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Linear Algebra Proof

  1. May 14, 2008 #1
    1. The problem statement, all variables and given/known data

    A = A(t), B = constant

    [tex]\frac{d}{dt} \left[A \cdot (\dot{A} \times B) \right] = \frac{d}{dt} \left[A \cdot (\ddot{A} \times B) \right] [/tex]

    3. The attempt at a solution

    In Einstein notation I get

    [tex]LHS = \frac{d}{dt} \left [ A_i (\dot{A} \times B)_i \right] = \frac{d}{dt} \left [ \epsilon_{ilm} A_i \dot{A}_l B_m \right] = \epsilon_{ilm} B_m \frac{d}{dt} \left [A_i \dot{A}_l \right] [/tex]

    Is this right so far? Product rule on A and A-dot from here?
     
    Last edited: May 14, 2008
  2. jcsd
  3. May 14, 2008 #2

    Dick

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    I think you have an extra d/dt on the right side of what you are trying to prove. But yes, now product rule. Then what?
     
  4. May 14, 2008 #3
    I think I got it:

    After using product rule I get

    [tex]A \cdot ( \ddot{A} \times B) + \dot{A} \cdot ( \dot{A} \times B)[/tex]

    Where the second term would be 0

    So the second term must expand as (A . A) x ( A . B) ?
     
  5. May 14, 2008 #4
    Yeah the RHS time derivative shouldn't be there.
     
  6. May 14, 2008 #5

    Dick

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    Yes, but you don't expand it like that A.A and A.B are scalars. How can you cross them? In terms of vectors axb is perpendicular to a and b. In terms of your tensor expansion the product of the two A dots is symmetric, the corresponding indices of the epsilon tensor are antisymmetric. So?
     
  7. May 14, 2008 #6
    Ah. This way just seemed quicker but I see where it makes no sense.

    Thanks for your help.
     
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