## 物理代写|热力学代写thermodynamics代考|MECH 3720

2022年7月20日

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## 物理代写|热力学代写thermodynamics代考|Photonic Baths

The electromagnetic (EM) bath operator that is derived from the canonical Lagrangian quantization in a big box of the volume $\mathcal{V}$ is the vector potential (in the Gaussian units)
$$\boldsymbol{B}(\boldsymbol{x}) \equiv \boldsymbol{A}(\boldsymbol{x})=\sum_{k, \lambda}\left(\frac{2 \pi \hbar c}{k \mathcal{V}}\right)^{1 / 2}\left(a_{k \lambda} \epsilon_{k \lambda} e^{i k \cdot x}+a_{k \lambda}^{\dagger} \epsilon_{k \lambda}^{*} e^{-i k \cdot x}\right)$$
where $A$ is the vector potential, $c$ is the light velocity in vacuum, and $\epsilon_{k \lambda}$ is a unit polarization vector. The three-dimensional (3d) plane-wave decomposition of this bath operator is appropriate only for free-space EM baths.
Cavity Baths
In the case of a high- $Q$ closed cavity, where $Q$ is the quality factor of the cavity, it is natural to decompose the EM bath in terms of standing-wave discrete modes $\phi_{i j}(\boldsymbol{x})$ labeled by two indices. For an open cavity, by contrast, most of the modes have low $Q$ and merge into a continuum. If the cavity is very large compared with the wavelength, as typically happens in optics, a few modes may have low losses, due to the directionality of the radiation.

In the standard optical Fabry-Pérot cavity, light bounces back and forth between two parallel flat mirrors that are made slightly transparent, to allow a coupling with external space. To obtain the mode frequencies of the high- $Q$ modes of the open Fabry-Pérot cavity, the latter can be approximately considered as a closed cavity. For a cavity of length $d$ (along the $z$-axis) and a square cross section $(2 a)^{2}$ (in the $x y$-plane), the mode frequencies have the form $$\omega=\frac{\pi c}{n_{\mathrm{r}}}\left[\left(\frac{q}{d}\right)^{2}+\left(\frac{r}{2 a}\right)^{2}+\left(\frac{s}{2 a}\right)^{2}\right]^{\frac{1}{2}}$$
where $q, r$, and $s$ are positive integers and $n_{r}$ is the refractive index of the medium in the cavity. Each mode corresponds to a plane wave bouncing back and forth between the walls and is characterized by the direction cosines. For waves that propagate at a small angle with the cavity axis, $q$ is approximately the total number of half wavelengths contained in $d$, a very large value, whereas $r$ and $s$ are small. Then, if the ratio $2 a / d$ is not too small, (3.6) can be written as
$$\omega=\frac{\pi c}{n_{\mathrm{r}}}\left(\frac{q}{d}+\frac{r^{2} d}{8 q a^{2}}+\frac{s^{2} d}{8 q a^{2}}\right) .$$

## 物理代写|热力学代写thermodynamics代考|Waveguide Baths

A photonic waveguide supports a $1 \mathrm{~d}$ continuum of modes along its axis $z$. The field modes are confined in the directions $(x, y)$ transverse to the axis $z$. There are two different types of photonic waveguides in terms of their mode structure:
(a) A hollow waveguide supports electric field modes with normalized mode functions
$$\boldsymbol{u}{m n k}^{\lambda}=\frac{1}{\sqrt{L}} \boldsymbol{E}{m n k}^{\lambda}(x, y) e^{i k z} .$$
Here the quantization length along $z$ is $L, \lambda$ is the polarization index, and $m$ and $n$ are the transverse-mode indices. The transverse modes obey the dispersion relation
$$\omega_{m n k}=c \sqrt{k_{m n}^{2}+k^{2}},$$
where $c k_{m n}$ is the lower-cutoff frequency of the mode. There are two possible polarizations for the transverse modes: transverse electric (TE) and transverse magnetic (TM).

As an example, we consider the rectangular hollow metal waveguide (the lengths of the sides being $a$ and $b$ ) whose mode frequencies are much lower than the plasma frequency of the metal, so that it is nearly lossless. By normalizing their transverse profiles, that is, demanding
$$\int_{0}^{a} d x \int_{0}^{b} d y \boldsymbol{E}{m n k}^{\lambda *}(x, y) \cdot \boldsymbol{E}{m n k}^{\lambda}(x, y)=1,$$
we obtain the mode functions; see Table $3.1$, where $A=a b$ is the transverse area of the waveguide. Note that the transverse profile of the TE modes does not depend on the wave number $k$ (or on frequency). The mode cutoff wave numbers are
$$k_{m n}=\pi \sqrt{(m / a)^{2}+(n / b)^{2}},$$
where $m$ and $n$ are nonnegative integers such that $m+n \geq 1$ or $m+n \geq 2$ for TE or TM modes, respectively. The lowest mode is $\mathrm{TE}{01}$ or $\mathrm{TE}{10}$. The lowest $\mathrm{TM}$ mode is $\mathrm{TM}_{11}$; it has a higher cutoff than the lowest TE mode.

# 热力学代写

## 物理代写|热力学代写thermodynamics代考|Photonic Baths

$$\boldsymbol{B}(\boldsymbol{x}) \equiv \boldsymbol{A}(\boldsymbol{x})=\sum_{k, \lambda}\left(\frac{2 \pi \hbar c}{k \mathcal{V}}\right)^{1 / 2}\left(a_{k \lambda} \epsilon_{k \lambda} e^{i k \cdot x}+a_{k \lambda}^{\dagger} \epsilon_{k \lambda}^{*} e^{-i k \cdot x}\right)$$

$$\omega=\frac{\pi c}{n_{\mathrm{r}}}\left[\left(\frac{q}{d}\right)^{2}+\left(\frac{r}{2 a}\right)^{2}+\left(\frac{s}{2 a}\right)^{2}\right]^{\frac{1}{2}}$$

$$\omega=\frac{\pi c}{n_{\mathrm{r}}}\left(\frac{q}{d}+\frac{r^{2} d}{8 q a^{2}}+\frac{s^{2} d}{8 q a^{2}}\right)$$

## 物理代写|热力学代写thermodynamics代考|Waveguide Baths

（a）中空波导支持具有归一化模式函数的电场模式
$$\boldsymbol{u} m n k^{\lambda}=\frac{1}{\sqrt{L}} \boldsymbol{E} m n k^{\lambda}(x, y) e^{i k z}$$

$$\omega_{m n k}=c \sqrt{k_{m n}^{2}+k^{2}},$$

$$\int_{0}^{a} d x \int_{0}^{b} d y \boldsymbol{E} m n k^{\lambda *}(x, y) \cdot \boldsymbol{E m n k}^{\lambda}(x, y)=1$$

$$k_{m n}=\pi \sqrt{(m / a)^{2}+(n / b)^{2}},$$

## 有限元方法代写

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## MATLAB代写

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