Linear Algebra Proof: Proving A=A(t), B=constant

[cscott]

Homework Statement

A = A(t), B = constant

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

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?

 

[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?

 

[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) ?

 

[cscott]

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.

 

[Dick]

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) ?

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?

 

[cscott]

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

Thanks for your help.