Posted on Categories:General Relativity, 广义相对论, 物理代写

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An important case occurs when the source varies harmonically in time $S=$ $4 T^{\mu \nu}=j_\omega\left[\vec{r}^{\prime}\right] \exp \left[-i \omega t^{\prime}\right]$. With the source near the origin, the observation point $\vec{r}$ may be in one of three zones: near, intermediate, and far. Each zone allows for different approximations. The far zone, where $d \ll \lambda \ll r$, is of interest to us. A violent event, triggering a gravitational wave, is likely to occur far from us. Here $d$ is the source size, and is much smaller than the wavelength of the radiation. That, in turn, is much less than the radial coordinate of the observation point. Then,
\begin{aligned} \left|\vec{r}-\vec{r}^{\prime}\right| & =\left(r^2+r^{\prime 2}-2 \vec{r} \cdot \vec{r}^{\prime}\right)^{1 / 2}=r\left(1-2 \vec{r} \cdot \vec{r}^{\prime} / r^2+\left(r^{\prime} / r\right)^2\right)^{1 / 2} \ & \approx r\left(1-\vec{r} \cdot \vec{r}^{\prime} / r^2\right)=r-\hat{e}r \cdot \vec{r}^{\prime} \ \bar{h}^{\mu \nu}[\vec{r}, t] & =\int{\infty} d V^{\prime} j_\omega\left[\vec{r}^{\prime}\right] \frac{\exp \left[-i \omega\left(t-\left[r-\hat{e_r} \cdot \vec{r}^{\prime}\right]\right)\right]}{r-\hat{e}r \cdot \vec{r}^{\prime}} \ & \approx \frac{\exp [i(k r-\omega t)]}{r} \int{\infty} d V^{\prime} j_\omega\left[\vec{r}^{\prime}\right] \exp \left[-i k \hat{e}r \cdot \vec{r}^{\prime}\right] \ & \approx \frac{\exp [i(k r-\omega t)]}{r} \int{\infty} d V^{\prime} j_\omega\left[\vec{r}^{\prime}\right], \text { to lowest order, } \ & =\frac{4}{r} \int_{\infty} d V^{\prime} T^{\mu \nu}\left[\vec{r}^{\prime}, t-r\right] . \end{aligned}

The far zone is approximately locally inertial because it is so far from the source. In natural units $2 \pi / \lambda=k=\omega$, but $k r$ and $\omega t$ are written so that the equations look familiar. For the harmonic dependence, the solution Eq. (7.24) looks like an outgoing spherical wave with amplitude given by the integral. Recall that $g_{\mu \nu}=\eta_{\mu \nu}+h_{\mu \nu}$, thus the solution for $h^{\mu \nu}$ and $\bar{h}^{\mu \nu}$ are expressed in terms of the rectangular coordinates. Raising and lowering indices is done by $\eta^{\mu \nu}$ and $\eta_{\mu \nu}$. After these manipulations are carried out, the amplitudes can be expressed in other coordinate systems.
From energy conservation, the lowest order approximation yields
\begin{aligned} 0 & =T^{\mu \nu}{ }{,}{ }\nu=T^{\mu \nu}{ }{,}{ }\nu+\Gamma_{\xi \nu}^\mu T^{\xi \nu}+\Gamma_{\nu \xi}^\nu T^{\mu \xi} \approx T^{\mu \nu}{ }{,} \ & =T^{0 \nu}{ }\nu=T^{00}{ }{, 0}+T^{0 k}{ }{, k} \ & =\left(T^{00}{ }0+T^{0 k}{ }{, k}\right){, 0}=T^{00}{ }{, 0}, 0+T^{0 k}{ }{, k}, 0, \ T^{00}{ }{, 0}, 0 & =\left(-T^{0 k}{ }{, 0}\right){{ }k}=-\left(-T^{j k}{ }{, j}\right){, k}=T^{j k}{ }{, j}, k, \ \int_{\infty} d V x^i x^n T^{00}{ }{, 0}, 0 & =\int{\infty} d V x^i x^n T^{j k}{ }_j,{ }_k . \end{aligned}

## 物理代写|广义相对论代写General Relativity代考|Gravity Wave Flux and Power

The wave flux is its energy/area/time. In natural units it is just $\mathrm{m}^{-2}$. This quantity is integrated over the area of a sphere. That determines the power $P$ or the wave luminosity $L$. As the source loses energy, it changes, and that observation can be used to detect the wave. The result from electromagnetic waves cannot be taken over directly as their amplitudes are tensors of rank 1, while a gravity wave amplitude is a tensor of rank 2 . Thus, while the flux is still proportional to the absolute square of the amplitude, the all important proportionality factor is different. In order to calculate it, the approach of B. Schutz (2009) is followed.

Consider a plane transverse traceless wave moving in the $z$-direction. The flux transferred to an approximately continuous array of oscillators, elemental springs, is calculated. The springs are aligned along the $x$-direction in the plane $z=0$. The springs have natural length $l_0$, equal masses $m$, small spring constant $m \omega_0^2 / 2$, and small damping constant $m \gamma$. The number of springs per unit area is $\frac{d n}{d A}=\alpha$. As the oscillators acquire energy, the wave loses energy, and its amplitude decreases. The relationship between flux and amplitude is found, not to depend on the springs, but is a property of the wave. The springs are just used as calculation facilitators.

Let the origin be at a spring’s center with the masses at $x_{1,2}$. In flat space, the equations of motion for the masses are as follows:
\begin{aligned} \frac{d^2 x_2}{d t^2} & =-\omega_0^2\left(x_2-x_1-l_0\right) / 2-\gamma \frac{d\left(x_2-x_1\right)}{d t}, \ \frac{d^2 x_1}{d t^2} & =\omega_0^2\left(x_2-x_1-l_0\right) / 2+\gamma \frac{d\left(x_2-x_1\right)}{d t}, \ \frac{d^2\left(x_2-x_1-l_0\right)}{d t^2} & =-\omega_0^2\left(x_2-x_1-l_0\right)-2 \gamma \frac{d\left(x_2-x_1-l_0\right)}{d t} . \end{aligned}
One element of the wave ${ }^{T T} \bar{h}{x x}$ is considered. The other elements, and there must be other elements, since the trace is zero, would be handled in the same manner. When the wave is encountered, Eq. (7.17) yields the proper length between the masses, \begin{aligned} l & =\left(x_2-x_1\right)\left(1+{ }^{T T} \bar{h}{x x} / 2\right) \ & \approx x_2-x_1+l_0 T T \bar{h}{x x} / 2, \ x_2-x_1 & =l-l_0 T T \bar{h}{x x} / 2 . \end{aligned}

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。