Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I have following form for wave vector $k_2=n_2 \omega/c_0$. Now because $\omega=2 \pi c/ \lambda$, then $k_2=n_2 \omega/c_0=\frac{n_2 2 \pi c_0}{c_0 \lambda}=\frac{n_2 2 \pi}{\lambda}$. But problem is that $k$ (is magnitude of wave vector~ wave number) and according to wikipedia $k=\frac{2 \pi}{\lambda}$. So where is mistake?

share|improve this question
    
Where did the $n_2$ come from in your original definition of the wave vector $k_2 = n_2w/c_0$? –  John Rennie Apr 18 '12 at 14:50
    
I don't know where it comes from but particular formula is related to nonlinear optics and according to the lecture notes $k_2=n_2 \omega/c_0$ is the wave vector of the Second harmonic generation wave. –  laovultai Apr 18 '12 at 15:00
1  
Mentioning "nonlinear optics" was very helpful here. Now we know where to begin our search. –  Pygmalion Apr 18 '12 at 15:04
    
If $\omega$ is the frequency of the fundamental then it's wave vector would be $\omega / c_0$, the wave vector of the first harmonic would be $2\omega / c_0$ because it's twice the frequency. The second harmonic would be $3\omega / c_0$ and so on. –  John Rennie Apr 18 '12 at 15:06
    
Please include reference and link for the lecture notes if possible. –  Qmechanic Apr 22 '12 at 15:30

2 Answers 2

up vote 0 down vote accepted

It looks like your mistake lies in the assumption that the various $\lambda$ are the same. It might help to write:

$$k_2 = n_2\frac{2\pi}{\lambda_0}$$

share|improve this answer

This is just a different convention. Some people use k to mean radians per meter, and $\omega $ to mean radians per second (angular wavenumber and frequency) while other people use it to mean cycles per meter or per second.

If you are reading a text that defined it in some way, then that is what it means in that text. Each author is free to define any symbols any way he likes. Wikipedia is not some sort of absolute authority on notation.

share|improve this answer

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.