# Understanding the Coriolis Force

1. Oct 24, 2014

### cooev769

I'm having a little trouble understanding why the Coriolis force is proportional to the velocity vector of the object in the rotating frame. It seems to me if you had a spinning cd for example and let a ball go on it, if the ball was moving slower from the inside to the outside it would divert and go in circles way more than if the ball was going really fast from the centre to the outside in which case it would look to have diverted from its path almost not at all. Am I misunderstanding this?

2. Oct 25, 2014

### Staff: Mentor

Remember that the velocity for determining the Coriolis force is measured in the rotating frame, not the inertial frame as you described. A slow ball in the inertial frame will have a tangential velocity with a centripetal Coriolis force whereas a fast ball in the inertial frame will have a radial velocity with a tangential Coriolis force.

3. Oct 25, 2014

### A.T.

This is correct, the faster ball will have a less curved path in the rotating frame. That's because the Coriolis force is proportional to velocity, while centripetal force is proportional to velocity squared.

4. Oct 25, 2014

### vanhees71

The Coriolis and centrifugal forces appear when looking at Newton's 2nd Law in a rotating frame of reference. To derive it, it's most convenient to use Hamilton's principle. Let $\vec{x}=(x_1,x_2,x_3)$ be the coordinates of a particle with respect to Cartesian coordinates of an inertial reference frame. The Lagrangian with a force that can be derived from a scalar potential field $V(\vec{x})$, is given by
$$L=\frac{m}{2} \dot{\vec{x}}^2-V(\vec{x}),$$
and the equations of motion by the Euler-Lagrange equations
$$\frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L}{\partial \dot{\vec{x}}}=\frac{\partial L}{\partial \vec{x}},$$
i.e., with the above given Lagrangian
$$m \ddot{\vec{x}}=-\vec{\nabla} V(\vec{x})=\vec{F}(\vec{x}),$$
which is Newton's 2nd law written in an inertial frame.

Now we consider a frame of reference, which rotates with respect to the inertial frame around the $z$ axis. The Cartesian coordinates with respect to the rotating frame are $(x',y',z')$. The inertial coordinates read in terms of the rotating ones
$$x=x' \cos(\omega t) - y' \sin (\omega t), \quad y=x' \sin(\omega t) +y' \cos(\omega t), \quad z=z'.$$
Now we have to express the Lagrangian in terms of the rotating coordinates. For that we need
$$\dot{x}=\dot{x}' \cos(\omega t) - x' \omega \sin(\omega t) - \dot{y}' \sin \omega t-y' \omega \cos(\omega t),$$
$$\dot{y}=\dot{x}' \sin(\omega t) + x' \omega \cos(\omega t) +\dot{y}' \cos(\omega t) -y' \omega \sin(\omega t).$$
Now you have to evaluate the kinetic energy. It turns out that after the dust has settled you can write
$$T=\frac{m}{2} \dot{x}^2=\frac{m}{2} (\dot{\vec{x}}'{}+\vec{\omega} \times \vec{x}')^2,$$
where
$$\vec{\omega}=(0,0,\omega).$$
Evaluating the Euler-Lagrange equations in the new coordinates finally leads to
m [\ddot{\vec{x}}'+2 \vec{\omega} \times \dot{\vec{x}'} + \vec{\omega} \times (\vec{\omega} \times \vec{x}')]=\vec{F}'(\vec{x}').[/tex]
Since $\vec{F}(\vec{x})$ is a vector field, you have
$$\vec{F}'(\vec{x})'=\hat{D} \vec{F}(\vec{x})=\hat{D} \vec{F}(\hat{D}^{-1} \vec{x}'),$$
where $\hat{D}$ denotes the rotation matrix
$$\hat{D}=\begin{pmatrix} \cos(\omega t) & \sin(\omega t) &0 \\ -\sin(\omega t) & \cos \omega(t) &0 \\ 0 & 0 & 1 \end{pmatrix}.$$
Now you rewrite this equation as
$$m \ddot{\vec{x}}'=\vec{F}'(\vec{x}')-2 m \vec{\omega} \times \dot{\vec{x}}' - m \vec{\omega} \times (\vec{\omega} \times \vec{x}'.$$
You get an equation of motion which looks like Newtons 2nd Law in an inertial frame but with two additional forces, the Coriolis force and the centrifugal force. They are "inertial forces", only appearing because we are looking at the motion from the point of view of an observer at rest wrt. a non-inertial frame of reference.

5. Oct 25, 2014

### haushofer

I have the feeling these algebraic derivations are rather complicated. Just consider Newton's second law and perform the coordinate transfo x'=R(t)x where R(t) is a time dependent element of SO(3). Newton's 2nd law, having a second time derivative, then produces two extra terms in the primed system, one being proportional to the velocity. That's the Coriolis term. Of course, this doesn't directly clarify the conceptual understanding :P

6. Oct 25, 2014

### A.T.

Neither derivation for the rotating frame addresses the OP's problem, which is not specific to the Coriolis force. You have a similar issue with the Lorentz force: Increasing the velocity will increase the sideways force, but the curvature of the path will still decrease. That's because keeping the curvature constant would require a centripetal force proportional to velocity squared, while both Coriolis and Lorentz are just proportional to velocity.

7. Oct 25, 2014

### NTW

This video is from 'Die Knoff Hoff Show', with a Spanish voice-over:

8. Oct 25, 2014

### vanhees71

I don't understand what you are saying. Of course, you can directly do the two time derivatives as stated in #5. The result is the same as with the Lagrange formalism, but a bit more cumbersome for my taste.

Also, I don't see, why these derivations do not answer the OP's question. They very precisely answer them. What else than concise mathematical derivation should do that?

The Lorentz force is a completely different business. To understand it, you need relativistic classical electrodynamics and the empirical fact that the electromagnetic field is a massless vector field. From this the Lorentz force follows from the minimal-coupling assumption and, of course, is not proportional to the velocity. The correct equation of motion reads (in non-covariant form)
$$m \frac{\mathrm{d} \vec{p}}{\mathrm{d} t}=q \left (\vec{E} +\frac{1}{c} \vec{v} \times \vec{B} \right )$$
with
$$\vec{p}=\frac{m \vec{v}}{\sqrt{1-\vec{v}^2/c^2}}.$$
As you see, the Lorentz force is much more complicated to conceptually understand from first principles than the Coriolis force.