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fluidistic
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Homework Statement
Hi guys, I don't reach the correct answer to an exercise. I'm following Ashcroft's book.
I must find that the reciprocal of the bcc Bravais lattice is a fcc one and the reciprocal of the fcc Bravais lattice is a bcc one.
Homework Equations
If a_1, a_2 and a_3 are vectors spaning the Bravais lattice, then the reciprocal lattice is spanned by the vectors [itex]b_1= 2 \pi \frac{a_2 \times a_3}{a_1 \cdot (a_2 \times a_3)}[/itex], [itex]b_2= 2 \pi \frac{a_3 \times a_1}{a_1 \cdot (a_2 \times a_3)}[/itex] and [itex]b_3= 2 \pi \frac{a_1 \times a_2}{a_1 \cdot (a_2 \times a_3)}[/itex]
The Attempt at a Solution
For a bcc lattice: [itex]a_1=\frac{a}{2}(\hat y +\hat z-\hat x )[/itex], [itex]a_2=\frac{a}{2}(\hat z +\hat x-\hat y )[/itex] and [itex]a_3=\frac{a}{2}(\hat x +\hat y-\hat z )[/itex].
While for a fcc lattice, [itex]a_1=\frac{a}{2}(\hat y +\hat z)[/itex], [itex]a_2=\frac{a}{2}(\hat z +\hat x )[/itex] and [itex]a_3=\frac{a}{2}(\hat y +\hat x )[/itex].
I first tried to show that the reciprocal of the fcc lattice is a bcc lattice. So I just used the formulae given but didn't reach what I should have.
[itex]b_1=2\pi \frac{[ \frac{a}{2}(\hat z + \hat x )] \times [ \frac{a}{2} (\hat x + \hat y ) ]}{[\frac{a}{2} (\hat y + \hat x )] \cdot [ \frac{a}{2} (\hat z + \hat x ) \times \frac{a}{2} (\hat x + \hat y ) ] }[/itex].
I calculated [itex](\hat z + \hat x ) \times (\hat x + \hat y )[/itex] to be worth [itex]\hat y + \hat z[/itex]. So that I reached that [itex]b_1=\frac{4 \pi }{a} (\hat y + \hat z )[/itex]. This is already wrong, according to the book I should have reached [itex]b_1=\frac{4 \pi }{a} \cdot \frac{1}{2} (\hat y + \hat z -\hat x )[/itex].
I have no clue on what I've done wrong.