# Starting point for a derivation of fictitious forces

I came across this expression at the start of a derivation of fictitious forces:

$$(dA/dt)_L = (dA/dt)_R + \omega \times A$$

Where the $L$ subscript refers to the laboratory (inertial) reference frame and the $R$ subscript is the rotating frame. $\omega$ is the angular velocity and $A$ is a vector.

This equation makes intuitive sense to me, the change in the laboratory frame has to be some combination of the change in the rotating frame as well as being augmented by the angular velocity. However, I would like to see an actual derivation of this.

• Just do the time-dependent transformation into the rotating frame, the $\omega\times A$ should appear by applying the chain rule. Commented Mar 23, 2015 at 20:51
• Commented Mar 23, 2015 at 21:53

## 1 Answer

You haven't given enough information: What you are asking for is that missing bit. I think I can guess just what that is: If $\vec{A}$ is a position vector (e.g. points from the origin to a point that is represented in both the inertial and the non-inertial frame of reference) and your non-inertial frame of reference rotates at a constant angular velocity $\left| \vec{\omega} \right|$ around an axis $\vec{\omega}$ through the origin, it all happens to make sense because then $-\vec{\omega} \times \vec{A}$ is the difference in the time-derivative $\frac{d}{dt} \left( \vec{A}_R - \vec{A}_L \right)$ where I use the subscript L and R the way you did.

To make this obvious, it would be neat to go to the previous (zeroth?) step, writing an equation relating $\vec{A}_L$ and $\vec{A}_R$ before taking the time-derivative. You probably have to do it yourself, because there is more than one way to express it, and to see what is happening, you should chose a representation that you are sufficiently at ease with to do the vectorial differentiation.