Making a pellet with correct density and thickness: As we know the mean free path of a pellet is: lambda=4*R/3/rou, where R is the particle radius, rou is volume density. We want to make a pellet because our sing crystal sample is too thick to transmit light. Then we want to grind sample to small particles that can transmit light and put them into pellets. The keys are: 1) pellet thickness d should be slightly larger than the mean free path: d > lambda =4*R/3/rou 2) effective thickness (deff) should be small enough so that light can penetrate: deff =d*rou < 1/alpha, where alpha is the absorption coefficient. Thus the correct thickness is: 4R/3/rou < d < 1/alpha/rou is the real condition that we want to meet. There is a implicit condition here: R < 3/alpha/4, which is very important. In conclusion, to make a good pellet: 1) particle size has to be R < 3/alpha/4 2) sample thickness 4R/3/rou < d < 1/alpha/rou, which can be tuned by the volume concentration rou. Example: If we have alpha=1e7 m^{-1}, then the condition of a good pellet is : 1) R < 75 nm. 2) d < 1e7/rou This condition can not be realized because the particle size needs to be very small.

where is the most color of the world coming from?

All the colors of insulators are decided by transmittance instead of reflectance. This has been a huge confusion that I just make it clear myself not so long ago. In most insulators, the refractive index n is between 1 and 3, therefore the reflectance R=(1-n)²/(1+n)² is between 10% and 20%. This is not a small amount of light, however, this part of light is normally not so colorful either. That's because n is not rapidly frequency dependent. We can see this from our everyday experience. Any flat surface will reflect white light, no matter what color the surface is.

If everything reflects white light, how come we see a colorful world? The reason is that surface roughness disperses reflection, making this part of light only important when you have a flat surface and you are looking at a appropriate angle. So, the colorful light we see on normal stuff is actually from double transmittance, because it does not care about the surface roughness. Remember, this part of light can be as much as 80-90% of the total light illuminated on the material. It acts like reflectance because it is transmitted twice. It is colorful because the color is filtered by the media. (As we know, absorption is truly a strong frequency dependent quantity.)

The question immediately follows is that if what we see is transmittance, why most of the stuff is not transparent at all. This is another confusing phenomenon that took me a long time to understand. Actually this has to do with scattering again. If you have a big piece of NaCl single crystal, it should be transparent. But if you grind it into powder, it will keep the color, but become not transparent due to too much scattering at the powder surface. In everyday life, the stuff we see all contain so called pigments, which are colorful transparent small particles. They act like powders, which keeps the color, but not the transparency. If you look everything under optical microscopy, you can see that. Another easy way to verify this is that there is no opaque and colorful (not black) single crystal.

Back to the transition metal oxides. Their band gap is often large enough to leave all the visible range in the gap. In that case, in principle, all the visible lights should go through, making these oxides white or colorless. However this is only true for Sc₂O₃ and TiO₂ when there is not 3d electrons left on the metal ion. Other oxides are normally black and their powder shows colors. The reason is there is the d-d transition. For example, NiO, the electronic configuration is 3d_⁸, (5t_2g,3e_g), which can be excited to 3d⁸* (4t_2g, 3e_g). The energy is in the visible range. The oscillator strength (the absolute value of the absorption) is however much smaller (normally 2 to 3 order of magnitude) than the O-p to Ni-d charge transfer excitation, simply because these kind of excitations are symmetry forbidden.

If it is so small, should it be so important? The answer is yes, because the absorption does not need to be large to make a 1 mm thick sample colorful or even black. We can easily estimate this: to make a 1mm thick sample black, the absorption coeffcient α> 1/1mm = 1000 1/m. (definition of alpha is that transmitted light intensity I=I₀*exp(-α*d), where d is the thickness). If we take one step further, we can estimate extinction coefficient: k=α*c/(ω*2), where c is the speed of light, ω is the angular frequency. For a light with wave length 500 nm, α = 100 /m corresponds to k= 4e-3. This is so small that you can not it in a figure that is scaled to a charge transfer excitations. In fact, you have to use logarithmic scale to bring this detail up. This also explains why the powder is colorful. Because the transmittance depends on the absorption coefficient alpha and sample thickness. As long as the sample is small enough, it is not black anymore. Instead it will show the intrinsic color which corresponds to the small excitation in the visible range. After all, green color for NiO, red color for Fe₂O₃ (the rust) are their intrinsic properties.

bandgap_and _colors

Color and band gap of 3d transition metal oxides

mateials	Band gap (eV)	IBulk	Film (100 nm)	Powder (10 micron)
alpha Fe2O3 (Hametite)	2.34	black
Fe3O4	0.05	black
BiFeO3	2.7	black	yellow
NiO	3.4	black
TiO2	3.03	colorless

People often use interference between the reflected light from front and back surface to measure the thickness of a film. However, there is this a very important formula:

d=nu/(2*n)*), where nu is the frequency difference (difference of 1/lambda ), n is the refractive index, and d is the film thickness.

The caveat is that, we should watch out for the absorption (or the extinction coefficient k).

The formula comes from interference between reflection of from and back surface of a material. However, it does not consider the phase shift of the reflectance, and besides, it does not consider the phaseshift of the transmission of the back-surface reflection.

The problem only goes away when k is very small.

Because the phaseshift dependce on

tan(theta)=epsilon2/epsilon1=2nk/(n^2-k^2),

where theta is the phase shift.

If k<<n, theta~0. Then there is no problem, fortunately, it is the case most of the time.

Interesting color

If the material is has a single band (band gap Eg), the color depends on that only.

Eg	Tranmission/ color	reflection/ color	color of powder
>3.2 eV	large transmission / no color	small reflection / no color	white!
1.6eV< <3.2 eV	depends on thickness	depends on thickness	change color with powder size
<1.6 eV	none	metallic lusters may have different color (e.g. copper: red)	never change color with size

The color is very interesting and confusing. Actually, all the color we see is the color of transmittance color!!

Some may say, there are a lot of non-transparent colorful stuff. That's illusion of color. Let's use paint as an example. The paint contains many small transparent small particles (which is also called pigments). It is the reflection between interfaces makes the light not able to go all the way through, and in turn make the paint not transparent. The real color we normally see is the double transmitted light from the back surface.

Relation between optical functions

We measure reflection R in the expreiment.

Using R we can calculation the phase shift:

θ(ω) = ω/π ∫ log(√(R/R₀))/(ω₀²-ω²)dω₀

Given R and θ as functions of ω, one can calculate all the optical functions.

1) n and k (refractive index and extinction factor):

because R=((n-1)²+k²)/((n+1)²+k²)

k²=4n/(1-R)-(n+1)²

therefore:

n=(1-R)/(1+R+2*√(R)cos(θ))

hence:

k=√(R(n+1)²-(n-1)²)/√(1-R)

2) other functions:

ε₁=n²-k²;

ε₂=2nk

σ₁=ε₂*ω*ε₀

α=k*ω/c, where c is the speed of light

Mean free path as a function of density in volume and particle size:

assuming a unit volume for KCl, then the density will be:

rou=V0, where V0 is the volume of the sample.

Number of particles will be:

Np=V0/V, where V=4pi/3*R^3 is the volume of the particles,

then Np=V0/(4pi/3*R^3)

The mean free path will be:

lambda=1/simga/Np, where sigma=pi*R^2 is the crossection of the particle.

Therefore lambda=4R/3/rou

Example:

rou=0.01, R=1e-6 meter

then the mean free path is:

lambda=1.3e-4 meter which is 0.13 mm.

Xpy log

2007/10/18

1) uiplot
X 2) dataoperationframe: we need to add reset group, reset all. and put save at the beginning.
X 3) write the part to load two files to calculate ratio
X 4) export to ascii
X 7) data explorer for everything
X 8) put "group" operation to the ui
9) load data with time
10) calculate the time constant
X 11) put file and datainfo on printed pages
X 12) dataoperation need to be able to rego.
X 13) make XpyFigure a DataFigure
14) copy and paste
Bugs to fix
* too slow opening data explorer
v * memory leak (looks like it is still leaking, but better)
X * data saving can not update


==========================================
2007/10/25
I decided to switch to python2.5 in order to fix the memory leak problem.
Python2.5 package I have:

python-2.5.1.msi
scipy-0.6.0.win32-py2.5.exe
PIL-1.1.6.win32-py2.5.exe
numpy-1.0.3.1.win32-py2.5.exe
matplotlib-0.90.1-py2.5-win32.egg
matplotlib-0.90.1.win32-py2.5.exe
ipython-0.8.1.win32.exe
basemap-0.9.5.win32-py2.5.exe
==========================================


2007/10/26

I started the programming on Sep 16 2007, it has been 40 days and I finally have version 1.0 released. This version is not perfect at all. But for the time being, it is already very useful for data analysis. There are a lot of to do including bugs to fix, more gui tools to make and make data analysis functions to add. However, the infrastructure is built. So all we have to do is to add on stuff.

To do	add date	finish date	comments
functions
uiplot	2007/10/26	N/A any more since I have separated plotcontrol
progress bar	2007/10/26
copy and paste	2007/10/26	11/02/2007
xpyfigure hold on/off	2007/10/26	done
new numeric/string new arbitrary member		can be done using copy/paste
exlporer window as subwin		N/A
to be able to explore unlimited number of layers (two listboxes, like window style)		10/28/2007
peak analysis		11/03 partly done
xyz plot		image done 11/19
py2exe		11/19/2007	1) cd current version dicrectory 2) import xpygui 3) cd .. 4) change setup.py to match current version 5)copy image folder to dist folder 6) run setup py2exe
separate figure control window		11/23/2007
balloon help		11/22/2007
pmw and py2exe		11/23/2007	change bundlepmw: regsub.gsub to re.sub run bundlepmw copy pmw.py pmwblt.py pmwcolor.py to xpy1.26
put on internet space to be downloaded		11/27/2007
log operation of dataobject		11/29/2007
fix label to support real inline latex			this is not trivial at all. I tried twice, this time I failed again. What happens is Pylab only accept the entire text as Latex string. You can not embed some Latex in normal text. I guess I can live without this inline Latex support.
complete support for physical units
plot numsheet?
add single analysis to plotfig	idea: make ageneric analysis procedure that can take a spectrum and some input and plot the result
unit of plotting as a check box in plotcontrolwindow	done
make list working as well as dataobject
Tix progressbar
Tix tree
tktable
wxpython investigation
new object menu in the data explorer
command window (shell)
spread sheet for displaying data
new object in the right click menu
fix the loading problem
bug fixing or efficiency improvment
too slow opening explorer	2007/10/26	fixed for new explorer
memory leak	2007/10/26	the memory leak may come from pylab	partly fixed by switching to python 2.51
newfig after saving
date missing		done 10/31
compress data		done using pick and gzip
slow garbage collection
explorer for figure doesn't have context menu

Xpy to do list

Learn
	Python
		Package
	wxPython
		root window (console)
Do
	GUI
		Shell
		Spreadsheet
	Data analysis	physical units, physical quantity representation
	Data organization
	Data visualization	com for microsoft spreadsheet
		contextmenu and explorermenu for an object