- #1

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* 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.