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Non-inertial Reference Frames

  1. Jun 4, 2008 #1
    [SOLVED] Non-inertial Reference Frames

    1. The problem statement, all variables and given/known data
    State Newton's Second Law of motion in a non-inertial reference frame.

    2. Relevant equations
    In an inertial frame F=ma
    If S and S' are two reference frames, the same point is related by
    r=A(t)r'+b(t) where A is an orthogonal matrix. (Galilean Relativity).

    3. The attempt at a solution
    Let r be a point in an inertial frame S, and r' be the same point in the frame S'.
    r=A(t) r' + b(t), A orthogonal.
    Differentiate twice with respect to time
    Where a dot represents a time derivative. Now it would be handy to get this in a more familiar form, in terms of angular velocity.

    For any vector Q in an infinitesimal circular rotation
    [tex]Q=Q+w\times Q dt[/tex]
    Where x is the cross product, and w is the angular velocity. (This is easier to see with a diagram).
    So this gives us a rotation from identity: for a general rotation we can think of it as lots of infinitesimal rotations one after another
    [tex]A=(I+\frac{t\vec{w}\times}{N})^N[/tex] in the limit N approaches infinity.
    [tex]A=exp(t \vec{w} \times)[/tex]
    Ok, so let's look at the special case when the axes of S and S' are aligned at t=0:
    [tex]A(t)=I+t \vec{w}\times + \frac{t^2}{2} \vec{w}\times(\vec{w}\times)[/tex] + higher order terms
    Now we in general allow w to vary
    [tex]\dot{A(t)}=\vec{w}+t\dot{\vec{w}}\times+t \vec{w}\times(\vec{w}\times)[/tex] + higher order terms

    And here lies the problem, plugging into my original equation I get a term of the form:
    [tex]2\dot{\vec{w}}\times r'[/tex]
    Which is TWICE the angular velocity term I should have.

    Possible solutions:
    If I ignored the second term in the first time derivative of A this problem would vanish, but it would be inconsistent.
    I'm wondering if perhaps in deriving my rotation matrix ignoring the variation in w, even though it is second order, leads to an issue.

    If someone could give me a hint where I'm going wrong it would be very much appreciated.
  2. jcsd
  3. Jun 4, 2008 #2

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    You're derivation of the rotation matrix and its time derivative is the problem.

    Here is how I prefer to look at things.

    Some nomenclature first:
    • [tex]\mathbf q[/tex] A vector, sans any concept of representation.

    • [tex]\mathbf q_F[/tex] The vector [tex]\mathbf q[/tex] as expressed in reference frame F.

    • [tex]\frac{d\mathbf q}{dt_F}[/tex] The derivative of some vector quantity q as seen by an observer attached to (i.e., not moving in) reference frame F.
    • [tex]\frac{d}{dt_F}[/tex] The above frame-dependent derivative written in operator form.

    With this notation, the derivative of some vector quantity q from the perspective of an inertial frame I and a rotating frame R are related via

    [tex]\frac{d\mathbf q}{dt_I} = \frac{d\mathbf q}{dt_R} + \mathbf {\omega}\times \mathbf q[/tex]

    where [itex]\mathbf {\omega}[/itex] is the rotation rate of frame R with respect to frame I. In operator form, this becomes

    [tex]\frac{d}{dt_I} = \frac{d}{dt_R} + \mathbf {\omega}\times[/tex]

    Back to this vector quantity q. The frame-dependent representations of the vector are related via

    [tex]\mathbf q_I = \mathbf A_{R\to I} \mathbf q_R[/tex]

    Differentiating with respect to time,

    [tex]{\boldmath {\dot q}}_I =
    {\boldmath {\dot A}}_{R\to I} \mathbf q_R +
    \mathbf A_{R\to I} {\boldmath {\dot q}}_R[/tex]

    The first form of the vector derivative addressed the frame of the observer but not the frame in which the vector is expressed. The second form simply differentiated a bunch of scalars (the individual components of a matrix and a vector are just scalars.) We need to reconcile the two forms. Differentiating the frame-dependent components of some vector yields the frame-dependent derivative of that vector. Thus the left-hand side of the first form results from calculating the derivative of q as expressed inertially while the right-hand side results from calculating the derivative in the rotating frame:

    [tex]{\boldmath {\dot q}}_I =
    \mathbf A_{R\to I}(
    {\boldmath {\dot q}}_R + \mathbf {\omega}_R\times \mathbf q_R)[/tex]

    Equating this form with the second form,

    [tex]\mathbf A_{R\to I}(
    {\boldmath {\dot q}}_R + \mathbf {\omega}_R\times \mathbf q_R) =
    {\boldmath {\dot A}}_{R\to I} \mathbf q_R +
    \mathbf A_{R\to I} {\boldmath {\dot q}}_R[/tex]

    and thus

    [tex]{\boldmath {\dot A}}_{R\to I} \mathbf q_R =
    \mathbf A_{R\to I}(\mathbf {\omega}_R\times \mathbf q_R)[/tex]

    In operator form,

    [tex]{\boldmath {\dot A}}_{R\to I} =
    \mathbf A_{R\to I}\mathbf {\omega}_R\times[/tex]

    Taking the derivative of this operator,

    [tex]{\boldmath {\ddot A}}_{R\to I} =
    \mathbf A_{R\to I}(
    \mathbf {\omega}_R\times \mathbf {\omega}_R\times +
    {\boldmath {\dot \omega}}_R)[/tex]
    Last edited: Jun 4, 2008
  4. Jun 4, 2008 #3
    Awesome answer, thanks.

    I think I need to spend a bit of time getting my ideas straight - I think the changing unit vectors, and how to view them in the non-inertial reference frame are confusing me.

    Thanks again for your clear exposition.
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