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Reciprocal Lattice Simple Question

by TheDestroyer
Tags: lattice, reciprocal, simple
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Apr15-06, 04:04 AM
P: 397
If we are studying FCC in the direct lattice, Why does the length of the cube side in the reciprocal lattice equal to 4*Pi/a Where a is the lattice constant,

a*=|G|=2*Pi/a Sqrt(4) = 4*Pi/a

Where a* is the length of the cube site in reciprocal lattice
Note: this thing is repeated in 2 problems and i wouldn't be able to know the reason.

Prefessor is writing it like this but i can't understand (LOL, he also doesn't know to answer me when i asked him, he's just reading from papers, Silliy Professors)

Any one can explain? thanks
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Apr15-06, 04:54 AM
P: 576
the factor 4 is usually left there to make it clear that fccs reciprocal lattice is the bcc lattice with a lattice constant of 4pi/a.
Apr15-06, 06:53 AM
P: 397
Can you explain it in mathemtical way? cuz this explanation is refused when it's said like that!!


Apr15-06, 07:50 AM
P: 576
Reciprocal Lattice Simple Question

allright. let's look at the primitive reciprocal lattice vector b1 of the fcc lattice. the a vectors are the direct space primitive lattice vectors.

[tex]b_1=2\pi \frac{a_2 \times a_3}{a_1 \cdot (a_2 \times a_3}[/tex]

Just plug in the fcc vectors and do the cross products and you'll get

[tex]b_1=\frac{4\pi}{a} 1/2(y+z-x)[/tex]

which is the a1 for bcc with a lattice constant of 4pi/a.
Apr17-06, 04:31 PM
P: 397
OK! Why did you put 1/2?

I know the reciprocal of fcc is 2Pi/a (-x+y+z)

Why did you multiply and device by 1/2????


Apr17-06, 05:00 PM
P: 576
The half is there just to show the connection between fcc and bcc in direct and reciprocal space.
Apr17-06, 07:41 PM
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Gokul43201's Avatar
P: 11,155
The basis vectors of an FCC in a symmetric form are :

[tex]a_1=\frac{a}{2}(\hat{x} + \hat{y}) [/tex]

[tex]a_2=\frac{a}{2}(\hat{y} + \hat{z}) [/tex]

[tex]a_3=\frac{a}{2}(\hat{z} + \hat{x}) [/tex]

If you plug these into the equation provided by inha in post#4 for the reciprocal lattice vectors, you get :

[tex]b_1=\frac{2\pi}{a}(\hat{x} + \hat{y} - \hat{z}) [/tex]

[tex]b_2=\frac{2\pi}{a}(\hat{y} + \hat{z} - \hat{x}) [/tex]

[tex]b_3=\frac{2\pi}{a}(\hat{z} + \hat{x} - \hat{y}) [/tex]

(also, as posted by inha in post #4)

The trick, next, is to recall that the basis vectors for a BCC, in symmetric form are :

[tex]a_1=\frac{a'}{2}(\hat{x} + \hat{y} - \hat{z}) [/tex]

[tex]a_2=\frac{a'}{2}(\hat{y} + \hat{z} - \hat{x}) [/tex]

[tex]a_3=\frac{a'}{2}(\hat{z} + \hat{x} - \hat{y}) [/tex]

where a' is the BCC lattice parameter (or cube edge).

Since these have the same form as the reciprocal vectors of the FCC, we understand that the reciprocal lattice of the FCC is in fact, a BCC.

Secondly, comparing coefficients, we find that :

[tex]\frac{2\pi}{a} = \frac{a'}{2} [/tex]

[tex]\implies a' = \frac{4\pi}{a} [/tex]
Apr18-06, 12:06 PM
P: 397
Thanks, I got it
Nov1-11, 03:17 AM
P: 1
Quote Quote by TheDestroyer View Post
Thanks, I got it
I think it is quite easy. By normally calculate we can get b=4*pi/a for fcc and bcc

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