Linear Algebra Proof

1. May 14, 2008

cscott

1. The problem statement, all variables and given/known data

A = A(t), B = constant

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

3. The attempt at a solution

In Einstein notation I get

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

Is this right so far? Product rule on A and A-dot from here?

Last edited: May 14, 2008
2. May 14, 2008

Dick

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. May 14, 2008

cscott

I think I got it:

After using product rule I get

$$A \cdot ( \ddot{A} \times B) + \dot{A} \cdot ( \dot{A} \times B)$$

Where the second term would be 0

So the second term must expand as (A . A) x ( A . B) ?

4. May 14, 2008

cscott

Yeah the RHS time derivative shouldn't be there.

5. May 14, 2008

Dick

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. May 14, 2008

cscott

Ah. This way just seemed quicker but I see where it makes no sense.