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I Who can identify a Bravais Lattice?

  1. Sep 20, 2016 #1
    I have a real hard time trying to imagine why a face centered cubic cell originates a Bravais lattice. Could you try to explain it? I have also been trying to figure out if a side-centered cell is a Bravais lattice as well. This cell is a simple cubic cell with additional point at the midpoints of the lines faces of cubic cell. Any help will be really appreciated!
  2. jcsd
  3. Sep 20, 2016 #2
    No, yours is a lattice with basis. I recommend you look at Ziman or Ashcroft and Mermin. Essentially a Bravais lattice is a point lattice which in 1-d is created by repetitive application of a single 1-d vector; in 2-d it is created by repeated application of 2 independent vectors, and in 3-d by repeated application of 3 independent vectors. If you play with this, you will find that there are an exactly fixed number of Bravais lattices in 1, 2, 3, 4, ... dimensions (modulo irrelevant length scalings in lower symmetry cases).
  4. Sep 20, 2016 #3
    No, the OP is right. FCC is a proper Bravais lattice with a 1 atom basis.
    It is quite easy to see in 2D. Just draw a (face) centered 2D square lattice, then draw a square with face center on a piece of transparent foil...
  5. Sep 21, 2016 #4
    But, how did you figure out that the side centered cell is not a Bravais Lattice?
  6. Sep 21, 2016 #5
    How does that prove that the thing is a Bravais Lattice? I am trying to imagine whether or not all points are equivalent. For instance, if the point in the middle of some face can be seen as the point of a corner? Do you understand? Do you think that if i always use this method, I will be able to figure out if the lattice is a Bravais lattice or not?
  7. Sep 21, 2016 #6
    Of course the FCC lattice is Bravais. I w as referring to his lattice with edge-centered atoms.
  8. Sep 21, 2016 #7
    It takes a little group theory to really work out how many lattices are Bravais. But it is quite straightforward to show whether a given lattice is one of them. The simplest way is to create a primitive unit cell using 3 (not generally orthogonal) independent vectors that has only one atom in the cell. That atom is often made up of pieces of several atoms that are shared with neighboring cells.Y ou then create a lattice of arbitrary size by stacking the cells. For the FCC lattice the simplest cell is made by using the vectors v1 = 0.5i +0.5j +0k; v2 = 0.5i +0j +0.5k; and v3 = 0i + 0.5j + 0.5k . Here i, j, and k are the unit vectors in the x, y, and z directions respectively. Then starting with an atom at the origin (0,0,0), you can create any lattice point on the FCC lattice as a integer linear combination of v1, v2, and v3: R = N1×v1 + N2×v2 + N3×v3. Note that each of these vectors takes a corner atom to a face atom and takes a face atom to a corner atom; so you can see how to build the lattice step by step with them. Now, the cubic FCC cell can be used as a building block; but it is not primitive because it does not contain only 1 atom. Why? Well it has eight corner atoms and 6 face atoms. Each face atom is shared between 2 cells; there are 6/2 =3 face atoms per cubic cell. Each corner atom is shared among 8 cells. So there is 8/8 = 1 corner atoms. Thus each cubic cell has 4 atoms total. The cell made up of the origin and the three face atoms at the end of the three primitive vectors above, v1, v2, and v3, is in fact primitive. It forms a rhombohedral unit cell made by using two diagonally opposite corners and the using the primitive vectors to connect the two corners through the adjacent face centered locations. If you can draw three-d figures well you can use simple counting arguments to see that there is only one atom per unit cell.
    Hope this helps.
  9. Sep 21, 2016 #8
    Given that "side" is a synonym for "face" in this context, it would be Bravais. But the edge centered case is not. You can tell because you cannot generate the entire lattice using the same three independent vectors repeatedly. By the way "independent" here means that you cannot find any way to form one of the vectors as a weighted sum of the other two vectors. Try it with the bcc vectors I gave you. You will see that it is not possible.
  10. Sep 22, 2016 #9
    Even in 2D the edge-centered square is not a Bravais lattice. Move from the corner to the edge center, and one of the edge-centers moves to the middle of the square - where there is no lattice point.
    If you add that point, then you just create a simple square lattice with half the lattice parameter.
    You can draw the 3D case by using different colors for the z=0 and z=1/2 planes.
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