics.stackexchange.com/q/68002 – joshphysics Jun 15 '13 at 5:33

Okay, here is my (hopefully rigorous) demonstration of the origin of these forces here, from first principles. I've tried to be pretty clear what's happening with the maths. Bear with me, it's a bit lengthy!

### Angular velocity vector

Let us start with the principal equation defining angular velocity in three dimensions,

$$\dot{\mathbf{r}} = \mathbf{\omega} \times \mathbf{r}\; .$$

(This can be derived roughly by considering a centripetal force acting on a particle. Note that this equation applies symmetrically in inertial and rotating reference frames.)

Notice that we can in fact generalise this statement in terms of $r$ for an arbitrary vector $a$ that is known to be fixed in the rotating body.

### Transformation between inertial and rotating frames

Now consider a vector $a$, which we can write in Cartesian coordinates (fixed within the body) as

$$\mathbf{a} = a_x \mathbf{\hat i} + a_y \mathbf{\hat j} + a_z \mathbf{\hat k}\; .$$

In Newtonian mechanics, scalar quantities must be invariant for any given choice of frame, so we can say

$$\left.\frac{\mathrm da_x}{\mathrm dt}\right|_I = \left.\frac{\mathrm da_x}{\mathrm dt}\right|_R$$

where $I$ indicates the value is for the inertial frame, and $R$ that the value is for the rotating frame. Equivalent statements apply for $a_y$ and $a_z$, of course. Hence, any transformation of $a$ between frames must be due to changes in the unit vectors of the basis.

Now by the product rule,

\begin{align}\left.\frac{\mathrm d\mathbf{a}}{\mathrm dt}\right|_I &= \frac{\mathrm d}{\mathrm dt} \left( a_x \mathbf{\hat i} + a_y \mathbf{\hat j} + a_z \mathbf{\hat k} \right) \\& = \left( \frac{\mathrm da_x}{\mathrm dt} \mathbf{\hat i} + \frac{\mathrm da_y}{\mathrm dt} \mathbf{\hat j} + \frac{\mathrm da_z}{\mathrm dt} \mathbf{\hat k} \right) + \left( a_x \frac{\mathrm d\mathbf{\hat i}}{\mathrm dt} + a_y \frac{\mathrm d\mathbf{\hat j}}{\mathrm dt} + a_z \frac{\mathrm d\mathbf{\hat k}}{\mathrm dt} \right) .\end{align}

Using the previous equation for angular velocity, we then have

\begin{align}\left.\frac{\mathrm d\mathbf{a}}{\mathrm dt}\right|_I &= \left( \frac{\mathrm da_x}{\mathrm dt} \mathbf{\hat i} + \frac{\mathrm da_y}{\mathrm dt} \mathbf{\hat j} + \frac{\mathrm da_z}{\mathrm dt} \mathbf{\hat k} \right) + \left( a_x \mathbf{\omega} \times \mathbf{\hat i} + a_y \mathbf{\omega} \times \mathbf{\hat j} + a_z \mathbf{\omega} \times \mathbf{\hat k} \right) \\&= \left.\frac{\mathrm d\mathrm{a}}{\mathrm dt}\right|_R + \mathbf{\omega} \times \mathbf{a} \;.\end{align}

Now consider a position vector on the surface of a rotating body. We can write

$$\mathbf{v}_I = \left.\frac{\mathrm d\mathbf{r}}{\mathrm dt}\right|_I = \left.\frac{\mathrm d\mathbf{r}}{\mathbf dt}\right|_R + \mathbf{\omega} \times \mathbf{r} ,$$

and similarly for $\mathbf{a} = \mathbf{v}_I$,

\begin{align}\left.\frac{\mathrm d^2\mathbf{r}}{\mathrm dt^2}\right|_I &= \left( \left.\frac{\mathrm d}{\mathrm dt}\right|_R + \mathrm{\omega} \times \right)^2 \mathbf{r} \\&= \left.\frac{\mathrm d^2\mathbf{r}}{\mathrm dt^2}\right|_R + 2\mathbf{\omega} \times \left.\frac{\mathrm d\mathbf{r}}{\mathrm dt}\right|_R + \mathbf{\omega} \times (\mathbf{\omega} \times \mathbf{r}) \;.\end{align}

### Forces on body in rotating frame

Now consider a force acting on an object at position $\mathbf{r}$ (for example, gravity). Newton's third law states

$$\mathbf{F} = m \left.\frac{\mathrm d^2\mathbf{r}}{\mathrm dt^2}\right|_I .$$

And so substituting this into the previous equation for $\left.\frac{\mathrm d^2\mathbf{r}}{\mathrm dt^2}\right|_I$ and rearranging we get

\begin{align}\mathbf{F}_\textrm{net} &= m \left.\frac{\mathrm d^2\mathbf{r}}{\mathrm dt^2}\right|_R \\&= \mathbf{F} - 2m \mathbf{\omega} \times \mathbf{v}_R - m \mathbf{\omega} \times (\mathbf{\omega} \times \mathbf{r})\\ &= \mathbf{F} - 2m \mathbf{\omega} \times \mathbf{v}_R + m \mathbf{\omega}^2 \mathbf{r}.\end{align}

And here we have it. The second term on the right is the Coriolis force, and the third term is the centrifugal force (clearly pointing away from the centre of rotation). Any interpretation of the Coriolis and centrifugal forces then follow naturally from this single important equation.

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(Apologies for the long lines, unfortunately the LaTeX rendering implementation here doesn't support line breaks.) Oh, and feel free to point out any small errors or typos. It's rather a lot of maths. – Noldorin Nov 9 '10 at 0:07
+1 nice derivation ;-) You put a plus sign instead of an equals in your formula for $\vec{v}_I = d\vec{r}/dt$, and I think you're missing an exponent of 2 on the operator $\left.d/dt\right|_R+\vec\omega\times$ in the next equation. – David Z Nov 9 '10 at 2:26
@David: Thanks, you're right on both accounts. I knew a typo or two would find its way in. ;) – Noldorin Nov 9 '10 at 3:17
@Noldorin: In the expression for $\vec{a} = a_x \vec{i} + a_y \vec{j} + a_z \vec{k}$ you meant $\vec{a} = a_x \hat{i} + a_y \hat{j} + a_z \hat{k}$, right? – Robert Smith Nov 9 '10 at 6:48
@Robert: What's the difference ? – Cedric H. Nov 9 '10 at 11:29

## protected by Qmechanic♦Dec 29 '13 at 12:30

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