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Linear Algebra Proof

  • Thread starter cscott
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  • #1
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Homework Statement



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]

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:

Answers and Replies

  • #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?
 
  • #3
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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) ?
 
  • #4
782
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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?
Yeah the RHS time derivative shouldn't be there.
 
  • #5
Dick
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Homework Helper
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618
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) ?
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?
 
  • #6
782
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Ah. This way just seemed quicker but I see where it makes no sense.

Thanks for your help.
 

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