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I write stuff sometimes on a pseudo-regular basis. You can find them here.
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Motivation So far the simulation utilises Dirichlet boundary conditions, which gets annoying because of wave reflectance. Hence, want to make an absorbing boundary, known as a perfectly matched layer (PML). Perfectly Absorbing Boundary A perfectly absorbing boundary works by absorbing a certai...
Derivation Skipping the tedious part of the derivation - very similar to the 1D, except we set $\mathbf E=(0,0,E_z)$ and $\mathbf{\tilde{H}}=(\tilde H_x,\tilde H_y,0)$, since these turn out to be the coupled equations (the other components become a ‘$H_z$’ mode). The curl equations reduce there...
Derivation of Update Equations Expanding the two curl equations: $$ \frac{\partial E_z}{\partial y} - \frac{\partial E_y}{\partial z} = -\frac{1}{c_{0}} \left( \mu_{xx}\frac{\partial \tilde H_x}{\partial t} + \mu_{xy}\frac{\partial \tilde H_y}{\partial t} + \mu_{xz}\frac{\partial \tilde H_z}...
Maxwell’s Equations The microscopic formulation of Maxwell’s equations are fairly well-known: $$\nabla \cdot \mathbf E = \frac{\rho}{\varepsilon_0}\qquad\text{(Gauss's Law)}$$ $$\nabla \cdot \mathbf B = 0\qquad \text{(Gauss's Law \#2?)}$$ $$\nabla \times \mathbf E = -\frac{ \partial \mat...
Expressions In relativistic dynamics, $$\mathbf p=\gamma_v m\mathbf v,\, E=\gamma mc^2$$ where $m$ is the rest mass. These quantities are conserved in all collisions. These have been empirically proven to be true. A rationalisation can be made through 4-vectors. This also implies a relation...
Rapidity Define $\phi$ as rapidity where $\tanh \phi=v$. Note that the velocity addition formula then reduces to $$\tanh\phi'=\tanh (\phi_1+\phi_2)$$ Similarly, scaling by $\gamma$ results in nice properties: $$\gamma = \frac{1}{\sqrt{1-v^2}}=\frac{1}{\sqrt{1-\tanh^2 \phi}}=\cosh\phi\impli...
Minkowski Distance Distance on the Minkowski diagram (using $x$ and $ct$ axes) is given by $$(\Delta s)^2=(c\Delta t)^2-(\Delta x)^2$$ and is invariant on hyperbolic rotation. An analogy is the Euclidean metric, where distance is measured as $$(\Delta s)^2=(\Delta x)^2+(\Delta y)^2$$ and is ...
Explanation Literally just geometry. Like draw crap and good luck. These don’t solve new problems, but give a nicer visualisation. Here and Now In a Minkowski diagram when boosting into $S'$, draw the $x'$ and $t'$ axes superimposed on the $S$ basis; allows for understanding: $t'$ axis make...
Derivation Consider a frame $S'$ moving at some speed $v$ in the positive x direction relative to another frame $S$. We aim to find a transformation such that we can write coordinates $(x,t)$ in $S$ in terms of $(x',t')$ in $S'$. Specifically, $$\Delta x=f_x(\Delta x',\Delta t'),\, \Delta t=f_t...
Toy Problem In some frame $S$, there exists a frame $S'$ moving at speed $v$. In frame $S'$, some object moves at speed $u$. The problem is to find the measured speed of this object in $S$. Derivation As a concretisation, let a train (frame $S'$) move at a speed $v$ rightwards as observed by t...
Breaking Simultaneity Most intuitive effect. Consider a setup with two lightbulbs on either side of a person $A$, all at rest relative to an inertial frame moving in the direction of one of the lightbulbs. An observer $B$ ‘outside’ of the frame sees all objects moving at speed $v$. In $A$’s fra...
About This is the first post in a series of posts on special relativity, asked for by a friend. Is this a good learning resource? Not really. I would recommend actually going through the textbooks, but I guess this is a decent suppelement/reading guide. Materials Using a combination of Wang/Ri...
Theory Consider the TDSE: $$i\hbar \frac{d}{dt} | \psi \rangle = \hat H |\psi\rangle$$ where $|\psi\rangle \coloneqq \psi(x,y,t)$ for this problem. Expanding the Hamiltonian for non-relativistic particle: $$i\hbar\frac{d}{dt}\psi(x,y,t)=\left(-\frac{\hbar^2}{2m}\nabla^2+V(x,y,t)\right)\ps...
Theory Similar to the 2D TDSE setup: $$i\hbar\frac{d}{dt}\ket{\psi}=\hat{H}\ket{\psi}$$ Doing much the same theoretical setup ($\hbar=2m=1$): $$i\frac{d}{dt} \psi(x,t) =(-\nabla^2+V(x)) \psi(x,t) \implies i\frac{d}{dt} \psi= - \frac{\partial^2 u}{\partial x^2}+V\psi$$ And for this proble...