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This is Eric's logbook for Nano Japan project from 2012.6.3-2012.7.27.
| Time and Day | Mon | Tue | Wed | Thu | Fri |
| 09:00-10:00 | Sato | Nugraha | Tapsanit | Tatsumi | Nugraha |
| 13:30-14:30 | Saito | Simon | Hasdeo | Simon | Tapsanit |
Hey guys! So I believe this is where I should post questions. I don't have any specific questions today, but was just trying to understand Brillouin zones and Bloch's theorem, which appear a lot in Saito-sensei's book on Carbon Nanotubes.
From google I have an ok idea of what Brillouin zones are, but am still trying to figure out how they relate to the energy gap of the cell such as on page 28 of the CN book. I was also wondering how on page 47 the Brillouin zone can be a long line segment instead of a polygon.
If anybody has any quick words of advice for how to understand Bloch's theorem that would be great as well! It's a little bit intimidating and I'm not sure if there's a simple way to begin to understand it.
Thanks so much and see you guys in a couple of weeks!
Eric
Please consider my answers one by one in order to understand the Brillouin zone.
Can I use a copy machine or printer?
Yes, and they are across the hall from my room.
I would like to know where the library is and how to use it.
The library is on the 7th floor. To check out a book I write information on a special wooden card and place it where the book had been. I also give the slip inside of the book to the librarian, and I return the book to the librarian.
This part is basically written by Eric-san. Any other people can add this. Here the information should be from new to old so that we do not need to scroll.
This morning Tapsanit-san taught me about the refractive index and the complex refractice index. Materials have a refelctivity R=I&tex(): Error! cURL: Could not resolve host: chart.apis.google.com;/I&tex(): Error! cURL: Could not resolve host: chart.apis.google.com;, where I&tex(): Error! cURL: Could not resolve host: chart.apis.google.com; is the intensity of the reflected light. They have a similar transmission coefficient T=I&tex(): Error! cURL: Could not resolve host: chart.apis.google.com;/I&tex(): Error! cURL: Could not resolve host: chart.apis.google.com;. Light slows down in a meduium, and we define the refractice index n=c/v. We define the absorption coefficient &tex(): Error! cURL: Could not resolve host: chart.apis.google.com; so that intensity I(z)=I&tex(): Error! cURL: Could not resolve host: chart.apis.google.com;e^(-&tex(): Error! cURL: Could not resolve host: chart.apis.google.com;z).
We defined the complex refractive index n~ = n +iK, where iK is the extinction coefficient, and then we derived that &tex(): Error! cURL: Could not resolve host: chart.apis.google.com; = 2Kw/c = 4&tex(): Error! cURL: Could not resolve host: chart.apis.google.com; K/ &tex(): Error! cURL: Could not resolve host: chart.apis.google.com;. K is the extinction coefficient and alpha is the absorption coefficient.
This morning Nugraha-san taught me more about computer programming in Fortran. He walked me through the process of making a program to calculate the optical and acoustic modes of a 1D phonon. He then showed me how to make a .dat file to plot the data in table form, and how to graph the function in xmgrace. Later in the morning Saito-sensei taught me how to modify the axis of the graph so that characters (like Pi/a) are displayed instead of units. The way to do this is to make the graph, go to plot, axis properties, special. Change a tab to 'numbers and characters', and insert characters in the appropriate window. For the Greek letters there is a special code, which I cannot remember but could be looked up online.
Also in the morning I compared a gaussian force (f=Ae^(-t^2/(a^2)) to a polynomial force (f=t^2(t-2b)^2) and graphed these two functions to see how similar they were. I integrated each the total area of the function and set that equal to 1. I then calculated the relationship between the coefficients. The graph of these two functions side by side on the interval from 0 to 2b were very similar. The polynomial function is easier to work with because it yields a much simpler solution to the equation of motion x'' + w^2x = f(t).
Next Saito-sensei taught me about the equation for a wave which is Qxx - c^2*Qtt= f(x,t). f(x,t) is the applied force and Q is a function of x and t. In the afternoon I solved the wave equation for f(x,t)=0 (with help from online notes from the University of British Columbia math department.) However, I also need to solve for the coefficients of the solution, which can be done using the initial conditions of the problem.
In the afternoon Florian-san also taught me more about differential equations relating to springs, and photocopied some of his notes so that I may use them as reference. He also taught me a lot about how to use POV-Ray animation software, and now I've seen how to make objects move in animations. Florian has a lot of code and information compiled on his website and I may use this in order to help me create my own animations.
To Do:
Today Saito-sensei taught me how to find a solution for the differential equation of a Gaussian. I learned a lot today about Gaussian functions, and about using LaTeX and Mathematica.
Saito-sensei taught me how to use the equation feature of LaTeX so that my equations will be numbered sequentially. He also found a function in Mathematica that converts an equation in Mathematica into LaTeX code, which will really help speed things up in the future.
The Mathematica function is: TeXForm[f(t)] Make sure to always use a capital letter X. Insert the function inside the brackets and LaTeX code will be created. Also, TeXForm[%] may be used if the TeXForm command is in the same section as the function with the % sign taking the place of the function.
On LaTeX I created a summary of the solution we found and include the recursion formulas that we found for the coefficients:
Here we have the equation of motion for an applied force, which we set to be a Gaussian: \[ \ddot{Q} - k\dot{Q}+\omega^2Q = F(t) \] \[ \ddot{Q} - \omega^2Q = A e^{-\frac{t^2}{\alpha^2}} \]
We Assume a solution of form: \[ \Sigma_{n}^{\infty} C_{n}t^{n}e^{-\frac{t^2}{\alpha^2}}+A_{n}\sin (\omega t)+B_{n}\cos(\omega t) \]
For Odd Coefficients we have where m=2n+1, m grater than or equal to 1: \[ C_{2m+1}=\frac{1}{4m^{2}+2m}[C_{2m-1}(\frac{8m-2}{\alpha^2})-\omega^2C_{2m-1}-\frac{4}{\alpha^2}C_{2m-3}] \]
And for Even Coefficients where m=2n, m greater than or equal to 2: \[ C_{2m}=\frac{1}{4m^{2}-2m}[C_{2m-2}(\frac{8m-6}{\alpha^2})-\omega^2C_{2m-2}-\frac{4}{\alpha^2}C_{2m-4}] \]
Saito-sensei and I also talked about how best to model the applied force. A Gaussian force is ideal but very complicated mathematically and also takes a long time to approach its peak. (It never truly starts and zero, and the closer to zero is starts the longer it takes to reach its peak.)
Saito-sensei suggested modeling the force as either a polynomic equation of degree 4, or something of the form f(t)=t*E^(-t^2). We talked about these approaches and I will study these functions more. Additionally, we began to think about the wave as a function Q(x,t) which satisfies the wave equation.
To Do:
This morning Nugraha-san came and we talked a little bit about the reciproical lattice. We then attacked the problem of the differential equation given by a Gaussian force. After working for a while we did not find a satisfying solution, and it's possible that for this equation a simple solution doesn't exists. If we had more boundary conditions it's possible that the solution would be simpler.
We then talked a little bit about Fortran and Nugraha-san showed me a website from Boston University that had a lot of good Frotran tutorials.
In the afternoon Tapsanit-san came in and we constructed the Brillioun zone of both the square lattice and of graphene. We obtained the symmetry points (Gamma, M, K and K') in the Brillioun zone. We compared the area of the Brillioun zone with that of the square lattice.
Finally, we looked at the Fortran task given by Hasdeo-san and we created a program that gives a cascading array of integers, as desired. Tapsanit-san printed out some pages on optical processes that I can read this weekend if I have time.
To Do:
In the morning Tatsumi-san and Ominato-san explained to me the reciprocal lattice of graphene. First we derived the unit cell of graphene. We derived the reciprocal lattice and found then found the Brillouin zone of graphene, which is also a hexagon but rotated by 90 degrees.
To somewhat review what Tapsanit-san taught me yesterday we then talked about the wavevector (k) in the reciprocal lattice, and frequency as a function of k. We showed that this satisfied Bloch's theorem and described the periodic boundary conditions. Then, we talked about the acoustic mode vs the optical mode. We showed that the number of modes will increase if we increase the number of atoms in a unit cell. In the problem where two different types of masses are connected with identical springs, the lower valued function is the 'acoustic mode', where all the atoms vibrate in the same direction at the same value of time. The higher valued function is the optical mode, where some of the atoms are pushed towards each other while other atoms are pushed away from each other.
At 10:30 I accompanied Hasdeo-san to Japanese class where I learned the ta form of verbs. For instance: "Kyoto e itta koto ga arimaska?" = "Have you ever been to Kyoto?". Iie, arimasen. Watashi wa Kyoto ni ikitai.
In the afternoon I went to the group meeting where Simon-san presented his animations and some differential equations. His POV-Ray animations were very cool and also really helped to illustrate the acoustic vs. optical vibrational modes. Tomorrow we made plans to discuss some of the differential equations he had been working with, because I need to know them as well.
Later in the afternoon, Ominato-san and Tatsumi-san returned to teach me more physics. Ominato-san had written a long sequence of very good problems some of which we worked through. We defined the Drude Model of electron conduction, and wrote the equation of electron motion including a frictional term: \[ m\frac{dv}{dt}=e(E+\frac{1}{c}v X H)-(\frac{m}{T})v \]
We derived v(t) when H=0, and then for the steady state solution (dv/dt=0) We then Derived &tex(): Error! cURL: Could not resolve host: chart.apis.google.com; (the conductivity) based on the relation: \[ j=nev= \sigma E \]
Next, we derived v(t) for a more general case and put into matrix form the equation: \[ j= \sigma E \]
Ominato-san explained the classical Hall effect and we breifly talked about special relativity. The energy of a particle with mass is given by E=+/- sqrt((cp)^2+(mc^2)^2) were E is a parabolic function of p. However in graphene electron behave as if they have no mass and E=+/- cp.
To Do:
I learned a tremendous amount of theory today. My understanding of the lattice has made good progress over the last several days.
In the afternoon Hasdeo-san explained further about Fortran programming language. After working through a program he made he gave me several assignments to do. Hasdeo-san also helped me learn to navigate the directories of command prompt.
To Do: 1. Get a correctly formatted w(k) graph to Nugraha-san AND learn about/explain which is the optical mode and which is the acoustic mode. 2. Solve Saito-sensei's differential equation by learning about Fourier transforms. 3. Solve Hasdeo-san's problems:
- plot the cascading list of integers - graph two fxns (different parabolas) using Fortran - output fxn data in column form
Also, Florian-san showed me how to use the POV-Ray software. We installed it on the desktop machine and I worked through the basic tutorial.
Also, solve the diff eq for a Gaussian.
If I have time consider the infinite 1D series of spring mass problems, but in the case that all masses are the same, but the spring constants are different.
Solution is: f(t)=(b/(aw^2))(t-(1/w)sin(wt)) Could look into the eqn and new initial conditions for t>a
to do:
\vec {a}_1 = \left(\frac{\sqrt{3}}{2}a,\frac{a}{2}\right), \quad
\vec {a}_2 = \left(\frac{\sqrt{3}}{2}a,-\frac{a}{2}\right)
\]\[ \vec{a}_1\cdot\vec{b}_1 = a_{1x}b_{1x}+a_{1y}b_{1y}=b_{1x}\frac{a\sqrt{3}}{2}+b_{1y}\frac{a}{2}=2\pi \] \[ \vec{a}_2\cdot\vec{b}_1 = a_{2x}b_{1x}+a_{2y}b_{1x}=b_{1x}\frac{a\sqrt{3}}{2}-b_{1y}\frac{a}{2}=0 \]
\[b_{1y}\frac{a}{2}=b_{1x}\frac{a\sqrt{3}}{2} \] \[b_{1x}\frac{a\sqrt{3}}{2}+b_{1x}\frac{a\sqrt{3}}{2}=2\pi \]
\[b_{1x}=\frac{2\pi}{a\sqrt{3}} \] \[b_{1y}=\frac{2\pi}{a} \]
\[
\vec {b}_1 = \left(\frac{2\pi}{a\sqrt{3}},\frac{2\pi}{a}\right), \quad
\vec {b}_2 = \left(\frac{2\pi}{a\sqrt{3}},\frac{-2\pi}{a}\right)
\]
