Last Modified 17 December 1999

The ``Static'' Tight-Binding Program: Example X

Orthorhombic strain to find C11-C12

Theory

Skip to the computational details

So far we've looked at the energy of the hcp Titanium lattice as a function of a and c (or V and c/a), keeping the full P63/mmc symmetry of the hcp lattice. This allowed us to over-determine the elastic constants

However, there are five independent elastic constants for a hexagonal crystal. Remaining are This page is devoted to C11 - C12, we'll do C44 later.

Just as in Example 4, finding C11 - C12 requires varying e1 and e2. Like Example 4, we can do this strain and keep volume conservation. Unlike Example 4, we cannot choose a strain such that E(-x) = E(x). Looking at the pre-defined strains table we see that static as options ``21'' and ``-21'' which handle this. We'll use option 21:

e1 = [(1+x)/(1-x)]½ - 1
e2 = [(1-x)/(1+x)]½ - 1      (1)
e3-6 =0

The strained hcp lattice then has the form

a1 = ( ½ a (1+e1), - ½ 3½ a (1+e2), 0)
a2 = ( ½ a (1+e1), - ½ 3½ a (1+e2), 0)   .       (2)
a3 = ( 0 , 0 , c )

Since C22 = C11 in an hcp lattice, the energy-strain relationship is thus

E(x) = E0 + V0 (C11-C12) x2 + O[x3]   .     (3)

The procedure should now be familiar. We change x, construct a new lattice (2), calculate the energy, repeat, and fit the results to (3). Unfortunately, one little problem emerges from all of this: (2) is not a hexagonal lattice. Instead, it is now a base-centered orthorhombic lattice. If we plug in the equilibrium positions of the basis vectors,

B1 = 1/3 a1 + 2/3 a2 + 1/4 a3 = ( ½ (1+e1) a ,   ½ 3-½ (1+e2) , ¼ c )
B2 = 2/3 a1 + 1/3 a2 + 3/4 a3 = ( ½ (1+e1) a , -½ 3-½ (1+e2) , ¾ c )   ,     (4)
then we find that the lattice has symmetry Cmcm, Lattice #63 in the International Tables, with the atoms located on the 4m sites. In this regard the lattice now looks much like the B33 (CrB) lattice, except that only the Chromium sites are occupied.

What makes life difficult is that atoms on the 4m sites are not fixed in space. The Crystallographic tables say that the atoms have positions (0,±y,±¼), where y is free to change. In terms of the basis vectors (4) this translates into

B1 = (1/3 - u) a1 + (2/3 + u) a2 + 1/4 a3 = ( ½ (1+e1) a ,   ½ 3-½ (1+e2) (1+2u), ¼ c )
B2 = 2/3 a1 + 1/3 a2 + 3/4 a3 = ( ½ (1+e1) a , -½ 3-½ (1+e2) (1+2u), ¾ c )   ,     (5)
where u is a free parameter. Note that when e1=e2=0, we only get back to the hcp lattice when we also have u=0.

In order to find the elastic constant C11-C12, then, we must first find the minimum energy E(x,u) at fixed x, and use this value as the E(x) in (3). This is very reminiscent of the minimization procedure required to obtain the equation of state for hcp Titanium. We'll pursue this in all its computational glory next.

Go on to the computational details.

Go back to the Example X home page.

Look at other examples.

Get other parameters from the Tight-binding periodic table.

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