$\left(\frac{d}{d t} \mathbf{L} \right)$ in a rotating coordinate system

Question

• We've a vector $\mathbf{A}$ lying in space and changing according to some rule.

• We introduce an inertial frame and find $\left(\frac{d}{d t} \mathbf{A} \right)_{i n}$ in it.

• We also introduce a co located frame rotating with $\mathbf{\omega}$. In this rotating frame I find $\left(\frac{d}{d t} \mathbf{A} \right)_{rot}$

• There exists a relationship between the two time rates as $\left ( \frac{d\mathbf A}{dt} \right )_\text{inertial} = \left ( \frac{d\mathbf A}{dt} \right )_\text{rot} + \boldsymbol \omega \times \mathbf A$

• In all of this derivation it was assumed that the vector $\mathbf{A}$ was independent of the coordinate system. We merely observed the vector in two frames. The vector is independent of the coordinate system.

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• Can we use the above equation on angular momentum vector $\mathbf{L}$ i.e $\left ( \frac{d\mathbf L}{dt} \right )_\text{inertial} = \left ( \frac{d\mathbf L}{dt} \right )_\text{rot} + \boldsymbol \omega \times \mathbf L$?

I think no we can't.

In the derivation of the $\left ( \frac{d\mathbf A}{dt} \right )_\text{inertial} = \left ( \frac{d\mathbf A}{dt} \right )_\text{rot} + \boldsymbol \omega \times \mathbf A$ we assumed that the vector $\mathbf{A}$ was independent of the coordinate system, its lengths and direction in space is independent of the coordinate system.

However for $\mathbf{L}$ that isn't the case. $\mathbf{L}$ has a different length in a stationary frame than in a rotating one. So the derivation doesn't apply.