Lecture 2 1. Look at homework problem 2. Look at coordinate systems—Chapter 2 in Massa

Homework Problem 1

Optical Crystallography

Polarizing Microscope

Vectors Vectors have length and direction We will use bold v to represent a vector |V| is the magnitude (length) of a vector The dot product of two vectors is as follows v 1 · v 2 = |v 1 | x |v 2 | x cos θ where theta is the angle between the vectors

Coordinates and Basis Vectors A coordinate system is composed of three basis vectors call a, b, and c. The angles between these vectors are given by α (alpha) the angle between b and c β (beta) the angle between a and c γ (gamma) the angle between a and b

Right Handed Coordinates

Cartesian Coordinates |a|, |b|, and |c| all equal 1 α, β, and γ all 90º A reminder cos(0)=1 cos (90)=0 sin(0)=0 sin(90)=1 The symbols x, y, and z represent positions along a, b, and c A general vector is given by v=xa + yb + zc Since the basis vectors magnitude is 1 they are ignored!

Some Useful Facts v 1 = (x 1,y 1,z 1 ) v 1 · v 2 = |v 1 | x |v 2 | x cos(θ)=x 1 x 2 +y 1 y 2 +z 1 z 2 v · v = |v| x |v| x cos(0)=x 2 +y 2 +z 2 |v| = (x 2 +y 2 +z 2 ) 1/2

Non-Cartesian Coordinates

Let's make the basis vectors the length, width, and height of the cargo compartment. Therefore |a|=121 |b|=19 and |c|=13.5 The basis vectors are orthogonal Place the origin at the front left of the plane. Now the coordinates inside the plane are fractional. v=xa + yb + zc

Non-orthogonal Coordinates If the basis vectors have a magnitude of 1 then a·a = b·b = c·c = 1 a·b = cos(γ)‏ b·c = cos(α)‏ a·c = cos(β)‏

Crystallographic Coordinate These can provide the worst of all possible worlds. They frequently are non-orthogonal The do not have unit vectors as the basis vectors. The coordinates system is defined by the edges of the unit cell.

Dot Product in random coordinates Assume |a|, |b|, and |c| not equal to one. Assume the vectors are not orthogonal a·b=x 1 x 2 |a| 2 +y 1 x 2 |a||b|cosγ+z 1 x 2 |a||c|cosβ+ x 1 y 2 |a||b|cosγ+y 1 y 2 |b| 2 +z 1 y 2 |b||c|cosα+ x 1 z 2 |a||c|cosβ+y 1 z 2 |b||c|cosα+z 1 z 2 |c| 2

Magnitude of a Vector This can be derived from the dot product formula by making x1 and x2 into x, etc. |v| 2 =x 2 |a| 2 +y 2 |b| 2 +z 2 |c| 2 +2xy|a||b|cosγ+ 2xz|a||c|cosβ+2yz|b||c|cosα

Homework Problem #2