How Does Compression Transform FCC Lattice to BCC and Affect Miller Indices?

In summary, the conversation discussed the process of compressing a face-centered cubic (FCC) lattice to a body-centered cubic (BCC) lattice. The rules for X-ray reflection in both FCC and BCC structures were also mentioned, with the key difference being that in FCC, the indices must be all even or all odd, while in BCC, the sum of the indices must be even. Finally, the conversation touched on the effects of compression on the Miller indices, with the conclusion that while the distance between lattice points may change, the indices themselves remain the same.
  • #1
unscientific
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Homework Statement


[/B]
(a) Show Compression of FCC leads to BCC.
(b) State rules for X-rays reflection in FCC.
(c) What are the new Miller Indices after compression?

25po47k.png

Homework Equations

The Attempt at a Solution



Part(a)[/B]
I'm quite confused as to what they mean by 'principal axes'.
For an FCC lattice, the basis is given by ##[0,0,0]## and ##[\frac{1}{2},\frac{1}{2},0]## and ##[\frac{1}{2},0,\frac{1}{2}]## and ##[0,\frac{1}{2},\frac{1}{2}]##..
The primitive lattice vectors are ##\left( \frac{1}{2} \hat x + \frac{1}{2}\hat y + 0 \hat z \right)## and ##\left( \frac{1}{2} \hat x + 0 \hat y + \frac{1}{2} \hat z \right)## and ##\left( 0 \hat x + \frac{1}{2} \hat y + \frac{1}{2} \hat z \right)##.
Are the principal axes the ##x,y,z## axes?

I'm not sure how the Bain compression changes FCC to BCC:
ksy6a.png


Part(b)
Structure Factor for FCC is ##S_{(hkl)} = f_{fcc} \left[1 + (-1)^{h+k} + (-1)^{h+l} + (-1)^{k+l} \right]##. For reflection to occur, rule is (hkl) must be all even or all odd.
Structure Factor for BCC is ##S_{(hkl)} = f_{bcc} \left[1 + (-1)^{h+k+l} \right]##. For reflection to occur, rule is (h+k+l) must be even.

Part(c)
How does compression change the miller indices?
 
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  • #2
bumpp
 
  • #3
bumpp
 
  • #4
bumpp
 
  • #5
bumpp on bain compression
 
  • #6
bumpppp
 
  • #7
bumpp anyone?
 
  • #8
Couldn't find much helpful material on bain compression. Any help?
 
  • #9
bumpp on bain compression
 
  • #10
bain bumpp
 
  • #11
I am not sure on bain compression, but I have taken some condensed state physics so I might be able to help out. Note that when the lattice parameters goes from
##(a_1, a_2, a_3) \to (\frac{a_1}{\sqrt{2}}, a_2, a_3)##, then the distance between two lattice points at ##(0,0,0)## and ##(a/2,a/2,0)## turns into the distance between two lattice points at ##(0,0,0)## and ##(a/\sqrt{8},a/2,0)##, which means the two points are precicely a distance ##a\sqrt{\frac{3}{8}} <a/\sqrt{2} = a\sqrt{\frac{4}{8}}## appart. This distance is too small to be stable, as the minimal distance between lattice points prior to compression was ##a/\sqrt{2}##. Thus this lattice point must shift to ##(a\sqrt{8},a/2,a/2)## where the distance between it and its surrounding lattice points becomes the distance between ##(0,0,0)## and ##(a/\sqrt{8},a/2,a/2)##, which is a distance of ##a\sqrt{\frac{5}{8}}>a\sqrt{\frac{4}{8}}## and thus stable. This is then a BCC.

I am not sure this explanation is correct, but I felt like someone should give it a try since you have not had a responce yet and this question has been out for a while.

With regard to question (c), the miller distance has not changed, but your basis vectors in your reciprocal lattice have.
i.e. ##(a_1,a_2,a_3) \to (a_1',a_2,a_3) \Rightarrow (b_1,b_2,b_3) \to (b_1',b_2',b_3')##
Therefore, we will have different interplanar distances corresponding to miller indicies, as ##d_{(a,b,c)} = \frac{2\pi}{|\boldsymbol{G}_{(a,b,c)}|}##
 
  • #12
Brage said:
I am not sure on bain compression, but I have taken some condensed state physics so I might be able to help out. Note that when the lattice parameters goes from
##(a_1, a_2, a_3) \to (\frac{a_1}{\sqrt{2}}, a_2, a_3)##, then the distance between two lattice points at ##(0,0,0)## and ##(a/2,a/2,0)## turns into the distance between two lattice points at ##(0,0,0)## and ##(a/\sqrt{8},a/2,0)##, which means the two points are precicely a distance ##a\sqrt{\frac{3}{8}} <a/\sqrt{2} = a\sqrt{\frac{4}{8}}## appart. This distance is too small to be stable, as the minimal distance between lattice points prior to compression was ##a/\sqrt{2}##. Thus this lattice point must shift to ##(a\sqrt{8},a/2,a/2)## where the distance between it and its surrounding lattice points becomes the distance between ##(0,0,0)## and ##(a/\sqrt{8},a/2,a/2)##, which is a distance of ##a\sqrt{\frac{5}{8}}>a\sqrt{\frac{4}{8}}## and thus stable. This is then a BCC.

I am not sure this explanation is correct, but I felt like someone should give it a try since you have not had a responce yet and this question has been out for a while.

Thanks for replying. I see your point, but then wouldn't compression of ##\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \cdots## also lead to a BCC lattice as well?

Brage said:
With regard to question (c), the miller distance has not changed, but your basis vectors in your reciprocal lattice have.
i.e. ##(a_1,a_2,a_3) \to (a_1',a_2,a_3) \Rightarrow (b_1,b_2,b_3) \to (b_1',b_2',b_3')##
Therefore, we will have different interplanar distances corresponding to miller indicies, as ##d_{(a,b,c)} = \frac{2\pi}{|\boldsymbol{G}_{(a,b,c)}|}##

I can calculate the new reciprocal lattice vectors ##b_1, b_2, b_3##. How do I use that to find the new miller indices? Can i simply equate the product of old and new?

[tex]h' \left|\vec b_1'\right| = h\left|\vec b_1\right| [/tex]
[tex]k' \left|\vec b_2'\right| = k\left|\vec b_2\right| [/tex]
[tex]l' \left|\vec b _3'\right| = l\left|\vec b_3\right| [/tex]
 
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  • #13
bumpp
 
  • #14
Hey again, I don't know if compression to more than ##\frac{1}{3}## would make the bonding distances between the centered lattice point and the corner lattice points too small to be possible, but you should be able to check that using the same procedure I used above. As regard with the miller indices, if origionally you have ##\vec{G}=n_1\vec{b}_1+n_2\vec{b}_2+n_3\vec{b}_3 \to \vec{G}' =n_1\vec{b}_1'+n_2\vec{b}_2'+n_3\vec{b}_3'##. In other words literally just keep the indeces, but change the basis vectors. Then change these basis vectors to the basis vectors for your lattice (not reciprocal lattice), by ##\vec{b}_i' \cdotp \vec{a}_i =2\pi\delta_{ij}##, and find out what your translational vector ##T=m_1\vec{a}_1+m_2\vec{a}_2+m_3\vec{a}_3##. Check your results by considering ##\vec{T}\cdotp\vec{G}=2\pi Z## where Z is an integer.
 
  • #15
Brage said:
Hey again, I don't know if compression to more than ##\frac{1}{3}## would make the bonding distances between the centered lattice point and the corner lattice points too small to be possible, but you should be able to check that using the same procedure I used above. As regard with the miller indices, if origionally you have ##\vec{G}=n_1\vec{b}_1+n_2\vec{b}_2+n_3\vec{b}_3 \to \vec{G}' =n_1\vec{b}_1'+n_2\vec{b}_2'+n_3\vec{b}_3'##. In other words literally just keep the indeces, but change the basis vectors. Then change these basis vectors to the basis vectors for your lattice (not reciprocal lattice), by ##\vec{b}_i' \cdotp \vec{a}_i =2\pi\delta_{ij}##, and find out what your translational vector ##T=m_1\vec{a}_1+m_2\vec{a}_2+m_3\vec{a}_3##. Check your results by considering ##\vec{T}\cdotp\vec{G}=2\pi Z## where Z is an integer.

Thanks again for replying.

I thought ##\vec a_1, \vec a_2, \vec a_3## are lattice vectors in real space while ##\vec b_1, \vec b_2, \vec b_3## are lattice vectors in reciprocal space. The formula to get these reciprocal vectors is:

eqn004.gif


Do I chuck in the new lattice vectors after compression ##\vec a_1', \vec a_2', \vec a_3'## to get the new reciprocal vectors ##\vec b_1', \vec b_2', \vec b_3'##?
 
  • #16
unscientific said:
Thanks again for replying.
I thought ##\vec a_1, \vec a_2, \vec a_3## are lattice vectors in real space while ##\vec b_1, \vec b_2, \vec b_3## are lattice vectors in reciprocal space.
By "reciprocal space" you actually mean momentum space, as we are doing a Fourier transform of the position vectors, which turns into momentum positions. This is also often referred to as K-space, which is the same as the wave vector K is directly related to the momentum of a wave.

unscientific said:
Thanks again for replying.
Do I chuck in the new lattice vectors after compression ##\vec a_1', \vec a_2', \vec a_3'## to get the new reciprocal vectors ##\vec b_1', \vec b_2', \vec b_3'##?
Yes i believe so, and then remember the process for finding miller indices can be reversed to find the axial intersections of the planes under consideration.
After this you can just use the bragg law for diffraction to answer the last part of the question ##2\vec{k}\cdotp\vec{G}=\vec{G}^2##
 
  • #17
Brage said:
By "reciprocal space" you actually mean momentum space, as we are doing a Fourier transform of the position vectors, which turns into momentum positions. This is also often referred to as K-space, which is the same as the wave vector K is directly related to the momentum of a wave.Yes i believe so, and then remember the process for finding miller indices can be reversed to find the axial intersections of the planes under consideration.
After this you can just use the bragg law for diffraction to answer the last part of the question ##2\vec{k}\cdotp\vec{G}=\vec{G}^2##

So after compression, the structure changes into a BCC. Thus, the new miller indices are: ##(110), (200), (211), (220)##? Then the answer they are looking for is simply ##h \vec b_1' + k \vec b_2' + l \vec b_3'## using the new vectors in momentum space?

I'm still not sure about the stability argument in part (a) though, about why it becomes an BCC when the atoms become too close to one another.
 
  • #18
As far as I understand, they do not want you to calculate new miller indices, but rather use the same miller indices for the new lattices, to find out what translational vector that corresponds to.
 
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FAQ: How Does Compression Transform FCC Lattice to BCC and Affect Miller Indices?

What are Miller indices and why are they important in FCC and BCC structures?

Miller indices are a notation system used to describe the orientation of crystal planes in a crystal structure. In FCC (face-centered cubic) and BCC (body-centered cubic) structures, Miller indices are important because they allow us to identify and characterize the different crystal planes present in the structure.

How do you determine the Miller indices for a crystal plane in an FCC or BCC structure?

To determine the Miller indices for a crystal plane in an FCC or BCC structure, you need to first identify the intercepts of the plane on the three axes (a, b, and c). Then, take the reciprocals of these intercepts and reduce them to the smallest possible integers. These integers are the Miller indices for the plane.

What is the significance of a crystal plane having a Miller index of zero in FCC and BCC structures?

A crystal plane with a Miller index of zero in FCC or BCC structures means that the plane is parallel to that particular axis. This is significant because it indicates that there is no intercept on that axis, and therefore the plane extends infinitely in that direction.

What does a higher Miller index mean in FCC and BCC structures?

In FCC and BCC structures, a higher Miller index indicates that the crystal plane is closer to being parallel to the corresponding axis. For example, a plane with Miller indices (0,0,1) is parallel to the c-axis, while a plane with Miller indices (0,0,2) is even closer to being parallel to the c-axis.

How can Miller indices be used to determine the density of a crystal structure?

Miller indices can be used to determine the density of a crystal structure by calculating the distance between adjacent planes of atoms in a crystal structure. This distance, known as the interplanar spacing, can be used in the Bragg's Law equation to calculate the density of the crystal structure.

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