Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

equation 1:

$-i\omega\underline \xi - 2i(\underline k \cdot \underline \Omega )\underline u = 0$

with:

$\underline \xi = i \underline k \times \underline u$

$\underline k \cdot \underline u = 0$

taking:

$i\underline k \times \text{(1)}$

derive:

$\omega = \pm 2\frac{(\underline k \cdot \underline \Omega )}{|\underline k|}$

I can get pretty close and can post my working if required - but I don't do anything complicated - any help = many thanks

share|improve this question
    
Can you write a couple of sentences describing what you want to do or to know? –  Mariano Suárez-Alvarez Feb 7 '11 at 20:31
    
Derive the bottom equation by taking ik cross product with equation 1 –  user6757 Feb 7 '11 at 20:51

1 Answer 1

You have $$\underline 0=i\underline k\times(-i\omega\underline \xi - 2i(\underline k \cdot \underline \Omega )\underline u))=-i^2\omega(\underline k\times \underline \xi)-2i^2(\underline k \cdot \underline \Omega)(\underline k\times\underline u)$$

But by the triple cross product expansion formula

$$\underline k\times \underline \xi=\underline k\times (i\underline k\times \underline u)=i|k|^2\underline u$$ since $\underline u\cdot\underline k=0$.

Thus the first equation can be restated as

$$-i^3\omega |k|^2\underline u=2i^2(\underline k \cdot \underline \Omega)(\underline k\times\underline u)$$

Taking norms, you will get what you want.

Remember that $|\underline k\times\underline u|=|\underline k||\underline u|$ because $\underline u\cdot\underline k=0$ (see this).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.