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

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

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

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

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

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

good call!
 
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