PROBLEM 1 Define the function we will be using for the forward differencing as \[u=ge^{ikx_j}\] \[}{}=iku\] and the forward differencing equation as \[\Delta_x^{'}f_j=-f_j}{\Delta}\] where are function f will be our functuon u. Putting this is we get \[\Delta_x^{'}u=}-ge^{ikx_j}}{\Delta}\] define xj + 1 = xj + Δ then the above equation becomes \[\Delta_x^{'}u=-ge^{ikx_j}}{\Delta}\] Factoring we get \[\Delta_x^{'}u=(e^{ik\Delta}-1)}{\Delta}\] and taylor expanding we get \[\approx(1+ik\Delta+{2}-1)}{\Delta}\] Then factoring \[=(1+{2})}{\Delta}\]. then \[\Delta_x^{'}u=}{}(1+{2})\] So \[\Delta_x^{'}=(1+{2})}{}\]

INTRODUCTION Hedrick Casimir first introduced the idea of the vacuum field in 1948. Interest started when Casimir and Polder earlier in 1948 described the interaction between q perfectly conducting plate and an atom by using the Van Der Waals-London forces, then correcting it for retardation effects in the limit of large distances. Then the next issue was to show the effects between the interaction between two perfectly conducting plates using this Van Der Waals retarded interaction. It is useful to understand the properties of the paper that lead to the Casimir effect. Casimir considers a cubic cavity of a volume L³ bounded by perfectly conducting square plate with side L be parallel to the XY face and investigates the situation where plate is at a small distance from XY face and situation where the distance between is a distance ${2}$ [1]. Looking at the resonant frequencies of this system, $ {2} \sum \hbar \omega $, the sum being over all the resonant frequencies, notice the problem that this goes to infinity and is not physical. The differences between the frequencies is well defined for the two situations. Labeling the situations of small interaction and long interaction, differences are \[{2}(\sum \hbar \omega )_I -{2}(\sum \hbar \omega)_{II}\] and the value given is the interaction between the plate and the XY face. Cavity will have dimensions \[0 \leq X \leq L, 0 \leq Y \leq L,0 \leq Z \leq a \] with the wavenumbers given by \[ k_i={L}, k_z={a} \] where i=x,y and a is the small distance to be used between plate and XY face. The total K is given by $K==$. For every kx, ky, kz, there is two be two standing waves unless ni is zero, which case there is one standing wave. kx, ky can be regarded as continuous variables for very large L, then in polar coordinates \[ {2} \sum \hbar \omega = {\pi^2 2} ^{\infty}^{\infty}{a^2}+x^2} x dx\] where the notation of (0)1 means the n=0 term is to be multiplied by ${2}$. When a becomes large, the summation can be turned into an integral. Now the interaction energy between plates for the situations above is \[{2} \sum \hbar \omega = {\pi^2 2} ^{\infty}^{\infty}{a^2}+x^2} x dx-^{\infty} ^{\infty} x dx({\pi}) \] For sake of brevity, the result is given for a force per cm^2 \[F={240 a^4} \] Casimir concluded that there exists a force between the two plates which is independent of material and interpreted as a zero point pressure of electromagnetic waves, this result is now known as the Casimir effect. This result is hard to understand without a knowledge of quantum field theory or quantum electrodynamics. A brief anecdote on QFT will be given to try to understand the Casimir effect and the vacuum field. Quantum Field Theory states that all the fundamental fields are quantized at each and every point in space. A field can be seen as space being filled with interconnected vibrating balls and springs. Vibrations in a field then propagate and are governed by the appropriate wave equation for the field in question. The second quantization of quantum field theory or canonical quantization requires that each ball and spring combination be quantized. The field at each point in space is a simple harmonic oscillator and its quantization puts a quantum harmonic oscillator at each point. The vacuum we assume to be empty is not so empty after all. As it would be, the vacuum has all the properties a particle may have such as spin, energy but on average these cancel out. If these properties cancel shouldn’t the vacuum be empty? The exception is the vacuum energy. Since the vacuum field’s quantization is a quantum oscillator, we know that the ground state of a quantum harmonic oscillator is $E_0={2} \hbar \omega$. For the vacuum field, this is called the zero point energy. This is the same energy Casimir uses when calculating the force between the two plates. [2]

Graphene has garnered a lot of interest for its unique electronic properties and its potential applications. Graphene is a two-dimensional form of carbon in which atoms are arranged in a honeycomb lattice. It is a zero-gap semiconductor, this allows the ability to dope graphene to have high concentrations by applying a voltage. This property of electric gating is also used to control conductivity and the dielectric constant of graphene. There is extensive research on the many properties of this monolayer of carbon but I will focus on conductivty, surface waves supported by graphene, and research into applications with optical modulators. There are two absorption properties that are involved in light-graphene interaction, interband and intraband absorption for small signals. This is described by the complex conductivity, \[\sigma_g=(\omega,\mu_c,\Gamma,T)+(\omega,\mu_c,\Gamma,T)\], where ω is the angular frequency of light, μc is the chemical potential, Γ is scattering rate, and T is the temperature [1].Whats interesting is that the chemical potential can be tuned with electrical gating, thus the conductivity can be controlled by the same process. Chemical potential also controls what absorption process is occurring.Interband corresponding to absorption from the valence band to conduction band and intraband absoprtion corresponding to absoprtion from a semiconductor like optical property to a metal like optical property [2]. For incoming light with energy ℏω, interband absoprtion dominates when $\mu_c < {2}$ and intraband dominates when $\mu_c > {2}$ and it is theoretically predicted that intraband absoprtion would be dominate when $\mu_c \approx {1.67}$ [2]. Zhaolin Lu et al investigates graphene conductivty at T = 300K with a scattering rate ℏΓ = 5meV. Figure 1 plots real and imaginary parts of conductivity vs chemical potential and the dielectric constant as a funciton of chemical potential. Notice the sensitivity of conductivity with little change in chemical potential. Here $=1-{i\omega\epsilon_0\Delta}$ where Δ = 0.7nm is the effective thickness of graphene. Notice the dip in in the magnitude of the effective dielectric constant, it is near zero at μt. Physically this is the marking of the transition from a dielectric type graphene to a metal type graphene. The transition chemical potential is μt = 0.515eV for this experiment. Theoretical value from [1] predicts μt = 0.479eV which has about 7 % error. Since conductivity can be tuned by electrical gating so too can the effective dielectric constant.

LECTURES AND NOTES week 8 notes review Taylors Theorem says if f is analytic on {z : |z − z₀|<r} and continuous on the domain that includes the boundary, then $f(z)=^{\infty}f^{(n)}(z_0){n!}$ and this series converges absolutely. Cauchys inequality says that if is analytic in {z ∈ 𝕔 : |z − z₀|<r} and |f(z)| ≤ c in the disk then the function converges absolutely. Isolated Singularities A function which is analytic on the punctured disk {z ∈ 𝕔 : 0 < |z − z₀|<r} has an isolated singularity at z₀.There are three examples of isolated singularities. 1. removable singularity: where f(z) is bounded for some r > 0 on {0 < |z − z₀|<r}, remains bounded as z → z₀ 2. poles: where limz → z₀|f(z)| = ∞ 3. essential singularity: when 1 or 2 dont apply lemma: if f has a removable singularity at z₀ then the limz → z₀f(z) exists and extends f to an analytic function at z₀.

Revisting the initial value problem of \[ (z)\\ (z)\\ = F(z;z_0) (z_0)\\ (z_0) \\ \] The question is to solve this subject to the initial condition of when power is launched only into the mode a at z₀ = 0. This means $(0)=0$ and $(0)\neq 0$. \[ F(z;z_0)= {\beta_c}e^{i \delta (z+z_0)} & }{\beta_c}sin\beta_c(z-z_0)e^{i \delta (z+z_0)}\\ }{\beta_c}sin\beta_c(z-z_0)e^{-i \delta (z+z_0)} &{\beta_c}e^{-i \delta (z+z_0)} \\ \] Subject to our initial conditions then \[ (z)\\ (z)\\ = F(z;0) (0)\\ 0 \\ \] Carrying the matrix multiplication step by step \[ (z)\\ (z)\\ = {\beta_c}e^{i \delta (z)} & }{\beta_c}sin\beta_c(z)e^{i \delta (z)}\\ }{\beta_c}sin\beta_c(z)e^{-i \delta (z)} &{\beta_c}e^{-i \delta (z)} \\ (0)\\ 0 \\ \] \[ (z)\\ (z)\\ = (0) \beta_c cos \beta_c(z)-i\delta sin \beta_c(z)}{\beta_c}e^{i \delta (z)} & 0\\ (0) }{\beta_c}sin\beta_c(z)e^{-i \delta (z)} & 0\\ \] This leaves us then with \[(z)=(0)\left( cos \beta_c z -{\beta_c} sin \beta_c z \right) e^{i \delta z} \] \[(z)=(0)\left(}{\beta_c} sin \beta_c z \right) e^{-i \delta z} \] Looking back at my slides I wrote that $(z)=(0)\left(}{\beta_c} sin \beta_c z \right) e^{-i \delta z} $, which given my initial conditions would be wrong since $(0)=0$. Thankfully, this error doesn’t propagate through the rest of the presentation as that error only shows up on slide 8 and the rest of the figures I made are still accurate. I want to point out another error I made in the paper in regards to figure 1 here. The error is that in the figure in the paper, I wrote the same error for the power, corresponding to the blue line in figure one. I wrote ${P_b(o)}$, when in fact it should be ${P_a(o)}$, It was corrected for the presentation and here.

Time between pulses of SL was found to be 3.069 × 10−5 ± 0.01 seconds, width of an SL pulse was found to be 1.925 ∗ 10−6 ± 0.04 seconds and Sonoluminescence was observed.

PROBLEM 1 Part C Plots done using mathematica, figure 1 using equation 8.25 excluding the ϕ dependence and using the appropriate l values for each mode.