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Lecture 2

1. Look at homework problem 2. Look at coordinate systems—Chapter 2 in Massa. Lecture 2. Optical Goniometer. Polarizing Microscope. I have been frequently asked in the past how to work with crystallographic coordinates.

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Lecture 2

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

  2. Optical Goniometer

  3. Polarizing Microscope

  4. I have been frequently asked in the past how to work with crystallographic coordinates. If the results of a crystal structure are to be used for further studies than it is important to be able to do this This is also important in deriving bond distances and angles from the atomic positions. An Introduction to Crystal Coordiantes

  5. 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 v1 · v2 = |v1| x |v2| x cos θ where theta is the angle between the vectors. The dot product is a scalar.

  6. Right Handed Coordinates

  7. Cartesian Coordinates • |a|, |b|, and |c| all equal 1 • α, β, and γall 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!

  8. v1=(x1,y1,z1) [shorthand for x1a+y1b+z1c] v1· v2 = |v1| x |v2| x cos(θ)=x1x2 +y1y2+z1z2 v · v= |v| x |v| x cos(0)=x2 +y2+z2 |v| = (x2 +y2+z2)1/2 Some Useful Facts

  9. Non-Cartesian Coordinates INSIDE LENGTH 39'5” INSIDE WIDTH 7'8'' INSIDE HEIGHT 7'10”

  10. Let's make the basis vectors the length, width, and height of the container. • Therefore |a|=39'5” |b|=7'8” and |c|=7'10” • The basis vectors are orthogonal • Place the origin at the door bottom left • Now the coordinates inside the box are fractional. • v=xa + yb + zc

  11. Non-orthogonal Coordinates There is no rule requiring the basis vectors make right angles! It is just very convenient if they do 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)‏ • If the basis vectors are orthogonal than the dot products are all zero.

  12. Crystallographic Coordinates • These 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.

  13. Assume |a|, |b|, and |c| not equal to one. Assume the vectors are not orthogonal a·b=x1x2|a|2+y1x2|a||b|cosγ+z1x2|a||c|cosβ+ x1y2|a||b|cosγ+y1y2|b|2+z1y2|b||c|cosα+ x1z2|a||c|cosβ+y1z2|b||c|cosα+z1z2|c|2 Dot Product in random coordinates

  14. This can be derived from the dot product formula by making x1 and x2 into x, etc. |v|2=x2|a|2+y2|b|2+z2|c|2+2xy|a||b|cosγ+ 2xz|a||c|cosβ+2yz|b||c|cosα Magnitude of a Vector

  15. Angles • The dot product contains the angle between the vectors V·W=|V||W|cos(Θ) then • Θ=cos-1(V·W/|V||W|) • Note to do this we need to take three dot products: V·VW·W and V·W • This is a good deal of arithmetic for non-orthogonal coordinates and much easier for Cartesian coordinates.

  16. Homework Problem #2

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