标签： PHYS4654

物理代写|广义相对论代写General relativity代考|PHY475

Posted on 2023年7月19日2023年7月19日 by statistics-lab

如果你也在怎样代写广义相对论General Relativity 这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。广义相对论General RelativityGR)于1915年发表，它包含了万有引力和加速度。有关英文翻译，请参见爱因斯坦(1905)。后一种理论预测了光在大质量天体(如太阳)附近的偏转。第一次世界大战结束后不久，由a·s·爱丁顿领导的一个英国小组证实了这一惊人的预言。这使爱因斯坦举世闻名，甚至在那些对科学没有特别兴趣的人中间也是如此。

广义相对论General Relativity现在——至少——是主流物理学的一部分。报道内容相当传统;在概述了需要一个引力理论来取代牛顿的理论之后，有两章专门讨论微分几何，包括微分形式和无坐标矢量的现代公式，然后是爱因斯坦场方程，史瓦西解，透镜-蒂林效应(最近观测证实)，黑洞，克尔解，引力辐射和宇宙学。这本书以场论一章结束，描述了广义相对论和粒子物理规范理论、黎曼时空中的狄拉克方程和卡鲁扎-克莱因理论之间的相似之处

statistics-lab™ 为您的留学生涯保驾护航在代写广义相对论General relativity方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写广义相对论General relativity代写方面经验极为丰富，各种代写广义相对论General relativity相关的作业也就用不着说。

物理代写|广义相对论代写General relativity代考|Remarks on the algebra of $\rho$-forms

Let us revert to considering 2-forms; and first show that if $\boldsymbol{\omega}$ and $\boldsymbol{\sigma}$ are both 1 -forms
$$
\omega=a_i \boldsymbol{\theta}^i, \quad \boldsymbol{\sigma}=b_k \boldsymbol{\theta}^k,
$$
then their wedge product is a 2 -form:
$$
\begin{aligned}
\omega \wedge \boldsymbol{\sigma} & =a_i b_k \boldsymbol{\theta}^i \wedge \boldsymbol{\theta}^k \
& =1 / 2\left(a_i b_k-a_k b_i\right) \boldsymbol{\theta}^i \wedge \boldsymbol{\theta}^k \
& =c_{i k} \boldsymbol{\theta}^i \wedge \boldsymbol{\theta}^k
\end{aligned}
$$
with
$$
c_{i k}=-c_{k i}=1 / 2\left(c_{i k}-c_{k i}\right)=c_{[i k]}
$$
the coefficients $c_{i k}$ are the components of an antisymmetric $\left(\begin{array}{l}0 \ 2\end{array}\right)$ tensor. It is then clear that $\boldsymbol{\omega} \wedge \boldsymbol{\sigma}$ is a 2 -form, and also that
$$
\omega \wedge \sigma=-\sigma \wedge \omega .
$$
Analogous relations hold for general wedge products. Let $\boldsymbol{\alpha}$ be a $p$-form and $\boldsymbol{\beta}$ a $q$-form, so that
$$
\boldsymbol{\alpha}=a_{k_1 \ldots k_p} \boldsymbol{\theta}^{k_1} \wedge \cdots \wedge \boldsymbol{\theta}^{k_p}=a_{\left[k_1 \ldots k_p\right]} \boldsymbol{\theta}^{k_1} \wedge \cdots \wedge \boldsymbol{\theta}^{k_p}
$$
and the coefficients $a_{\left[k_1 \ldots k_p\right]}$ are the components of a totally antisymmetric $\left(\begin{array}{c}0 \ p\end{array}\right)$ tensor. A similar formula holds for $\boldsymbol{\beta}$ and it then follows, by manipulations similar to those which lead to (3.79), that
$$
\alpha \wedge \boldsymbol{\beta}=(-1)^{p q} \beta \wedge \boldsymbol{\alpha}
$$

物理代写|广义相对论代写General relativity代考|A note on orientation

We saw above that the area $A$ of a parallelogram defined by the vectors $\mathbf{v}, \mathbf{w}$ is $\pm\left|\begin{array}{cc}v_x & v_y \ w_x & w_y\end{array}\right|$ (and $A$ is always taken to be positive). For simplicity take $\mathbf{v}$ and $\mathbf{w}$ to be at right angles, and let us assume that in some local Cartesian coordinate system $\mathbf{v}$ is in the $+x$ direction and $\mathbf{w}$ in the $+y$ direction; then
$$
A=\left|\begin{array}{cc}
v_x & v_y \
w_x & w_y
\end{array}\right|=\left|\begin{array}{cc}
v_x & 0 \
0 & w_y
\end{array}\right|>0 .
$$
If, however, the $x$ and $y$ axes are interchanged (see Fig. 3.8), we find that
$$
\left|\begin{array}{cc}
v_x & v_y \
w_x & w_y
\end{array}\right|=\left|\begin{array}{cc}
0 & v_y \
w_x & 0
\end{array}\right|<0 . $$ If a space has the property that it is possible to define $A>0$ consistently over the whole space, the space is called orientable. Otherwise it is non orientable. In an orientable space the existence of two distinguishable classes $A>0$ and $A<0$ allows a global distinction between right-handed and left-handed coordinate systems over the space, but this will not hold in a non-orientable one. An example of a non-orientable (2-dimensional) space is the Möbius strip, illustrated in Fig. 3.9. A coordinate system is set up at $P$, and it is seen that after transporting it round the band to $Q$ the $x$ and $y$ axes have been interchanged, so a consistent definition of the sign of $A$ over the surface is not possible. The Möbius strip is actually also an example of a fibre bundle. This is seen as follows: a cylinder is made by drawing a rectangle and joining together the edges marked with arrows, so that the arrows are aligned, as in Fig. 3.10(a). Coordinatising the rectangle, this means that the point $(1, y)$ becomes identified with the point $(1, y)$. If, however, one of the arrows on the rectangle is inverted, then joining the edges in such a way that the arrows are still aligned results in a Möbius strip, as in Fig. 3.10(b). This corresponds to the identification of the points $(1, y)$ and $(1, y)$ in the original rectangle. Now compare the rectangles in (a) and (b). Moving from the point $(1, y)$, keeping $y$ constant but with $x$ decreasing, describes a journey on the cylinder where we eventually return to the original point – so that $x$ has completed a circuit and $y$ has remained unchanged. But on the Möbius strip after $x$ has completed a circuit, from $x=1$ to $x=1, y$ has changed. This means it is not possible to define a Cartesian coordinate system over the whole space; the space may be coordinatised by $(x, y)$, but this does not represent a Cartesian product of $x$ and $y$. Calling the $x$ axis the base space and the $y$ axis the fibre, a closed circuit in the base space results in a motion along the fibre. (The reader will recall that it was argued at the beginning of this chapter that space-time is a fibre bundle.)

广义相对论代考

物理代写|广义相对论代写General relativity代考|Remarks on the algebra of $\rho$-forms

让我们回到考虑两种形式;首先证明$\boldsymbol{\omega}$和$\boldsymbol{\sigma}$都是1 -型
$$
\omega=a_i \boldsymbol{\theta}^i, \quad \boldsymbol{\sigma}=b_k \boldsymbol{\theta}^k,
$$
那么它们的楔形积是一个2型:
$$
\begin{aligned}
\omega \wedge \boldsymbol{\sigma} & =a_i b_k \boldsymbol{\theta}^i \wedge \boldsymbol{\theta}^k \
& =1 / 2\left(a_i b_k-a_k b_i\right) \boldsymbol{\theta}^i \wedge \boldsymbol{\theta}^k \
& =c_{i k} \boldsymbol{\theta}^i \wedge \boldsymbol{\theta}^k
\end{aligned}
$$
有
$$
c_{i k}=-c_{k i}=1 / 2\left(c_{i k}-c_{k i}\right)=c_{[i k]}
$$
系数$c_{i k}$是一个反对称的$\left(\begin{array}{l}0 \ 2\end{array}\right)$张量的分量。很明显，$\boldsymbol{\omega} \wedge \boldsymbol{\sigma}$是2 -form，而且
$$
\omega \wedge \sigma=-\sigma \wedge \omega .
$$
类似的关系适用于一般的楔形积。设$\boldsymbol{\alpha}$为$p$ -form, $\boldsymbol{\beta}$为$q$ -form，这样
$$
\boldsymbol{\alpha}=a_{k_1 \ldots k_p} \boldsymbol{\theta}^{k_1} \wedge \cdots \wedge \boldsymbol{\theta}^{k_p}=a_{\left[k_1 \ldots k_p\right]} \boldsymbol{\theta}^{k_1} \wedge \cdots \wedge \boldsymbol{\theta}^{k_p}
$$
系数$a_{\left[k_1 \ldots k_p\right]}$是一个完全反对称的$\left(\begin{array}{c}0 \ p\end{array}\right)$张量的分量。类似的公式适用于$\boldsymbol{\beta}$，然后通过类似于导致式(3.79)的操作，得到
$$
\alpha \wedge \boldsymbol{\beta}=(-1)^{p q} \beta \wedge \boldsymbol{\alpha}
$$

物理代写|广义相对论代写General relativity代考|A note on orientation

我们在上面看到，由向量$\mathbf{v}, \mathbf{w}$定义的平行四边形的面积$A$等于$\pm\left|\begin{array}{cc}v_x & v_y \ w_x & w_y\end{array}\right|$ ($A$总是被认为是正的)。为简单起见，假设$\mathbf{v}$和$\mathbf{w}$成直角，并假设在某个局部笛卡尔坐标系中$\mathbf{v}$在$+x$方向，$\mathbf{w}$在$+y$方向;然后
$$
A=\left|\begin{array}{cc}
v_x & v_y \
w_x & w_y
\end{array}\right|=\left|\begin{array}{cc}
v_x & 0 \
0 & w_y
\end{array}\right|>0 .
$$
然而，如果$x$和$y$轴互换(见图3.8)，我们发现
$$
\left|\begin{array}{cc}
v_x & v_y \
w_x & w_y
\end{array}\right|=\left|\begin{array}{cc}
0 & v_y \
w_x & 0
\end{array}\right|<0 . $$如果一个空间有可能在整个空间中一致地定义$A>0$，那么这个空间就被称为可定向空间。否则它是不可定向的。在可定向空间中，两个可区分的类$A>0$和$A<0$的存在允许在空间上区分右手和左手坐标系，但这在不可定向空间中不成立。不可定向(2维)空间的一个例子是Möbius条，如图3.9所示。在$P$处建立了一个坐标系统，可以看到，在将其沿带移动到$Q$后，$x$和$y$轴被交换了，因此不可能在表面上对$A$的符号进行一致的定义。Möbius条实际上也是纤维束的一个例子。如图3.10(a)所示，通过绘制矩形并将箭头标记的边缘连接在一起，使箭头对齐，从而形成一个圆柱体。协调矩形，这意味着点$(1, y)$与点$(1, y)$相等。然而，如果矩形上的一个箭头是倒置的，那么以箭头仍然对齐的方式连接边缘会产生Möbius条，如图3.10(b)所示。这对应于原始矩形中点$(1, y)$和$(1, y)$的标识。现在比较(a)和(b)中的矩形。从点$(1, y)$开始移动，保持$y$不变，但$x$减小，描述了在圆柱体上的旅程，我们最终返回到原始点-因此$x$完成了一个回路，$y$保持不变。但是在Möbius条上，当$x$完成一条赛道后，从$x=1$到$x=1, y$的位置发生了变化。这意味着在整个空间上定义笛卡尔坐标系是不可能的;空间可以通过$(x, y)$进行协调，但这并不表示$x$和$y$的笛卡尔积。将$x$轴称为基空间，将$y$轴称为纤维，基空间中的闭合回路导致沿纤维运动。(读者应该还记得，在本章的开头，我们曾讨论过时空是一个纤维束。)

物理代写|广义相对论代写General relativity代考请认准statistics-lab™

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

金融工程是使用数学技术来解决金融问题。金融工程使用计算机科学、统计学、经济学和应用数学领域的工具和知识来解决当前的金融问题，以及设计新的和创新的金融产品。

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

术语广义线性模型（GLM）通常是指给定连续和/或分类预测因素的连续响应变量的常规线性回归模型。它包括多元线性回归，以及方差分析和方差分析（仅含固定效应）。

有限元方法代写

有限元方法（FEM）是一种流行的方法，用于数值解决工程和数学建模中出现的微分方程。典型的问题领域包括结构分析、传热、流体流动、质量运输和电磁势等传统领域。

有限元是一种通用的数值方法，用于解决两个或三个空间变量的偏微分方程（即一些边界值问题）。为了解决一个问题，有限元将一个大系统细分为更小、更简单的部分，称为有限元。这是通过在空间维度上的特定空间离散化来实现的，它是通过构建对象的网格来实现的：用于求解的数值域，它有有限数量的点。边界值问题的有限元方法表述最终导致一个代数方程组。该方法在域上对未知函数进行逼近。[1] 然后将模拟这些有限元的简单方程组合成一个更大的方程系统，以模拟整个问题。然后，有限元通过变化微积分使相关的误差函数最小化来逼近一个解决方案。

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

随机分析代写

随机微积分是数学的一个分支，对随机过程进行操作。它允许为随机过程的积分定义一个关于随机过程的一致的积分理论。这个领域是由日本数学家伊藤清在第二次世界大战期间创建并开始的。

时间序列分析代写

随机过程，是依赖于参数的一组随机变量的全体，参数通常是时间。随机变量是随机现象的数量表现，其时间序列是一组按照时间发生先后顺序进行排列的数据点序列。通常一组时间序列的时间间隔为一恒定值（如1秒，5分钟，12小时，7天，1年），因此时间序列可以作为离散时间数据进行分析处理。研究时间序列数据的意义在于现实中，往往需要研究某个事物其随时间发展变化的规律。这就需要通过研究该事物过去发展的历史记录，以得到其自身发展的规律。

回归分析代写

多元回归分析渐进（Multiple Regression Analysis Asymptotics）属于计量经济学领域，主要是一种数学上的统计分析方法，可以分析复杂情况下各影响因素的数学关系，在自然科学、社会和经济学等多个领域内应用广泛。

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，工具箱允许您学习和应用专业技术。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。

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

物理代写|广义相对论代写General relativity代考|A remark on inertial mass

Posted on 2023年6月29日2023年6月29日 by statistics-lab

如果你也在怎样代写广义相对论General Relativity 这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。广义相对论General Relativity又称广义相对论和爱因斯坦引力理论，是爱因斯坦在1915年发表的引力几何理论，是目前现代物理学中对引力的描述。广义相对论概括了狭义相对论并完善了牛顿的万有引力定律，将引力统一描述为空间和时间或四维时空的几何属性。特别是，时空的曲率与任何物质和辐射的能量和动量直接相关。这种关系是由爱因斯坦场方程规定的，这是一个二阶偏微分方程系统。

广义相对论General Relativity描述经典引力的牛顿万有引力定律，可以看作是广义相对论对静止质量分布周围几乎平坦的时空几何的预测。然而，广义相对论的一些预言却超出了经典物理学中牛顿的万有引力定律。这些预言涉及时间的流逝、空间的几何、自由落体的运动和光的传播，包括引力时间膨胀、引力透镜、光的引力红移、夏皮罗时间延迟和奇点/黑洞。到目前为止，对广义相对论的所有测试都被证明与该理论一致。广义相对论的时间相关解使我们能够谈论宇宙的历史，并为宇宙学提供了现代框架，从而导致了大爆炸和宇宙微波背景辐射的发现。尽管引入了一些替代理论，广义相对论仍然是与实验数据一致的最简单的理论。然而，广义相对论与量子物理学定律的协调仍然是一个问题，因为缺乏一个自洽的量子引力理论；以及引力如何与三种非引力–强、弱和电磁力统一起来。

物理代写|广义相对论代写General relativity代考|A remark on inertial mass

The Equivalence Principle states the equality of gravitational and inertial mass, as we have just seen above. It is worthwhile, however, making the following remark. The inertial mass of a particle refers to its mass (deduced, for example, from its behaviour analysed according to Newton’s laws) when it undergoes non-uniform, or non-inertial, motion. There are, however, two different types of such motion; it may for instance be acceleration in a straight line, or circular motion with constant speed. In the first case the magnitude of the velocity vector changes but its direction remains constant, while in the second case the magnitude is constant but the direction changes. In each of these cases the motion is non-inertial, but there is a conceptual distinction to be made. To be precise we should observe this distinction and denote the two types of mass $m_{\mathrm{i}, \text { acc }}$ and $m_{\mathrm{i}, \text { rot }}$. We believe, without, as far as I know, proper evidence, that they are equal
$$
m_{\mathrm{i}, \mathrm{acc}}=m_{\mathrm{i}, \mathrm{rot}}
$$
The interesting thing is that Einstein’s formulation of the Equivalence Principle referred to inertial mass measured in an accelerating frame, $m_{\mathrm{i}, \mathrm{acc}}$, whereas the Eötvös experiment, described above, establishes the equality (to within the stated bounds) of $m_{\mathrm{i}, \text { rot }}$ and the gravitational mass. The question is: can an experiment be devised to test the equality of $m_{\mathrm{i}, \text { acc }}$ and $m_{\mathrm{g}}$ ? Or even to test (1.19)?

物理代写|广义相对论代写General relativity代考|Tidal forces

The Principle of Equivalence is a local principle. To see this, consider the Einstein box in the gravitational field of the Earth, as in Fig. 1.3. If the box descends over a large distance towards the centre of the Earth, it is clear that two test bodies in the box will approach one another, so over this extended journey it is clear that they are in a genuine gravitational field, and not in an accelerating frame (in which they would stay the same distance apart). In other words, the Equivalence Principle has broken down. We conclude that this principle is only valid as a local principle. Over small distances a gravitational field is equivalent to an acceleration, but over larger distances this equivalence breaks down. The effect is known as a tidal effect, and ultimately is due to the curvature produced by a real gravitational field.
Another way of stating the situation is to note that an object in free fall is in an inertial frame. The effect of the gravitational field has been cancelled by the acceleration of the elevator (the ‘acceleration due to gravity’). The accelerations required to annul the gravitational fields of the two test bodies, however, are slightly different, because they are directed along the radius vectors. So the inertial frames of the two bodies differ slightly. The frames are ‘locally inertial’. The Equivalence Principle treats a gravitational field at a single point as equivalent to an acceleration, but it is clear that no gravitational fields encountered in nature give rise to a uniform acceleration. Most real gravitational fields are produced by more or less spherical objects like the Earth, so the equivalence in question is only a local one.

We may find an expression for the tidal forces which result from this non-locality. Figure 1.4 shows the forces exerted on the two test bodies – call them $\mathrm{A}$ and $\mathrm{B}$ – in the gravitational field of a body at $\mathrm{O}$. They both experience a force towards $\mathrm{O}$ of magnitude where $m$ is the mass of $\mathrm{A}$ and $\mathrm{B}, M$ is the mass of the Earth and $r$ the distance of $\mathrm{A}$ and $\mathrm{B}$ from its centre. In addition, let the distance between $\mathrm{A}$ and $\mathrm{B}$ be $x$. Consider the frame in which $\mathrm{A}$ is at rest. This frame is realised by applying a force equal and opposite to $F_{\mathrm{A}}$, to both $\mathrm{A}$ and $\mathrm{B}$, as shown in Fig. 1.4. In this frame, B experiences a force $F$, directed towards A, which is the vector sum of $F_{\mathrm{B}}$ and $F_{\mathrm{A}}$ :
$$
F=2 F_{\mathrm{A}} \sin \alpha=2 F_{\mathrm{A}} \cdot \frac{x}{2 r}=\frac{m M G}{r^3} x .
$$
A then observes B to be accelerating towards him with an acceleration given by $F=m \mathrm{~d}^2 x /$ $\mathrm{d} t^2$, i.e.
$$
\frac{\mathrm{d}^2 x}{\mathrm{~d} t^2}=-\frac{M G}{r^3} x
$$
The $1 / r^3$ behaviour is characteristic of tidal forces.

广义相对论代考

物理代写|广义相对论代写General relativity代考|A remark on inertial mass

等效原理说明了引力质量和惯性质量是相等的，就像我们刚才看到的那样。然而，提出以下评论是值得的。一个粒子的惯性质量是指它在进行非均匀或非惯性运动时的质量(例如，根据牛顿定律分析它的行为推导出来的)。然而，这种运动有两种不同的类型;例如，它可以是直线加速，也可以是匀速圆周运动。在第一种情况下，速度矢量的大小改变，但方向保持不变，而在第二种情况下，速度矢量的大小不变，但方向改变。在每一种情况下，运动都是非惯性的，但是要做一个概念上的区别。确切地说，我们应该注意到这种区别，并表示两种类型的质量$m_{\mathrm{i}, \text { acc }}$和$m_{\mathrm{i}, \text { rot }}$。据我所知，在没有适当证据的情况下，我们相信它们是平等的
$$
m_{\mathrm{i}, \mathrm{acc}}=m_{\mathrm{i}, \mathrm{rot}}
$$
有趣的是，爱因斯坦的等效原理的公式指的是在加速坐标系中测量的惯性质量，$m_{\mathrm{i}, \mathrm{acc}}$，而上面描述的Eötvös实验，建立了$m_{\mathrm{i}, \text { rot }}$和引力质量的等式(在规定的范围内)。问题是:可以设计一个实验来测试$m_{\mathrm{i}, \text { acc }}$和$m_{\mathrm{g}}$的相等性吗?或者甚至测试(1.19)?

物理代写|广义相对论代写General relativity代考|Tidal forces

等效原理是一种局部原理。要了解这一点，请考虑地球引力场中的爱因斯坦盒子，如图1.3所示。如果盒子向地球中心下降了一段很长的距离，很明显盒子里的两个测试物体会相互靠近，所以在这段漫长的旅程中，很明显它们处于真正的引力场中，而不是在加速坐标系中(在加速坐标系中，它们之间的距离会保持不变)。换句话说，等效原理失效了。我们得出结论，这一原则仅作为局部原则有效。在较短的距离上，引力场相当于加速度，但在较长的距离上，这个等价就失效了。这种效应被称为潮汐效应，最终是由于实际引力场产生的曲率。
说明这种情况的另一种方法是注意到自由落体的物体处于惯性系中。引力场的影响已经被电梯的加速度抵消了(“重力加速度”)。然而，使两个测试体的引力场失效所需的加速度略有不同，因为它们是沿半径矢量方向的。所以这两个物体的惯性系稍有不同。坐标系是“局部惯性的”。等效原理将单点处的引力场等效为加速度，但很明显，在自然界中遇到的引力场不会产生均匀加速度。大多数真实的引力场或多或少都是由像地球这样的球形物体产生的，因此所讨论的等效性只是局部的。

我们可以找到这种非定域性所产生的潮汐力的表达式。图1.4显示了施加在两个测试体上的力-称之为它们 $\mathrm{A}$ 和 $\mathrm{B}$ -在物体的引力场中 $\mathrm{O}$．它们都经历了朝向的力 $\mathrm{O}$ 的大小 $m$ 的质量 $\mathrm{A}$ 和 $\mathrm{B}, M$ 地球的质量是多少 $r$ 的距离 $\mathrm{A}$ 和 $\mathrm{B}$ 从它的中心。此外，让距离之间 $\mathrm{A}$ 和 $\mathrm{B}$ 他 $x$．考虑一下这个框架 $\mathrm{A}$ 是静止的。这个坐标系是通过施加一个相等且相反的力来实现的 $F_{\mathrm{A}}$，两者都有 $\mathrm{A}$ 和 $\mathrm{B}$如图1.4所示。在这个坐标系中，B受到一个力 $F$，指向A，它是向量的和 $F_{\mathrm{B}}$ 和 $F_{\mathrm{A}}$ ：
$$
F=2 F_{\mathrm{A}} \sin \alpha=2 F_{\mathrm{A}} \cdot \frac{x}{2 r}=\frac{m M G}{r^3} x .
$$
然后A观察到B向他加速，加速度为 $F=m \mathrm{~d}^2 x /$ $\mathrm{d} t^2$，即:
$$
\frac{\mathrm{d}^2 x}{\mathrm{~d} t^2}=-\frac{M G}{r^3} x
$$
The $1 / r^3$ 潮汐力的特性。

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

物理代写|广义相对论代写General relativity代考|Fluids

Posted on 2023年6月29日2023年6月29日 by statistics-lab

物理代写|广义相对论代写General relativity代考|Fluids

In order to solve Einstein’s equation (5.9), the energy-momentum tensor $T_{\mu \nu}$ is needed. Also, even though gravity is weak, GR is required. For example, if the universe had a uniform mass density $\bar{\rho}$, the quantity $M / r=4 \pi \bar{\rho} r^2 / 3>1$, at some $r$. In order to make headway, the many bodies making up the universe are subject to a simplifying assumption, namely that they constitute a perfect fluid. Then an momentum-energy tensor, that makes testable predictions, can be obtained.

In the SR frame $\mathrm{O}$ where all the particles are at rest the fluid is called dust. In this frame, the number density of dust particles is $N / V=n \mathrm{~m}^{-3}$. In another SR frame $\mathrm{O}^{\prime}$, where the particles are moving with velocity $\vec{v}^{\prime}$, the volume element is contracted by a factor $1 / \gamma=\left(1-\left|\vec{v}^{\prime}\right|^2\right)^{1 / 2}$. In $\mathrm{O}^{\prime}$ $n^{\prime}=\gamma n$.

The flux of particles across a surface is the number crossing the surface in the direction of the normal to the surface per unit area per unit time. Thus, all particles within a distance $v^{\bar{i}^{\prime}} \Delta t^{\prime}$ of the surface, where $v^{\bar{i}^{\prime}}$ is the speed in the normal direction and within area $\Delta A$ that defines the size of the surface, will cross the surface in time period $\Delta t^{\prime}$,
$$
\begin{aligned}
f & =\left(n^{\prime} v^{\bar{i}^{\prime}} \Delta t^{\prime} \Delta A\right) /\left(\Delta t^{\prime} \Delta A\right) \
& =\gamma n v^{i^{\prime}} .
\end{aligned}
$$

Note that the vector $N^{\bar{\mu}^{\prime}}=n U^{\bar{\mu}^{\prime}}$ combines both the flux and number density,
$$
\begin{aligned}
N^{\overline{0}^{\prime}} & =n U^{\overline{0}^{\prime}}=\gamma n, \
N^{\bar{i}^{\prime}} & =n U^{\bar{i}^{\prime}}=\gamma n v^{\bar{i}^{\prime}}, \
N^{\bar{\mu}^{\prime}} N_{\bar{\mu}^{\prime}} & =\gamma^2 n^2\left(-1+\left(v^{\prime}\right)^2\right)=-n^2 .
\end{aligned}
$$

物理代写|广义相对论代写General relativity代考|Robertson–Walker Einstein Dynamics

The nonzero C symbols, Ricci tensor and scalar, and Einstein tensor for the Robertson-Walker metric were evaluated in Problems 4.7 and 4.8. The following results were obtained for the $\mathrm{C}$ symbols,
$$
\begin{aligned}
\Gamma_{i i}^0 & =g_{i i} \frac{1}{Q} \frac{d Q}{d t}, \quad \Gamma_{0 i}^i=\frac{1}{Q} \frac{d Q}{d t}, \
\Gamma_{r r}^r & =\frac{k r}{1-k r^2}, \sin ^2 \theta \Gamma_{\theta \theta}^r=-r\left(1-k r^2\right) \sin ^2 \theta=\Gamma_{\phi \phi}^r, \
\Gamma_{\theta r}^\theta & =r^{-1}, \Gamma_{\phi \phi}^\theta=-\sin \theta \cos \theta, \quad \Gamma_{\phi r}^\phi=r^{-1}, \Gamma_{\phi \theta}^\phi=\cot \theta .
\end{aligned}
$$
These allowed the calculation of the Ricci tensor and scalar,
$$
\begin{aligned}
R_{00} & ==-3 \frac{1}{Q} \frac{d^2 Q}{d t^2}=R^{00}, \
R_{i i} & =g_{i i} \frac{1}{Q^2}\left[2\left(k+\left(\frac{d Q}{d t}\right)^2\right)+Q \frac{d^2 Q}{d t^2}\right]=g_{i i}^2 R^{i i}=g_{i i} R^{i i} / g^{i i}, \
R & =g^{\mu \nu} R_{\mu \nu}=g^{\mu \mu} R_{\mu \mu}=6 \frac{1}{Q^2}\left[Q \frac{d^2 Q}{d t^2}+\left(\frac{d Q}{d t}\right)^2+k\right],
\end{aligned}
$$
and the Einstein tensor elements,
$$
\begin{aligned}
G_{\mu \nu} & =R_{\mu \nu}-g_{\mu \nu} R / 2, \quad G^{\mu \nu}=R^{\mu \nu}-g^{\mu \nu} R / 2, \
G_{00} & =3 \frac{1}{Q^2}\left[\left(\frac{d Q}{d t}\right)^2+k\right], \
G_{i i} & =-g_{i i} \frac{1}{Q^2}\left[2 Q \frac{d^2 Q}{d t^2}+\left(\frac{d Q}{d t}\right)^2+k\right] .
\end{aligned}
$$

广义相对论代考

物理代写|广义相对论代写General relativity代考|Fluids

为了解爱因斯坦方程(5.9)，需要能量动量张量$T_{\mu \nu}$。此外，即使重力很弱，GR也是必需的。例如，如果宇宙有一个均匀的质量密度$\bar{\rho}$，量$M / r=4 \pi \bar{\rho} r^2 / 3>1$，在$r$。为了取得进展，构成宇宙的许多物体服从于一个简化的假设，即它们构成一种完美的流体。然后可以得到一个动量-能量张量，它可以进行可测试的预测。

在SR框架$\mathrm{O}$中，所有的颗粒都处于静止状态，流体被称为尘埃。在这个框架中，尘埃粒子的数量密度为$N / V=n \mathrm{~m}^{-3}$。在另一个SR坐标系$\mathrm{O}^{\prime}$中，粒子以速度$\vec{v}^{\prime}$运动，体积元被一个因子$1 / \gamma=\left(1-\left|\vec{v}^{\prime}\right|^2\right)^{1 / 2}$收缩。在$\mathrm{O}^{\prime}$$n^{\prime}=\gamma n$。

通过表面的粒子通量是单位时间内单位面积上沿表面法线方向通过表面的粒子数。因此，在距离表面$v^{\bar{i}^{\prime}} \Delta t^{\prime}$的范围内的所有粒子，其中$v^{\bar{i}^{\prime}}$是法线方向上的速度，并且在定义表面大小的面积$\Delta A$内，将在时间周期$\Delta t^{\prime}$内穿过表面。
$$
\begin{aligned}
f & =\left(n^{\prime} v^{\bar{i}^{\prime}} \Delta t^{\prime} \Delta A\right) /\left(\Delta t^{\prime} \Delta A\right) \
& =\gamma n v^{i^{\prime}} .
\end{aligned}
$$

注意，向量$N^{\bar{\mu}^{\prime}}=n U^{\bar{\mu}^{\prime}}$结合了通量和数密度，
$$
\begin{aligned}
N^{\overline{0}^{\prime}} & =n U^{\overline{0}^{\prime}}=\gamma n, \
N^{\bar{i}^{\prime}} & =n U^{\bar{i}^{\prime}}=\gamma n v^{\bar{i}^{\prime}}, \
N^{\bar{\mu}^{\prime}} N_{\bar{\mu}^{\prime}} & =\gamma^2 n^2\left(-1+\left(v^{\prime}\right)^2\right)=-n^2 .
\end{aligned}
$$

物理代写|广义相对论代写General relativity代考|Robertson–Walker Einstein Dynamics

在问题4.7和4.8中对Robertson-Walker度规的非零C符号、Ricci张量和标量以及Einstein张量进行了评估。对$\mathrm{C}$符号得到如下结果:
$$
\begin{aligned}
\Gamma_{i i}^0 & =g_{i i} \frac{1}{Q} \frac{d Q}{d t}, \quad \Gamma_{0 i}^i=\frac{1}{Q} \frac{d Q}{d t}, \
\Gamma_{r r}^r & =\frac{k r}{1-k r^2}, \sin ^2 \theta \Gamma_{\theta \theta}^r=-r\left(1-k r^2\right) \sin ^2 \theta=\Gamma_{\phi \phi}^r, \
\Gamma_{\theta r}^\theta & =r^{-1}, \Gamma_{\phi \phi}^\theta=-\sin \theta \cos \theta, \quad \Gamma_{\phi r}^\phi=r^{-1}, \Gamma_{\phi \theta}^\phi=\cot \theta .
\end{aligned}
$$
这样就可以计算里奇张量和标量，
$$
\begin{aligned}
R_{00} & ==-3 \frac{1}{Q} \frac{d^2 Q}{d t^2}=R^{00}, \
R_{i i} & =g_{i i} \frac{1}{Q^2}\left[2\left(k+\left(\frac{d Q}{d t}\right)^2\right)+Q \frac{d^2 Q}{d t^2}\right]=g_{i i}^2 R^{i i}=g_{i i} R^{i i} / g^{i i}, \
R & =g^{\mu \nu} R_{\mu \nu}=g^{\mu \mu} R_{\mu \mu}=6 \frac{1}{Q^2}\left[Q \frac{d^2 Q}{d t^2}+\left(\frac{d Q}{d t}\right)^2+k\right],
\end{aligned}
$$
和爱因斯坦张量元素，
$$
\begin{aligned}
G_{\mu \nu} & =R_{\mu \nu}-g_{\mu \nu} R / 2, \quad G^{\mu \nu}=R^{\mu \nu}-g^{\mu \nu} R / 2, \
G_{00} & =3 \frac{1}{Q^2}\left[\left(\frac{d Q}{d t}\right)^2+k\right], \
G_{i i} & =-g_{i i} \frac{1}{Q^2}\left[2 Q \frac{d^2 Q}{d t^2}+\left(\frac{d Q}{d t}\right)^2+k\right] .
\end{aligned}
$$

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

物理代写|广义相对论代写General relativity代考|Equations of Motion

Posted on 2023年6月29日2023年6月29日 by statistics-lab

物理代写|广义相对论代写General relativity代考|Equations of Motion

Motion in the equatorial plane $\theta=\pi / 2$ of a black hole is considered. Problem 7 shows this is a possible solution, even in Kerr space. The equations of motion for objects with finite rest mass $m$ are most easily obtained by noting $0=g_{\mu \nu, 0}=g_{\mu \nu, 2}$. So $U_0=P_0 / m$ and $U_2=P_2 / m$ are constant. The constants will be chosen so that the equations of motion match Eqs. (6.2)(6.5), with Eq. (6.6), when $a=0$. Thus,
$$
\begin{aligned}
\frac{d \phi}{d \tau} & =U^2=g^{2 \nu} U_\nu \
& =g^{20} U_0+g^{22} U_2=\frac{g_{20} U_0-g_{00} U_2}{\Lambda} .
\end{aligned}
$$
When $a=0$,
$$
\frac{d \phi}{d \tau}=U_2 \frac{-g_{00}}{\Lambda}=U_2 \frac{(1-R / r)}{r^2(1-R / r)}=\frac{U_2}{r^2}
$$
Equations (6.2) and (6.6) yield
$$
\begin{aligned}
\frac{d \phi}{d \tau} & =\frac{J}{r^2 E^{\prime 1 / 2}}, \text { so choose, } \
U_2 & =\frac{J}{E^{\prime 1 / 2}} .
\end{aligned}
$$
Carrying on,
$$
\begin{aligned}
\frac{d t}{d \tau} & =U^0=g^{0 \nu} U_\nu \
& =g^{00} U_0+g^{02} U_2=\frac{-g_{22} U_0+g_{02} U_2}{\Lambda} .
\end{aligned}
$$

物理代写|广义相对论代写General relativity代考|Circular Motion

First the circular orbits of photons are considered. It is convenient to calculate $\left(\frac{d r}{d q}\right)^2$ and then take $E^{\prime}=0$ :
$$
\begin{aligned}
\left(\frac{d r}{d q}\right)^2 & =-E^{\prime} \frac{\Lambda}{r^2}+1-\frac{J^2-a^2}{r^2}+\frac{(J-a)^2 R}{r^3} \
& =1-\frac{J^2-a^2}{r^2}+\frac{(J-a)^2 R}{r^3} \
1 & =\left(\frac{d r}{d q}\right)^2+\frac{J^2-a^2}{r^2}-\frac{(J-a)^2 R}{r^3}
\end{aligned}
$$
Per unit mass, the left-hand side of the above equation is a constant energy and on the right-hand side is a kinetic energy term plus an equivalent gravitational potential energy term $\equiv V_p[r]$. For circular motion $r=R_c$, $J=J_c, \frac{d r}{d q}=0$, and $\frac{d V_p}{d r}=0$. For stable orbits, $\frac{d^2 V_p}{d r^2}>0$, while for unstable orbits, $\frac{d^2 V_p}{d r^2}<0$. In this case,
$$
\begin{aligned}
0 & =1+\left(J_c-a\right)^2 R / R_c^3-\left(J_c^2-a^2\right) / R_c^2, \
\left.\frac{d V_p}{d r}\right|{R_c} & =0=3\left(J_c-a\right)^2 R / R_c-2\left(J_c^2-a^2\right), \ \left.\frac{d^2 V_p}{d r^2}\right|{R_c} & =-3\left(J_c-a\right)^2 R / R_c^5<0 .
\end{aligned}
$$
From Eq. (8.75), it is seen that photon circular orbits are unstable. From Eq. (8.74),
$$
R_c=\frac{3 R}{2} \frac{\left(J_c-a\right)^2}{J_c^2-a^2}=\frac{3 R}{2} \frac{J_c-a}{J_c+a}
$$

广义相对论代考

物理代写|广义相对论代写General relativity代考|Equations of Motion

考虑黑洞在赤道平面$\theta=\pi / 2$上的运动。问题7表明这是一个可能的解决方案，即使在克尔空间中也是如此。静止质量有限的物体$m$的运动方程最容易通过记$0=g_{\mu \nu, 0}=g_{\mu \nu, 2}$得到。所以$U_0=P_0 / m$和$U_2=P_2 / m$是常数。常数的选择将使运动方程与方程相匹配。(6.2)(6.5)，与式(6.6)，当$a=0$。因此，
$$
\begin{aligned}
\frac{d \phi}{d \tau} & =U^2=g^{2 \nu} U_\nu \
& =g^{20} U_0+g^{22} U_2=\frac{g_{20} U_0-g_{00} U_2}{\Lambda} .
\end{aligned}
$$
当$a=0$，
$$
\frac{d \phi}{d \tau}=U_2 \frac{-g_{00}}{\Lambda}=U_2 \frac{(1-R / r)}{r^2(1-R / r)}=\frac{U_2}{r^2}
$$
式(6.2)和式(6.6)得
$$
\begin{aligned}
\frac{d \phi}{d \tau} & =\frac{J}{r^2 E^{\prime 1 / 2}}, \text { so choose, } \
U_2 & =\frac{J}{E^{\prime 1 / 2}} .
\end{aligned}
$$
继续，
$$
\begin{aligned}
\frac{d t}{d \tau} & =U^0=g^{0 \nu} U_\nu \
& =g^{00} U_0+g^{02} U_2=\frac{-g_{22} U_0+g_{02} U_2}{\Lambda} .
\end{aligned}
$$

物理代写|广义相对论代写General relativity代考|Circular Motion

首先考虑光子的圆形轨道。方便计算$\left(\frac{d r}{d q}\right)^2$，然后取$E^{\prime}=0$:
$$
\begin{aligned}
\left(\frac{d r}{d q}\right)^2 & =-E^{\prime} \frac{\Lambda}{r^2}+1-\frac{J^2-a^2}{r^2}+\frac{(J-a)^2 R}{r^3} \
& =1-\frac{J^2-a^2}{r^2}+\frac{(J-a)^2 R}{r^3} \
1 & =\left(\frac{d r}{d q}\right)^2+\frac{J^2-a^2}{r^2}-\frac{(J-a)^2 R}{r^3}
\end{aligned}
$$
每单位质量，上面方程的左边是一个恒定的能量右边是动能项加上一个等效的重力势能项$\equiv V_p[r]$。对于圆周运动$r=R_c$, $J=J_c, \frac{d r}{d q}=0$和$\frac{d V_p}{d r}=0$。对于稳定轨道，$\frac{d^2 V_p}{d r^2}>0$，对于不稳定轨道，$\frac{d^2 V_p}{d r^2}<0$。在这种情况下，
$$
\begin{aligned}
0 & =1+\left(J_c-a\right)^2 R / R_c^3-\left(J_c^2-a^2\right) / R_c^2, \
\left.\frac{d V_p}{d r}\right|{R_c} & =0=3\left(J_c-a\right)^2 R / R_c-2\left(J_c^2-a^2\right), \ \left.\frac{d^2 V_p}{d r^2}\right|{R_c} & =-3\left(J_c-a\right)^2 R / R_c^5<0 .
\end{aligned}
$$
由式(8.75)可知，光子圆轨道是不稳定的。由式(8.74)可知，
$$
R_c=\frac{3 R}{2} \frac{\left(J_c-a\right)^2}{J_c^2-a^2}=\frac{3 R}{2} \frac{J_c-a}{J_c+a}
$$

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

物理代写|广义相对论代写General relativity代考|Binary Neutron Star System Radiation

Posted on 2023年6月9日2023年6月9日 by statistics-lab

物理代写|广义相对论代写General relativity代考|Binary Neutron Star System Radiation

The binary pulsar PSR B1913+16, shown schematically in Fig. 7.3, was discovered by R. Hulse and J. H. Taylor (Hulse, 1975). This occurred in 1973 at the Arecibo radio telescope. It is in a gravitational bound state with an unseen neutron star. The pulsar is a magnetized neutron star, whose rapid rotation generates a plasma, the source of beamed radio waves. They are seen at earth, as periodic pulses, every $0.059 \mathrm{~s}$. This is because the radio waves are beamed along the magnetic axis, but that axis rotates about the spin axis of the star. The rotation period of such a massive compact body is very stable against external perturbations. It is actually a very accurate clock. Modern timing devices can measure the period with high precision. Search the WWW for “pulsar” and you’ll find some wonderful images. Neutron stars are highly compact massive objects. They are supported against gravitational collapse by neutron degeneracy, a purely quantum effect. For a mass of $1.4 M_s$, the neutron star surface radial coordinate is $\approx 10-20 \mathrm{~km}$.

Upon discovery, it was noted that the pulsing rate varied. This was interpreted as due to the pulsar traveling in a changing gravitational field, as it orbited an unseen neighbor. The pulsar was tracked for decades, and the parameters of the orbit were obtained from the slight changes in the pulsing rate. The orbiting is a source of a gravitational wave. The wave carries away energy, reflected in changes in the orbit. Since the pulsar is a radio emitter, the experimenters have to remove the distortion, due to the index of refraction of the intergalactic medium. An optical pulsar would have been simpler.

The discoverers were joined by J. M. Weisberg who performed much of the data analysis. Their paper Weisberg (2010) and references therein, describe the intricacies of extracting the orbit parameters. They found that this is a wonderful system with which to test GR. For example, the advance of the periastron is $\approx 35,000$ times that of the perihelion of Mercury. The periastron is the distance of closest approach to its unseen neighbor. However, the prize here is the detection of a gravitational wave carrying energy away from the system.

物理代写|广义相对论代写General relativity代考|Static Black Holes

In the era before SR and GR, there was speculation about possible compact spherically symmetric objects of large mass $M^{\prime}$ and radius $R$ from which even light could not escape. The incorrect argument was made on the basis of the escape speed $v_E$, of an object of mass $M$, using Newtonian mechanics. The escape speed condition is that the total energy vanishes at $r=R$. Then the object stops at $r=\infty$. This yields
$$
M\left(v_E\right)^2 / 2-M M^{\prime} / R=0, \quad v_E=\left(2 M^{\prime} / R\right)^{1 / 2} .
$$
So when $R=2 M^{\prime}, v_E=1$, and light would be bound to the compact object.

The connection to GR is easy to see. It is just where the Schwarzschild metric has a singularity other than $r=0$. As spherical coordinates will be used, let $r^p \equiv(r)^p$. The metric is
$$
\begin{aligned}
(d \tau)^2= & \left(1-2 M^{\prime} / r\right)(d t)^2-\left(1-2 M^{\prime} / r\right)^{-1}(d r)^2 \
& -r^2\left[(d \theta)^2+\sin ^2 \theta(d \phi)^2\right] \
= & (1-R / r)(d t)^2-(1-R / r)^{-1}(d r)^2-r^2\left[(d \theta)^2+\sin ^2 \theta(d \phi)^2\right] .
\end{aligned}
$$
Thus,
$$
1-R / r=0, \quad(1-R / r)^{-1}=\infty \quad \text { when } r=R,
$$
where $R$ is called the Schwarzschild radius. For an object with the mass of the sun, it has a very small value $R=2 M_s=2.968 \times 10^3 \mathrm{~m}$. In the case of the sun, such a radius is well within the sun’s radius, and wouldn’t contain much of the sun’s mass. A black hole, however, is a real singularity at $r=0$, and $R$ is external to it. In the region accessible to observation $R / r<1$, the applications of the Schwarzschild metric found in Chapter 6 apply. However, to emphasize that black holes are spoken of, $2 M^{\prime}$ will be replaced by $R$ for the rest of this chapter.

The apparent singularity at $R$ is not real and is due to the choice of coordinates. This can be seen by recalling that in deriving the Schwarzschild metric, the Ricci tensor $R_{\mu \nu}$ vanished in vacuum. Thus, there can’t be a real singularity at the vacuum point $r=R$. One can seek other coordinates that make the apparent singularity disappear. The following ones, known as the Kruskal coordinates (Kruskal, 1960) $\left(r^{\prime}, t^{\prime}\right)$, do the trick:
$$
\begin{aligned}
r^{\prime 2}-t^{\prime 2} & =K^2(r / R-1) \exp [r / R] \
2 r^{\prime} t^{\prime} /\left(r^{\prime 2}+t^{\prime 2}\right) & =\tanh [t / R], t=R \tanh ^{-1}\left[2 r^{\prime} t^{\prime} /\left(r^{\prime 2}+t^{\prime 2}\right)\right]
\end{aligned}
$$

广义相对论代考

物理代写|广义相对论代写General relativity代考|Binary Neutron Star System Radiation

双星脉冲星PSR B1913+16如图7.3所示，是由R. Hulse和J. H. Taylor (Hulse, 1975)发现的。这发生在1973年的阿雷西博射电望远镜上。它与一颗看不见的中子星处于引力束缚状态。脉冲星是一颗磁化的中子星，它的快速旋转产生等离子体，这是射电电波的来源。它们在地球上被视为周期性脉冲，每$0.059 \mathrm{~s}$一次。这是因为无线电波是沿着磁轴发射的，但磁轴是围绕恒星的自转轴旋转的。这样一个大质量致密体的旋转周期对外部扰动是非常稳定的。它实际上是一个非常精确的钟。现代计时装置可以高精度地测量周期。在WWW上搜索“脉冲星”，你会发现一些美妙的图像。中子星是高度致密的大质量物体。中子简并(一种纯粹的量子效应)支持它们对抗引力坍缩。对于质量为$1.4 M_s$的中子星，其表面径向坐标为$\approx 10-20 \mathrm{~km}$。

发现后，人们注意到脉冲速率是变化的。这被解释为由于脉冲星在一个变化的引力场中运行，因为它绕着一个看不见的邻居运行。这颗脉冲星被跟踪了几十年，轨道的参数是从脉冲速率的微小变化中得到的。轨道是引力波的来源。波带走能量，反映在轨道的变化上。由于脉冲星是射电发射器，实验人员必须消除由于星系间介质折射率造成的畸变。光学脉冲星会更简单。

J. M. Weisberg加入了发现者的行列，他进行了大量的数据分析。他们的论文Weisberg(2010)和其中的参考文献描述了提取轨道参数的复杂性。他们发现这是一个很好的测试GR的系统。例如，近日点的速度是水星近日点速度的$\approx 35,000$倍。近日点是它与看不见的邻居最接近的距离。然而，这里的奖励是探测到引力波携带能量离开系统。

物理代写|广义相对论代写General relativity代考|Static Black Holes

在SR和GR之前的时代，有人推测可能存在致密的球对称物体，其质量$M^{\prime}$和半径$R$大，甚至光也无法逃脱。错误的论点是基于逃逸速度$v_E$，质量物体$M$，使用牛顿力学。逃逸速度条件是总能量在$r=R$处消失。然后对象停在$r=\infty$。这产生了
$$
M\left(v_E\right)^2 / 2-M M^{\prime} / R=0, \quad v_E=\left(2 M^{\prime} / R\right)^{1 / 2} .
$$
所以当$R=2 M^{\prime}, v_E=1$，光会被束缚在致密的物体上。

与GR的联系很容易看到。就是史瓦西度规有一个奇点而不是$r=0$的地方。由于将使用球坐标，设$r^p \equiv(r)^p$。度规是
$$
\begin{aligned}
(d \tau)^2= & \left(1-2 M^{\prime} / r\right)(d t)^2-\left(1-2 M^{\prime} / r\right)^{-1}(d r)^2 \
& -r^2\left[(d \theta)^2+\sin ^2 \theta(d \phi)^2\right] \
= & (1-R / r)(d t)^2-(1-R / r)^{-1}(d r)^2-r^2\left[(d \theta)^2+\sin ^2 \theta(d \phi)^2\right] .
\end{aligned}
$$
因此，
$$
1-R / r=0, \quad(1-R / r)^{-1}=\infty \quad \text { when } r=R,
$$
$R$被称为史瓦西半径。对于太阳质量的物体，它的值非常小$R=2 M_s=2.968 \times 10^3 \mathrm{~m}$。以太阳为例，这样的半径正好在太阳的半径之内，不会包含太多太阳的质量。然而，黑洞在$r=0$是一个真正的奇点，而$R$在它的外部。在可观测的$R / r<1$区域，适用于第6章中发现的史瓦西度规的应用。然而，为了强调黑洞的存在，在本章的其余部分，$2 M^{\prime}$将被$R$所取代。

在$R$的明显奇点是不真实的，是由于坐标的选择。回想一下，在推导史瓦西度规时，里奇张量$R_{\mu \nu}$在真空中消失了，就可以看出这一点。因此，在真空点$r=R$不可能有真正的奇点。我们可以寻找其他的坐标使奇点消失。下面这些，被称为Kruskal坐标(Kruskal, 1960) $\left(r^{\prime}, t^{\prime}\right)$，可以做到这一点:
$$
\begin{aligned}
r^{\prime 2}-t^{\prime 2} & =K^2(r / R-1) \exp [r / R] \
2 r^{\prime} t^{\prime} /\left(r^{\prime 2}+t^{\prime 2}\right) & =\tanh [t / R], t=R \tanh ^{-1}\left[2 r^{\prime} t^{\prime} /\left(r^{\prime 2}+t^{\prime 2}\right)\right]
\end{aligned}
$$

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

物理代写|广义相对论代写General relativity代考|Gravity Wave Detection: LIGO Experiment

Posted on 2023年6月9日2023年6月9日 by statistics-lab

物理代写|广义相对论代写General relativity代考|Gravity Wave Detection: LIGO Experiment

In order to see how the wave affects free particles, consider a particle initially at rest in a Lorentz frame. The velocity components are $U^0=1$ and $U^i=0$. As previously shown, in the TT gauge this velocity forced ${ }^{T T} h_{\mu 0}=0$. The following equations of motion for the particle’s coordinates are obtained
$$
\begin{aligned}
0 & =\frac{d U^\mu}{d \tau}+\Gamma_{\nu \chi}^\mu U^\nu U^\chi \
\frac{d U^\mu}{d \tau}[x=0] & =-\Gamma_{00}^\mu \
& =-\eta^{\mu \sigma}\left({ }^{T T} h_{\sigma 0,0}+{ }^{T T} h_{0 \sigma, 0}-{ }^{T T} h_{00, \sigma}\right) / 2=0 .
\end{aligned}
$$
So initially the acceleration vanishes. This means that the particle will be at rest at a later time, and by the same argument, the acceleration will be zero at a later time. Thus, the particle can remain at its coordinate in this gauge. This has no invariant geometrical meaning. However, suppose particles at $x=0, \epsilon$ experience a wave. The proper distance between them changes to
$$
\begin{aligned}
l & =\int_0^\epsilon\left(g_{x x}\right)^{1 / 2} d x \approx \epsilon\left(g_{x x}[x=0]\right)^{1 / 2} \
& =\epsilon\left(1+{ }^{T T} h_{x x}[x=0]\right)^{1 / 2} \
& \approx \epsilon\left(1+{ }^{T T} h_{x x}[x=0] / 2\right) .
\end{aligned}
$$
Thus, the wave can be detected because the proper distance between two objects will wiggle in its presence.

物理代写|广义相对论代写General relativity代考|Wave Equation Solution With Sources

If there are no sources, plane wave solutions are allowed. However, without knowledge of the source, the wave amplitudes and the power carried by the wave, cannot be calculated. When scientists build a detector-like LIGO, they know the minimum signals that can be observed. Before many hundreds of millions of dollars are spent, one would like to be confident, that the sought for gravitational waves, carry enough power to be detected. Thus solution of the wave equation with sources is a necessary part of the program. Such a signal is distinctive from the background noise, even though the two may have the same power level.
Most readers have solved the wave equation with sources in electrodynamics courses. They may skip this section without loss. It is included for completeness, and as a handy reference. The method of Fourier analysis is used to solve Eq. (7.6). This analysis makes use of the following theorem concerning the Dirac delta function:
$$
\delta\left(t^{\prime}-t\right)=\frac{1}{2 \pi} \int_{-\infty}^{\infty} d \omega \exp \left[i \omega\left(t^{\prime}-t\right)\right]
$$
One sees that Eq. (7.6) looks like,
$$
\square \Psi[\vec{r}, t]=\left(\nabla^2-\frac{\partial^2}{\partial t^2}\right) \Psi[\vec{r}, t]=-4 \pi S[\vec{r}, t],
$$
where $\Psi=\bar{h}^{\mu \nu}$ is the wave function and $S=4 T^{\mu \nu}$ is the source. Using Fourier transforms, one can write
$$
\Psi[\vec{r}, t]=\frac{1}{2 \pi} \int_{-\infty}^{\infty} d \omega \Psi[\vec{r}, \omega] \exp [-i \omega t]
$$
where
$$
\Psi[\vec{r}, \omega]=\int_{-\infty}^{\infty} d t^{\prime} \Psi\left[\vec{r}, t^{\prime}\right] \exp \left[i \omega t^{\prime}\right]
$$
since
$$
\begin{aligned}
\Psi[\vec{r}, t] & =\frac{1}{2 \pi} \int_{-\infty}^{\infty} d \omega \int_{-\infty}^{\infty} d t^{\prime} \Psi\left[\vec{r}, t^{\prime}\right] \exp \left[i \omega\left(t^{\prime}-t\right)\right] \
& \left.=\int_{-\infty}^{\infty} d t^{\prime} \Psi\left[\vec{r}, t^{\prime}\right] \frac{1}{2 \pi} \int_{-\infty}^{\infty} d \omega \exp \left[i \omega\left(t^{\prime}-t\right)\right]\right) \
& =\int_{-\infty}^{\infty} d t^{\prime} \Psi\left[\vec{r}, t^{\prime}\right] \delta\left(t^{\prime}-t\right)=\Psi[\vec{r}, t] .
\end{aligned}
$$

广义相对论代考

物理代写|广义相对论代写General relativity代考|Gravity Wave Detection: LIGO Experiment

为了了解波如何影响自由粒子，考虑一个粒子最初在洛伦兹坐标系中静止。速度分量为$U^0=1$和$U^i=0$。如前所示，在TT测量中，此速度强制${ }^{T T} h_{\mu 0}=0$。得到了粒子坐标的运动方程
$$
\begin{aligned}
0 & =\frac{d U^\mu}{d \tau}+\Gamma_{\nu \chi}^\mu U^\nu U^\chi \
\frac{d U^\mu}{d \tau}[x=0] & =-\Gamma_{00}^\mu \
& =-\eta^{\mu \sigma}\left({ }^{T T} h_{\sigma 0,0}+{ }^{T T} h_{0 \sigma, 0}-{ }^{T T} h_{00, \sigma}\right) / 2=0 .
\end{aligned}
$$
所以一开始加速度消失了。这意味着粒子将在稍后的时间处于静止状态，同样的道理，加速度将在稍后的时间为零。因此，粒子可以保持在这个量规的坐标上。这没有不变的几何意义。然而，假设在$x=0, \epsilon$处的粒子经历了波。它们之间的固有距离变为
$$
\begin{aligned}
l & =\int_0^\epsilon\left(g_{x x}\right)^{1 / 2} d x \approx \epsilon\left(g_{x x}[x=0]\right)^{1 / 2} \
& =\epsilon\left(1+{ }^{T T} h_{x x}[x=0]\right)^{1 / 2} \
& \approx \epsilon\left(1+{ }^{T T} h_{x x}[x=0] / 2\right) .
\end{aligned}
$$
因此，这个波可以被探测到，因为两个物体之间的适当距离会在它的存在下摆动。

物理代写|广义相对论代写General relativity代考|Wave Equation Solution With Sources

如果没有源，则允许采用平面波解。然而，如果不知道震源，就无法计算出波的振幅和波所携带的能量。当科学家们建造类似LIGO的探测器时，他们知道可以观测到的最小信号。在花费数亿美元之前，人们希望有信心，所寻找的引力波具有足够的能量来被探测到。因此，求解带源的波动方程是程序的必要部分。这样的信号与背景噪声不同，尽管两者可能具有相同的功率水平。
大多数读者在电动力学课程中都解过带源的波动方程。他们可以跳过这一节而不会有损失。包括它是为了完整性，并作为一个方便的参考。采用傅立叶分析方法求解式(7.6)。这种分析利用了狄拉克函数的下列定理:
$$
\delta\left(t^{\prime}-t\right)=\frac{1}{2 \pi} \int_{-\infty}^{\infty} d \omega \exp \left[i \omega\left(t^{\prime}-t\right)\right]
$$
方程(7.6)是这样的，
$$
\square \Psi[\vec{r}, t]=\left(\nabla^2-\frac{\partial^2}{\partial t^2}\right) \Psi[\vec{r}, t]=-4 \pi S[\vec{r}, t],
$$
其中$\Psi=\bar{h}^{\mu \nu}$为波函数，$S=4 T^{\mu \nu}$为源。用傅里叶变换，可以写成
$$
\Psi[\vec{r}, t]=\frac{1}{2 \pi} \int_{-\infty}^{\infty} d \omega \Psi[\vec{r}, \omega] \exp [-i \omega t]
$$
在哪里
$$
\Psi[\vec{r}, \omega]=\int_{-\infty}^{\infty} d t^{\prime} \Psi\left[\vec{r}, t^{\prime}\right] \exp \left[i \omega t^{\prime}\right]
$$
自从
$$
\begin{aligned}
\Psi[\vec{r}, t] & =\frac{1}{2 \pi} \int_{-\infty}^{\infty} d \omega \int_{-\infty}^{\infty} d t^{\prime} \Psi\left[\vec{r}, t^{\prime}\right] \exp \left[i \omega\left(t^{\prime}-t\right)\right] \
& \left.=\int_{-\infty}^{\infty} d t^{\prime} \Psi\left[\vec{r}, t^{\prime}\right] \frac{1}{2 \pi} \int_{-\infty}^{\infty} d \omega \exp \left[i \omega\left(t^{\prime}-t\right)\right]\right) \
& =\int_{-\infty}^{\infty} d t^{\prime} \Psi\left[\vec{r}, t^{\prime}\right] \delta\left(t^{\prime}-t\right)=\Psi[\vec{r}, t] .
\end{aligned}
$$

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

物理代写|广义相对论代写General relativity代考|Equations of Motion

Posted on 2023年6月9日2023年6月9日 by statistics-lab

物理代写|广义相对论代写General relativity代考|Equations of Motion

The metric in the solar system is due mainly to the relativistic mass of the sun $M^{\prime}=M_s$. The general two-body problem in GR has not been solved analytically, so the sun’s mass will be taken as much larger than that of the planet considered, and the gravitational effects of other planets will be neglected. Also the sun’s rotation will be neglected. Under these conditions, the metric is that obtained by Schwarzschild. In this chapter, spherical coordinates will be used and $r^p \equiv(r)^p$. The equations of motion involve the C symbols, and from Eqs. (5.16)-(5.18),
$$
\begin{aligned}
\exp [-2 \Delta] & =1-2 M^{\prime} / r, \quad-2 \Delta=\ln \left[1-2 M^{\prime} / r\right] \
\exp [2 \Phi] & =1-2 M^{\prime} / r, \quad 2 \Phi=\ln \left[1-2 M^{\prime} / r\right] \
\Gamma_{r t}^t & =\Phi,r=\left(1-2 M^{\prime} / r\right)^{-1}\left(M^{\prime} / r^2\right), \ \Gamma{r r}^r & =\Delta, r=-\left(1-2 M^{\prime} / r\right)^{-1}\left(M^{\prime} / r^2\right), \
\Gamma_{t t}^r & =\exp [2 \Phi] \exp [-2 \Delta] \Phi, r=\left(1-2 M^{\prime} / r\right)\left(M^{\prime} / r^2\right), \
\sin ^2 \theta \Gamma_{\theta \theta}^r & =-r \sin ^2 \theta\left(1-2 M^{\prime} / r\right)=\Gamma_{\phi \phi}^r
\end{aligned}
$$
Two constants of the motion are expected for massive particles, since $g_{\mu \nu, 0}$ and $g_{\mu \nu, 2}=0$. However, the motion of both photons and massive particles is desired. So the equations of motion (3.25) are written in terms of an affine parameter $q$,
$$
0=\frac{d^2 x^\mu}{d q^2}+\Gamma_{\chi \nu}^\mu \frac{d x^\chi}{d q} \frac{d x^\nu}{d q}
$$

物理代写|广义相对论代写General relativity代考|Orbit Equations

The orbit equations are obtained from Eqs. (6.2), (6.3), (6.5) and (6.6), e.g.,
$$
\begin{gathered}
\frac{d \phi}{d r}=\frac{d \phi}{d q} / \frac{d r}{d q}= \pm \frac{J}{r^2\left(1-2 M^{\prime} / r\right)^{1 / 2}} \frac{1}{\left(\left[\left(1-2 M^{\prime} / r\right)^{-1}-E^{\prime}\right]-[J / r]^2\right)^{1 / 2}}, \
D[r] \equiv \frac{1}{\left(\left[\left(1-2 M^{\prime} / r\right)^{-1}-E^{\prime}\right] / J^2-1 / r^2\right)^{1 / 2}} .
\end{gathered}
$$

Thus,
$$
\begin{gathered}
d \phi= \pm \frac{d r}{r^2\left(1-2 M^{\prime} / r\right)^{1 / 2}} D[r], \
d t= \pm \frac{d r}{J\left(1-2 M^{\prime} / r\right)^{3 / 2}} D[r], \
d \tau= \pm \frac{d r E^{\prime 1 / 2}}{J\left(1-2 M^{\prime} / r\right)^{1 / 2}} D[r] .
\end{gathered}
$$
It should be noted that $t$ is the time according to a faraway at-rest observer, while $\tau$ is the time on a clock attached to the particle. Integration of the above equations will give $r$ as a function of $\phi, t$, or $\tau$. The correct sign is determined by the result of the integration. Each integral is an elliptic integral, and could be worked out numerically. However, if $M^{\prime} / r \ll 1$ everywhere, then to first order in $M^{\prime} / r$, the integration can be done analytically. In order to accomplish this, one expands $\left(1-2 M^{\prime} / r\right)^n=$ $1-n\left(2 M^{\prime} / r\right)+[n(n-1) / 2]\left(2 M^{\prime} / r\right)^2+\cdots$, and keeps the lowest order term. In some of the mathematical manipulations below, I have followed Weinberg (1972), and filled in some steps.

广义相对论代考

物理代写|广义相对论代写General relativity代考|Equations of Motion

太阳系中的公制主要是由于太阳的相对论性质量$M^{\prime}=M_s$。广义相对论中一般的二体问题还没有解析解决，所以太阳的质量将被认为比所考虑的行星的质量大得多，而其他行星的引力效应将被忽略。太阳的自转也将被忽略。在这些条件下，度规就是史瓦西得到的度规。在本章中，将使用球坐标和$r^p \equiv(r)^p$。运动方程包含C符号，从方程。(5.16)-(5.18);
$$
\begin{aligned}
\exp [-2 \Delta] & =1-2 M^{\prime} / r, \quad-2 \Delta=\ln \left[1-2 M^{\prime} / r\right] \
\exp [2 \Phi] & =1-2 M^{\prime} / r, \quad 2 \Phi=\ln \left[1-2 M^{\prime} / r\right] \
\Gamma_{r t}^t & =\Phi,r=\left(1-2 M^{\prime} / r\right)^{-1}\left(M^{\prime} / r^2\right), \ \Gamma{r r}^r & =\Delta, r=-\left(1-2 M^{\prime} / r\right)^{-1}\left(M^{\prime} / r^2\right), \
\Gamma_{t t}^r & =\exp [2 \Phi] \exp [-2 \Delta] \Phi, r=\left(1-2 M^{\prime} / r\right)\left(M^{\prime} / r^2\right), \
\sin ^2 \theta \Gamma_{\theta \theta}^r & =-r \sin ^2 \theta\left(1-2 M^{\prime} / r\right)=\Gamma_{\phi \phi}^r
\end{aligned}
$$
对于大质量粒子，有两个运动常数，因为$g_{\mu \nu, 0}$和$g_{\mu \nu, 2}=0$。然而，光子和大质量粒子的运动是需要的。所以运动方程(3.25)是用仿射参数$q$表示的，
$$
0=\frac{d^2 x^\mu}{d q^2}+\Gamma_{\chi \nu}^\mu \frac{d x^\chi}{d q} \frac{d x^\nu}{d q}
$$

物理代写|广义相对论代写General relativity代考|Orbit Equations

轨道方程由方程得到。(6.2)、(6.3)、(6.5)和(6.6)，例如:
$$
\begin{gathered}
\frac{d \phi}{d r}=\frac{d \phi}{d q} / \frac{d r}{d q}= \pm \frac{J}{r^2\left(1-2 M^{\prime} / r\right)^{1 / 2}} \frac{1}{\left(\left[\left(1-2 M^{\prime} / r\right)^{-1}-E^{\prime}\right]-[J / r]^2\right)^{1 / 2}}, \
D[r] \equiv \frac{1}{\left(\left[\left(1-2 M^{\prime} / r\right)^{-1}-E^{\prime}\right] / J^2-1 / r^2\right)^{1 / 2}} .
\end{gathered}
$$

因此，
$$
\begin{gathered}
d \phi= \pm \frac{d r}{r^2\left(1-2 M^{\prime} / r\right)^{1 / 2}} D[r], \
d t= \pm \frac{d r}{J\left(1-2 M^{\prime} / r\right)^{3 / 2}} D[r], \
d \tau= \pm \frac{d r E^{\prime 1 / 2}}{J\left(1-2 M^{\prime} / r\right)^{1 / 2}} D[r] .
\end{gathered}
$$
应该指出，$t$是远处静止的观察者的时间，而$\tau$是附着在粒子上的时钟上的时间。以上方程的积分将得到$r$作为$\phi, t$或$\tau$的函数。正确的符号是由积分结果决定的。每个积分都是一个椭圆积分，可以用数值计算。然而，如果$M^{\prime} / r \ll 1$无处不在，那么到$M^{\prime} / r$的一阶，积分可以解析地完成。为了实现这一点，展开$\left(1-2 M^{\prime} / r\right)^n=$$1-n\left(2 M^{\prime} / r\right)+[n(n-1) / 2]\left(2 M^{\prime} / r\right)^2+\cdots$，并保留最低阶项。在下面的一些数学操作中，我遵循了Weinberg(1972)，并填写了一些步骤。

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

物理代写|广义相对论代写General relativity代考|Parallel Transport

Posted on 2023年5月12日2023年5月12日 by statistics-lab

物理代写|广义相对论代写General relativity代考|Parallel Transport

The characterization of curvature starts with the concept of parallel transport. On a flat surface, as in Fig. 4.2, draw an arbitrary closed path $A B C A$. Here, a circle is used, so that at various places along the path, some tangent vectors $\vec{W}$, that can point in all directions, are shown. For parallel transport start at $A$, draw on the surface, a small parallel transport vector $\vec{V}$ in any direction. Proceed to a neighboring point on the path, draw on the surface a small transport vector, as parallel as possible to the $\vec{V}$ previously drawn. On a flat surface, it is possible to draw the vector exactly parallel. When once again at $A$, the identical vectors would be redrawn. In this sense, a flat surface has no intrinsic curvature. A cylinder can be constructed by rolling a flat sheet, and so has no intrinsic curvature.

A sphere cannot be made from a flat sheet. It has intrinsic curvature. One can find at least one path on the sphere’s surface, as in Fig. 4.3, for which the vectors $\vec{V}$, would not repeat. Pick the path $A B C A$, such that $B$ and $C$ are on the equator, and $A$ is at a pole. At $A$, start with a vector $\vec{V}$ on the sphere’s surface, that is tangent to an arc of longitude. As one proceeds to $B$, along the longitude, a new parallel transport vector cannot be drawn on the surface, exactly parallel to $\vec{V}$. The best one can do is draw that vector along the tangent vector. At $B$ that vector is perpendicular to the equator, and remains so as one proceeds to $C$. From there the return to $A$ is again along a longitude. The parallel transport vectors on the surface will be opposite the tangent vectors. Upon reaching $A$, the final parallel transport vector is different from the initial one.

In spacetime, these vectors have four components $V^\mu, W^\mu$. At any point $P$, one can go to a locally inertial frame. In a small enough neighborhood of $P$, as you proceed along the curve specified by affine parameter $q$ and $W^{\bar{\nu}}=\frac{d x^{\bar{\nu}}}{d q}$, the vector $V^{\bar{\mu}}$ is constant. This leads to a tensor equation, that is taken as the frame invariant definition of parallel transport of $V^\mu$ along $W^\nu$
$$
0=\left.\frac{d V^{\bar{\mu}}}{d q}\right|P=V{, \bar{\nu}}^{\bar{\nu}} \frac{d x^{\bar{\nu}}}{d q}=W^{\bar{\nu}} V_{, \bar{\nu}}^{\bar{\mu}}=W^{\bar{\nu}} V^{\bar{\mu}} ; \bar{\nu}=W^\nu V^\mu ;_\nu
$$

物理代写|广义相对论代写General relativity代考|Curvature Tensors

When gravity is present and there are no boundary surfaces, there is no reason to allow just $x^{1,2}$ to vary, so let them be replaced by generalized coordinates, $x^{\gamma, \lambda}$. Then, the above quantity in parentheses is defined as the Riemann curvature tensor,
$$
R_{\beta \gamma \lambda}^\mu=\Gamma_{\lambda \beta}^\nu \Gamma_{\gamma \nu}^\mu-\Gamma_{\gamma \beta}^\mu, \lambda-\Gamma_{\gamma \beta}^\nu \Gamma_{\lambda \nu}^\mu+\Gamma_{\lambda \beta}^\mu,\gamma $$ The proof that $R{\beta \gamma \lambda}^\mu$ is a tensor was carried out in Problem 3.9 , where the following results were obtained for vector $V^\mu$ :
$$
V^\mu ;\lambda ; V^\mu{ }{; \gamma} ; \lambda=R_{\beta \lambda \gamma}^\mu V^\beta, R_{\beta \lambda \gamma}^\mu=-R_{\beta \gamma \lambda}^\mu
$$
Since the covariant derivative of a tensor is a tensor, the left-hand side of Eq. (4.6) is a tensor of rank 3. The right-hand side of the first equality must be a tensor. Since $V^\beta$ is a tensor of rank $1, R_{\beta \lambda \gamma}^\mu$ must be a tensor of rank 4.

This tensor simplifies for rectangular coordinates in a locally inertial frame because the $\mathrm{C}$ symbols, but not their partial derivatives vanish, and
$$
\begin{aligned}
g^{\bar{\mu} \bar{\nu}} ;{\bar{\chi}} & =g^{\bar{\mu} \bar{\nu}}, \bar{\chi}=\eta^{\mu \nu}, \bar{\chi}=0 \ \Gamma{\bar{\lambda} \bar{\beta}}^{\bar{\mu}}, \bar{\gamma} & =\left[\left(g^{\bar{\alpha} \bar{\mu}}\left[g_{\bar{\beta} \bar{\alpha}, \bar{\lambda}}+g_{\bar{\lambda} \bar{\alpha}, \bar{\beta}}-g_{\bar{\beta} \bar{\lambda}, \bar{\alpha}}\right]\right), \bar{\gamma}\right] / 2 \
& =\left(g^{\bar{\alpha} \bar{\mu}}, \bar{\gamma}\left[g_{\bar{\beta} \bar{\alpha}, \bar{\lambda}}+g_{\bar{\lambda} \bar{\alpha}, \bar{\beta}}-g_{\bar{\beta} \bar{\lambda}, \bar{\alpha}}\right]+g^{\bar{\alpha} \bar{\mu}}\left[g_{\bar{\beta} \bar{\alpha}, \bar{\lambda}}+g_{\bar{\lambda} \bar{\alpha}, \bar{\beta}}-g_{\bar{\beta} \bar{\lambda}, \bar{\alpha}}\right], \bar{\gamma}\right) / 2 \
& =\left(g^{\bar{\alpha} \bar{\mu}}\left[g_{\bar{\beta} \bar{\alpha}}, \bar{\lambda}+g_{\bar{\lambda} \bar{\alpha}}, \bar{\beta}-g_{\bar{\beta} \bar{\lambda}}, \bar{\alpha}\right], \bar{\gamma}\right) / 2 \
& =g^{\bar{\alpha} \bar{\mu}}\left(g_{\bar{\beta} \bar{\alpha}}, \bar{\lambda}, \bar{\gamma}+g_{\bar{\lambda} \bar{\alpha}, \bar{\beta}, \bar{\gamma}}-g_{\bar{\beta} \bar{\lambda}, \bar{\alpha}, \bar{\gamma}}\right) / 2
\end{aligned}
$$

广义相对论代考

物理代写|广义相对论代写General relativity代考|Parallel Transport

曲率的表征始于平行移动的概念。在平面上，如图4.2所示，绘制任意闭合路径$A B C A$。这里用了一个圆，所以在路径的不同地方，一些切向量$\vec{W}$，可以指向所有的方向。对于从$A$开始的平行移动，在表面上画一个小的平行移动矢量$\vec{V}$在任何方向上。继续到路径上的邻近点，在表面上绘制一个小的传输矢量，尽可能平行于先前绘制的$\vec{V}$。在一个平面上，可以画出完全平行的矢量。当再次在$A$时，相同的向量将被重新绘制。从这个意义上说，一个平面没有内在曲率。一个圆柱体可以通过滚动一个平板来构造，因此它没有固有曲率。

平坦的薄片不能做成球体。它有固有曲率。我们可以在球面上找到至少一条路径，如图4.3所示，对于这条路径，向量$\vec{V}$不会重复。选择路径$A B C A$，这样$B$和$C$在赤道上，$A$在极点上。在$A$，从球面上的矢量$\vec{V}$开始，它与经度弧线相切。当一个人沿着经度到达$B$时，不能在表面上画一个新的平行移动向量，与$\vec{V}$完全平行。最好的办法就是沿着切向量画这个向量。在$B$处，这个矢量垂直于赤道，一直到$C$处都是如此。从那里回到$A$还是沿着一条经度。表面上的平行移动向量与切向量相对。到达$A$后，最终的平行移动向量与初始的不同。

在时空中，这些向量有四个分量$V^\mu, W^\mu$。在任意一点$P$，都可以进入局部惯性系。在$P$的一个足够小的邻域内，当您沿着仿射参数$q$和$W^{\bar{\nu}}=\frac{d x^{\bar{\nu}}}{d q}$指定的曲线前进时，矢量$V^{\bar{\mu}}$是常数。这就得到了一个张量方程，作为$V^\mu$沿轴平行移动的坐标系不变定义 $W^\nu$
$$
0=\left.\frac{d V^{\bar{\mu}}}{d q}\right|P=V{, \bar{\nu}}^{\bar{\nu}} \frac{d x^{\bar{\nu}}}{d q}=W^{\bar{\nu}} V_{, \bar{\nu}}^{\bar{\mu}}=W^{\bar{\nu}} V^{\bar{\mu}} ; \bar{\nu}=W^\nu V^\mu ;_\nu
$$

物理代写|广义相对论代写General relativity代考|Curvature Tensors

当重力存在并且没有边界面时，没有理由允许$x^{1,2}$变化，所以让它们被广义坐标$x^{\gamma, \lambda}$代替。那么，上面括号中的量定义为黎曼曲率张量，
$$
R_{\beta \gamma \lambda}^\mu=\Gamma_{\lambda \beta}^\nu \Gamma_{\gamma \nu}^\mu-\Gamma_{\gamma \beta}^\mu, \lambda-\Gamma_{\gamma \beta}^\nu \Gamma_{\lambda \nu}^\mu+\Gamma_{\lambda \beta}^\mu,\gamma $$在问题3.9中证明了$R{\beta \gamma \lambda}^\mu$是张量，对于向量$V^\mu$得到如下结果:
$$
V^\mu ;\lambda ; V^\mu{ }{; \gamma} ; \lambda=R_{\beta \lambda \gamma}^\mu V^\beta, R_{\beta \lambda \gamma}^\mu=-R_{\beta \gamma \lambda}^\mu
$$
因为张量的协变导数是一个张量，所以方程(4.6)的左边是一个秩为3的张量。第一个等式的右边必须是一个张量。因为$V^\beta$是一个秩张量$1, R_{\beta \lambda \gamma}^\mu$一定是一个秩4张量。

这个张量在局部惯性坐标系中简化为直角坐标因为$\mathrm{C}$符号，而不是它们的偏导数消失了，并且
$$
\begin{aligned}
g^{\bar{\mu} \bar{\nu}} ;{\bar{\chi}} & =g^{\bar{\mu} \bar{\nu}}, \bar{\chi}=\eta^{\mu \nu}, \bar{\chi}=0 \ \Gamma{\bar{\lambda} \bar{\beta}}^{\bar{\mu}}, \bar{\gamma} & =\left[\left(g^{\bar{\alpha} \bar{\mu}}\left[g_{\bar{\beta} \bar{\alpha}, \bar{\lambda}}+g_{\bar{\lambda} \bar{\alpha}, \bar{\beta}}-g_{\bar{\beta} \bar{\lambda}, \bar{\alpha}}\right]\right), \bar{\gamma}\right] / 2 \
& =\left(g^{\bar{\alpha} \bar{\mu}}, \bar{\gamma}\left[g_{\bar{\beta} \bar{\alpha}, \bar{\lambda}}+g_{\bar{\lambda} \bar{\alpha}, \bar{\beta}}-g_{\bar{\beta} \bar{\lambda}, \bar{\alpha}}\right]+g^{\bar{\alpha} \bar{\mu}}\left[g_{\bar{\beta} \bar{\alpha}, \bar{\lambda}}+g_{\bar{\lambda} \bar{\alpha}, \bar{\beta}}-g_{\bar{\beta} \bar{\lambda}, \bar{\alpha}}\right], \bar{\gamma}\right) / 2 \
& =\left(g^{\bar{\alpha} \bar{\mu}}\left[g_{\bar{\beta} \bar{\alpha}}, \bar{\lambda}+g_{\bar{\lambda} \bar{\alpha}}, \bar{\beta}-g_{\bar{\beta} \bar{\lambda}}, \bar{\alpha}\right], \bar{\gamma}\right) / 2 \
& =g^{\bar{\alpha} \bar{\mu}}\left(g_{\bar{\beta} \bar{\alpha}}, \bar{\lambda}, \bar{\gamma}+g_{\bar{\lambda} \bar{\alpha}, \bar{\beta}, \bar{\gamma}}-g_{\bar{\beta} \bar{\lambda}, \bar{\alpha}, \bar{\gamma}}\right) / 2
\end{aligned}
$$

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

物理代写|广义相对论代写General relativity代考|Differentiation of Invariants and Vectors

Posted on 2023年5月12日2023年5月12日 by statistics-lab

物理代写|广义相对论代写General relativity代考|Differentiation of Invariants and Vectors

In the previous chapters, the importance of tensors in spacetime was stressed. Any time a new quantity is encountered, it will have to be checked to see if it is a tensor. If it isn’t, its transformation properties are not obvious. Construction of new tensors has, so far, taken the form of products of known tensors or total differentiation with respect to $\tau$. For example, $(d \tau)^2=d r_\mu d r^\mu, g^{\mu \nu} g_{\xi \nu}=\delta_{\xi}^\mu=\delta_{\xi}^\mu$, or $U^\mu=\frac{d r^\mu}{d \tau}$. From studies of the calculus of 3 -vectors, one recalls that partial differentiation with respect to the coordinates produces new 3-vectors and scalars through the gradient and divergence operations. In spacetime, such partial differentiation also leads to important new tensors.

Consider an invariant that is a function of position, $\Phi=\Phi\left(x^\mu\right)=$ $\Phi\left(x^{\mu^{\prime}}\right)$, e.g., $d \tau$. It has no index associated with it. Taking the partial derivative with respect to a coordinate yields
$$
\Phi,{ }{,}=x^{\xi^{\prime}}, \Phi{, \xi^{\prime}}
$$
However, this is the rule for the transformation of a covariant vector and so another vector is added to our arsenal.

The gradient of a scalar $\Phi$ is given by $g^{\mu \nu} \Phi,{ }_\nu$ because in an inertial frame the expected results for the spatial components are obtained
$$
\begin{aligned}
\nabla^{\bar{\mu}} \Phi & \equiv g^{\bar{\mu} \bar{\nu}} \Phi, \bar{\nu}=\eta^{\mu \nu} \Phi, \bar{\nu} \
\vec{\nabla} \Phi & =\Phi,{ }_x \hat{e}_x+\Phi{ }_y \hat{e}_y+\Phi, z \hat{e}_z
\end{aligned}
$$
ligned}
$$

物理代写|广义相对论代写General relativity代考|Differentiation of Tensors

Given two vectors $V$ and $W$, the product, $V^\mu W_\nu$, transforms like a mixed tensor of rank 2 , and its covariant derivative yields
$$
\begin{aligned}
T_\nu^\mu ;\alpha & =\left(V^\mu W\nu\right) ;\alpha=V^\mu ;\alpha W_\nu+V^\mu W_\nu ; \alpha_\alpha \
& =\left(V^\mu{ }\alpha+\Gamma{\beta \alpha}^\mu V^\beta\right) W_\nu+V^\mu\left(W_\nu,\alpha-\Gamma{\nu \alpha}^\beta W_\beta\right) \
& =\left(V^\mu W_\nu\right){ }\alpha+\Gamma{\beta \alpha}^\mu V^\beta W_\nu-\Gamma_{\nu \alpha}^\beta V^\mu W_\beta \
& =T_\nu^\mu{ }\alpha+\Gamma{\beta \alpha}^\mu T_\nu^\beta-\Gamma_{\nu \alpha}^\beta T_\beta^\mu,
\end{aligned}
$$
yielding a mixed tensor of rank 3 . The contravariant index requires a positive sign, while the covariant index requires a negative sign for the $\mathrm{C}$ symbol. In a similar manner, one obtains the covariant derivatives of a covariant or contravariant tensor of rank 2 . If the rank is higher, say $n$, then $n \mathrm{C}$ symbols with appropriate signs are needed. In the case of the metric tensor,
$$
\begin{gathered}
g^{\mu \nu} ;\alpha=g^{\mu \nu}{ }\alpha+\Gamma_{\beta \alpha}^\mu g^{\beta \nu}+\Gamma_{\alpha \beta}^\nu g^{\mu \beta}=0, \
g_{\mu \nu} ;\alpha=g{\mu \nu},\alpha-\Gamma{\mu \alpha}^\beta g_{\beta \nu}-\Gamma_{\alpha \nu}^\beta g_{\mu \beta}=0 .
\end{gathered}
$$
The reason the above tensors are zero is that in an inertial frame $g_{\bar{\mu} \bar{\nu}} ; \bar{\alpha}=$ $\eta_{\mu \nu} ; \bar{\alpha}=\eta_{\mu \nu}, \bar{\alpha}=0$. As this is a tensor equation, it holds in all frames, and leads to the more useful form for $\Gamma_{\mu \nu}^\lambda$,
$$
\begin{aligned}
& 0=g_{\mu \nu} ; \alpha+g_{\mu \alpha} ;{ }\nu-g{\alpha \nu} ; \mu \
& =g_{\mu \nu},\alpha+g{\mu \alpha}, \nu-g_{\alpha \nu},{ }\mu-\Gamma{\mu \alpha}^\beta g_{\beta \nu}-\Gamma_{\alpha \nu}^\beta g_{\mu \beta} \
& -\Gamma_{\mu \nu}^\beta g_{\beta \alpha}-\Gamma_{\alpha \nu}^\beta g_{\mu \beta}+\Gamma_{\mu \alpha}^\beta g_{\beta \nu}+\Gamma_{\mu \nu}^\beta g_{\alpha \beta}, \
& 2 g_{\mu \beta} \Gamma_{\alpha \nu}^\beta=\left(g_{\mu \nu},{ }\alpha+g{\mu \alpha, \nu}-g_{\alpha \nu}, \mu\right) \text {, } \
& 2 g^{\mu \lambda} g_{\mu \beta} \Gamma_{\alpha \nu}^\beta=2 \delta_\beta^\lambda \Gamma_{\alpha \nu}^\beta=g^{\mu \lambda}\left(g_{\mu \nu}, \alpha+g_{\mu \alpha},\nu-g{\alpha \nu}, \mu\right) \text {, } \
& \Gamma_{\alpha \nu}^\lambda=g^{\mu \lambda}\left(g_{\mu \nu},\alpha+g{\mu \alpha, \nu}-g_{\alpha \nu, \mu}\right) / 2 \text {. } \
&
\end{aligned}
$$

广义相对论代考

物理代写|广义相对论代写General relativity代考|Differentiation of Invariants and Vectors

时候遇到一个新的量，都必须检查它是否是张量。如果不是，它的变换性质就不明显。到目前为止，新张量的构造都是已知张量的积或关于$\tau$的全微分的形式。例如,$ (d \τ)^ 2 = d r_ \ d r ^ \μμg ^{\μ\ν}g_ {\ xi \ν}= \ delta_ {\ xi} ^ \μ= \ delta_ {\ xi} ^ \μ美元,美元U ^ \μ= \压裂{d r ^ \μ}{d \τ}$。从3向量微积分的研究中，我们可以回忆起关于坐标的偏微分通过梯度和散度运算产生新的3向量和标量。在时空中，这种偏微分也导致了重要的新张量。

考虑一个位置函数的不变量，$\Phi=\Phi\left(x^\mu\right)=$ $\Phi\left(x^{\mu^{\prime}}\right)$，例如$d \tau$。它没有与之相关的索引。对坐标求偏导数
$ $
\φ,{}{}= x ^ {\ xi ^{\ ‘}},φ\ \ {,xi ^ {\ ‘}}
$ $
然而，这是协变向量变换的规则所以另一个向量加入了我们的武器库。

标量$\Phi$的梯度由$g^{\mu \nu} \Phi，{}_\nu$给出，因为在惯性系中得到了空间分量的期望结果
$ $
开始{对齐}
酒吧\微分算符^{{\μ}}\φ& \枚g ^{\酒吧{\μ}\酒吧{\ν}}\φ,{\ν}= \ \酒吧埃塔^{μ\ }\φ,{\ν}\ \酒吧
vec{\微分算符}\ \ &φ= \φ,{}φ值\帽子}{e值+ \ {}_y吗\帽子{e} _y吗+ \φ,z \ {e} _z帽子
结束{对齐}
$ $
线}
$ $

物理代写|广义相对论代写General relativity代考|Differentiation of Tensors

给定两个向量$V$和$W$，乘积$V^\mu W_\nu$像一个2阶的混合张量一样变换，它的协变导数是
$$
\begin{aligned}
T_\nu^\mu ;\alpha & =\left(V^\mu W\nu\right) ;\alpha=V^\mu ;\alpha W_\nu+V^\mu W_\nu ; \alpha_\alpha \
& =\left(V^\mu{ }\alpha+\Gamma{\beta \alpha}^\mu V^\beta\right) W_\nu+V^\mu\left(W_\nu,\alpha-\Gamma{\nu \alpha}^\beta W_\beta\right) \
& =\left(V^\mu W_\nu\right){ }\alpha+\Gamma{\beta \alpha}^\mu V^\beta W_\nu-\Gamma_{\nu \alpha}^\beta V^\mu W_\beta \
& =T_\nu^\mu{ }\alpha+\Gamma{\beta \alpha}^\mu T_\nu^\beta-\Gamma_{\nu \alpha}^\beta T_\beta^\mu,
\end{aligned}
$$
得到一个3阶的混合张量。逆变指标需要一个正号，而协变指标需要一个负号来表示$\mathrm{C}$符号。用类似的方法，我们可以得到秩为2的协变张量或逆变张量的协变导数。如果排名较高，例如$n$，则需要使用带有适当符号的$n \mathrm{C}$符号。在度规张量的情况下，
$$
\begin{gathered}
g^{\mu \nu} ;\alpha=g^{\mu \nu}{ }\alpha+\Gamma_{\beta \alpha}^\mu g^{\beta \nu}+\Gamma_{\alpha \beta}^\nu g^{\mu \beta}=0, \
g_{\mu \nu} ;\alpha=g{\mu \nu},\alpha-\Gamma{\mu \alpha}^\beta g_{\beta \nu}-\Gamma_{\alpha \nu}^\beta g_{\mu \beta}=0 .
\end{gathered}
$$
以上张量为零的原因是在惯性系$g_{\bar{\mu} \bar{\nu}} ; \bar{\alpha}=$$\eta_{\mu \nu} ; \bar{\alpha}=\eta_{\mu \nu}, \bar{\alpha}=0$中。因为这是一个张量方程，它适用于所有坐标系，并引出了$\Gamma_{\mu \nu}^\lambda$更有用的形式，
$$
\begin{aligned}
& 0=g_{\mu \nu} ; \alpha+g_{\mu \alpha} ;{ }\nu-g{\alpha \nu} ; \mu \
& =g_{\mu \nu},\alpha+g{\mu \alpha}, \nu-g_{\alpha \nu},{ }\mu-\Gamma{\mu \alpha}^\beta g_{\beta \nu}-\Gamma_{\alpha \nu}^\beta g_{\mu \beta} \
& -\Gamma_{\mu \nu}^\beta g_{\beta \alpha}-\Gamma_{\alpha \nu}^\beta g_{\mu \beta}+\Gamma_{\mu \alpha}^\beta g_{\beta \nu}+\Gamma_{\mu \nu}^\beta g_{\alpha \beta}, \
& 2 g_{\mu \beta} \Gamma_{\alpha \nu}^\beta=\left(g_{\mu \nu},{ }\alpha+g{\mu \alpha, \nu}-g_{\alpha \nu}, \mu\right) \text {, } \
& 2 g^{\mu \lambda} g_{\mu \beta} \Gamma_{\alpha \nu}^\beta=2 \delta_\beta^\lambda \Gamma_{\alpha \nu}^\beta=g^{\mu \lambda}\left(g_{\mu \nu}, \alpha+g_{\mu \alpha},\nu-g{\alpha \nu}, \mu\right) \text {, } \
& \Gamma_{\alpha \nu}^\lambda=g^{\mu \lambda}\left(g_{\mu \nu},\alpha+g{\mu \alpha, \nu}-g_{\alpha \nu, \mu}\right) / 2 \text {. } \
&
\end{aligned}
$$

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

物理代写|广义相对论代写General relativity代考|Tensor Transforms

Posted on 2023年5月12日2023年5月12日 by statistics-lab

物理代写|广义相对论代写General relativity代考|Tensor Transforms

To study the transformation properties of tensors of rank $>1$, start by constructing a quantity that depends on two indexes that have a known transformation, e.g., $V^\mu W_\nu, V^\mu W^\nu$ and $V_\mu W_\nu$, where $V$ and $W$ are vectors. Such quantities are defined to transform as tensors, and it is easy to see how higher ranked tensors must transform. Using Eqs. (2.6) and (2.7), we have
$$
\begin{gathered}
\left(V^\mu W_\nu\right)=x^\mu, \psi^{\prime} V^{\psi^{\prime}} x^{\xi^{\prime}},{ }\nu W{\xi^{\prime}}=x^\mu, \psi^{\prime} x^{\xi^{\prime}}, \nu\left(V^{\psi^{\prime}} W_{\xi^{\prime}}\right) \
\left(V^\mu W^\nu\right)=x^\mu, \psi^{\prime} V^{\psi^{\prime}} x^\nu, \xi^{\prime} W^{\xi^{\prime}}=x^\mu, \psi^{\prime} x^\nu, \xi^{\prime}\left(V^{\psi^{\prime}} W^{\xi^{\prime}}\right) \
\left(V_\mu W_\nu\right)=x^{\psi^{\prime}},{ }\mu V{\psi^{\prime}} x^{\xi^{\prime}},{ }\nu W{\xi^{\prime}}=x^{\psi^{\prime}, \mu} x^{\xi^{\prime}}, \nu\left(V_{\psi^{\prime}} W_{\xi^{\prime}}\right)
\end{gathered}
$$
One then declares that the above multiplications of the two vectors produce tensors of rank 2 , such as $T_\nu^\mu, T^{\mu \nu}$, and $T_{\mu \nu}$. The above equations are the transformation rule, for any tensor of rank 2 . However, this must be tested on two quantities asserted to be tensors of rank 2 , the metric and Kronecker delta tensors. Let $V, W$ be vectors. If $g_{\mu \nu}$ is a tensor, then
$$
\begin{aligned}
g_{\mu \nu} V^\mu W^\nu & =x^{\xi^{\prime}},{ }\mu x^{\chi^{\prime}}, \nu g{\xi^{\prime} \chi^{\prime}} x^\mu, \alpha^{\prime} V^{\alpha^{\prime}} x^\nu, \beta^{\prime} W^{\beta^{\prime}} \
& =x^{\xi^{\prime}},{ }\mu x^\mu, \alpha^{\prime} x^{\chi^{\prime}}, \nu x^\nu, \beta^{\prime} g{\xi^{\prime} \chi^{\prime}} V^{\alpha^{\prime}} W^{\beta^{\prime}} \
& =x^{\xi^{\prime}}, \alpha^{\prime} x^{\chi^{\prime}}, \beta^{\prime} g_{\xi^{\prime} \chi^{\prime}} V^{\alpha^{\prime}} W^{\beta^{\prime}} \
& =\delta^{\xi^{\prime}}{ }{\alpha^{\prime}} \delta{\beta^{\prime}}^{\chi^{\prime}} g_{\xi^{\prime} \chi^{\prime}} V^{\alpha^{\prime}} W^{\beta^{\prime}}=g_{\xi^{\prime} \chi^{\prime}} V^{\xi^{\prime}} W^{\chi^{\prime}} .
\end{aligned}
$$

物理代写|广义相对论代写General relativity代考|Forming Other Vectors

Since $d r^\mu$ is a vector and $d \tau$ is an invariant, the quantity $U^\mu=\frac{d r^\mu}{d \tau}$ is another vector with units of velocity. This is not true for photons because $d \tau=0$. Suppose SR observer $\mathrm{O}^{\prime}$ says an object has 3 -velocity rectangular components $\frac{d r^{\bar{i}^{\prime}}}{d t^{\prime}}=v^{\bar{i}^{\prime}}=v_{\bar{i}^{\prime}}$. An observer moving with the object says that $d \tau$ has elapsed. Chapter 1 results yielded $d \tau=d t^{\prime} / \bar{\gamma}^{\prime}$, where $d t^{\prime}$ is the time elapsed according to $O^{\prime}$. Here $\gamma^{\prime}=\left(1-\left|v^{\prime}\right|^2\right)^{-1 / 2}$ and $\left|v^{\prime}\right|^2=v^{\bar{i}^{\prime}} v_{\bar{i}^{\prime}}$. Then,
$$
\begin{aligned}
& U^{\bar{i}^{\prime}}=\gamma^{\prime} \frac{d r^{i^{\prime}}}{d t^{\prime}}=\gamma^{\prime} v^{\bar{v}^{\prime}}=U_{\bar{i}^{\prime}}, \
& U^{\overline{0}^{\prime}}=\gamma^{\prime} \frac{d r^{0^{\prime}}}{d t^{\prime}}=\gamma^{\prime}=-U_{\overline{0}^{\prime}}, \
&-\eta_{\mu^{\prime} \nu^{\prime}} U^{\bar{\mu}^{\prime}} U^{\bar{\nu}^{\prime}}=-\gamma^{\prime 2}\left(-1+\left|v^{\prime}\right|^2\right)=1, \text { invariant. }
\end{aligned}
$$
There are similar equations for the $U^{\bar{\mu}}$ in $\mathrm{O}$ – just remove the primes.
The 3 -velocity transforms come from applying the Lorentz transform to these vectors. If $\mathrm{O}^{\prime}$ moves relative to $\mathrm{O}$, with speed $V$ in the $z$ direction, let $\gamma[V]=\left(1-|V|^2\right)^{-1 / 2}$. Then,
$$
\begin{aligned}
U^{\overline{0}} & =\gamma=\gamma[V]\left(U^{\overline{0}^{\prime}}+V U^{\overline{3}^{\prime}}\right)=\gamma[V] \gamma^{\prime}\left(1+V v^{\overline{3}^{\prime}}\right), \
U^{\overline{3}} & =\gamma v^{\overline{3}}=\gamma[V]\left(U^{\overline{3}^{\prime}}+V U^{\overline{0}^{\prime}}\right)=\gamma[V] \gamma^{\prime}\left(v^{\overline{3}^{\prime}}+V\right), \
v^{\overline{3}} & =\frac{v^{\overline{3}^{\prime}}+V}{1+V v^{\overline{3}^{\prime}}}=\frac{\gamma[V]\left(d x^{\overline{3}^{\prime}}+V d x^{\overline{0}^{\prime}}\right)}{\gamma[V]\left(d x^{\overline{0}^{\prime}}+V d x^{3^{\prime}}\right)}=\frac{d x^{\overline{3}}}{d t}, \
\gamma v^{\overline{1}, \overline{2}} & =\gamma^{\prime} v^{\overline{\overline{ }^{\prime}}, \overline{2}^{\prime}}, \
v^{\overline{1}, \overline{2}} & =\frac{v^{\overline{1}^{\prime}, \overline{2}^{\prime}}}{\gamma[V]\left(1+V \bar{v}^{3^{\prime}}\right)}=\frac{d x^{\overline{1}^{\prime}, \overline{2}^{\prime}}}{\gamma[V]\left(d x^{\overline{0}^{\prime}}+V d x^{\overline{3}^{\prime}}\right)}=\frac{d x^{\overline{1}, \overline{2}}}{d t} .
\end{aligned}
$$
The quantity $d \tau$ has vanished from Eqs. (2.19) and (2.20). Thus, they also hold for photons, e.g., if $v^{\overline{1}^{\prime}}=1, v^{\overline{2}^{\prime}}=v^{\overline{3}^{\prime}}=0$, then $v_{\overline{1}} v^{\overline{1}}=(\gamma[V])^{-2}, v^{\overline{2}}=$ $0, v_{\overline{3}} v^{\overline{3}}=|V|^2$ and $|v|^2=v_{\bar{i}} v^{\bar{i}}=1$. All observers see the same speed for light. For a slowly moving object like the earth, under the gravitational influence of the sun, $\gamma \approx 1, U^0 \approx 1$ and $U^i \approx 0$.

广义相对论代考

物理代写|广义相对论代写General relativity代考|Tensor Transforms

为了研究秩为$>1$的张量的变换性质，首先构造一个依赖于两个具有已知变换的指标的量，例如$V^\mu W_\nu, V^\mu W^\nu$和$V_\mu W_\nu$，其中$V$和$W$是向量。这样的量被定义为张量的变换，很容易看出高阶张量必须如何变换。使用等式。(2.6)和(2.7)，我们有
$$
\begin{gathered}
\left(V^\mu W_\nu\right)=x^\mu, \psi^{\prime} V^{\psi^{\prime}} x^{\xi^{\prime}},{ }\nu W{\xi^{\prime}}=x^\mu, \psi^{\prime} x^{\xi^{\prime}}, \nu\left(V^{\psi^{\prime}} W_{\xi^{\prime}}\right) \
\left(V^\mu W^\nu\right)=x^\mu, \psi^{\prime} V^{\psi^{\prime}} x^\nu, \xi^{\prime} W^{\xi^{\prime}}=x^\mu, \psi^{\prime} x^\nu, \xi^{\prime}\left(V^{\psi^{\prime}} W^{\xi^{\prime}}\right) \
\left(V_\mu W_\nu\right)=x^{\psi^{\prime}},{ }\mu V{\psi^{\prime}} x^{\xi^{\prime}},{ }\nu W{\xi^{\prime}}=x^{\psi^{\prime}, \mu} x^{\xi^{\prime}}, \nu\left(V_{\psi^{\prime}} W_{\xi^{\prime}}\right)
\end{gathered}
$$
然后声明上述两个向量的乘法产生秩为2的张量，如$T_\nu^\mu, T^{\mu \nu}$和$T_{\mu \nu}$。以上方程是变换规则，适用于任何秩为2的张量。然而，这必须在两个被断言为2阶张量的量上检验，度规张量和克罗内克张量。设$V, W$为向量。如果$g_{\mu \nu}$是张量，那么
$$
\begin{aligned}
g_{\mu \nu} V^\mu W^\nu & =x^{\xi^{\prime}},{ }\mu x^{\chi^{\prime}}, \nu g{\xi^{\prime} \chi^{\prime}} x^\mu, \alpha^{\prime} V^{\alpha^{\prime}} x^\nu, \beta^{\prime} W^{\beta^{\prime}} \
& =x^{\xi^{\prime}},{ }\mu x^\mu, \alpha^{\prime} x^{\chi^{\prime}}, \nu x^\nu, \beta^{\prime} g{\xi^{\prime} \chi^{\prime}} V^{\alpha^{\prime}} W^{\beta^{\prime}} \
& =x^{\xi^{\prime}}, \alpha^{\prime} x^{\chi^{\prime}}, \beta^{\prime} g_{\xi^{\prime} \chi^{\prime}} V^{\alpha^{\prime}} W^{\beta^{\prime}} \
& =\delta^{\xi^{\prime}}{ }{\alpha^{\prime}} \delta{\beta^{\prime}}^{\chi^{\prime}} g_{\xi^{\prime} \chi^{\prime}} V^{\alpha^{\prime}} W^{\beta^{\prime}}=g_{\xi^{\prime} \chi^{\prime}} V^{\xi^{\prime}} W^{\chi^{\prime}} .
\end{aligned}
$$

物理代写|广义相对论代写General relativity代考|Forming Other Vectors

因为$d r^\mu$是矢量，$d \tau$是不变量，所以$U^\mu=\frac{d r^\mu}{d \tau}$是另一个以速度为单位的矢量。这对光子是不成立的，因为$d \tau=0$。假设SR观察者$\mathrm{O}^{\prime}$说一个物体有3速度的矩形分量$\frac{d r^{\bar{i}^{\prime}}}{d t^{\prime}}=v^{\bar{i}^{\prime}}=v_{\bar{i}^{\prime}}$。随着物体移动的观察者说$d \tau$已经过去了。第1章的结果为$d \tau=d t^{\prime} / \bar{\gamma}^{\prime}$，其中$d t^{\prime}$是根据$O^{\prime}$所经过的时间。这里是$\gamma^{\prime}=\left(1-\left|v^{\prime}\right|^2\right)^{-1 / 2}$和$\left|v^{\prime}\right|^2=v^{\bar{i}^{\prime}} v_{\bar{i}^{\prime}}$。然后，
$$
\begin{aligned}
& U^{\bar{i}^{\prime}}=\gamma^{\prime} \frac{d r^{i^{\prime}}}{d t^{\prime}}=\gamma^{\prime} v^{\bar{v}^{\prime}}=U_{\bar{i}^{\prime}}, \
& U^{\overline{0}^{\prime}}=\gamma^{\prime} \frac{d r^{0^{\prime}}}{d t^{\prime}}=\gamma^{\prime}=-U_{\overline{0}^{\prime}}, \
&-\eta_{\mu^{\prime} \nu^{\prime}} U^{\bar{\mu}^{\prime}} U^{\bar{\nu}^{\prime}}=-\gamma^{\prime 2}\left(-1+\left|v^{\prime}\right|^2\right)=1, \text { invariant. }
\end{aligned}
$$
$\mathrm{O}$中也有类似的$U^{\bar{\mu}}$方程——只是去掉质数。
三速度变换来自于对这些向量应用洛伦兹变换。如果$\mathrm{O}^{\prime}$相对于$\mathrm{O}$移动，速度$V$在$z$方向，让$\gamma[V]=\left(1-|V|^2\right)^{-1 / 2}$。然后，
$$
\begin{aligned}
U^{\overline{0}} & =\gamma=\gamma[V]\left(U^{\overline{0}^{\prime}}+V U^{\overline{3}^{\prime}}\right)=\gamma[V] \gamma^{\prime}\left(1+V v^{\overline{3}^{\prime}}\right), \
U^{\overline{3}} & =\gamma v^{\overline{3}}=\gamma[V]\left(U^{\overline{3}^{\prime}}+V U^{\overline{0}^{\prime}}\right)=\gamma[V] \gamma^{\prime}\left(v^{\overline{3}^{\prime}}+V\right), \
v^{\overline{3}} & =\frac{v^{\overline{3}^{\prime}}+V}{1+V v^{\overline{3}^{\prime}}}=\frac{\gamma[V]\left(d x^{\overline{3}^{\prime}}+V d x^{\overline{0}^{\prime}}\right)}{\gamma[V]\left(d x^{\overline{0}^{\prime}}+V d x^{3^{\prime}}\right)}=\frac{d x^{\overline{3}}}{d t}, \
\gamma v^{\overline{1}, \overline{2}} & =\gamma^{\prime} v^{\overline{\overline{ }^{\prime}}, \overline{2}^{\prime}}, \
v^{\overline{1}, \overline{2}} & =\frac{v^{\overline{1}^{\prime}, \overline{2}^{\prime}}}{\gamma[V]\left(1+V \bar{v}^{3^{\prime}}\right)}=\frac{d x^{\overline{1}^{\prime}, \overline{2}^{\prime}}}{\gamma[V]\left(d x^{\overline{0}^{\prime}}+V d x^{\overline{3}^{\prime}}\right)}=\frac{d x^{\overline{1}, \overline{2}}}{d t} .
\end{aligned}
$$
量$d \tau$从等式中消失了。(2.19)和

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写