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I understand that it is possible to express a vector $\textbf{A}$ in two different frames such that $\textbf{A}_{I} = A_{x, I}\hat{\textbf{i}}_{I} + A_{y, I}\hat{\textbf{j}}_{I} + A_{z, I}\hat{\textbf{k}}_{I}$ or $\textbf{A}_{R} = A_{x, R}\hat{\textbf{i}}_{R} + A_{y, R}\hat{\textbf{j}}_{R} + A_{z, R}\hat{\textbf{k}}_{R}$. Then differentiating both in terms of the inertial frame $I$ and applying product rule to the rotating frame gives: $\frac{d \textbf{A}_{I}}{dt}\big|_{I} = \frac{d \textbf{A}_{R}}{dt}\big|_{I} = \big(\frac{dA_{x, R}}{dt} \hat{\textbf{i}}_{R} + \frac{dA_{y, R}}{dt} \hat{\textbf{j}}_{R} + \frac{dA_{z, R}}{dt} \hat{\textbf{k}}_{R}\big) + \big( A_{x, R}\frac{d\hat{\textbf{i}}_{R}}{dt} + A_{y, R}\frac{d\hat{\textbf{j}}_{R}}{dt} + A_{z, R}\frac{d\hat{\textbf{k}}_{R}}{dt} \big) = \frac{d\textbf{A}_{R}}{dt}\big|_{R} + \mathbf{\omega}\times \textbf{A}_{R}$.

If this is then applied to a velocity vector such that $\frac{d\textbf{v}_{I}}{dt}\big|_{I} = \frac{d\textbf{v}_{R}}{dt}\big|_{R} + \mathbf{\omega}\times \textbf{v}_{R}$, how is the usual form for acceleration in the inertial frame derived from this ? (also in proofs I have read online most say that $\textbf{A}_{I} = \textbf{A}_{R}$ but why not $\textbf{A}_{I} = R \textbf{A}_{R}$ where R is the rotation between the two bases?)

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1 Answer 1

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This is a common confusion in non-inertial dynamics. $A_R$ and $A_I$ are the same vectors, just expressed in a different basis, so $A_R = A_I$. Now for your other question:

$\frac{d\textbf{v}_{R}}{dt}\big|_{I}= \frac{d^2\textbf{x}_{R}}{dt^2}\big|_{I}= \frac{d}{dt}\left(\frac{d\textbf{x}_{R}}{dt}\big|_{I}\right)\big|_{I}=\left(\frac{d}{dt}\big|_{R} + \omega \times\right)\left(\frac{d\textbf{x}_{R}}{dt}\big|_{R} + \mathbf{\omega}\times \textbf{x}_{R}\right) =\frac{d^2\textbf{x}_{R}}{dt^2}\big|_{R} + \omega \times \frac{d\textbf{x}_{R}}{dt}\big|_{R} + \omega \times (\omega \times x_R) + \frac{d}{dt}\left(\omega \times x_R\right)\big|_{R} = \frac{d^2\textbf{x}_{R}}{dt^2}\big|_{R} + 2\omega \times \frac{d\textbf{x}_{R}}{dt}\big|_{R} + \omega \times (\omega \times x_R) + \frac{d\omega}{dt}\big|_{R} \times x_R $

If the origin of the non-intertial frame is also moving with respect to the fixed frame, then we get:

$\frac{d\textbf{v}_{R}}{dt}\big|_{I}= \frac{d^2R}{dt^2}\big|_{I} +\frac{d^2\textbf{x}_{R}}{dt^2}\big|_{R} + 2\omega \times \frac{d\textbf{x}_{R}}{dt}\big|_{R} + \omega \times (\omega \times x_R) + \frac{d\omega}{dt}\big|_{R} \times x_R $,

where $\frac{d^2R}{dt^2}\big|_{I}$ is the acceleration of the origin of the non-intertial frame with respect to the fixed frame.

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  • $\begingroup$ Thank you very much for your reply! What I am still struggling however is how to reconcile (assuming the origins are shared and fixed, and $\omega$ is constant): $\frac{dv_{I}}{dt}\big|_{I} = \frac{dv_{R}}{dt}\big|_{I}= \frac{dv_{R}}{dt}\big|_{R} + \omega \times v_{R}$ using $A_{I} = v_{I} = v_{R} = A_{R}$ with the form above where, instead the transformation is applied twice as a second derivative to give instead $\frac{dv_{I}}{dt}\big|_{I} = frac{dv_{R}}{dt}\big|_{I}= \frac{dv_{R}}{dt}\big|_{R} + 2\omega\times v_{R} + \omega \times \omega \times x_{R}$? $\endgroup$
    – Hello
    May 5, 2019 at 14:15
  • $\begingroup$ I think you are assuming that $v_R= \frac{dx_R}{dt}\big|_R$, which is not the case. $\endgroup$
    – Shamaz
    May 5, 2019 at 21:53
  • $\begingroup$ Oh, I thought $\textbf{v}_{R} = v_{x, R}\textbf{i}_{R} + v_{y, R}\textbf{j}_{R} + v_{z, R}\textbf{k}_{R}$ and $\frac{d\textbf{x}_{R}}{dt}\big|_{R} = \frac{dx_{x, R}}{dt}\big|_{R}\textbf{i}_{R} + \frac{dx_{y, R}}{dt}\big|_{R}\textbf{j}_{R} + \frac{dx_{z, R}}{dt}\big|_{R} \textbf{k}_{R}$ (and then product rule also differentiates the basis vectors but these are fixed in the rotating frame giving zero). What is the form of $v_{R}$ in component form? Is $\frac{dx_{x, R}}{dt}\big|_{R} = v_{x, R}$ not the case? $\endgroup$
    – Hello
    May 5, 2019 at 22:31
  • $\begingroup$ No, $v_{x,R} = \frac{dx_{x, R}}{dt}\big|_{R} + \left(\omega \times x_R\right)_x$ $\endgroup$
    – Shamaz
    May 5, 2019 at 22:40
  • $\begingroup$ So $\textbf{v}_{R} = \frac{d\textbf{x}_{R}}{dt}\big|_{I} = \frac{d\textbf{x}_{R}}{dt}\big|_{R} + \omega \times \textbf_{x}_{R}$? What does $v_{R}$ represents physically and compared to $v_{I}$? Is velocity vector as seen by a observer fixed in the rotating frame $\frac{d\textbf{x}_{R}}{dt}\big|_{R}$? (Thank you very much for the help, this has always confused me) $\endgroup$
    – Hello
    May 5, 2019 at 22:52

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