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You need to construct its dielectric function ε(ω) in the frequency range of interest (physical part), and supply it somehow to comsol (software part). For the physical part, I tried to collect various experimental sources to build ε(ω) spectra for common materials, including GaAs: my dielectric spectra page. You can find the Lorentz-Drude models under a ...

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With forward Euler, you're simply out of luck. Because of the way the scheme is constructed, you always make an error in momentum in the same direction, which then compounds, leading to exponential behavior. You need a symplectic integrator, the simplest being leapfrog, or any other Verlet integrator. These still won't conserve momentum on a timestep to ...

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TL;DR: Yes. Although in reality you don't have unlimited floating point precision, and this will almost always break time-reversibility. I should point out that not all integrators are time-reversible. For example, predictor-corrector schemes, and most schemes that deal with constraints. The Verlet method, however, is time-reversible, even for large ...

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I don't quite understand what you are asking. In your diagram you have \begin{aligned} a_{tangential} & = \dot{\omega} \ell \\ a_{centripetal} & = \omega^2 \ell -g \end{aligned} what else to do want to know? If the angle is changing (and therefore your acceleration are not aligned with X and Y) then use \begin{aligned} a_X & = ... 1 For a 2D planar simulation with zero friction do the following Definitions Each body has 3 degrees of freedom. These are (x_1,y_1,\theta_1) and (x_2,y_2,\theta_2) defined at the center of mass. Each body has mass and mass moment of inertia. These are m_1, m_2 and Iz_1, Iz_2. The contact is at point A with coordinates (x_A,y_A) and with ... 0 No. Momentum is still conserved. In particular, the component of momentum parallel to the ground is conserved. So if the ball is going to the right before hitting the ground, it will continue going to the right after. The formula you refer to is for one-dimensional collisions. That applies only if the elements are arranged so that there actually is ... 5 I assume you're talking about the numerical instabilities that arise from having an infinite potential at r=0. Here are three common solutions: Use a soft-core potential that behaves like 1/r except very close to r=0 where it levels off to a finite value. For example, 1/\sqrt{\epsilon+r^2} instead of 1/r is common. Add hard sphere collision ... 0 The heat balance of this system can be written as (in \mathrm{W/s}):\dot{Q}=\dot{Q_1}+\dot{Q_2}+\dot{Q_3} Where $\dot{Q_2}$ is the heat energy consumed by the machine tool and $\dot{Q_1},\dot{Q_2}$ heat losses resp. in the feed pipe and the return pipe. Assuming constant mass flow $\dot{m}$ and specific heat capacity $c_p$ of the heat fluid, ...

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