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I Correction term while switching from inertial to body fixed

  1. Mar 5, 2016 #1
    Suppose we have an equation in inertial frame A.
    [tex]\begin{equation}\frac{{}^Ad\bf{H}_C}{dt} = \bf{M}_C\end{equation}[/tex]
    Now we want to switch to body fixed frame B. For this need to employ correction factor [itex]{}^A\bf\omega^B\times\bf{H}_B[/itex]. Why do we have this correction factor? How to derive this correction factor?
    [tex]\begin{equation}\frac{{}^Bd\bf{H}_C}{dt} + {}^A\bf{\omega}^B\times\bf{H}_C = \bf{M}_C\end{equation}[/tex]
     
    Last edited: Mar 5, 2016
  2. jcsd
  3. Mar 5, 2016 #2

    Simon Bridge

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    What happens if you don't use a correction factor?
    ... start with the frames you are interested in and write out the rules for converting between them.
     
  4. Mar 5, 2016 #3
    The rule to convert between frames is multiply by rotation matrix. But here we add some term, which is strange for me. I can very well understand this in case of measuring angular momentum about different points... Oooo, maybe here they think of different frames also as diferent points to measure angular momentum?
     
  5. Mar 5, 2016 #4

    vanhees71

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    Let ##\Sigma## be an inertial frame and ##\Sigma'## one, which rotates arbitrarily against ##\Sigma##. Then there's some rotation matrix ##\hat{D}(t) \in \mathrm{SO}(3)## for the components of vectors:
    $$\vec{V}=\hat{D} \vec{V}',$$
    where ##\vec{V}## are the components of an arbitrary vector wrt. the orthonormal basis at rest in ##\Sigma## and ##\vec{V}'## that wrt. to the one in ##\Sigma'##.

    For the time derivative of ##\vec{V}## you get
    $$\vec{A}:=\dot{\vec{V}}=\dot{\hat{D}} \vec{V}'+\hat{D} \dot{\vec{V}}'.$$
    For the components of ##\vec{A}## wrt. ##\Sigma'## it follows
    $$\vec{A}'=\hat{D}^{-1} \vec{A}=\hat{D}^{T} \vec{A}=\hat{D}^T \dot{\hat{D}} \vec{V}'+\dot{\vec{V}}'.$$
    Now since ##\hat{D} \in \mathrm{SO}(3)## we have
    $$\hat{D}^T \hat{D}=1 \; \Rightarrow \; \hat{D}^T \dot{\hat{D}}=-\dot{\hat{D}}^T \hat{D}=-(\hat{D}^T \dot{\hat{T}})^T.$$
    i.e., ##\hat{D}^T \dot{\hat{D}}## is antisymmetric, and thus you can introduce an axial vector ##\vec{\omega}## such that
    $$(\hat{D}^T \dot{\hat{D}})_{ij}=-\epsilon_{ijk} \omega_k.$$
    So you get
    $$A_i'=\dot{V}_i'-\epsilon_{ijk} \omega_k V_j = \dot{V}_i' + \epsilon_{ikj} \omega_k V_j'.$$
    In vector notation that reads
    $$\vec{A}''=\dot{V}'+\vec{\omega} \times \vec{V}',$$
    which proves your formula.
     
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