14

The key is Weyl's famous observation that electrodynamics is really (classical) $U(1)$-gauge theory. Concretely: You generalise the global $1$-form $\mathcal{A}$ on $M$ to a connection $\nabla$ on a Hermitian line bundle $\mathcal{L} \to M$, which can locally be written as $d + \mathcal{A}$ for $\mathcal{A}$ the so-called connection $1$-form. The ...


13

The answer is yes to both questions. If you cast Maxwell's Equations in cylindrical coordinates for a fiber optic cable, and you take birefringence into account, you get the coupled nonlinear Schrödinger equations. You can then solve those by means of the Inverse Scattering Transform, which takes the original system of nonlinear pde's (nonlinear because of ...


5

Electric fields due to $Q_1$ and $Q_2$: $$\vec{E}_1=\frac{Q_1}{4\pi\epsilon_0}\frac{x\hat{x}+y\hat{y}+z\hat{z}}{\left(x^2+y^2+z^2\right)^{3/2}},$$ $$\vec{E}_2=\frac{Q_2}{4\pi\epsilon_0}\frac{x\hat{x}+y\hat{y}+(z-R)\hat{z}}{\left(x^2+y^2+(z-R)^2\right)^{3/2}}.$$ Dot product of electric fields in spherical coordinates: $$\begin{align} \vec{E}_1\cdot\vec{E}...


5

If the curl were an arbitrary vector field, it would “contain as much information as the field itself”. (I'm using scare quotes because this is all on the same hand-waving level of rigorosity that your arguments address.) However, the curl satisfies $\nabla\cdot(\nabla\times E)=0$. Thus, we have one scalar constraint on the curl, which reduces ...


5

It means that the differential operator $$ \mathbf{A} \cdot \nabla = (A_x,A_y,A_z) \cdot (\partial_x,\partial_y,\partial_z) = A_x \partial_x + A_y \partial_y + A_z \partial_z $$ acts componentwise on the vector $\mathbf{B}$.


4

The answer to the question if whether a solution $\chi$ to the the following equation exists $$ -\frac{1}{c^{2}}\square=\left(\Delta - \frac{1}{c^2}\frac{\partial^2}{\partial t^2}\right)\chi=f\;\text{ in }\;\Bbb R^4\equiv \Bbb R^3\times \Bbb R \label{w}\tag{W} $$ under mild smoothness requirements on the datum $f$ is yes: I explain below why it is so in a ...


4

Given that $\dot v_x = \dfrac{dv_x}{dt} = \dfrac{qB_0}{m}v_y, \tag 1$ and $\dot v_y = \dfrac{dv_y}{dt} = -\dfrac{qB_0}{m}v_x, \tag 2$ since $q$, $B_0$, and $m$ are constant, we may differentiate each equation with respect to $t$ and so obtain $\ddot v_x = \dfrac{d^2v_x}{dt^2} = \dfrac{qB_0}{m} \dot v_y, \tag 3$ $\ddot v_y = \dfrac{d^2v_y}{dt^2} = -\...


3

Yes, solving for $\epsilon_0$ first is a good idea: $$V = \frac{1}{4\pi\epsilon_0}\frac{Q}{r} \\ \epsilon_0 = \frac{1}{4\pi V}\frac{Q}{r} \\ [\epsilon_0] = \frac{1}{\textrm{V}}\frac{\textrm{C}}{\textrm{m}} = \textrm{C} \textrm{V}^{-1}\textrm{m}^{-1}$$ Read out, $\epsilon_0$ has units of "Coulombs per volt per meter".


3

The presence of a curl term in the Helmholtz decomposition is very much a three-dimensional phenomenon. While it is reasonable to define the curl of a vector field tangent to a surface, what you get is either a field normal to the sphere or a scalar field, so it's not going to help us decompose tangent vector fields. The Hodge decomposition tells us that ...


3

There is this (very brute-force type of calculations) way to do it. We let that $\hat{n} = \sin(\theta)\cos(\phi)\hat{x} + \sin(\theta)\sin(\phi)\hat{y}+\cos(\theta)\hat{z}$ and $\vec{B} = B_x\hat{x}+B_y\hat{y}+B_z\hat{z}$ so we will have that $$\dot{\hat{n}} = \left(\begin{matrix}\color{blue}{\dot\theta} \cos(\theta)\cos(\phi) - \color{red}{\dot \phi \sin(...


3

Note that with $G=C\frac{e^{-j\beta r}}{r}$, heuristically (or in distribution) we have for any $R>0$ $$\begin{align} \int_{|\vec r|\le R}\left(\nabla^2 G+\beta^2G\right)\,dV&= \oint_{|\vec r|=R}\nabla G\cdot \hat n\,\,dS+\beta^2\,\int_{|\vec r|\le R} G\,dV\\\\ &=\int_0^{2\pi}\int_0^\pi C \left(-j\beta\, \frac{e^{-j\beta R}}{\epsilon}-\frac{e^{-j\...


3

Your result gives the electric field for a polygon whose $n$ sides have length $2L$ and has then a radius $R=L/\sin(\pi/n)\to\infty$, no surprise then if $E\to0$. You should instead keep $R$ constant and plug $L=R\sin(\pi/n)$ into your formula for a polygon with $n$ sides: $$ E = \frac{1}{4\pi\varepsilon_0} \frac{2\lambda Rz\ n\sin(\pi/n) } {\big(z^2 + R^2\...


3

Unfortunately not. The reason is that the problem is not well-posed. The reason for this is that you are trying to find two functions $J_x$ and $J_y$ (because $\bf J$ is a vector with two components) but you have only one equation that $J_x$ and $J_y$ should satisfy. It is possible to demonstrate that you have infinite solutions. In particular, if ${\bf J}(...


2

Starting with your expression (which after cancelling factors can be written) $$dE = \frac{\sigma z}{2 \epsilon_0} \frac{ada}{(z^2+a^2)^{3/2}}$$ Now integrate to get $$E = \frac{\sigma z}{2 \epsilon_0} \int_0^R\frac{ada}{(z^2+ a^2)^{3/2}}$$ To solve the integral let $w = z^2 + a^2$ then $$E = \frac{\sigma z}{4 \epsilon_0} \int_{z^2}^{z^2+R^2}\frac{dw}{w^...


2

Consider the surface $S$ defined by the Ampèrian loop $\partial S$. Then, a vector normal to this surface and with magnitude equal to the area of the surface is $\vec a$. This is commonly referred to as the vector area of $S$. Then, the unit vector $\hat a$ is a vector parallel to $\vec a$, but with unit length. The total current flowing through $S$ is ...


2

multiply the first denominator by $$R-iwL$$ and the second one by $$1-iwRC$$ and you will get $$\frac{L^2Rw^2}{R^2+w^2L^2}+\frac{R}{1+w^2R^2C^2}+i\left(\frac{LR^2w}{R^2+w^2L^2}-\frac{CR^2w}{1+w^2R^2C^2}\right)$$ and you can solve your problem


2

NOTE: Herein, we use the time convention $\displaystyle e^{i\omega t}$ If $u(x,y)=Ae^{-i(k_1x_1+k_2x_2)}+Be^{-i(k_1x_1-k_2y_2 )}$, then on the boundary we have $$A+B+\beta(A(-ik_2)-B(-ik_2))=0$$ whence we have $$\begin{align} B&=-\frac{1-ik_2\beta}{1+ik_2\beta}A\\\\ &=-\left(\frac{\left(1+k_2\text{Im}(\beta)\right)-i\left(k_2\text{Re}(\beta)\...


2

We will show that one need not calculate the energy stored in the electrostatic field by integration. To that end, we proceed. We start with the electrostatic energy $W$ stored as given by $$W=\frac{\epsilon_0}{2}\int_V \vec E\cdot \vec E \,dV \tag 1$$ where $V$ is all of space. We now assume that the field is induced by a charge distribution ...


2

We begin with the expression $$\frac1{\epsilon_0}\int_V\phi_1\rho_2\,dV+\oint_{\partial V}\phi_1\nabla\phi_2\cdot \hat n \,dS=\frac1{\epsilon_0}\int_V\phi_2\rho_1\,dV+\oint_{\partial V}\phi_2\nabla\phi_1 \cdot\hat n\, dS \tag 1$$ GIVENS: We are given that $\phi_1=\Phi$ on the surface $|\vec x|=a$ and $\rho_1=0$ for $|\vec x|<a$. We take $\phi_2=\frac{...


2

The primed coordinates are just "dummy" coordinates over which we integrate the contributions to the electric field at a fixed observation point $\vec r$ from a differential charge, $dq=\rho(\vec r')\,dV'$, at $\vec r'$. The distance between to fixed observation point $\vec r$ and the differential charge at $\vec r'$ is $|\vec r-\vec r'|$ and the unit ...


2

Kolmogorov and Fomin's Introductory Real Analysis does a nice job introducing the ideas with the rigor of distributions in their section on Generalized Functions. This book is also a nice introduction to functional analysis. Another place you might look, for a more general, in-depth, and measure theoretic perspective, is Folland's Real Analysis, chapter 3.


2

You could just say that $$ \langle \frac{\partial}{\partial r}, \frac{1}{r}\frac{\partial}{\partial \theta}, \frac{1}{r\sin\theta}\frac{\partial}{\partial \phi}\rangle $$ is what the gradient operator looks like in spherical polar coordinates. Here's an outline of a derivation: Starting with ${\bf E} = -\nabla V$, imagine an (arbitrary) infinitesimal ...


2

Hint: if $\theta$ is the angle betwee the $z$ axis and the line passing thorough the origin and the point $P=(x,y,z)$ and $r$ is the distance $\overline{OP}$, than The distance from $P$ to the point $(0,0,a)$ is: $$ \sqrt{x^2+y^2+(z-a)^2}=\sqrt{x^2+y^2+z^2+a^2-2za}=\sqrt{r^2+a^2-2ar\cos \theta} $$ (because $z=r\cos \theta$). Analogously you can find the ...


2

The electric field at $\vec r=\hat z a$, due to a disk of uniform charge density $\sigma$ with radius $\sqrt 3 a$ and located at $\rho <\sqrt 3 a$, $z=0$ is given by $$\begin{align} \vec E(0,0,a)&=\frac{\sigma}{4\pi \epsilon_0}\int_0^{2\pi}\int_0^{a\sqrt 3} \frac{\hat z a-\hat\rho'\rho'}{\left(a^2+\rho'^2\right)^{3/2}}\,\rho'\,d\rho'\,d\phi'\\\\ &...


2

The lower and upper boundaries for $y$ in $S_1$ should be $-r$ and $r$ instead of $0$ and $r$. I was thinking in polar coordinates instead of in Cartesian coordinates... I'll edit the original post accordingly


2

Adrian has given an interesting answer already, but I think it is worth pointing out two key points which were necessary for his soliton situation. Firstly, it was necessary to impose some specific form of initial-boundary data (to constrain the waves to inside the fibre optic cable), and secondly it was necessary to impose physical assumptions on the medium ...


2

The way I remembered the rule for how magnetic fields around wires work is the following. Take your right hand, and point your thumb in such a way so that it points in the same direction as the current, and curl the rest of your fingers. The direction your fingers curl represents the direction of the magnetic field. So, when you use this rule, you'd see ...


2

Hint: $$\int \frac{1}{(x^2+\alpha^2)^{3/2}} dx= \frac{x}{\alpha^2\sqrt{x^2+\alpha^2}}+c$$


2

In general, $r \theta$ is the length of the arc of the circle of radius r subtended by a central angle $\theta$. This comes straight from the definition of radian. In the picture, it seems they are approximating the "straight" leg of the little triangle, with the "curved" arc swept out by the arm of length $r$ through angle $d\theta$. This is reasonable ...


2

Good question! The answer, at least to me, lies in that the integral form holds for arbitrary surfaces $\Sigma$. This can be interpreted in two ways: For a given closed loop $\ell\in\mathbb{R}^3$, there are infinitely many smooth surfaces $\Sigma$ such that $\partial\Sigma=\ell$; The closed loop $\ell$ itself could also be arbitrarily specified. Therefore,...


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