You may also like

problem icon

Ball Bearings

If a is the radius of the axle, b the radius of each ball-bearing, and c the radius of the hub, why does the number of ball bearings n determine the ratio c/a? Find a formula for c/a in terms of n.

problem icon

Air Routes

Find the distance of the shortest air route at an altitude of 6000 metres between London and Cape Town given the latitudes and longitudes. A simple application of scalar products of vectors.

problem icon

Epidemic Modelling

Use the computer to model an epidemic. Try out public health policies to control the spread of the epidemic, to minimise the number of sick days and deaths.

Crystal Symmetry

Stage: 5 Challenge Level: Challenge Level:2 Challenge Level:2

 

Caesium Chloride assumes a 'body centred cubic structure' in which each caesium ion is surrounded by $8$ chlorine ions located at the vertices of a cube, and vice-versa.

Mathematically, we can choose cartesian coordinates such that the ions lie on the integer lattice comprising the points $(l, m, n)$ with $l, m,n$ integers. The caesium ions are located at points with $l+m+n$ even and the chlorine ions at points with $l-m-n$ odd.

To start, make sure that you understand why the lattice points represent correctly a body centred cubic structure.

I can transform the points ${\bf v}$ in a lattice by multiplying by a constant matrix $M$ or adding a constant vector ${\bf c}$ through
$$
{\bf v} \rightarrow {\bf v} +{\bf c}\text{ or } {\bf v} \rightarrow M{\bf v}
$$

Which of the following vectors and matrices preserve exactly the structure of the caesium chloride when they transform the lattice?

$$M =\pmatrix{1&0&0\cr 0&0&-1\cr 0&1&0}\,,\pmatrix{3&0&0\cr 0&1&0\cr 0&0&2}\,,\pmatrix{1&1&-1\cr -1&1&1\cr 1&-1&1}$$

$$ {\bf c} = \pmatrix{1\cr 0\cr 0}\,,\pmatrix{1\cr 1\cr 0}\,,\pmatrix{2\cr 2\cr 2}\,,\pmatrix{4\cr -2\cr -8}$$

In each case prove why the crystal structure is preserved, or explain what goes wrong.

Using your geometrical intuition as a guide, investigate the types of transformations which will leave the crystal invariant.