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# 物理代写|电动力学作业代写Electrodynamics代考|PHYS102 Wave packets

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## 物理代写|电动力学作业代写Electrodynamics代考|Wave packets

In real physical conditions a quasi-monochromatic wave with a frequency $\nu$ is represented by wave packets (wave train), such a wave may be generated by charges oscillating with frequency $\nu$ (see section 3.8). For example, if a generator is switched on in a moment $t=0$ and switched off in a moment $\tau \gg \nu^{-1}$, the process of switching may be described by a modulating function.

Speaking again about waves in a vacuum we can start with harmonic waves (3.22) (nonzero in whole space $R^{3}$ ), but combining its Fourier superposition from $k$ to $k+\delta k$.
$3-5$
Practical Electrodynamics with Advanced Applications
Suppose that in the initial moment the $x$-component of the electric field is given by
$$E_{x}(\vec{r}, 0)=\phi_{x}(x, y, z)=\int \mathrm{d} k_{x} \mathrm{~d} k_{y} \mathrm{~d} k_{z} E_{x}\left(k_{x}, k_{y}, k_{z}\right),$$

• the Fourier integral for $E_{x}\left(k_{x}, k_{y}, k_{z}\right)=E_{0 x}\left(k_{x}, k_{y}, k_{z}\right) e^{i(\vec{k} \vec{r})}+c . c$. Recall that the dispersion relation (3.21) root with the sign ‘ $+$ ‘ links $\omega=c k$. The amplitude function $E_{0 x}\left(k_{x}, k_{y}, k_{z}\right)$ is calculated as the inverse Fourier transform.
The form of Fourier integral in three dimensions:
$$\vec{E}(\vec{r}, t)=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{+\infty} \mathrm{d} \vec{k} \vec{E}(\vec{k}) e^{i(\vec{k} \vec{r}-c k t)},$$
gives the general solution of a Cauchy problem for the homogeneous wave equation. The inverse transformation
$$E_{j}(\vec{k})=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{+\infty} \mathrm{d} \vec{r} \phi_{j}(x, y, z) e^{-i(\vec{k} \vec{r})},$$
gives the expression for all components $j=1,2,3$ amplitudes $E_{j}(\vec{k})$ via the initial conditions $\phi_{j}(x, y, z)$. Remember the I Maxwell equation as a link between electric field components. For illustration we plot in figure $3.3$ the electric field component $E=E_{x}$ with zero others, so that is
$$E=\exp \left[-\frac{(x-c t)^{2}}{3}\right] \cos (n(x-c t))$$

## 物理代写|电动力学作业代写Electrodynamics代考|Cauchy Problem in 1 + 1 space–time

Our starting point is the Maxwell equations for vacuum with $\rho=0, \vec{j}=0$, in the Lorentz-Heaviside’s unit system. For didactic reasons we start with the onedimensional model as of [1], the $x$-axis chosen as the direction of a pulse propagation. The first Maxwell equation (Coulomb’s law) (2.17) in this one-dimensional case means the simple form $\partial E_{x} / \partial x=0$, hence, following the finite-energy field condition we assume that $E_{x}=0$ and, similarly by (2.18) $B_{x}=0$ taking into account the only polarization of electromagnetic wave, or $E_{z}=0, B_{y}=0$, denoting for brevity $E_{y}=E, B_{z}=B$. This allows us to rewrite the Maxwell equations as
\begin{aligned} &\frac{1}{c} \frac{\partial E}{\partial t}=\frac{\partial B}{\partial x} \ &\frac{1}{c} \frac{\partial B}{\partial t}=\frac{\partial E}{\partial x} \end{aligned}
In this case we rewrite the system (3.27) as the matrix equation
$$\Psi_{t}=L \Psi,$$
where the field vector (3.27), denoting $\partial_{x}=\frac{\partial}{\partial x}$,
$$\Psi=\left(\begin{array}{l} E \ B \end{array}\right) \text { and the matrix operator } L=\left(\begin{array}{cc} 0 & c \partial_{x} \ c \partial_{x} & 0 \end{array}\right),$$
that enter the matrix operator equation (3.27) as
$$\left(\begin{array}{l} \frac{\partial}{\partial t} E \ \frac{\partial}{\partial t} B \end{array}\right)=\left(\begin{array}{cc} 0 & c \partial_{x} \ c \partial_{x} & 0 \end{array}\right)\left(\begin{array}{l} E \ B \end{array}\right)=\left(\begin{array}{c} c \frac{\partial}{\partial x} B \ c \frac{\partial}{\partial x} E \end{array}\right)$$
Let us formulate now a Cauchy problem for the system equation (3.28) on $x \in(-\infty, \infty)$. Besides the equations (3.28) the problem contains two initial conditions
$$\Psi_{0}=\left(\begin{array}{l} E(x, 0) \ B(x, 0) \end{array}\right)=\left(\begin{array}{l} \phi(x) \ \chi(x) \end{array}\right) .$$
It is easy to verify that the problem formulation has a unique solution, that is given and investigated in the next subsection.

## 物理代写|电动力学作业代写Electrodynamics代考| Wave packets

$3-5$

$$E_{x}(\vec{r}, 0)=\phi_{x}(x, y, z)=\int \mathrm{d} k_{x} \mathrm{~d} k_{y} \mathrm{~d} k_{z} E_{x}\left(k_{x}, k_{y}, k_{z}\right),$$

$$\vec{E}(\vec{r}, t)=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{+\infty} \mathrm{d} \vec{k} \vec{E}(\vec{k}) e^{i(\vec{k} \vec{r}-c k t)},$$

$$E_{j}(\vec{k})=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{+\infty} \mathrm{d} \vec{r} \phi_{j}(x, y, z) e^{-i(\vec{k} \vec{r})},$$

$$E=\exp \left[-\frac{(x-c t)^{2}}{3}\right] \cos (n(x-c t))$$

## 物理代写|电动力学作业代写 Electrodynamics代考|Cauchy Problem in $1+1$ space-time

$$\frac{1}{c} \frac{\partial E}{\partial t}=\frac{\partial B}{\partial x} \quad \frac{1}{c} \frac{\partial B}{\partial t}=\frac{\partial E}{\partial x}$$

$$\Psi_{t}=L \Psi,$$

$\Psi=(E B)$ and the matrix operator $L=\left(\begin{array}{lll}0 & c \partial_{x} & c \partial_{x}\end{array}\right.$

$$\left(\frac{\partial}{\partial t} E \frac{\partial}{\partial t} B\right)=\left(\begin{array}{lll} 0 & c \partial_{x} c \partial_{x} & 0 \end{array}\right)(E B)=\left(c \frac{\partial}{\partial x} B c \frac{\partial}{\partial x} E\right)$$

$$\Psi_{0}=(E(x, 0) B(x, 0))=(\phi(x) \chi(x)) .$$

## MATLAB代写

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