Hello Everyone,
As I'm sure we have all noticed, the exact definitions of the relationships between Bravais Lattices and Reciprocal Lattices are not exactly firmly defined. Question 3 on the assessed assignment, for example, seems to ask us to prove a definition! In this case, I can share my collected thoughts: this question is asking us to prove that the crystallography definition of Miller indices (i.e. the picture of the direct lattice axes and a plane intersecting them, according to Ashcroft and Mermin) is equivalent to the Physics definition of Miller indices (the co-efficients of the normal vector to the (hkl) plane. Confusing it all is how the plane is called the hkl plane: according to Ash.&Merm. the reason the plane is called the hkl plane is because these are the co-efficients of the normal vector that defines the plane!
Another problem I've come across is Question 1 of the assessed problems, where you need to find the reciprocal lattice of a 2D Bravais lattice. Obviously, if you take the 3rd lattice vector to be zero, you get zeroes everywhere (also known as the Australian Zero Party :~) )! I found a site suggesting that the 3rd vector should be a unit vector in the 3rd dimension, which seems to make a bit of sense to me, and the problem seems to work out when you do this. I'm not sure of the proper mathematical basis for it though (no pun intended); perhaps I should revise the parallelepipeds that were considered 'unimportant' by my maths C teacher (along with eigenvalues...)? Does anyone else know? Specifically, anyone who has done lots of maths at uni?
In conclusion, I think that Ash.&Merm. again have shown their canny textbook writing skills, by writing the most unambiguous description of the relationship between Miller indices and direct lattices I have seen (CHEM3004 was clearer, but only about the diffraction pattern relationships to Miller indices, not the direct lattice relationships. Admittedly, you use software to do that bit for you using the structure factors)!
Hopefully, this is helpful.
From down the Hill,
Josh Harbort
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I beg to differ with your comment regarding Question 3. I see it as more exploring properties and consequences of taking such a definition.
ReplyDeleteI don't quite understand your comments regarding Question 1. Why would there be a third lattice (basis) vector when we take a 2D lattice? YOu should need exactly two vectors for a 2D lattice, should you not?
I do note that A&M is becoming more readable now, thankfully!
I can't help but notice that someone wishes to introduce a little politics into this...
Whenever I get confused between bravais and reciprocal lattices, I always go back to definition of the reciprocal lattice-discrete sets of k values,K that yield plane waves with periodicity of a Bravais lattice. Its good to remember that reciprocal lattice vector,K corresponds to a family of planes in real space lattice, a Miller index (klm) and a specific x-ray diffraction peak.
ReplyDeleteAs for question 1, only 2 primitive vectors are required for 2D lattice but what Josh meant is how what are we suppose to plug as the third vector in the three equations (5.3) A&M to generate the reciprocal lattice primitive vectors. I think it treating the third vector as a unit vector is the right idea.
I see what you mean by the third component.
ReplyDeletemakes physical sense to just have the third dimension as a unit vector in the third dimension.
Picturing our 2D lattice, we could picture this as if it was itself pulled out to the third dimension, i.e. all slices along that third dimension would give us that same 2D lattice.
Sort of like taking a triangle and working with triangular prisms. All slices are the triangle that we want.
Taking the projection from looking onto, we have the triangles.
Sidenote: Reminds me of Flatland...
We went over this in the tutorial on Wednesday, I remember the one issue I was having was determining which situation is modelled correctly. As in, for the tutorial and assignment question, it states that the 2D layers are stacked neatly one atop the other, so it can be assumed that there is this neat three dimensional object. Therefore you could construct a reciprocal lattice for this crystal structure.
ReplyDeleteThe way suggested in class was to return to the original definition of the reciprocal vectors, where we essentialy have exp^(iK.R) = 1. Which suggests K.R = 2*pi*some integer. Hence the components of K can be found in a 2D case by solving the above.
The thing is, these two both give the same b1 and b2 vectors (as they should). Effectively, we can consider having a1 = ax + by, a2 = cx + dy, and a3 = ez, so that a1 and a2 lie on the same plane, with a3 perpendicular to them. Thus to find b1, we first find the triple product of a1.(a2 x a3) (which is just a number). Then b1 is proportional to a2 x a3, which being perpendicular to a3 must lie on the plane spanned by a1 and a2. Similarly b2 is perpendicular to a3 x a1, so again lies on this plane always.
Finally, it doesn't matter whether it's a unit vector chosen or not, as a3 is both on the numerator and the denominator. Hence the magnitude can be taken outside of both the vectors, leaving just uni vectors in the vector products. Hence you'll be left with the correct b1 and b2 vectors, no matter the choice of vector a3 (as long as it's perpendicular to the plane).
Anyway, that's my wishy washy proof about why they're equivalent! It's just another way of getting to the same answer though. In terms of whether a 2D or 3D reciprocal lattice would be a better answer, it was suggested by Paul that if only values for two of the dimensions are given in this case, it'd be a safer bet to give the reciprocal answer in just 2D (done by simply omitting the arbitrary b3 vector from K).
Yes, we did cover this in tutorial...quite productive!
ReplyDeleteWe have a formula generalised to 3D...convenient in that we live in 3D space, of which 1 and 2D is subset. This 3D generailsation has cross products.
The issue comes with our definition of cross product (and I suppose also scaler triple product).
We define cross products to exist only in 3 (or 7) dimensions.
Thus, to use the generalised formula for 2D problems, we need to bring our 3D grid to 3D-- in whatever way as Dale suggests.
So insteading of pretending our 2D lattice is 3D, just to find the reciprocal lattice, we could just work strictly in 2D, e.g. with the definition of exp(iK.R).
It's trippy in that we really need to think about every assumption, formula and consequence regarding everything we write. (perhaps we've forgotten to do this a bit...)
Sidenote: When do we work with 7 dimensions?