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I am watching a youtube video regarding time derivatives in a rotating reference frame. The stationary reference frame is S and the rotating one is S'.

Unfortunately I don't understand why the red underlined formula is valid. Is it because there is no external force acting?

enter image description here

Because with two reference systems, where one is stationary and the other moves with a constant velocity, it is the case, if a force acts on a test particle, that the force is the same in both reference systems, but not the measured velocity.

Qmechanic
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Lambda
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1 Answers1

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The vector $\vec{A}$ has different components depending on which direction vectors you want to use.

In the example they declare the components of $\vec{A}$ in terms of the frame S' as $$\vec{A} =\pmatrix{A_x' \\ A_y' \\ A_z'}$$

such that $$\vec{A} = A_x' \hat{i}' +A_y' \hat{j}' + A_z' \hat{k}' \tag{1}$$

This is done because $\vec{A}$ is a vector defined in S' and riding along the moving frame. But the components are measured from S. This is a bit confusing with this nomenclature.

Now let's see how the components vary with time. We call this

$$\left. \frac{{\rm d}\vec{A}}{{\rm d}t} \right|_S \tag{2}$$

because we are sitting on S measuring the components of $\vec{A}$

Now the first line just substitutes (1) into (2).

$$\left. \frac{{\rm d}\vec{A}}{{\rm d}t} \right|_S = \left. \frac{{\rm d}}{{\rm d}t} \right|_S \left( A_x' \hat{i}' +A_y' \hat{j}' + A_z' \hat{k}' \right) \tag{3}$$

John Alexiou
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