What is the equivalence of the Bragg condition in vectorial form?

[thefireman]

Hello, I have a quick question regarding the bragg condition.

I know that it is most often stated as [tex]2dSin \theta=n\lambda[/tex]

But I have come across a case (Kittel chp9 pg 255, where it is written as ([tex]\vec{k}[/tex]+[tex]\vec{G}[/tex])[tex]^{2} = k[/tex]

I cannot really see how the vectorial case is the same as the simpler former one.
Could someone elucidate?

Thanks

 

[malawi_glenn]

if you know what k and G are, it is simple to show it. It is called the Laue-equation:

http://en.wikipedia.org/wiki/Laue_equation

 

[nasu]

Or look in the same book (Kittel -Chapter 2) where the equivalence of the two formulas is discussed explicitly.

 

[malawi_glenn]

Or look in the same book (Kittel -Chapter 2) where the equivalence of the two formulas is discussed explicitly.

good call!