[Schrödinger]

Hey guys

I have some doubts regarding solid state. I've tried to find the answers, but I haven't had much success. Some of my doubts might even be silly, so please don't get offended.

First : Why are there only 14 possible Bravais lattices? For example, I read that hexagonal crystals can only be primitive. Why can't we have face-centred or body-centred or end-centred hexagonal crystal systems? 

[cth]

Because the other lattices can be described using the 14 Bravais lattices by changing the vectors a, b and c.


Let's take an example easy to visualise: cubic C.
Cubic lattices can only be P (primitive), I (body centered) and F (face centered).

For cubic C, imagine several cubes next to one another, their faces up and down being the ones that are centered. Let's consider the plane with the centered faces, you would have a pattern looking a bit like that:

¤-------¤-------¤-------¤
|          |          |          |
|    ¤    |    ¤     |    ¤    |
|          |          |          |
¤-------¤-------¤-------¤
|          |          |          |
|    ¤    |    ¤     |    ¤    |
|          |          |          |
¤-------¤-------¤-------¤

But, you can have smaller squares without moving the points:

¤-------¤-------¤-------¤
|          |          |          |
|    ¤    |    ¤     |    ¤    |
|          | /     \  |          |
¤-------¤-------¤-------¤
|          | \     /  |          |
|    ¤    |    ¤     |    ¤    |
|          |          |          |
¤-------¤-------¤-------¤
The new squares are half the size of the previous ones. And now, you can move back to the 3D lattice. You realise that the cubic C is actually equivalent to a tetragonal P with half a unit cell volume.

[Schrödinger]

Thanks for putting all that effort to explain . But I regret to say that I haven't understood completely (or maybe anything at all) 

So, I hope you can help me again.

For cubic C, imagine several cubes next to one another, their faces up and down being the ones that are centered.

What do up and down mean here?

[cth]

Sure. When I said "faces up and down", I meant:
A cube has six faces. For a cubic C, it would have two faces centered, and four faces not centered. I just supposed that the centered faces were lying above and below the cube, and therefore the lateral faces are not centered.
Probably the choice of word "up and down" was not optimum. Replace it by "above and below".


To illustrate it, take a pack of dices. They all have their faces numbered from 1 to 6. Now, let's consider that faces #1 and 6 are centered. So, faces # 2, 3, 4 and 5 are not centered. Put the dices on a table and order them in a way that:
-face #1 is touching the table
-face #6 is facing the sky
-face #2 is touching face #5 of an adjacent dice
-face #3 is touching face #4 of an adjacent dice
Now, if you look at the many dices ordered like that, you should see a tiling of faces #6 (which are the one centered) which look like:
¤-------¤-------¤-------¤
|          |          |          |
|    ¤    |    ¤     |    ¤    |
|          |          |          |
¤-------¤-------¤-------¤
|          |          |          |
|    ¤    |    ¤     |    ¤    |
|          |          |          |
¤-------¤-------¤-------¤
in red is highlighted one of the dices.
You can add more layers on top of this one to build up the 3D lattice.

Now that you have done this, you realise that you can join up all the points using a smaller unit cell while keeping the translation symmetry condition:
¤-------¤-------¤-------¤
|          |          |          |
|    ¤    |    ¤     |    ¤    |
|          | /     \  |          |
¤-------¤-------¤-------¤
|          | \     /  |          |
|    ¤    |    ¤     |    ¤    |
|          |          |          |
¤-------¤-------¤-------¤
New cell highlighted in red.
From a cubic C, you end up with a tetragonal P.

[Schrödinger]

Sure. When I said "faces up and down", I meant:
A cube has six faces. For a cubic C, it would have two faces centered, and four faces not centered. I just supposed that the centered faces were lying above and below the cube, and therefore the lateral faces are not centered.
Probably the choice of word "up and down" was not optimum. Replace it by "above and below".

I have 2 questions :
1. By centered, do you mean that atoms are present at the centre of the face?
    i.e., if side A is centered, then does it mean that an atom is present at the centre of side A?

2. If so, then shouldn't all sides of the cube be cenetred? (fcc)
   Why do you say that 4 faces will not be centered?

[cth]

I have 2 questions :
1. By centered, do you mean that atoms are present at the centre of the face?
    i.e., if side A is centered, then does it mean that an atom is present at the centre of side A?

Yes. C centered means a node is placed at the center of the face delimited by the vectors a and b.
       A centered means a node is placed at the center of the face delimited by the vectors b and c.
       B centered means a node is placed at the center of the face delimited by the vectors a and c.
The vectors I mention are the unit cell parameters: a, b, c with the angles α, β and γ between them.

A small remark: I talk about nodes, which are not only atoms but can be as well molecules or groups of molecules.
To give you an example: the wallpaper in your kitchen. The nodes in this case probably forms a rectangular pattern. You have the group of "molecules", let's imagine in this case: a table with a flower pot on it, a chair and a cat. If you take this group and apply it at each node of the pattern, you can reconstruct the entire wallpaper.

2. If so, then shouldn't all sides of the cube be cenetred? (fcc)
   Why do you say that 4 faces will not be centered?

Yes, if you're talking about a cubic F, then all the faces are centered. And cubic F are possible, no problem.
In my previous posts, I was talking about a cubic C, and explaining why it doesn't exist because it is equivalent to a tetragonal P.
You asked why some bravais lattices, like hexagonal I or F, don't exist. And I answered by taking an easy example: cubic C.

Cubic C, if it existed, would have 2 faces centered and 4 faces not centered.

[Schrödinger]

That cleared up a lot of things. I didn't know you were using the symbol 'C' to denote the type of lattice. I thought it was just some kind of name you had given to the cube.

Now that you have done this, you realise that you can join up all the points using a smaller unit cell while keeping the translation symmetry condition:
¤-------¤-------¤-------¤
|          |          |          |
|    ¤    |    ¤     |    ¤    |
|          | /     \  |          |
¤-------¤-------¤-------¤
|          | \     /  |          |
|    ¤    |    ¤     |    ¤    |
|          |          |          |
¤-------¤-------¤-------¤
New cell highlighted in red.
From a cubic C, you end up with a tetragonal P.

I have understood that the new cube is actually tertagonal (because the side into the page is longer than the ones on the plane (the diagonals that you have drawn))

I also realise that since you have derived another lattice (tetragonal P) out of this one, we should 'invalidate' this one (cubic C), because it simply contradicts the definition of unit cell.


1. I tried to use this technique, but failed miserably. I proved that a cubic F can't exist!!! All I assumed was that the cubes you had drawn were cubic F. After drawing diagonals like you did, I ended up with a tetragonal I. What am I doing wrong?

2. What is this translational symmetry you are talking about? (I'm sorry, but I have no   
   insight into group theory) I tried wikipedia, but it was all Greek and Latin to me.

[cth]

1. I tried to use this technique, but failed miserably. I proved that a cubic F can't exist!!! All I assumed was that the cubes you had drawn were cubic F. After drawing diagonals like you did, I ended up with a tetragonal I. What am I doing wrong?

You're right with cubic F, well done. What I explained is too crude and has some problems. So, I went to talk with the crystallographer at the university about that. Here is his explanation:

Cubic C is impossible because if you have 2 faces centered and 4 faces non-centered, you break the cubic symmetry (which is 4 3-fold rotation axises passing by the corners). So, it is not a cube anymore and it falls into a lower symmetry lattice which is tetragonal. To explain it a bit differently, if you have one face centered in a cube, then all the other faces are necessarily centered as well because of the symmetry of the cube (all the faces of the cube are the same), or else it is not a cube.

The same is true with hexagonal. To have an hexagonal lattice, one requires to have a 6-fold rotation axis. Hexagonal I is impossible because adding a central node breaks the 6-fold symmetry. So, it is not hexagonal anymore. Hexagonal I is impossible. The same is true with hexagonal F: if you have centered faces, then the 6-fold symmetry is not valid anymore.

Cubic F cannot actually be considered tetragonal I, because when you do that you loose the cubic symmetry and the 3-fold rotation axises.

2. What is this translational symmetry you are talking about? (I'm sorry, but I have no   
   insight into group theory) I tried wikipedia, but it was all Greek and Latin to me.

Translational symmetry simply means: identical by translation. In crystals, when you move from one unit cell to another one, the cell content is the same. For example, consider an idealised brickwall: the bricks are identical from one to the next. And you go from one brick to another brick by translating it ("pushing" it without any rotation).

[Schrödinger]

I'm kinda okay with everything else except the n-fold rotation axes. I have checked out wikipedia on what they mean, and worked out some examples (2D) . But in 3D, I am lost. I'm not able to visualize rotating these polyhedra, with the diagonals as axes. Can you help me?

[cth]

Sure. I found on Youtube a video where a guy, after a bit of talking, rotates a cube along one of the major axes http://www.youtube.com/watch?v=mvX2nS8xJDA.

You can also take a cube at home (a die, a paper box,...) and rotate it slowly in the same way. You'll see that when you rotate it by 120 degree, the cube comes back to the same position as it was at the start. If there is no inscription on that cube, you can't tell if it has been rotated by 120 degree or not. It looks exactly the same: it is symmetrical by that 3-fold rotation axis.

[Schrödinger]

Thanks for the link, cth... it was good. The only problem was that the cube was being rotated too fast... couldn't visualize properly. Thanks anyway