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2023年7月19日2023年7月19日 General Relativity, 广义相对论, 物理代写

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

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物理代写|广义相对论代写General Relativity代考|Exterior derivative operator: generalised Stokes’ theorem

We have seen that (in an $n$-dimensional space) we may introduce 1-forms, with a corresponding line integral, and 2-forms, with a corresponding area integral-as well as 3-forms, with a volume integral, and so on. Consider, however, Stokes’ theorem
$$
\int \mathbf{A} \cdot \mathrm{d} \mathbf{s}=\int \nabla \times \mathbf{A} \cdot \mathrm{d} \boldsymbol{\Sigma}
$$

The left hand side is a line integral – the integral of a 1-form, while the right hand side is an area integral – the integral of a 2 -form. The very existence of a theorem like Stokes’ theorem implies that there must be a relation between 1-forms and 2-forms. In a similar way, Gauss’s theorem relates an integral over an area to one over a volume, implying a relation between 2 -forms and 3-forms. We now investigate this relation and find a beautiful generalisation of Stokes’ theorem, applicable to general forms, which yields both Stokes’ theorem and Gauss’s theorem as special cases.

The key to finding the relation is to define an exterior derivative operator $\mathbf{d}$ which converts a $p$-form into a $(p+1)$-form. Let $\omega$ be the $p$-form
$$
\omega=a_{1, \ldots, p}(x) \mathbf{d} x^1 \wedge \cdots \wedge \mathbf{d} x^p .
$$
then
$$
\mathbf{d} \boldsymbol{\omega}=\frac{\partial a_{i_1 \cdots i_p}}{\partial x^k} \mathbf{d} x^k \wedge \mathbf{d} x^{i_1} \wedge \cdots \wedge \mathbf{d} x^{i_p} .
$$

物理代写|广义相对论代写General Relativity代考|Generalised Stokes’ theorem

Let $\omega$ be a $p$-form, and let $c$ be a $(p+1)$-chain. Define $\partial$, the boundary operator on chains, so that $\partial c$ is the boundary of $c ; \partial c$ is a $p$-chain. Two simple examples are drawn in Fig. 3.11; in (a) $c$ is an area (2-chain) which is bounded by the closed line $\partial c$, a 1-chain. In (b) $c$ is a volume (3-chain) bounded by the surface $\partial c$, a (closed) 2-chain. Note that in both these cases the boundary is itself $c l o s e d ; \partial c$ has no boundary, or
$$
\partial(\partial c)=\partial^2 c=0
$$
This is a general result for $p$-chains:
$$
\partial^2=0
$$
and may be understood as being a result ‘dual’ to the Poincaré lemma $\mathbf{d}^2=0,(3.90)$ above. The boundary operator $\partial$ is dual to the exterior derivative operator $\mathbf{d}$.

Having defined the boundary operator we are now in a position to state the generalised Stokes’ theorem, which is
$$
\int_{\partial c} \omega=\int_c d \omega .
$$
Stokes’ theorem holds in any space, but to illustrate it let us work in $R^3$; and first consider the case $p=1$. Then $\omega$ is a 1-form, of the type (3.85), and $c$ is an area, with boundary $\partial c$, as in Fig. 3.11(a). The 2-form d $\omega$ is given by (3.86), where, as remarked already, the coefficients are the components of $\nabla \times \mathbf{a}$. Then (3.93) gives
$$
\int_{\partial c} a \cdot \mathrm{d} l=\int_c(\nabla \times a) \cdot n \mathrm{~d} \Sigma
$$
where $\mathrm{d} \boldsymbol{\Sigma}$ is an element of surface area, with unit normal $\mathbf{n}$. This is clearly Stokes’ theorem. As a second example take the case $p=2$, so $\omega$ is a 2-form, and therefore of the form (3.87); $\mathbf{d} \boldsymbol{\omega}$ is the 3 -form given by (3.88). The 3-chain $c$ is a volume $V$ with boundary $\partial c=\partial V$ (Fig. 3.11(b)) and (3.92) then gives
$$
\int_{\partial V} b \cdot n \mathrm{~d} \Sigma=\int_V(\nabla \cdot b) \mathrm{d} V .
$$
The reader will recognise this as Gauss’s theorem.
It will be appreciated that the generalised Stokes’ theorem is a neat and powerful theorem. The reader will doubtless recall that the ‘usual’ formulation of Stokes’ and Gauss’s theorems requires the stipulation of ‘directional’ notions – the normal $\mathbf{n}$ is an outward, not an inward, normal; and in Stokes’ theorem the path round the closed boundary is taken in an anticlock wise sense. These notions are however automatically encoded in the present formulation based on the exterior derivative operator, which, as we have seen, antisymmetrises and differentiates at the same time.

广义相对论代写

物理代写|广义相对论代写General Relativity代考|Exterior derivative operator: generalised Stokes’ theorem

我们已经看到(在$n$维空间中)我们可以引入相应的线积分的1-形式，相应的面积积分的2-形式，以及体积积分的3-形式，等等。然而，考虑一下斯托克斯定理
$$
\int \mathbf{A} \cdot \mathrm{d} \mathbf{s}=\int \nabla \times \mathbf{A} \cdot \mathrm{d} \boldsymbol{\Sigma}
$$

左边是线积分——1型积分，而右边是面积积分——2型积分。像Stokes定理这样的定理的存在意味着一形式和二形式之间一定存在联系。类似地，高斯定理将面积上的积分与体积上的积分联系起来，暗示了二式和三式之间的关系。我们现在研究这一关系，并找到了适用于一般形式的Stokes定理的一个漂亮的推广，它产生了Stokes定理和高斯定理作为特殊情况。

找到这种关系的关键是定义一个外部导数运算符$\mathbf{d}$，它将$p$ -形式转换为$(p+1)$ -形式。设$\omega$为$p$ -形式
$$
\omega=a_{1, \ldots, p}(x) \mathbf{d} x^1 \wedge \cdots \wedge \mathbf{d} x^p .
$$
然后
$$
\mathbf{d} \boldsymbol{\omega}=\frac{\partial a_{i_1 \cdots i_p}}{\partial x^k} \mathbf{d} x^k \wedge \mathbf{d} x^{i_1} \wedge \cdots \wedge \mathbf{d} x^{i_p} .
$$

物理代写|广义相对论代写General Relativity代考|Generalised Stokes’ theorem

让$\omega$成为一个$p$形式，让$c$成为一个$(p+1)$链。定义链上的边界运算符$\partial$，使得$\partial c$是$c ; \partial c$是一个$p$ -链的边界。图3.11给出了两个简单的例子;在(a)中$c$是一个区域(2-chain)，它被封闭的线$\partial c$ (1-chain)包围。(b) $c$是一个以表面$\partial c$为界的体积(3链)，一个(封闭的)2链。注意，在这两种情况下，边界本身都是$c l o s e d ; \partial c$没有边界，或者
$$
\partial(\partial c)=\partial^2 c=0
$$
这是$p$ -chains的一般结果:
$$
\partial^2=0
$$
也可以理解为上面庞卡莱引理$\mathbf{d}^2=0,(3.90)$的对偶结果。边界算子$\partial$对偶于外导数算子$\mathbf{d}$。

定义了边界算符之后，我们现在可以陈述广义Stokes定理，它是
$$
\int_{\partial c} \omega=\int_c d \omega .
$$
斯托克斯定理适用于任何空间，但为了说明它，让我们用$R^3$;首先考虑$p=1$这个例子。则$\omega$为1型，类型为(3.85)，$c$为一个区域，边界为$\partial c$，如图3.11(a)所示。2-形式的d $\omega$由式(3.86)给出，其中，如前所述，系数是$\nabla \times \mathbf{a}$的分量。则(3.93)得到
$$
\int_{\partial c} a \cdot \mathrm{d} l=\int_c(\nabla \times a) \cdot n \mathrm{~d} \Sigma
$$
式中$\mathrm{d} \boldsymbol{\Sigma}$为表面积的元，单位法线为$\mathbf{n}$。这显然是Stokes定理。第二个例子以$p=2$为例，因此$\omega$是一个2-form，因此是(3.87);$\mathbf{d} \boldsymbol{\omega}$是式(3.88)给出的3 -形式。3链$c$是一个体积$V$，其边界为$\partial c=\partial V$(图3.11(b))，然后为(3.92)
$$
\int_{\partial V} b \cdot n \mathrm{~d} \Sigma=\int_V(\nabla \cdot b) \mathrm{d} V .
$$
读者会认出这就是高斯定理。
一般化的斯托克斯定理是一个简洁而有力的定理。读者无疑还记得，斯托克斯定理和高斯定理的“通常”表述需要规定“定向”概念——正态$\mathbf{n}$是向外的正态，而不是向内的正态;在Stokes定理中，绕封闭边界的路径是在逆时针的意义上取的。然而，这些概念在基于外部导数算子的当前公式中被自动编码，正如我们所看到的，它同时具有反对称和微分。

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微观经济学代写

微观经济学是主流经济学的一个分支，研究个人和企业在做出有关稀缺资源分配的决策时的行为以及这些个人和企业之间的相互作用。my-assignmentexpert™ 为您的留学生涯保驾护航在数学Mathematics作业代写方面已经树立了自己的口碑, 保证靠谱, 高质且原创的数学Mathematics代写服务。我们的专家在图论代写Graph Theory代写方面经验极为丰富，各种图论代写Graph Theory相关的作业也就用不着说。

线性代数代写

线性代数是数学的一个分支，涉及线性方程，如：线性图，如：以及它们在向量空间和通过矩阵的表示。线性代数是几乎所有数学领域的核心。

博弈论代写

现代博弈论始于约翰-冯-诺伊曼（John von Neumann）提出的两人零和博弈中的混合策略均衡的观点及其证明。冯-诺依曼的原始证明使用了关于连续映射到紧凑凸集的布劳威尔定点定理，这成为博弈论和数学经济学的标准方法。在他的论文之后，1944年，他与奥斯卡-莫根斯特恩（Oskar Morgenstern）共同撰写了《游戏和经济行为理论》一书，该书考虑了几个参与者的合作游戏。这本书的第二版提供了预期效用的公理理论，使数理统计学家和经济学家能够处理不确定性下的决策。

微积分代写

微积分，最初被称为无穷小微积分或 “无穷小的微积分”，是对连续变化的数学研究，就像几何学是对形状的研究，而代数是对算术运算的概括研究一样。

它有两个主要分支，微分和积分；微分涉及瞬时变化率和曲线的斜率，而积分涉及数量的累积，以及曲线下或曲线之间的面积。这两个分支通过微积分的基本定理相互联系，它们利用了无限序列和无限级数收敛到一个明确定义的极限的基本概念。

计量经济学代写

什么是计量经济学？
计量经济学是统计学和数学模型的定量应用，使用数据来发展理论或测试经济学中的现有假设，并根据历史数据预测未来趋势。它对现实世界的数据进行统计试验，然后将结果与被测试的理论进行比较和对比。

根据你是对测试现有理论感兴趣，还是对利用现有数据在这些观察的基础上提出新的假设感兴趣，计量经济学可以细分为两大类：理论和应用。那些经常从事这种实践的人通常被称为计量经济学家。

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。

2023年7月19日2023年7月19日 General Relativity, 广义相对论, 物理代写

物理代写|广义相对论代写General Relativity代考|MATHS565

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物理代写|广义相对论代写General Relativity代考|MATHS565

物理代写|广义相对论代写General Relativity代考|One-forms

The definition of vectors given above involved the specifying of basis vectors, which are in essence directions along coordinate lines $-\partial / \partial x, \partial / \partial r$ etc. This gives a sense to the idea that a vector is a quantity ‘with magnitude and direction’. But there is another type of ‘vector’ quantity defined in elementary calculus, which is the gradient of a (scalar) function. For example the function $f(x, y, z)$ has a gradient
$$
\nabla f=\frac{\partial f}{\partial x} \mathbf{i}+\frac{\partial f}{\partial y} \mathbf{j}+\frac{\partial f}{\partial z} \mathbf{k}
$$
As a simple illustration consider the surface $S^2$ (sphere) $r=$ const in $R^3$. Taking the normal to the surface we have
$$
\begin{aligned}
\nabla r & =\frac{\partial r}{\partial x} \mathbf{i}+\frac{\partial r}{\partial y} \mathbf{j}+\frac{\partial r}{\partial z} \mathbf{k}=\frac{x}{r} \mathbf{i}+\frac{y}{r} \mathbf{j}+\frac{z}{r} \mathbf{k}=\frac{1}{r} \mathbf{r}=\mathbf{e}r ; \ \nabla \theta & =\frac{\partial \theta}{\partial x} \mathbf{i}+\frac{\partial \theta}{\partial y} \mathbf{j}+\frac{\partial \theta}{\partial z} \mathbf{k}=\frac{\cos \theta \cos \phi}{r} \mathbf{i}+\frac{\cos \theta \sin \phi}{r} \mathbf{j}-\frac{\sin \theta}{r} \mathbf{k}=\frac{1}{r^2} \mathbf{e}\theta \
\nabla \phi & =\frac{\partial \phi}{\partial x} \mathbf{i}+\frac{\partial \phi}{\partial y} \mathbf{j}+\frac{\partial \phi}{\partial z} \mathbf{k}=-\frac{\sin \phi}{r \sin \theta} \mathbf{i}+\frac{\cos \phi}{r \sin \theta} \mathbf{j}=\frac{1}{r^2 \sin ^2 \theta} \mathbf{e}\phi . \end{aligned} $$ In this case the gradients are proportional to the vectors $\mathbf{e}_r, \mathbf{e}\theta, \mathbf{e}\phi$, with $\mathbf{e}\theta, \mathbf{e}_\phi$ spanning the tangent space to $S^2$. This, however, is not typical, since a spherical surface in a flat 3-dimensional space has a very high degree of symmetry. Conceptually, gradients and basis vectors are distinct; basis vectors are associated with coordinate lines, but gradients with ‘lines of steepest descent’ from one surface to another, as sketched in Fig. 3.5. This difference may be shown up by considering a non-orthogonal coordinate system in $R^3$; define $u, v$ and $w$ by
$$
x=u+v, \quad y=u-v, \quad z=2 u v+w
$$
with inverse
$$
u=1 / 2(x+y), \quad v=1 / 2(x-y), \quad w=z-1 / 2\left(x^2-y^2\right) .
$$

物理代写|广义相对论代写General Relativity代考|Transformation rules

One-forms, like vectors, have the same form in different coordinate systems. Under a transformation $x^\mu \rightarrow x^{\mu^{\prime}}$ the (coordinate) basis $\boldsymbol{\theta}^\mu$ transforms as
$$
\boldsymbol{\theta}^\mu \rightarrow \boldsymbol{\theta}^\mu=\frac{\partial x^{\prime \mu}}{\partial x^v} \boldsymbol{\theta}^v
$$
and the components $\omega_v$ as
$$
\omega_v \rightarrow \omega_v^{\prime}=\frac{\partial x^\lambda}{\partial x^{\prime v}} \omega_\lambda
$$
so that $\boldsymbol{\omega}=\omega_\mu \boldsymbol{\theta}^\mu$ is invariant:
$$
\boldsymbol{\omega} \rightarrow \omega_\mu^{\prime} \boldsymbol{\theta}^{\prime \mu}=\frac{\partial x^\lambda}{\partial x^{\prime \mu}} \frac{\partial x^{\prime \mu}}{\partial x^\kappa} \omega_\lambda \boldsymbol{\theta}^\kappa=\omega_\lambda \boldsymbol{\theta}^\lambda=\boldsymbol{\omega},
$$
in analogy with (3.17-3.19). In more old-fashioned language the transformation rules (3.17) and (3.36) are respectively the transformation rules for covariant and contravariant vector fields. What we now call vectors have components in a given coordinate system which transform ‘contravariantly’, and what we now call 1-forms have components in a given coordinate system which transform ‘covariantly’. For convenience we state the formulae again:
$$
\begin{array}{ll}
\text { covariant vector: } \quad V^\mu(x) & \rightarrow V^{\prime \mu}(x)=\frac{\partial x^{\prime \mu}}{\partial x^\lambda}(x) V^\lambda(x), \
\text { contravariant vector: } V_v(x) & \rightarrow V_v^{\prime}(x)=\frac{\partial x^\lambda}{\partial x^{\prime \nu}}(x) V_\lambda(x) .
\end{array}
$$
These formulae emphasise, by including the arguments $(x)$, that these vector fields are defined at a specific point $x$, and the transformation coefficients are evaluated the same point. The reader will appreciate that for non-linear transformations the coefficients $\frac{\partial x^{\prime \mu}}{\partial x^\lambda}$ and $\frac{\partial x^\lambda}{\partial x^{\prime v}}$ will themselves depend on $x$. The above transformation laws are specific to one particular point in the manifold.

广义相对论代写

物理代写|广义相对论代写General Relativity代考|One-forms

上面给出的向量的定义涉及基向量的指定，基向量本质上是沿坐标线$-\partial / \partial x, \partial / \partial r$等方向。这让我们理解了矢量是一个“有大小和方向”的量。但在初等微积分中还定义了另一种“向量”，即(标量)函数的梯度。例如，函数$f(x, y, z)$有一个梯度
$$
\nabla f=\frac{\partial f}{\partial x} \mathbf{i}+\frac{\partial f}{\partial y} \mathbf{j}+\frac{\partial f}{\partial z} \mathbf{k}
$$
作为一个简单的例子，考虑$R^3$中的表面$S^2$(球体)$r=$ const。取曲面的法线
$$
\begin{aligned}
\nabla r & =\frac{\partial r}{\partial x} \mathbf{i}+\frac{\partial r}{\partial y} \mathbf{j}+\frac{\partial r}{\partial z} \mathbf{k}=\frac{x}{r} \mathbf{i}+\frac{y}{r} \mathbf{j}+\frac{z}{r} \mathbf{k}=\frac{1}{r} \mathbf{r}=\mathbf{e}r ; \ \nabla \theta & =\frac{\partial \theta}{\partial x} \mathbf{i}+\frac{\partial \theta}{\partial y} \mathbf{j}+\frac{\partial \theta}{\partial z} \mathbf{k}=\frac{\cos \theta \cos \phi}{r} \mathbf{i}+\frac{\cos \theta \sin \phi}{r} \mathbf{j}-\frac{\sin \theta}{r} \mathbf{k}=\frac{1}{r^2} \mathbf{e}\theta \
\nabla \phi & =\frac{\partial \phi}{\partial x} \mathbf{i}+\frac{\partial \phi}{\partial y} \mathbf{j}+\frac{\partial \phi}{\partial z} \mathbf{k}=-\frac{\sin \phi}{r \sin \theta} \mathbf{i}+\frac{\cos \phi}{r \sin \theta} \mathbf{j}=\frac{1}{r^2 \sin ^2 \theta} \mathbf{e}\phi . \end{aligned} $$在这种情况下，梯度与向量$\mathbf{e}r, \mathbf{e}\theta, \mathbf{e}\phi$成正比，$\mathbf{e}\theta, \mathbf{e}\phi$张成切空间到$S^2$。然而，这并不典型，因为平面三维空间中的球面具有非常高的对称性。从概念上讲，梯度和基向量是不同的;基向量与坐标线相关联，但梯度与从一个表面到另一个表面的“最陡下降线”相关联，如图3.5所示。这种差异可以通过考虑$R^3$中的非正交坐标系来显示;定义$u, v$和$w$ by
$$
x=u+v, \quad y=u-v, \quad z=2 u v+w
$$
与逆
$$
u=1 / 2(x+y), \quad v=1 / 2(x-y), \quad w=z-1 / 2\left(x^2-y^2\right) .
$$

物理代写|广义相对论代写General Relativity代考|Transformation rules

一种形式，像向量一样，在不同的坐标系中有相同的形式。在转换$x^\mu \rightarrow x^{\mu^{\prime}}$下，(坐标)基$\boldsymbol{\theta}^\mu$转换为
$$
\boldsymbol{\theta}^\mu \rightarrow \boldsymbol{\theta}^\mu=\frac{\partial x^{\prime \mu}}{\partial x^v} \boldsymbol{\theta}^v
$$
分量$\omega_v$ as
$$
\omega_v \rightarrow \omega_v^{\prime}=\frac{\partial x^\lambda}{\partial x^{\prime v}} \omega_\lambda
$$
所以$\boldsymbol{\omega}=\omega_\mu \boldsymbol{\theta}^\mu$是不变的
$$
\boldsymbol{\omega} \rightarrow \omega_\mu^{\prime} \boldsymbol{\theta}^{\prime \mu}=\frac{\partial x^\lambda}{\partial x^{\prime \mu}} \frac{\partial x^{\prime \mu}}{\partial x^\kappa} \omega_\lambda \boldsymbol{\theta}^\kappa=\omega_\lambda \boldsymbol{\theta}^\lambda=\boldsymbol{\omega},
$$
类比于(3.17-3.19)。在更老式的语言中，变换规则(3.17)和式(3.36)分别是协变和逆变向量场的变换规则。我们现在所说的向量在给定坐标系中有逆变变换的分量，而我们现在所说的1-型在给定坐标系中有协变变换的分量。为方便起见，我们将公式重新表述一遍:
$$
\begin{array}{ll}
\text { covariant vector: } \quad V^\mu(x) & \rightarrow V^{\prime \mu}(x)=\frac{\partial x^{\prime \mu}}{\partial x^\lambda}(x) V^\lambda(x), \
\text { contravariant vector: } V_v(x) & \rightarrow V_v^{\prime}(x)=\frac{\partial x^\lambda}{\partial x^{\prime \nu}}(x) V_\lambda(x) .
\end{array}
$$
这些公式通过包含参数$(x)$来强调，这些向量场是在一个特定的点$x$定义的，并且变换系数是在同一点上计算的。读者会意识到，对于非线性变换，系数$\frac{\partial x^{\prime \mu}}{\partial x^\lambda}$和$\frac{\partial x^\lambda}{\partial x^{\prime v}}$本身取决于$x$。以上的变换定律只适用于流形中的某一点。

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2023年7月19日2023年7月19日 General Relativity, 广义相对论, 物理代写

物理代写|广义相对论代写General Relativity代考|ASTR3740

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物理代写|广义相对论代写General Relativity代考|ASTR3740

物理代写|广义相对论代写General Relativity代考|Twin paradox: accelerations

The so-called twin paradox is not a paradox. It is the following statement: if A and B are twins and A remains on Earth while B goes on a long trip, say to a distant star and back again, then on return B is younger than A. Suppose the star is a distance $l$ away and B travels with speed $v$ there and back. Then, as measured in A’s frame, B is away for a time $2 l / v$, and that is how much A has aged when B returns. When A looks at B’s clock, however, there is a time dilation factor of $\gamma=\left(\begin{array}{ll}1 & v^2 / c^2\end{array}\right)^{1 / 2}$, so B’s clock – including her biological clock – has only registered a passage of time $2 l / v v=(2 l / v)\left(1 \quad v^2 / c^2\right)^{1 / 2}$; on return, therefore, she is younger than A. This is the true situation. It appears paradoxical because one is tempted to think that ‘time is relative’, so that while A reckons B to be younger on return, as argued above, B should also reckon A to be younger; so in actual fact, one might think, they are the same age after the trip, just as before it. This, however, is wrong, and the reason is that while A remains in an inertial frame (or at least the approximately inertial frame of the Earth), B does not, since B has to reverse her velocity for the return trip, and that means she undergoes an acceleration. There is no reason why the twins should be the same age after B’s space trip, and they are not.

It may be useful to consider some numbers. Suppose the star is 15 light years away and B (Bianca) travels at speed $v=(3 / 5) c$. Then, measured by A (Andrei), Bianca reaches the star in $\frac{15}{3 / 5}=25$ years, so Andrei is 50 years older when Bianca returns (see Fig. 2.2). The time dilation factor is $1 / \gamma=\left(1 \quad v^2 / c^2\right)^{1 / 2}=4 / 5$, so, as seen by Andrei, Bianca takes a time $25 \times(4 / 5)=20$ years to reach the star, and will therefore be 40 years older when she returns. She will therefore be 10 years younger than Andrei after the trip. Of course, this is an approximation, since we have ignored the time taken for Bianca to change her velocity from $+v$ to $v$; this is indicated by B’s ‘smoothed out’ world-line near the star in Fig. 2.2. We now show, however, that this time may be as short as desired, if B is subjected to a large enough acceleration. We must therefore consider the treatment of accelerations in Minkowski space-time.
First define the 4-velocity $u^\mu$ :
$$
u^\mu=\frac{\mathrm{d} x^\mu}{\mathrm{d} \tau}=\left(c \frac{\mathrm{d} t}{\mathrm{~d} \tau}, \frac{\mathrm{d} x}{\mathrm{~d} \tau}, \frac{\mathrm{d} y}{\mathrm{~d} \tau}, \frac{\mathrm{d} z}{\mathrm{~d} \tau}\right) .
$$
In view of (2.16) and (2.17) we have
$$
\eta_{\mu v} u^\mu u^v=u^\mu u_\mu=-c^2
$$
the 4-velocity has constant length. Differentiating this gives $\left(\right.$ with $\left.\dot{u}^\mu=\frac{\mathrm{d} u^\mu}{\mathrm{d} \tau}\right)$
$$
\frac{\mathrm{d}}{\mathrm{d} \tau}\left(u^\mu u_\mu\right)=0=2 \dot{u}^\mu u_\mu
$$
or, defining the acceleration four vector $a^\mu=\dot{u}^\mu$,
$$
\eta_{\mu v} a^\mu u^v=a^\mu u_\mu=0 .
$$

物理代写|广义相对论代写General Relativity代考|Rotating frames: the Sagnac effect

The twin paradox arises when one of two observers (twins), but not the other one, undergoes an acceleration in Minkowski space-time; that is, motion in which $\frac{\mathrm{d} \mathbf{v}}{\mathrm{d} t} \neq 0$, where $\mathbf{v}$ is the relative velocity of the twins. Putting $\mathbf{v}=\mathbf{n} v$, however, we may distinguish in general two cases in which $\frac{\mathrm{d} \mathbf{v}}{\mathrm{d} t} \neq 0$, (i) $\frac{\mathrm{d} \mathbf{n}}{\mathrm{d} t}=0, \frac{\mathrm{d} v}{\mathrm{~d} t} \neq 0$; this is the case of acceleration in a straight line, (ii) $\frac{\mathrm{d} \mathbf{n}}{\mathrm{d} t} \neq 0, \frac{d v}{d t}=0$; this is the case of motion with changing direction but constant speed, for example, in a rotating frame. Let us now consider this case; we expect to find, and do find, some interesting new effects.

Let us suppose that $S^{\prime}$ rotates relative to $S$ around their common $z$ axis. It is convenient to use cylindrical coordinates, so that in $S$

$$
\mathrm{d} s^2=-c^2 \mathrm{~d} t^2+\mathrm{d} r^2+r^2 \mathrm{~d} \phi^2+\mathrm{d} z^2 .
$$
The coordinates in $S^{\prime}$ are related to those in $S$ by
$$
t^{\prime}=t, \quad r^{\prime}=r, \quad \phi^{\prime}=\phi-\omega t, \quad z^{\prime}=z
$$
and hence
$$
\begin{aligned}
\mathrm{d} s^{\prime 2} & =-c^2 \mathrm{~d} t^{\prime 2}+\mathrm{d} r^{\prime 2}+r^{\prime 2}\left(\mathrm{~d} \phi^{\prime}+\omega \mathrm{d} t^{\prime}\right)^2+\mathrm{d} z^{\prime 2} \
& =-\left(c^2-\omega^2 r^{\prime 2}\right) \mathrm{d} t^{\prime 2}+2 \omega r^{\prime 2} \mathrm{~d} \phi^{\prime} \mathrm{d} t^{\prime}+\mathrm{d} r^{\prime 2}+r^{\prime 2} \mathrm{~d} \phi^{\prime 2}+\mathrm{d} z^{\prime 2} .
\end{aligned}
$$
Dropping the primes we then have for the invariant space-time interval in a rotating frame
$$
\mathrm{d} s^2=-\left(c^2-\omega^2 r^2\right) \mathrm{d} t^2+2 \omega r^2 \mathrm{~d} \phi \mathrm{d} t+\mathrm{d} r^2+r^2 \mathrm{~d} \phi^2+\mathrm{d} z^2
$$
We write this in the form
$$
\mathrm{d} s^2=g_{\mu v} \mathrm{~d} x^\mu \mathrm{d} x^v=g_{00}\left(\mathrm{~d} x^0\right)^2+2 \mathrm{~g}{0 i} \mathrm{~d} x^0 \mathrm{~d} x^i+g{i k} \mathrm{~d} x^i \mathrm{~d} x^k,
$$
where, as usual, $i$ and $k$ are summed over spatial indices $1-3$ only. With $x^\mu=\left(x^0, x^1, x^2\right.$, $\left.x^3\right)=(c t, r, \phi, z), g_{\mu v}$ takes on the form
$$
g_{\mu v}=\left(\begin{array}{cccc}
-\left(1-\frac{\omega^2 r^2}{c^2}\right) & 0 & \frac{\omega r^2}{c} & 0 \
0 & 1 & 0 & 0 \
\frac{\omega r^2}{c} & 0 & r^2 & 0 \
0 & 0 & 0 & 1
\end{array}\right) .
$$

广义相对论代写

物理代写|广义相对论代写General Relativity代考|Twin paradox: accelerations

所谓的孪生悖论并不是一个悖论。它是这样的陈述:如果A和B是双胞胎，A留在地球上，而B进行一次长途旅行，比如说去一颗遥远的恒星并返回，那么返回时B比A年轻。假设这颗恒星距离我们$l$远，B以$v$的速度往返。然后，在A的坐标系中测量，B离开了一段时间$2 l / v$，这就是B回来时A的年龄。然而，当A看B的时钟时，有一个时间膨胀因子$\gamma=\left(\begin{array}{ll}1 & v^2 / c^2\end{array}\right)^{1 / 2}$，所以B的时钟——包括她的生物钟——只记录了一段时间$2 l / v v=(2 l / v)\left(1 \quad v^2 / c^2\right)^{1 / 2}$;因此，反过来，她比a年轻。这是真实的情况。这似乎是矛盾的，因为人们倾向于认为“时间是相对的”，因此，正如上文所述，当A认为B回来时更年轻时，B也应该认为A更年轻;所以事实上，有人可能会认为，他们在旅行后和旅行前是一样的年龄。然而，这是错误的，原因是当A保持在惯性系(或至少是地球的近似惯性系)时，B却没有，因为B在回程时必须反转她的速度，这意味着她经历了加速度。这对双胞胎没有理由在B的太空旅行后应该是相同的年龄，而且他们不是。

考虑一些数字可能会有用。假设这颗恒星离我们15光年远，B(比安卡)以$v=(3 / 5) c$的速度运行。那么，用A (Andrei)测量，比安卡到达恒星的时间是$\frac{15}{3 / 5}=25$年，所以当比安卡返回时，Andrei要老50岁(见图2.2)。时间膨胀系数是$1 / \gamma=\left(1 \quad v^2 / c^2\right)^{1 / 2}=4 / 5$，所以，正如安德烈所看到的，比安卡需要$25 \times(4 / 5)=20$年的时间才能到达这颗恒星，因此当她返回时，她将老40岁。因此，这次旅行结束后，她将比安德烈小10岁。当然，这只是一个近似值，因为我们忽略了比安卡将速度从$+v$改变到$v$所花费的时间;图2.2中B在恒星附近的“平滑”世界线表明了这一点。然而，我们现在表明，如果B受到足够大的加速度，这个时间可以像期望的那样短。因此我们必须考虑在闵可夫斯基时空中对加速度的处理。
首先定义4速$u^\mu$:
$$
u^\mu=\frac{\mathrm{d} x^\mu}{\mathrm{d} \tau}=\left(c \frac{\mathrm{d} t}{\mathrm{~d} \tau}, \frac{\mathrm{d} x}{\mathrm{~d} \tau}, \frac{\mathrm{d} y}{\mathrm{~d} \tau}, \frac{\mathrm{d} z}{\mathrm{~d} \tau}\right) .
$$
鉴于(2.16)和(2.17)，我们有
$$
\eta_{\mu v} u^\mu u^v=u^\mu u_\mu=-c^2
$$
速度的长度是恒定的。微分得到$\left(\right.$和$\left.\dot{u}^\mu=\frac{\mathrm{d} u^\mu}{\mathrm{d} \tau}\right)$
$$
\frac{\mathrm{d}}{\mathrm{d} \tau}\left(u^\mu u_\mu\right)=0=2 \dot{u}^\mu u_\mu
$$
或者定义加速度4向量$a^\mu=\dot{u}^\mu$，
$$
\eta_{\mu v} a^\mu u^v=a^\mu u_\mu=0 .
$$

物理代写|广义相对论代写General Relativity代考|Rotating frames: the Sagnac effect

两个观察者(双胞胎)中的一个而不是另一个在闵可夫斯基时空中经历加速度时，孪生悖论就出现了;也就是说，运动中$\frac{\mathrm{d} \mathbf{v}}{\mathrm{d} t} \neq 0$, $\mathbf{v}$是双胞胎的相对速度。然而，把$\mathbf{v}=\mathbf{n} v$放在一起，我们可以大体上区分出两种情况:$\frac{\mathrm{d} \mathbf{v}}{\mathrm{d} t} \neq 0$， (i) $\frac{\mathrm{d} \mathbf{n}}{\mathrm{d} t}=0, \frac{\mathrm{d} v}{\mathrm{~d} t} \neq 0$;这是直线加速度的情况，(ii) $\frac{\mathrm{d} \mathbf{n}}{\mathrm{d} t} \neq 0, \frac{d v}{d t}=0$;这是运动方向改变但速度不变的情况，例如，在旋转框架中。现在让我们考虑这种情况;我们期望发现，也确实发现了一些有趣的新效应。

让我们假设$S^{\prime}$相对于$S$围绕它们共同的$z$轴旋转。使用柱坐标很方便，所以在$S$

$$
\mathrm{d} s^2=-c^2 \mathrm{~d} t^2+\mathrm{d} r^2+r^2 \mathrm{~d} \phi^2+\mathrm{d} z^2 .
$$
$S^{\prime}$中的坐标与$S$中的坐标相关
$$
t^{\prime}=t, \quad r^{\prime}=r, \quad \phi^{\prime}=\phi-\omega t, \quad z^{\prime}=z
$$
因此
$$
\begin{aligned}
\mathrm{d} s^{\prime 2} & =-c^2 \mathrm{~d} t^{\prime 2}+\mathrm{d} r^{\prime 2}+r^{\prime 2}\left(\mathrm{~d} \phi^{\prime}+\omega \mathrm{d} t^{\prime}\right)^2+\mathrm{d} z^{\prime 2} \
& =-\left(c^2-\omega^2 r^{\prime 2}\right) \mathrm{d} t^{\prime 2}+2 \omega r^{\prime 2} \mathrm{~d} \phi^{\prime} \mathrm{d} t^{\prime}+\mathrm{d} r^{\prime 2}+r^{\prime 2} \mathrm{~d} \phi^{\prime 2}+\mathrm{d} z^{\prime 2} .
\end{aligned}
$$
去掉质数，我们就得到了旋转坐标系中不变时空区间
$$
\mathrm{d} s^2=-\left(c^2-\omega^2 r^2\right) \mathrm{d} t^2+2 \omega r^2 \mathrm{~d} \phi \mathrm{d} t+\mathrm{d} r^2+r^2 \mathrm{~d} \phi^2+\mathrm{d} z^2
$$
我们把它写成这样
$$
\mathrm{d} s^2=g_{\mu v} \mathrm{~d} x^\mu \mathrm{d} x^v=g_{00}\left(\mathrm{~d} x^0\right)^2+2 \mathrm{~g}{0 i} \mathrm{~d} x^0 \mathrm{~d} x^i+g{i k} \mathrm{~d} x^i \mathrm{~d} x^k,
$$
其中，像往常一样，$i$和$k$仅对空间指数$1-3$求和。对于$x^\mu=\left(x^0, x^1, x^2\right.$, $\left.x^3\right)=(c t, r, \phi, z), g_{\mu v}$采用了这种形式
$$
g_{\mu v}=\left(\begin{array}{cccc}
-\left(1-\frac{\omega^2 r^2}{c^2}\right) & 0 & \frac{\omega r^2}{c} & 0 \
0 & 1 & 0 & 0 \
\frac{\omega r^2}{c} & 0 & r^2 & 0 \
0 & 0 & 0 & 1
\end{array}\right) .
$$

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线性代数代写

线性代数是数学的一个分支，涉及线性方程，如：线性图，如：以及它们在向量空间和通过矩阵的表示。线性代数是几乎所有数学领域的核心。

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2023年6月29日2023年6月29日 General Relativity, 广义相对论, 物理代写

物理代写|广义相对论代写General Relativity代考|The need for a theory of gravity

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。

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物理代写|广义相对论代写General Relativity代考|The need for a theory of gravity

Newton’s theory of gravitation is a spectacularly successful theory. For centuries it has been used by astronomers to calculate the motions of the planets, with a staggering success rate. It has, however, the fatal flaw that it is inconsistent with Special Relativity. We begin by showing this.

As every reader of this book knows, Newton’s law of gravitation states that the force exerted on a mass $m$ by a mass $M$ is
$$
\mathbf{F}=-\frac{M m G}{r^3} \mathbf{r}
$$
Here $M$ and $m$ are not necessarily point masses; $r$ is the distance between their centres of mass. The vector $\mathbf{r}$ has a direction from $M$ to $m$. Now suppose that the mass $M$ depends on time. The above formula will then become
$$
\mathbf{F}(t)=-\frac{M(t) m G}{r^3} \mathbf{r}
$$
This means that the force felt by the mass $m$ at a time $t$ depends on the value of the mass $M$ at the same time t. There is no allowance for time delay, as Special Relativity would require. From our experience of advanced and retarded potentials in electrodynamics, we can say that Special Relativity would be satisfied if, in the above equation, $M(t)$ were modified to $M(t \quad r / c)$. This would reflect the fact that the force felt by the small mass at time $t$ depended on the value of the large mass at an earlier time $t \quad r / c$; assuming, that is, that the relevant gravitational ‘information’ travelled at the speed of light. But this would then not be Newton’s law. Newton’s law is Equation (1.2) which allows for no time delay, and therefore implicitly suggests that the information that the mass $M$ is changing travels with infinite velocity, since the effect of a changing $M$ is felt at the same instant by the mass $m$. Since Special Relativity implies that nothing can travel faster than light, Equations (1.1) and (1.2) are incompatible with it. If two theories are incompatible, at least one of them must be wrong. The only possible attitude to adopt is that Special Relativity must be kept intact, so Newton’s law has to be changed.

物理代写|广义相对论代写General Relativity代考|Gravitation and inertia: the Equivalence Principle in mechanics

Einstein’s new approach to gravity sprang from the work of Galileo (1564-1642; he was born in the same year as Shakespeare and died the year Newton was born). Galileo conducted a series of experiments rolling spheres down ramps. He varied the angle of inclination of the ramp and timed the spheres with a water clock. Physicists commonly portray Galileo as dropping masses from the Leaning Tower of Pisa and timing their descent to the ground. Historians cast doubt on whether this happened, but for our purposes it hardly matters whether it did or didn’t; what matters is the conclusion Galileo drew. By extrapolating to the limit in which the ramps down which the spheres rolled became vertical, and therefore that the spheres fell freely, he concluded that all bodies fall at the same rate in a gravitational field. This, for Einstein, was a crucially important finding. To investigate it further consider the following ‘thought-experiment’, which I refer to as ‘Einstein’s box’. A box is placed in a gravitational field, say on the Earth’s surface (Fig. 1.1(a)). An experimenter in the box releases two objects, made of different materials, from the same height, and measures the times of their fall in the gravitational field g. He finds, as Galileo found, that they reach the floor of the box at the same time. Now consider the box in free space, completely out of the reach of any gravitational influences of planets or stars, but subject to an acceleration a (Fig. 1.1(b)). Suppose an experimenter in this box also releases two objects at the same time and measures the time which elapses before they reach the floor. He will find, of course, that they take the same time to reach the floor; he must find this, because when the two objects are released, they are then subject to no force, because no acceleration, and it is the floor of the box that accelerates up to meet them. It clearly reaches them at the same time. We conclude that this experimenter, by releasing objects and timing their fall, will not be able to tell whether he is in a gravitational field or being accelerated through empty space. The experiments will give identical results. A gravitational field is therefore equivalent to an accelerating frame of reference – at least, as measured in this experiment. This, according to Einstein, is the significance of Galileo’s experiments, and it is known as the Equivalence Principle. Stated in a more general way, the Equivalence Principle says that no experiment in mechanics can distinguish between a gravitational field and an accel erating frame of reference. This formulation, the reader will note, already goes beyond Galileo’s experiments; the claim is made that all experiments in mechanics will yield the same results in an accelerating frame and in a gravitational field. Let us now analyse the consequences of this.

We begin by considering a particle subject to an acceleration a. According to Newton’s second law of motion, in order to make a particle accelerate it is necessary to apply a force to it. We write
$$
\mathbf{F}=m_{\mathrm{i}} \mathbf{a}
$$
Here $m_{\mathrm{i}}$ is the inertial mass of the particle. The above law states that the reason a particle needs a force to accelerate it is that the particles possesses inertia. A very closely related idea is that acceleration is absolute; (constant) velocity, on the other hand, is relative. Now consider a particle falling in a gravitational field g. It will experience a force (see (1.2) and (1.3) above) given by
$$
\mathbf{F}=m_{\mathrm{g}} \mathbf{g} .
$$

广义相对论代写

物理代写|广义相对论代写General Relativity代考|The need for a theory of gravity

牛顿的万有引力理论是一个非常成功的理论。几个世纪以来，天文学家一直用它来计算行星的运动，成功率惊人。然而，它有一个致命的缺陷，那就是它与狭义相对论不一致。我们从这个开始。

本书的每位读者都知道，牛顿的万有引力定律表明，施加在质量$m$上的力$M$是
$$
\mathbf{F}=-\frac{M m G}{r^3} \mathbf{r}
$$
这里$M$和$m$不一定是质点;$r$是它们质心之间的距离。向量$\mathbf{r}$的方向是从$M$到$m$。现在假设质量$M$与时间有关。上面的公式就变成
$$
\mathbf{F}(t)=-\frac{M(t) m G}{r^3} \mathbf{r}
$$
这意味着质量$m$在某一时刻感受到的力$t$取决于质量$M$在同一时刻t的值。不考虑时间延迟，正如狭义相对论所要求的那样。根据我们在电动力学中关于先进势和迟滞势的经验，我们可以说，如果把上式中的$M(t)$修改为$M(t \quad r / c)$，则狭义相对论是满足的。这将反映一个事实，即时间$t$的小质量所感受到的力取决于时间$t \quad r / c$的大质量的值;也就是说，假设相关的引力“信息”以光速传播。但这就不是牛顿定律了。牛顿定律是方程(1.2)，它不考虑时间延迟，因此隐含地表明，质量$M$变化的信息以无限大的速度传播，因为变化的$M$的影响在同一时刻被质量$m$感受到。由于狭义相对论暗示没有任何东西可以比光传播得更快，公式(1.1)和式(1.2)与之不相容。如果两个理论不相容，那么至少有一个是错的。唯一可能采取的态度是，狭义相对论必须保持不变，因此牛顿定律必须改变。

物理代写|广义相对论代写General Relativity代考|Gravitation and inertia: the Equivalence Principle in mechanics

爱因斯坦研究引力的新方法源于伽利略(1564-1642;他和莎士比亚同一年出生，死于牛顿出生的那一年)。伽利略进行了一系列让球体滚下斜坡的实验。他改变了斜坡的倾斜角度，并用一个水钟给球体计时。物理学家通常把伽利略描绘成从比萨斜塔上扔下物体，并计算它们下落到地面的时间。历史学家对这是否发生过表示怀疑，但就我们的目的而言，它是否发生并不重要;重要的是伽利略得出的结论。通过外推球体滚下的斜坡变为垂直的极限，因此球体自由下落，他得出结论，所有物体在引力场中都以相同的速度下落。对爱因斯坦来说，这是一个至关重要的发现。为了进一步研究它，请考虑下面的“思想实验”，我称之为“爱因斯坦的盒子”。一个盒子被放置在引力场中，比如地球表面(图1.1(A))。盒子里的实验者从相同的高度释放两个由不同材料制成的物体，并测量它们在引力场g中下落的时间。他发现，正如伽利略发现的那样，它们同时到达盒子的底部。现在考虑自由空间中的盒子，它完全不受行星或恒星的任何引力影响，但受到加速度a的影响(图1.1(b))。假设这个盒子里的实验者也同时释放两个物体，并测量它们到达地板之前所经过的时间。当然，他会发现它们到达地面的时间是一样的;他必须找到这个，因为当两个物体被释放时，它们就不会受到力的作用，因为没有加速度，是盒子的地板加速向上与它们相遇。很明显，它同时到达了他们。我们的结论是，通过释放物体并确定它们下落的时间，实验者将无法判断他是在引力场中还是在真空中被加速。这些实验将得出相同的结果。因此，引力场相当于一个加速参考系——至少在这个实验中是这样测量的。根据爱因斯坦的说法，这就是伽利略实验的意义所在，这就是所谓的等效原理。用更一般的方式表述，等效原理说，在力学中没有任何实验可以区分引力场和加速参照系。读者会注意到，这个公式已经超越了伽利略的实验;有人声称，所有力学实验在加速坐标系和引力场中都会得到相同的结果。现在让我们分析一下这种情况的后果。

我们首先考虑一个受到加速度a的粒子。根据牛顿第二运动定律，为了使一个粒子加速，必须对它施加力。我们写
$$
\mathbf{F}=m_{\mathrm{i}} \mathbf{a}
$$
这里$m_{\mathrm{i}}$是粒子的惯性质量。上述定律指出，一个粒子需要一个力来加速它的原因是粒子具有惯性。一个非常密切相关的观点是加速度是绝对的;另一方面，(恒定)速度是相对的。现在考虑一个粒子落在引力场g中。它将受到一个力(见上面的(1.2)和(1.3))
$$
\mathbf{F}=m_{\mathrm{g}} \mathbf{g} .
$$

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2023年6月29日2023年6月29日 General Relativity, 广义相对论, 物理代写

物理代写|广义相对论代写General Relativity代考|Determining Distance

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物理代写|广义相对论代写General Relativity代考|Determining Distance

For very close stars, the distance can be determined by parallax. As the earth orbits the sun, the star whose distance is to be measured, appears to move against the background of seemingly fixed distant stars, as indicated in Fig. 9.4. From the extreme wanderings that occur half a year apart, the parallax angle $p$ can be determined from the straight line light paths,
$$
b / d_{|}=b / d=\tan p \approx p, \quad d_{|}=b / p
$$
The origin is at the sun’s center, $b=1.496 \times 10^{11} \mathrm{~m}$ is the radial coordinate of earth, and $d$ is the radial coordinate of the star. In Newtonian physics, these coordinates are the distances. This formula is good enough because gravity is so weak, and only close stars are considered.

General Relativity (GR) requires taking gravity into consideration. The light paths are geodesics as indicated by the dashed curve in Fig. 9.4. Following the development of Weinberg (1972), light leaves the source at position $\vec{d}$, and eventually reaches us. In the coordinate system $x^{\mu^{\prime}}$, where the origin is at the light source, the tip of the ray path is at $\vec{r}^{\prime}=\hat{n} r^{\prime}$. Here $\hat{n}$ is a fixed unit vector and $r^{\prime}$ is a parameter describing positions along the path. In order to translate to the coordinate system in which the light source is at $\vec{d}$ and the origin is at the center of the sun, the quasi-translation Equation (9.4) must be used, so that the same metric holds for both observers. For this case, use $\vec{a}=\vec{d}$ and $x^i \leftrightarrow x^{i^{\prime}}$. Thus,
$$
\vec{r}\left(r^{\prime}\right)=r^{\prime} \hat{n}+\vec{d}\left[\left(1-k r^2\right)^{1 / 2}-\left[1-\left(1-k d^2\right)^{1 / 2}\right] \frac{r^{\prime} \hat{n} \cdot \hat{d}}{d}\right] .
$$

物理代写|广义相对论代写General Relativity代考|Red Shift Versus Distance Relation

Return to the conditions of Section 9.2 where the red shift was defined. An expansion of $Q$ about $t_0$ can provide a relation for $z$ in terms of the radiation travel time $u_1 \equiv t_0-t_1$ or the luminosity distance $d_L$. Assume, the expansion can neglect terms of third order or higher, in the expansion variable $u=t_0-t$
$$
\begin{aligned}
Q & =Q_0-u \frac{d Q_0}{d t}+\frac{u^2}{2} \frac{d^2 Q_0}{d t^2},\left.\quad \frac{d^2 Q_0}{d t^2} \equiv \frac{d^2 Q}{d t^2}\right|_{t_0} \
& =Q_0\left[1-\frac{u}{Q_0} \frac{d Q_0}{d t}+\frac{u^2}{2 Q_0} \frac{d^2 Q_0}{d t^2}\right] \
& \equiv Q_0\left[1-H_0 u-q_0 H_0^2 u^2 / 2\right]
\end{aligned}
$$
In the above equation,
$$
H=\frac{1}{Q} \frac{d Q}{d t},-q=\frac{1}{H^2 Q} \frac{d^2 Q}{d t^2}=1+\frac{1}{H^2} \frac{d H}{d t},
$$
where $H$ is the Hubble parameter with present value $H_0$ and $-q$ is the acceleration parameter with present value $-q_0$.

Evaluate Eq. (9.14) at time $t_1$, where $u=u_1$, and find the relation between the travel time and red shift,
$$
\begin{aligned}
1 & =\frac{Q_0}{Q_1}\left[1-H_0 u_1-H_0^2 q_0 u_1^2 / 2\right] \
& =(1+z)\left[1-H_0 u_1-H_0^2 q_0 u_1^2 / 2\right], \
z & =\frac{H_0 u_1+H_0^2 q_0 u_1^2 / 2}{1-H_0 u_1-H_0^2 q_0 u_1^2 / 2} \
& \approx\left[H_0 u_1+H_0^2 q_0 u_1^2 / 2\right]\left[1+H_0 u_1\right] \
& \approx H_0 u_1+H_0^2 u_1^2\left(1+q_0 / 2\right) .
\end{aligned}
$$

广义相对论代写

物理代写|广义相对论代写General Relativity代考|Determining Distance

对于距离很近的恒星，距离可以通过视差来确定。当地球绕太阳公转时，要测量的恒星在看似固定的遥远恒星的背景下似乎在移动，如图9.4所示。从相隔半年的极端漫游中，可以从直线光路确定视差角$p$，
$$
b / d_{|}=b / d=\tan p \approx p, \quad d_{|}=b / p
$$
原点在太阳的中心，$b=1.496 \times 10^{11} \mathrm{~m}$是地球的径向坐标，$d$是恒星的径向坐标。在牛顿物理学中，这些坐标就是距离。这个公式很好，因为引力很弱，而且只考虑距离很近的恒星。

广义相对论(GR)需要考虑重力。光路为测地线，如图9.4虚线所示。根据Weinberg(1972)的发展，光在$\vec{d}$位置离开光源，最终到达我们。在$x^{\mu^{\prime}}$坐标系中，原点在光源处，射线路径的尖端在$\vec{r}^{\prime}=\hat{n} r^{\prime}$。这里$\hat{n}$是一个固定的单位矢量，$r^{\prime}$是描述路径位置的参数。为了转换到光源在$\vec{d}$和原点在太阳中心的坐标系，必须使用准平移方程(9.4)，以便两个观测者保持相同的度规。对于这种情况，使用$\vec{a}=\vec{d}$和$x^i \leftrightarrow x^{i^{\prime}}$。因此，
$$
\vec{r}\left(r^{\prime}\right)=r^{\prime} \hat{n}+\vec{d}\left[\left(1-k r^2\right)^{1 / 2}-\left[1-\left(1-k d^2\right)^{1 / 2}\right] \frac{r^{\prime} \hat{n} \cdot \hat{d}}{d}\right] .
$$

物理代写|广义相对论代写General Relativity代考|Red Shift Versus Distance Relation

回到9.2节中定义红移的条件。对$t_0$进行$Q$的展开式，可以得到$z$在辐射传播时间$u_1 \equiv t_0-t_1$或光度距离$d_L$方面的关系。假设，在展开变量$u=t_0-t$中，可以忽略三阶或更高阶的项
$$
\begin{aligned}
Q & =Q_0-u \frac{d Q_0}{d t}+\frac{u^2}{2} \frac{d^2 Q_0}{d t^2},\left.\quad \frac{d^2 Q_0}{d t^2} \equiv \frac{d^2 Q}{d t^2}\right|_{t_0} \
& =Q_0\left[1-\frac{u}{Q_0} \frac{d Q_0}{d t}+\frac{u^2}{2 Q_0} \frac{d^2 Q_0}{d t^2}\right] \
& \equiv Q_0\left[1-H_0 u-q_0 H_0^2 u^2 / 2\right]
\end{aligned}
$$
在上式中，
$$
H=\frac{1}{Q} \frac{d Q}{d t},-q=\frac{1}{H^2 Q} \frac{d^2 Q}{d t^2}=1+\frac{1}{H^2} \frac{d H}{d t},
$$
其中$H$是具有现值的哈勃参数$H_0$, $-q$是具有现值的加速度参数$-q_0$。

在时间$t_1$ ($u=u_1$)处求式(9.14)，得到行程时间与红移之间的关系，
$$
\begin{aligned}
1 & =\frac{Q_0}{Q_1}\left[1-H_0 u_1-H_0^2 q_0 u_1^2 / 2\right] \
& =(1+z)\left[1-H_0 u_1-H_0^2 q_0 u_1^2 / 2\right], \
z & =\frac{H_0 u_1+H_0^2 q_0 u_1^2 / 2}{1-H_0 u_1-H_0^2 q_0 u_1^2 / 2} \
& \approx\left[H_0 u_1+H_0^2 q_0 u_1^2 / 2\right]\left[1+H_0 u_1\right] \
& \approx H_0 u_1+H_0^2 u_1^2\left(1+q_0 / 2\right) .
\end{aligned}
$$

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2023年6月29日2023年6月29日 General Relativity, 广义相对论, 物理代写

物理代写|广义相对论代写General Relativity代考|The C Symbols

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物理代写|广义相对论代写General Relativity代考|The C Symbols

The $\mathrm{C}$ symbols involve $g^{\mu \nu}$. They are also obtained as expansions in $\bar{v}^n$, that satisfy Eq. (8.21). From, $g^{\mu \xi} g_{\nu \xi}=\delta_\nu^\mu$,
$$
\begin{aligned}
1 & =g^{0 \xi} g_{0 \xi}=g^{00} g_{00}+g^{0 i} g_{0 i}=\left(\eta^{00}+{ }^2 h^{00}\right)\left(\eta_{00}+{ }^2 h_{00}\right) \
& =1-{ }^2 h^{00}-{ }^2 h_{00} \
{ }^2 h^{00} & =-{ }^2 h_{00}, \
0 & =g^{0 \xi} g_{i \xi}=g^{00} g_{i 0}+g^{0 j} g_{i j}={ }^3 h_{i 0} \eta^{00}+{ }^3 h^{0 j} \eta_{i j}, \
{ }^3 h^{i 0} & ={ }^3 h_{i 0}, \
\delta_j^i & =g^{i \xi} g_{j \xi}=g^{i 0} g_{j 0}+g^{i k} g_{j k}=\left(\eta^{i k}+{ }^2 h^{i k}\right)\left(\eta_{j k}+{ }^2 h_{j k}\right) \
& =\eta^{i k} \eta_{j k}+{ }^2 h_{j k} \eta^{i k}+{ }^2 h^{i k} \eta_{j k}=\delta^i{ }j+{ }^2 h{j k} \eta^{i k}+{ }^2 h^{i k} \eta_{j k}, \
{ }^2 h^{i j} & =-{ }^2 h_{j i} .
\end{aligned}
$$

The $\mathrm{C}$ symbols involve terms with a partial derivative with respect to time. It’s important to note that as for powers of $\bar{v}$,
$$
\frac{\partial}{\partial t} \propto \bar{v} / r
$$
This result and Eqs. (8.18)-(8.20) are required to make sure Eq. (8.21) is satisfied. The $\mathrm{C}$ symbols are obtained from
$$
\Gamma_{\mu \nu}^{\xi}=g^{\xi \chi}\left(g_{\mu \chi}, \nu+g_{\nu \chi}, \mu-g_{\mu \nu}, \chi\right) / 2 .
$$
Those with contravariant index 0 are as follows:
$$
\begin{aligned}
\Gamma_{00}^0 & =g^{0 \chi}\left(g_{0 \chi, 0}+g_{0 \chi}, 0-g_{00, \chi}\right) / 2 \
& =g^{00}\left(g_{00,0}\right) / 2+g^{0 i}\left(2 g_{0 i}, 0-g_{00, i}\right) / 2, \
{ }^3 \Gamma_{00}^0 & =\eta^{00}{ }^2 h_{00,0} / 2=-{ }^2 h_{00,0} / 2, \
\Gamma_{0 i}^0 & =g^{0 \chi}\left(g_{i \chi, 0}+g_{0 \chi, i}-g_{i 0, \chi}\right) / 2 \
& =g^{00}\left(g_{i 0,0}+g_{00, i}-g_{i 0,0}\right) / 2+g^{0 j}\left(g_{i j, 0}+g_{0 j, i}-g_{0 i, j}\right) / 2, \
{ }^2 \Gamma_{0 i}^0 & =\eta^{00}\left({ }^2 h_{00, i}\right) / 2=-{ }^2 h_{00, i} / 2 \
\Gamma_{i j}^0 & =g^{0 \chi}\left(g_{i \chi, j}+g_{j \chi, i}-g_{i j}, \chi\right) / 2 \
& =g^{00}\left(g_{i 0, j}+g_{j 0, i}-g_{i j, 0}\right) / 2+g^{0 k}\left(g_{i k, j}+g_{j k, i}-g_{i j, k}\right) / 2, \
{ }^3 \Gamma_{i j}^0 & =-\left({ }^3 h_{i 0, j}+{ }^3 h_{j 0, i}-{ }^2 h_{i j, 0}\right) / 2 .
\end{aligned}
$$

物理代写|广义相对论代写General Relativity代考|Ricci Tensor and Einstein Field Equations

In order to express the various expansion terms of a metric, as potentials in terms of the energy momentum tensor, just like ${ }^2 h_{00}=-2 \Psi_G$, the Einstein field equations are needed,
$$
\begin{aligned}
G_{\mu \xi} & =R_{\mu \xi}-g_{\mu \xi} R / 2=8 \pi T_{\mu \xi}, \
g^{\mu \xi}\left(R_{\mu \xi}-g_{\mu \xi} R / 2\right) & =8 \pi g^{\mu \xi} T_{\mu \xi}, \
R-2 R & =-R=8 \pi g^{\mu \xi} T_{\mu \xi}, \
R_{\mu \nu} & =8 \pi\left(T_{\mu \nu}-g_{\mu \nu} g^{\chi \xi} T_{\chi \xi} / 2\right) .
\end{aligned}
$$
The nonzero Ricci tensor elements $R_{\mu \nu}$ will be evaluated so that Eq. (8.21) is obeyed. The reader should note that $R_{\mu \nu}$ in this text has the opposite sign of $R_{\mu \nu}$ in Weinberg’s text. In some of the equations below, $\eta^{i i}$ is used to remind the reader to sum over $i$.
$$
R_{\mu \nu}=R_{\mu \xi \nu}^{\xi}=\Gamma_{\mu \nu}^\chi \Gamma_{\xi \chi}^{\xi}-\Gamma_{\xi \mu}^\chi \Gamma_{\nu \chi}^{\xi}+\Gamma_{\mu \nu, \xi}^{\xi}-\Gamma_{\xi \mu}^{\xi}, \nu
$$
However, the product $\Gamma \Gamma \propto \bar{v}^{>3}$ and may be neglected. Moreover,
$$
\begin{aligned}
R_{00} & =\Gamma_{00, \xi}^{\xi}-\Gamma_{\xi 0,0}^{\xi}=\Gamma_{00, i}^i-\Gamma_{i 0,0}^i, \
{ }^2 R_{00} & ={ }^2 \Gamma_{00, i}^i=-\eta^{i i}{ }^i h_{00, i},{ }i / 2=-\nabla^2{ }^2 h{00} / 2 .
\end{aligned}
$$

广义相对论代写

物理代写|广义相对论代写General Relativity代考|The C Symbols

$\mathrm{C}$符号包含$g^{\mu \nu}$。它们也可以在$\bar{v}^n$中展开，满足式(8.21)。来自，$g^{\mu \xi} g_{\nu \xi}=\delta_\nu^\mu$;
$$
\begin{aligned}
1 & =g^{0 \xi} g_{0 \xi}=g^{00} g_{00}+g^{0 i} g_{0 i}=\left(\eta^{00}+{ }^2 h^{00}\right)\left(\eta_{00}+{ }^2 h_{00}\right) \
& =1-{ }^2 h^{00}-{ }^2 h_{00} \
{ }^2 h^{00} & =-{ }^2 h_{00}, \
0 & =g^{0 \xi} g_{i \xi}=g^{00} g_{i 0}+g^{0 j} g_{i j}={ }^3 h_{i 0} \eta^{00}+{ }^3 h^{0 j} \eta_{i j}, \
{ }^3 h^{i 0} & ={ }^3 h_{i 0}, \
\delta_j^i & =g^{i \xi} g_{j \xi}=g^{i 0} g_{j 0}+g^{i k} g_{j k}=\left(\eta^{i k}+{ }^2 h^{i k}\right)\left(\eta_{j k}+{ }^2 h_{j k}\right) \
& =\eta^{i k} \eta_{j k}+{ }^2 h_{j k} \eta^{i k}+{ }^2 h^{i k} \eta_{j k}=\delta^i{ }j+{ }^2 h{j k} \eta^{i k}+{ }^2 h^{i k} \eta_{j k}, \
{ }^2 h^{i j} & =-{ }^2 h_{j i} .
\end{aligned}
$$

$\mathrm{C}$符号涉及对时间有偏导数的项。需要注意的是，对于$\bar{v}$的幂，
$$
\frac{\partial}{\partial t} \propto \bar{v} / r
$$
这个结果和等式。需要(8.18)-(8.20)来确保满足式(8.21)。$\mathrm{C}$符号取自
$$
\Gamma_{\mu \nu}^{\xi}=g^{\xi \chi}\left(g_{\mu \chi}, \nu+g_{\nu \chi}, \mu-g_{\mu \nu}, \chi\right) / 2 .
$$
逆变指标为0的有:
$$
\begin{aligned}
\Gamma_{00}^0 & =g^{0 \chi}\left(g_{0 \chi, 0}+g_{0 \chi}, 0-g_{00, \chi}\right) / 2 \
& =g^{00}\left(g_{00,0}\right) / 2+g^{0 i}\left(2 g_{0 i}, 0-g_{00, i}\right) / 2, \
{ }^3 \Gamma_{00}^0 & =\eta^{00}{ }^2 h_{00,0} / 2=-{ }^2 h_{00,0} / 2, \
\Gamma_{0 i}^0 & =g^{0 \chi}\left(g_{i \chi, 0}+g_{0 \chi, i}-g_{i 0, \chi}\right) / 2 \
& =g^{00}\left(g_{i 0,0}+g_{00, i}-g_{i 0,0}\right) / 2+g^{0 j}\left(g_{i j, 0}+g_{0 j, i}-g_{0 i, j}\right) / 2, \
{ }^2 \Gamma_{0 i}^0 & =\eta^{00}\left({ }^2 h_{00, i}\right) / 2=-{ }^2 h_{00, i} / 2 \
\Gamma_{i j}^0 & =g^{0 \chi}\left(g_{i \chi, j}+g_{j \chi, i}-g_{i j}, \chi\right) / 2 \
& =g^{00}\left(g_{i 0, j}+g_{j 0, i}-g_{i j, 0}\right) / 2+g^{0 k}\left(g_{i k, j}+g_{j k, i}-g_{i j, k}\right) / 2, \
{ }^3 \Gamma_{i j}^0 & =-\left({ }^3 h_{i 0, j}+{ }^3 h_{j 0, i}-{ }^2 h_{i j, 0}\right) / 2 .
\end{aligned}
$$

物理代写|广义相对论代写General Relativity代考|Ricci Tensor and Einstein Field Equations

为了表达度规的各种展开项，作为能量动量张量的势，就像${ }^2 h_{00}=-2 \Psi_G$，需要爱因斯坦场方程，
$$
\begin{aligned}
G_{\mu \xi} & =R_{\mu \xi}-g_{\mu \xi} R / 2=8 \pi T_{\mu \xi}, \
g^{\mu \xi}\left(R_{\mu \xi}-g_{\mu \xi} R / 2\right) & =8 \pi g^{\mu \xi} T_{\mu \xi}, \
R-2 R & =-R=8 \pi g^{\mu \xi} T_{\mu \xi}, \
R_{\mu \nu} & =8 \pi\left(T_{\mu \nu}-g_{\mu \nu} g^{\chi \xi} T_{\chi \xi} / 2\right) .
\end{aligned}
$$
将计算非零Ricci张量元素$R_{\mu \nu}$，使公式(8.21)得到遵守。读者应该注意到，$R_{\mu \nu}$在这篇文章中有相反的符号$R_{\mu \nu}$在温伯格的文本。在下面的一些方程中，$\eta^{i i}$被用来提醒读者对$i$求和。
$$
R_{\mu \nu}=R_{\mu \xi \nu}^{\xi}=\Gamma_{\mu \nu}^\chi \Gamma_{\xi \chi}^{\xi}-\Gamma_{\xi \mu}^\chi \Gamma_{\nu \chi}^{\xi}+\Gamma_{\mu \nu, \xi}^{\xi}-\Gamma_{\xi \mu}^{\xi}, \nu
$$
但是，产品$\Gamma \Gamma \propto \bar{v}^{>3}$和可能被忽略。而且，
$$
\begin{aligned}
R_{00} & =\Gamma_{00, \xi}^{\xi}-\Gamma_{\xi 0,0}^{\xi}=\Gamma_{00, i}^i-\Gamma_{i 0,0}^i, \
{ }^2 R_{00} & ={ }^2 \Gamma_{00, i}^i=-\eta^{i i}{ }^i h_{00, i},{ }i / 2=-\nabla^2{ }^2 h{00} / 2 .
\end{aligned}
$$

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2023年6月9日2023年6月9日 General Relativity, 广义相对论, 物理代写

物理代写|广义相对论代写General Relativity代考|Quadrupole Radiation

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物理代写|广义相对论代写General Relativity代考|Quadrupole Radiation

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, $$ \begin{aligned} l & =\left(x_2-x_1\right)\left(1+{ }^{T T} \bar{h}{x x} / 2\right) \
& \approx x_2-x_1+l_0 T T \bar{h}{x x} / 2, \ x_2-x_1 & =l-l_0 T T \bar{h}{x x} / 2 .
\end{aligned}
$$

广义相对论代写

物理代写|广义相对论代写General Relativity代考|Quadrupole Radiation

一个重要的情况发生在源随时间谐波变化$S=$$4 T^{\mu \nu}=j_\omega\left[\vec{r}^{\prime}\right] \exp \left[-i \omega t^{\prime}\right]$。当震源靠近原点时，观测点$\vec{r}$可能处于三个区域之一:近、中、远。每个区域允许不同的近似值。遥远的区域，$d \ll \lambda \ll r$，是我们感兴趣的。引发引力波的剧烈事件很可能发生在离我们很远的地方。这里$d$是源的大小，比辐射的波长小得多。这反过来又比观测点的径向坐标小得多。然后，
$$
\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}
$$

远区近似为局部惯性，因为它离源太远。用自然单位$2 \pi / \lambda=k=\omega$，但是$k r$和$\omega t$的写法让方程看起来很熟悉。对于谐波依赖，解Eq.(7.24)看起来像一个输出的球形波，其振幅由积分给出。回想一下$g_{\mu \nu}=\eta_{\mu \nu}+h_{\mu \nu}$，因此$h^{\mu \nu}$和$\bar{h}^{\mu \nu}$的解是用直角坐标表示的。通过$\eta^{\mu \nu}$和$\eta_{\mu \nu}$提高和降低指数。这些操作完成后，振幅可以在其他坐标系中表示。
从能量守恒，得到最低阶近似
$$
\begin{aligned}
0 & =T^{\mu \nu}{ }{,}{ }\nu=T^{\mu \nu}{ }{,}{ }\nu+\Gamma_{\xi \nu}^\mu T^{\xi \nu}+\Gamma_{\nu \xi}^\nu T^{\mu \xi} \approx T^{\mu \nu}{ }{,} \ & =T^{0 \nu}{ }\nu=T^{00}{ }{, 0}+T^{0 k}{ }{, k} \
& =\left(T^{00}{ }0+T^{0 k}{ }{, k}\right){, 0}=T^{00}{ }{, 0}, 0+T^{0 k}{ }{, k}, 0, \ T^{00}{ }{, 0}, 0 & =\left(-T^{0 k}{ }{, 0}\right){{ }k}=-\left(-T^{j k}{ }{, j}\right){, k}=T^{j k}{ }{, j}, k, \
\int_{\infty} d V x^i x^n T^{00}{ }{, 0}, 0 & =\int{\infty} d V x^i x^n T^{j k}{ }_j,{ }_k .
\end{aligned}
$$

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

波通量是它的能量/面积/时间。自然单位是$\mathrm{m}^{-2}$。这个量是对球面面积的积分。这决定了能量$P$或波的光度$L$。当震源失去能量时，它会发生变化，这种观察结果可以用来探测波。电磁波的结果不能直接接受，因为它们的振幅是1阶张量，而重力波的振幅是2阶张量。因此，虽然通量仍然与振幅的绝对平方成正比，但所有重要的比例因子都是不同的。为了计算它，遵循B. Schutz(2009)的方法。

考虑一个沿$z$ -方向运动的无迹横波。计算了传递到近似连续的振子阵列(元素弹簧)上的通量。弹簧在平面$z=0$上沿$x$ -方向排列。弹簧具有自然长度$l_0$、等质量$m$、小弹簧常数$m \omega_0^2 / 2$和小阻尼常数$m \gamma$。单位面积的弹簧数为$\frac{d n}{d A}=\alpha$。当振子获得能量时，波失去能量，其振幅减小。发现通量和振幅之间的关系不依赖于弹簧，而是波的性质。弹簧只是用作计算辅助工具。

设原点在弹簧的中心，质量在$x_{1,2}$。在平坦空间中，质量的运动方程如下:
$$
\begin{aligned}
\frac{d^2 x_2}{d t^2} & =-\omega_0^2\left(x_2-x_1-l_0\right) / 2-\gamma \frac{d\left(x_2-x_1\right)}{d t}, \
\frac{d^2 x_1}{d t^2} & =\omega_0^2\left(x_2-x_1-l_0\right) / 2+\gamma \frac{d\left(x_2-x_1\right)}{d t}, \
\frac{d^2\left(x_2-x_1-l_0\right)}{d t^2} & =-\omega_0^2\left(x_2-x_1-l_0\right)-2 \gamma \frac{d\left(x_2-x_1-l_0\right)}{d t} .
\end{aligned}
$$
考虑了${ }^{T T} \bar{h}{x x}$波的一个元素。其他元素(由于跟踪为零，因此必须有其他元素)将以相同的方式处理。当遇到波时，公式(7.17)给出质量之间的适当长度， $$ \begin{aligned} l & =\left(x_2-x_1\right)\left(1+{ }^{T T} \bar{h}{x x} / 2\right) \
& \approx x_2-x_1+l_0 T T \bar{h}{x x} / 2, \ x_2-x_1 & =l-l_0 T T \bar{h}{x x} / 2 .
\end{aligned}
$$

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线性代数是数学的一个分支，涉及线性方程，如：线性图，如：以及它们在向量空间和通过矩阵的表示。线性代数是几乎所有数学领域的核心。

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2023年6月9日2023年6月9日 General Relativity, 广义相对论, 物理代写

物理代写|广义相对论代写General Relativity代考|Plane Waves

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物理代写|广义相对论代写General Relativity代考|Plane Waves

In empty space there is no source, so plane wave solutions are possible. With enough plane waves of different wave vectors and accompanying amplitudes, any wave shape can be accommodated by superposition. In the case of a single wave vector $k_\chi$ with amplitude $A_{\mu \nu}$, a complex constant, the wave function in rectangular coordinates is the real part of $\bar{h}{\mu \nu}=A{\mu \nu} \exp \left(i k_\chi x^\chi\right)$. The phase factor is an invariant. It is easily shown that
$$
\begin{aligned}
& \bar{h}{\mu \nu, \beta}=i k\beta \bar{h}{\mu \nu}, \ & \square \bar{h}{\mu \nu}=-\left(k_\beta k^\beta\right) \bar{h}{\mu \nu}=0, \quad k\beta k^\beta=0 .
\end{aligned}
$$
As with an electromagnetic wave, the relation between frequency $k^0 \equiv \omega$ and wave 3 -vector $\vec{k}$ is identical. In free space, there is no dispersion, so the phase and group velocities are unity. The direction of $\vec{k}$ is the direction of wave travel. The gauge condition forced $\bar{h}\mu^\nu,{ }\nu=0$. Thus,
$$
k_\nu A_\mu^\nu=0
$$
This is another restriction, an orthogonality restriction on $A_\mu^\nu$.
A more useful solution can be obtained, by again applying a gauge transformation with vector $\xi_\alpha$. The vector satisfies $\square \xi_\alpha=0$. It can produce a solution ${ }^{T T} \bar{h}{\mu \nu}$, with amplitude $\bar{A}{\mu \nu}$, that is traceless ${ }^{T T} \bar{h}\mu^\mu=\bar{A}\mu^\mu=0$. Using $\xi_\mu=B_\mu \exp \left(i k_\chi r^\chi\right)$ and results from Problem 2 ,
$$
\begin{aligned}
\bar{h}{\mu^{\prime} \nu^{\prime}} & \equiv T T \bar{h}{\mu \nu}=\bar{h}{\mu \nu}-\xi\mu,{ }\nu-\xi{\nu,{ }\mu}+\eta{\mu \nu} \xi^\chi,{ }\chi, \ \bar{A}{\mu \nu} & =A_{\mu \nu}-i\left(B_\mu k_\nu+k_\mu B_\nu-\eta_{\mu \nu} B^\chi k_\chi\right), \
\bar{A}\mu^\alpha & =A\mu^\alpha-i\left(B_\mu k^\alpha+k_\mu B^\alpha-\delta_\mu^\alpha B^\chi k_\chi\right), \
0 & =\bar{A}\mu^\mu=A\mu^\mu-i\left(B_\mu k^\mu+k_\mu B^\mu-4 B^\chi k_\chi\right) \
& =A_\mu^\mu+2 i B_\mu k^\mu=A_\mu^\mu+2 i B^\mu k_\mu, \quad i B^\mu k_\mu=-A_\mu^\mu / 2 .
\end{aligned}
$$

物理代写|广义相对论代写General Relativity代考|The Graviton

For electromagnetic waves, the wave function of the vector field $A_\mu$ can describe all of the physics. When this field is quantized, the quanta are photons with spin $s=1$. In quantum electrodynamics, the interactions to lowest order are the exchange of virtual photons. In GR, the wave function of the field describing the physics is a tensor of rank $2 \bar{h}{\mu \nu}$. Thus, a quantum theory of gravity has a exchange particle of $\operatorname{spin} s=2$, with zero rest mass, called the graviton. A transparent way to see this is to consider what happens to a transverse electromagnetic or transverse, traceless gravitational plane wave amplitude, under rotation. If the plane wave is traveling in the 3-direction, the only nonzero amplitudes are $A_j$ for the electromagnetic wave and $\bar{A}{j k}$ for the gravitational wave. Here $(j, k) \neq 3$. One can rotate these wave functions by angle $\phi$ about the axis along the propagation direction, using the information in Fig. 1.1. Another set of rectangular axes, where basis and unit vectors are the same, is obtained. The nonzero elements of the rotation matrix $x^i, j^{\prime}$ are: $R{1^{\prime}}^1=R{2^{\prime}}^2=\cos \phi, R_{2^{\prime}}^1=-R_{1^{\prime}}^2=\sin \phi, R_{3^{\prime}}^3=1$.
Using the rotation $A_{j^{\prime}}=R_{j^{\prime}}^k A_k$, the electromagnetic amplitudes become
$$
\begin{aligned}
& A_{1^{\prime}}=R_{1^{\prime}}^1 A_1+R_{1^{\prime}}^2 A_2=\cos \phi A_1-\sin \phi A_2 \
& A_{2^{\prime}}=R_{2^{\prime}}^1 A_1+R_{2^{\prime}}^2 A_2=\sin \phi A_1+\cos \phi A_2
\end{aligned}
$$
These equations yield
$$
\begin{aligned}
A_{1^{\prime}} \pm i A_{2^{\prime}} & =(\cos \phi \pm i \sin \phi) A_1+(-\sin \phi \pm i \cos \phi) A_2 \
& =\exp ( \pm i \phi) A_1 \pm(\cos \phi \pm i \sin \phi) i A_2=\exp ( \pm i \phi)\left[A_1 \pm i A_2\right]
\end{aligned}
$$
Using the rotation $\bar{A}{j^{\prime} k^{\prime}}=R{j^{\prime}}^l R_{k^{\prime}}^n \bar{A}{l n}$ and Eq. (7.15), the gravitational amplitudes become $$ \begin{aligned} \bar{A}{1^{\prime} 1^{\prime}} & =R_{1^{\prime}}^1 R_{1^{\prime}}^1 \bar{A}{11}+R{1^{\prime}}^1 R_{1^{\prime}}^2 \bar{A}{12}+R{1^{\prime}}^2 R_{1^{\prime}}^1 \bar{A}{21}+R{1^{\prime}}^2 R_{1^{\prime}}^2 \bar{A}{22} \ & =\cos ^2 \phi \bar{A}{11}-2 \sin \phi \cos \phi \bar{A}{12}-\sin ^2 \phi \bar{A}{11} \
& =\cos 2 \phi \bar{A}{11}-\sin 2 \phi \bar{A}{12} \
\bar{A}{2^{\prime} 2^{\prime}} & =R{2^{\prime}}^1 R_{2^{\prime}}^1 \bar{A}{11}+R{2^{\prime}}^1 R_{2^{\prime}}^2 \bar{A}{12}+R{2^{\prime}}^2 R_{2^{\prime}}^1 \bar{A}{21}+R{2^{\prime}}^2 R_{2^{\prime}}^2 \bar{A}{22} \ & =\sin ^2 \phi \bar{A}{11}+2 \sin \phi \cos \phi \bar{A}{12}-\cos ^2 \phi \bar{A}{11} \
& =-\cos 2 \phi \bar{A}{11}+\sin 2 \phi \bar{A}{12} \
\bar{A}{1^{\prime} 2^{\prime}} & =R{1^{\prime}}^1 R_{2^{\prime}}^1 \bar{A}{11}+R{1^{\prime}}^1 R_{2^{\prime}}^2 \bar{A}{12}+R{1^{\prime}}^2 R_{2^{\prime}}^1 \bar{A}{21}+R{1^{\prime}}^2 R_{2^{\prime}}^2 \bar{A}{22} \ & =\sin \phi \cos \phi \bar{A}{11}+\left(\cos ^2 \phi-\sin ^2 \phi\right) \bar{A}{12}+\sin \phi \cos \phi \bar{A}{11} \
& =\sin 2 \phi \bar{A}{11}+\cos 2 \phi \bar{A}{12} .
\end{aligned}
$$

广义相对论代写

物理代写|广义相对论代写General Relativity代考|Plane Waves

在真空中没有源，所以平面波解是可能的。只要有足够多的不同波矢量和伴随振幅的平面波，就可以通过叠加容纳任何波形。在单波矢量$k_\chi$的情况下，振幅$A_{\mu \nu}$，一个复常数，波函数在直角坐标系是$\bar{h}{\mu \nu}=A{\mu \nu} \exp \left(i k_\chi x^\chi\right)$的实部。相位因子是不变量。这很容易证明 $$ \begin{aligned} & \bar{h}{\mu \nu, \beta}=i k\beta \bar{h}{\mu \nu}, \ & \square \bar{h}{\mu \nu}=-\left(k_\beta k^\beta\right) \bar{h}{\mu \nu}=0, \quad k\beta k^\beta=0 . \end{aligned} $$ 与电磁波一样，频率$k^0 \equiv \omega$与波3矢量$\vec{k}$之间的关系是相同的。在自由空间中，不存在色散，因此相速度和群速度是统一的。$\vec{k}$的方向是波的传播方向。压力表状态强制$\bar{h}\mu^\nu,{ }\nu=0$。因此， $$ k_\nu A_\mu^\nu=0 $$ 这是另一个限制，$A_\mu^\nu$的正交性限制。一个更有用的解可以得到，再次应用规范变换与向量$\xi_\alpha$。这个向量满足$\square \xi_\alpha=0$。它可以产生一个解${ }^{T T} \bar{h}{\mu \nu}$，振幅$\bar{A}{\mu \nu}$，无迹可循${ }^{T T} \bar{h}\mu^\mu=\bar{A}\mu^\mu=0$。利用$\xi_\mu=B_\mu \exp \left(i k_\chi r^\chi\right)$和问题2的结果， $$ \begin{aligned} \bar{h}{\mu^{\prime} \nu^{\prime}} & \equiv T T \bar{h}{\mu \nu}=\bar{h}{\mu \nu}-\xi\mu,{ }\nu-\xi{\nu,{ }\mu}+\eta{\mu \nu} \xi^\chi,{ }\chi, \ \bar{A}{\mu \nu} & =A_{\mu \nu}-i\left(B_\mu k_\nu+k_\mu B_\nu-\eta_{\mu \nu} B^\chi k_\chi\right), \ \bar{A}\mu^\alpha & =A\mu^\alpha-i\left(B_\mu k^\alpha+k_\mu B^\alpha-\delta_\mu^\alpha B^\chi k_\chi\right), \ 0 & =\bar{A}\mu^\mu=A\mu^\mu-i\left(B_\mu k^\mu+k_\mu B^\mu-4 B^\chi k_\chi\right) \ & =A_\mu^\mu+2 i B_\mu k^\mu=A_\mu^\mu+2 i B^\mu k_\mu, \quad i B^\mu k_\mu=-A_\mu^\mu / 2 . \end{aligned} $$

物理代写|广义相对论代写General Relativity代考|The Graviton

对于电磁波，向量场的波函数$A_\mu$可以描述所有的物理现象。当这个场被量子化时，量子就是自旋为$s=1$的光子。在量子电动力学中，最低阶的相互作用是虚光子的交换。在GR中，描述物理的场的波函数是阶为$2 \bar{h}{\mu \nu}$的张量。因此，量子引力理论有一个静止质量为零的交换粒子$\operatorname{spin} s=2$，称为引力子。看这个的一个透明的方法是考虑横向电磁或横向无迹引力平面波振幅在旋转下的变化。如果平面波在3方向上传播，那么电磁波的非零振幅为$A_j$，引力波的非零振幅为$\bar{A}{j k}$。这里$(j, k) \neq 3$。利用图1.1中的信息，可以沿传播方向绕轴以$\phi$角度旋转这些波函数。得到另一组矩形轴，其中基向量和单位向量是相同的。旋转矩阵$x^i, j^{\prime}$的非零元素为:$R{1^{\prime}}^1=R{2^{\prime}}^2=\cos \phi, R_{2^{\prime}}^1=-R_{1^{\prime}}^2=\sin \phi, R_{3^{\prime}}^3=1$。
通过旋转$A_{j^{\prime}}=R_{j^{\prime}}^k A_k$，电磁振幅变成
$$
\begin{aligned}
& A_{1^{\prime}}=R_{1^{\prime}}^1 A_1+R_{1^{\prime}}^2 A_2=\cos \phi A_1-\sin \phi A_2 \
& A_{2^{\prime}}=R_{2^{\prime}}^1 A_1+R_{2^{\prime}}^2 A_2=\sin \phi A_1+\cos \phi A_2
\end{aligned}
$$
这些方程得到
$$
\begin{aligned}
A_{1^{\prime}} \pm i A_{2^{\prime}} & =(\cos \phi \pm i \sin \phi) A_1+(-\sin \phi \pm i \cos \phi) A_2 \
& =\exp ( \pm i \phi) A_1 \pm(\cos \phi \pm i \sin \phi) i A_2=\exp ( \pm i \phi)\left[A_1 \pm i A_2\right]
\end{aligned}
$$
利用旋转$\bar{A}{j^{\prime} k^{\prime}}=R{j^{\prime}}^l R_{k^{\prime}}^n \bar{A}{l n}$和公式(7.15)，引力振幅变为 $$ \begin{aligned} \bar{A}{1^{\prime} 1^{\prime}} & =R_{1^{\prime}}^1 R_{1^{\prime}}^1 \bar{A}{11}+R{1^{\prime}}^1 R_{1^{\prime}}^2 \bar{A}{12}+R{1^{\prime}}^2 R_{1^{\prime}}^1 \bar{A}{21}+R{1^{\prime}}^2 R_{1^{\prime}}^2 \bar{A}{22} \ & =\cos ^2 \phi \bar{A}{11}-2 \sin \phi \cos \phi \bar{A}{12}-\sin ^2 \phi \bar{A}{11} \
& =\cos 2 \phi \bar{A}{11}-\sin 2 \phi \bar{A}{12} \
\bar{A}{2^{\prime} 2^{\prime}} & =R{2^{\prime}}^1 R_{2^{\prime}}^1 \bar{A}{11}+R{2^{\prime}}^1 R_{2^{\prime}}^2 \bar{A}{12}+R{2^{\prime}}^2 R_{2^{\prime}}^1 \bar{A}{21}+R{2^{\prime}}^2 R_{2^{\prime}}^2 \bar{A}{22} \ & =\sin ^2 \phi \bar{A}{11}+2 \sin \phi \cos \phi \bar{A}{12}-\cos ^2 \phi \bar{A}{11} \
& =-\cos 2 \phi \bar{A}{11}+\sin 2 \phi \bar{A}{12} \
\bar{A}{1^{\prime} 2^{\prime}} & =R{1^{\prime}}^1 R_{2^{\prime}}^1 \bar{A}{11}+R{1^{\prime}}^1 R_{2^{\prime}}^2 \bar{A}{12}+R{1^{\prime}}^2 R_{2^{\prime}}^1 \bar{A}{21}+R{1^{\prime}}^2 R_{2^{\prime}}^2 \bar{A}{22} \ & =\sin \phi \cos \phi \bar{A}{11}+\left(\cos ^2 \phi-\sin ^2 \phi\right) \bar{A}{12}+\sin \phi \cos \phi \bar{A}{11} \
& =\sin 2 \phi \bar{A}{11}+\cos 2 \phi \bar{A}{12} .
\end{aligned}
$$

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2023年6月9日2023年6月9日 General Relativity, 广义相对论, 物理代写

物理代写|广义相对论代写General Relativity代考|Schwarzschild Solution

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物理代写|广义相对论代写General Relativity代考|Schwarzschild Solution

Almost immediately after Einstein introduced the field equations, $K$. Schwarzschild found an exact solution. An English translation of his paper is available, see Schwarzschild (1916). The case in point was the one that was considered with weak gravity, the metric produced by a static, spherically symmetric, massive object in vacuum. The metric tensor won’t depend on $t$, but will depend on $\vec{r}$ and $d \vec{r}$, such that it has rotational invariance. Time independence leads to energy conservation, and rotational invariance leads to conservation of certain angular momentum components. Thus, it pays to work in spherical coordinates. So for the rest of this chapter, $r^p$ means $(r)^p$, and not the $p$ th component of the position vector.
Try the most general rotationally invariant form for the proper-time element,
$$
\begin{aligned}
(d \tau)^2= & -g_{\mu \nu} d x^\mu d x^\nu \
\equiv & A(r)(d t)^2-2 B(r)(\vec{r} \cdot d \vec{r}) d t-C(r)(\vec{r} \cdot d \vec{r})^2-D(r) d \vec{r} \cdot d \vec{r} \
= & A(r)(d t)^2-2 B(r) r d r d t-\left[C(r) r^2+D(r)\right](d r)^2 \
& -D(r) r^2\left[(d \theta)^2+(\sin \theta d \phi)^2\right]
\end{aligned}
$$
One can eliminate the $d t d r$ term with the following transformation:
$$
t \equiv t^{\prime}-E(r), \quad d E(r) \equiv-r d r B(r) / A(r)
$$
It is then easy to show, see Problem 4 , that this leads to
$$
\begin{aligned}
(d \tau)^2= & A(r)\left(d t^{\prime}\right)^2-\left[(r B(r))^2 / A(r)+C(r) r^2+D(r)\right](d r)^2 \
& -D(r) r^2\left[(d \theta)^2+\sin ^2 \theta(d \phi)^2\right] \
\equiv & A(r)\left(d t^{\prime}\right)^2-F(r)(d r)^2-r^2 D(r)\left[(d \theta)^2+(\sin \theta d \phi)^2\right]
\end{aligned}
$$
A final transform redefines $r$, and allows the proper time to be cast in a form where the metric tensor is diagonal,
$$
\begin{aligned}
r^{\prime 2} \equiv & r^2 D(r) \
(d \tau)^2 \equiv & \exp \left[2 \Phi\left(r^{\prime}\right)\right]\left(d t^{\prime}\right)^2-\exp \left[2 \Delta\left(r^{\prime}\right)\right]\left(d r^{\prime}\right)^2 \
& -r^{\prime 2}\left[(d \theta)^2+(\sin \theta d \phi)^2\right]
\end{aligned}
$$

物理代写|广义相对论代写General Relativity代考|Conserved Quantities: Massive Particles

Knowledge of the metric tells us what quantities, if any, are conserved. Such knowledge is very helpful in solving the equations of motion. Light has one constant of the motion, its speed. Equation (4.3) was obtained from parallel transport and for a particle of rest mass $m, d \tau \neq 0$ so it can be used instead of $d q$. Using $g_{\mu \alpha} ;\nu=0$, and renaming summed over indexes, $$ \begin{aligned} 0 & =W^\nu W^\mu ;\nu=U^\nu U^\mu ;\nu=P^\nu P^\mu ;{ }\nu \
& =g_{\mu \alpha} P^\nu P^\mu ;{ }\nu=P^\nu\left(g{\mu \alpha} P^\mu\right){{ }\nu} \
& =P^\nu P_\alpha ;\nu=P^\nu\left(P\alpha,{ }\nu-P\beta \Gamma_{\alpha \nu}^\beta\right) \
& =m \frac{d x^\nu}{d \tau} P_{\alpha, \nu}-P^\nu P_\beta \Gamma_{\alpha \nu}^\beta=m \frac{d P_\alpha}{d \tau}-P^\nu P_\beta \Gamma_{\alpha \nu}^\beta .
\end{aligned}
$$
Then,
$$
\begin{aligned}
\frac{d P_\alpha}{d \tau} & =P^\nu P_\beta \Gamma_{\alpha \nu}^\beta / m=P^\nu P_\beta g^{\beta \chi}\left(g_{\alpha \chi},{ }\nu+g{\chi \nu},{ }\alpha-g{\alpha \nu},\chi\right) /(2 m) \ & =\left(P^\nu P^\chi g{\alpha \chi},\nu+P^\chi P^\nu\left[g{\chi \nu},{ }\alpha-g{\alpha \nu}, \chi_\chi\right]\right) /(2 m) \
& =P^\chi P^\nu\left(g_{\alpha \nu},\chi+g{\chi \nu},{ }\alpha-g{\alpha \nu},\chi\right) /(2 m) \ & =P^\chi P^\nu g{\chi \nu},_\alpha /(2 m) .
\end{aligned}
$$

So if $g_{\chi \nu}, \alpha=0$, then $P_\alpha$ is constant along the geodesic. In a stationary metric $g_{\chi \nu, 0}=0$ and $P_0$ is constant. In the case of weak gravity and low speeds, the total energy is constant.
The constancy of energy can be illustrated to lowest order in small quantities. Use the fact that $|\vec{P}| \ll m$ so that $h_{i j} P^i P^j / m^2$ and $h_{i i}|\vec{P}|^2 / m^2$ can be neglected,
$$
\begin{aligned}
m^2 & =-g_{\mu \nu} P^\mu P^\nu \
& =-\left(-1+h_{00}\right)\left(P^0\right)^2-\left[\left(1+h_{i i}\right)|\vec{P}|^2+2 h_{i j} P^i P^j\right], \
1 & \approx\left(1-h_{00}\right)\left(P^0 / m\right)^2-(|\vec{P}| / m)^2 \
& =\left(1+2 M^{\prime} / r\right)\left(P^0 / m\right)^2-(|\vec{P}| / m)^2, \
P^0 / m & =\left(1+(|\vec{P}| / m)^2\right)^{1 / 2}\left(1+2 M^{\prime} / r\right)^{-1 / 2} \
& \approx\left(1+(|\vec{P}| / m)^2 / 2\right)\left(1-M^{\prime} / r\right), \
P^0 & \approx m-m M^{\prime} / r+|\vec{P}|^2 /(2 m)=R E+P E+K E=E, \
P_0 & =g_{00} P^0 \approx-P^0=-E, \text { constant. }
\end{aligned}
$$

广义相对论代写

物理代写|广义相对论代写General Relativity代考|Schwarzschild Solution

几乎是在爱因斯坦引入场方程之后，$K$。史瓦西找到了一个精确解。他的论文有英文译本，见Schwarzschild(1916)。最恰当的例子就是弱引力下的度规，它是由真空中一个静态的、球对称的大质量物体产生的。度规张量不依赖于$t$，但依赖于$\vec{r}$和$d \vec{r}$，因此它具有旋转不变性。时间无关性导致能量守恒，旋转不变性导致某些角动量分量守恒。因此，在球坐标下工作是值得的。因此，对于本章的其余部分，$r^p$表示$(r)^p$，而不是$p$位置向量的第一个分量。
试试固有时元素最一般的旋转不变形式，
$$
\begin{aligned}
(d \tau)^2= & -g_{\mu \nu} d x^\mu d x^\nu \
\equiv & A(r)(d t)^2-2 B(r)(\vec{r} \cdot d \vec{r}) d t-C(r)(\vec{r} \cdot d \vec{r})^2-D(r) d \vec{r} \cdot d \vec{r} \
= & A(r)(d t)^2-2 B(r) r d r d t-\left[C(r) r^2+D(r)\right](d r)^2 \
& -D(r) r^2\left[(d \theta)^2+(\sin \theta d \phi)^2\right]
\end{aligned}
$$
我们可以通过下面的变换消除$d t d r$项:
$$
t \equiv t^{\prime}-E(r), \quad d E(r) \equiv-r d r B(r) / A(r)
$$
这就很容易说明了，参见问题4，这导致了什么
$$
\begin{aligned}
(d \tau)^2= & A(r)\left(d t^{\prime}\right)^2-\left[(r B(r))^2 / A(r)+C(r) r^2+D(r)\right](d r)^2 \
& -D(r) r^2\left[(d \theta)^2+\sin ^2 \theta(d \phi)^2\right] \
\equiv & A(r)\left(d t^{\prime}\right)^2-F(r)(d r)^2-r^2 D(r)\left[(d \theta)^2+(\sin \theta d \phi)^2\right]
\end{aligned}
$$
最后一个变换重新定义$r$，并允许将固有时转换为度量张量是对角线的形式，
$$
\begin{aligned}
r^{\prime 2} \equiv & r^2 D(r) \
(d \tau)^2 \equiv & \exp \left[2 \Phi\left(r^{\prime}\right)\right]\left(d t^{\prime}\right)^2-\exp \left[2 \Delta\left(r^{\prime}\right)\right]\left(d r^{\prime}\right)^2 \
& -r^{\prime 2}\left[(d \theta)^2+(\sin \theta d \phi)^2\right]
\end{aligned}
$$

物理代写|广义相对论代写General Relativity代考|Conserved Quantities: Massive Particles

度规的知识告诉我们什么量是守恒的，如果有的话。这些知识对解运动方程很有帮助。光的运动有一个常数，它的速度。方程(4.3)是由平行输运得到的，对于静止质量的粒子$m, d \tau \neq 0$，因此可以用它来代替$d q$。使用$g_{\mu \alpha} ;\nu=0$，并将索引的总和重命名为$$ \begin{aligned} 0 & =W^\nu W^\mu ;\nu=U^\nu U^\mu ;\nu=P^\nu P^\mu ;{ }\nu \
& =g_{\mu \alpha} P^\nu P^\mu ;{ }\nu=P^\nu\left(g{\mu \alpha} P^\mu\right){{ }\nu} \
& =P^\nu P_\alpha ;\nu=P^\nu\left(P\alpha,{ }\nu-P\beta \Gamma_{\alpha \nu}^\beta\right) \
& =m \frac{d x^\nu}{d \tau} P_{\alpha, \nu}-P^\nu P_\beta \Gamma_{\alpha \nu}^\beta=m \frac{d P_\alpha}{d \tau}-P^\nu P_\beta \Gamma_{\alpha \nu}^\beta .
\end{aligned}
$$
然后，
$$
\begin{aligned}
\frac{d P_\alpha}{d \tau} & =P^\nu P_\beta \Gamma_{\alpha \nu}^\beta / m=P^\nu P_\beta g^{\beta \chi}\left(g_{\alpha \chi},{ }\nu+g{\chi \nu},{ }\alpha-g{\alpha \nu},\chi\right) /(2 m) \ & =\left(P^\nu P^\chi g{\alpha \chi},\nu+P^\chi P^\nu\left[g{\chi \nu},{ }\alpha-g{\alpha \nu}, \chi_\chi\right]\right) /(2 m) \
& =P^\chi P^\nu\left(g_{\alpha \nu},\chi+g{\chi \nu},{ }\alpha-g{\alpha \nu},\chi\right) /(2 m) \ & =P^\chi P^\nu g{\chi \nu},_\alpha /(2 m) .
\end{aligned}
$$

如果$g_{\chi \nu}, \alpha=0$，那么$P_\alpha$在测地线上是恒定的。在平稳度规中$g_{\chi \nu, 0}=0$和$P_0$是常数。在弱重力和低速的情况下，总能量是恒定的。
能量的恒常性可以用小的量表示到最低的数量级。利用$|\vec{P}| \ll m$这个事实，可以忽略$h_{i j} P^i P^j / m^2$和$h_{i i}|\vec{P}|^2 / m^2$，
$$
\begin{aligned}
m^2 & =-g_{\mu \nu} P^\mu P^\nu \
& =-\left(-1+h_{00}\right)\left(P^0\right)^2-\left[\left(1+h_{i i}\right)|\vec{P}|^2+2 h_{i j} P^i P^j\right], \
1 & \approx\left(1-h_{00}\right)\left(P^0 / m\right)^2-(|\vec{P}| / m)^2 \
& =\left(1+2 M^{\prime} / r\right)\left(P^0 / m\right)^2-(|\vec{P}| / m)^2, \
P^0 / m & =\left(1+(|\vec{P}| / m)^2\right)^{1 / 2}\left(1+2 M^{\prime} / r\right)^{-1 / 2} \
& \approx\left(1+(|\vec{P}| / m)^2 / 2\right)\left(1-M^{\prime} / r\right), \
P^0 & \approx m-m M^{\prime} / r+|\vec{P}|^2 /(2 m)=R E+P E+K E=E, \
P_0 & =g_{00} P^0 \approx-P^0=-E, \text { constant. }
\end{aligned}
$$

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微观经济学代写

线性代数代写

线性代数是数学的一个分支，涉及线性方程，如：线性图，如：以及它们在向量空间和通过矩阵的表示。线性代数是几乎所有数学领域的核心。

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2023年5月12日2023年5月12日 General Relativity, 广义相对论, 物理代写

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

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

In the locally inertial frame specified by $x^{\bar{\mu}}$, all objects travel in a straight line without acceleration. The equation of motion of a particle with rest mass is as follows:
$$
\frac{d^2 x^{\bar{\mu}}}{d \tau^2}=\frac{d U^{\bar{\mu}}}{d \tau}=0 .
$$

Using, Eq. (3.13), the above, in another frame becomes
$$
\begin{aligned}
0 & =\frac{d U^{\bar{\alpha}}}{d \tau}=\frac{d\left(x^{\bar{\alpha}}, \mu U^\mu\right)}{d \tau}=x^{\bar{\alpha}}, \frac{d U^\mu}{d \tau}+U^\mu \frac{d}{d \tau} x^{\bar{\alpha}}, \mu \
& =x^{\bar{\alpha}},\mu \frac{d U^\mu}{d \tau}+U^\mu x^{\bar{\alpha}}, \mu, \nu \frac{d x^\nu}{d \tau} \ & =x^{\bar{\alpha}},\mu \frac{d U^\mu}{d \tau}+U^\mu x^{\bar{\alpha}}, \mu, \nu U^\nu \
& =x^\beta,{\bar{\alpha}}\left(x^{\bar{\alpha}}, \mu \frac{d U^\mu}{d \tau}+U^\mu x^{\bar{\alpha}}, \mu, \nu U^\nu\right)=\delta^\beta \frac{d U^\mu}{d \tau}+U^\mu U^\nu \Gamma{\mu \nu}^\beta \
& =\frac{d U^\beta}{d \tau}+U^\mu U^\nu \Gamma_{\mu \nu}^\beta
\end{aligned}
$$
The equation of motion involves the $\mathrm{C}$ symbols, which depend on the metric tensor.

Since a photon has $d \tau=0$, the above cannot be used as its equation of motion. One must substitute another parameter, say $d q$. Allowable parameters are called affine parameters. For example, $d \tau=k d q$ yields an allowable $d q$, where for photons $k=0$. The parameter $q$ describes the path, such that as the photon moves $\frac{d x^\mu}{d q}=W^\mu$ is the tangent vector, with the property,
$$
\begin{aligned}
W_\mu W^\mu & =g_{\mu \nu} W^\mu W^\nu=g_{\mu \nu} \frac{d x^\mu}{d q} \frac{d x^\nu}{d q}=\frac{g_{\mu \nu} d x^\mu d x^\nu}{d q d q} \
& =\left(\frac{d \tau}{d q}\right)^2=0
\end{aligned}
$$

物理代写|广义相对论代写General Relativity代考|Geodesics

The thought experiments of Einstein, illustrated in Fig. 4.1, show that in the presence of gravity light moves in a curved path. The top of the figure shows two equivalent observers $\mathrm{O}$, one in gravity free space, and the other freely falling in a region of uniform gravity. They observe that a horizontally traveling light ray enters and exits their capsule a distance $L$ above the floor. The bottom of the figure shows an observer $\mathrm{O}^{\prime}$, not freely falling due to gravity. $\mathrm{O}^{\prime}$ also observes, that for the freely falling capsule, light entered and exited the same distance above the floor. However, according to $\mathrm{O}^{\prime}$ the exit point will have fallen in the time light crossed the capsule. Thus, the light also must have fallen or moved in a downward-curved path. The conclusion to be drawn is that gravity affects light, a break from Newtonian physics.

In the geometry of flat space, geodesics are the paths of minimum distance between two points, for motion with constant velocity, or the paths that minimize the travel time. Light in empty space certainly fits this case and before GR the phrase, “light travels in straight lines,” was often heard. This is because gravity is so weak, that the deviation from a straight line path, was too small to be observed. GR knowledgeable observers where gravity acts, but not freely falling, know that nothing can make the trip between two points faster than light. Thus, the “straight lines” or geodesics, are actually curved paths.

广义相对论代写

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

在$x^{\bar{\mu}}$指定的局部惯性系中，所有物体在没有加速度的情况下沿直线运动。静止质量粒子的运动方程为:
$$
\frac{d^2 x^{\bar{\mu}}}{d \tau^2}=\frac{d U^{\bar{\mu}}}{d \tau}=0 .
$$

使用上述的Eq.(3.13)，在另一个框架中得到
$$
\begin{aligned}
0 & =\frac{d U^{\bar{\alpha}}}{d \tau}=\frac{d\left(x^{\bar{\alpha}}, \mu U^\mu\right)}{d \tau}=x^{\bar{\alpha}}, \frac{d U^\mu}{d \tau}+U^\mu \frac{d}{d \tau} x^{\bar{\alpha}}, \mu \
& =x^{\bar{\alpha}},\mu \frac{d U^\mu}{d \tau}+U^\mu x^{\bar{\alpha}}, \mu, \nu \frac{d x^\nu}{d \tau} \ & =x^{\bar{\alpha}},\mu \frac{d U^\mu}{d \tau}+U^\mu x^{\bar{\alpha}}, \mu, \nu U^\nu \
& =x^\beta,{\bar{\alpha}}\left(x^{\bar{\alpha}}, \mu \frac{d U^\mu}{d \tau}+U^\mu x^{\bar{\alpha}}, \mu, \nu U^\nu\right)=\delta^\beta \frac{d U^\mu}{d \tau}+U^\mu U^\nu \Gamma{\mu \nu}^\beta \
& =\frac{d U^\beta}{d \tau}+U^\mu U^\nu \Gamma_{\mu \nu}^\beta
\end{aligned}
$$
运动方程包含$\mathrm{C}$符号，它依赖于度规张量。

因为光子有$d \tau=0$，所以不能用上面的方程作为它的运动方程。必须替换另一个参数，例如$d q$。允许参数称为仿射参数。例如，$d \tau=k d q$产生一个允许的$d q$，而对于光子$k=0$。参数$q$描述了路径，当光子移动时$\frac{d x^\mu}{d q}=W^\mu$是切矢量，其性质是，
$$
\begin{aligned}
W_\mu W^\mu & =g_{\mu \nu} W^\mu W^\nu=g_{\mu \nu} \frac{d x^\mu}{d q} \frac{d x^\nu}{d q}=\frac{g_{\mu \nu} d x^\mu d x^\nu}{d q d q} \
& =\left(\frac{d \tau}{d q}\right)^2=0
\end{aligned}
$$

物理代写|广义相对论代写General Relativity代考|Geodesics

如图4.1所示，爱因斯坦的思想实验表明，在重力存在的情况下，光以弯曲的路径运动。图的顶部显示了两个等效的观察者$\mathrm{O}$，一个在无重力空间中，另一个在均匀重力区域中自由下落。他们观察到一束水平传播的光线在距离地面$L$处进出他们的胶囊。图的底部显示了一个观察者$\mathrm{O}^{\prime}$，由于重力而没有自由下落。$\mathrm{O}^{\prime}$还观察到，对于自由落体的胶囊，光线进入和离开地面的距离相同。然而，根据$\mathrm{O}^{\prime}$的说法，在光穿过太空舱的时候，出口点将会下降。因此，光也必须以向下弯曲的路径落下或移动。由此得出的结论是，引力影响光，这是对牛顿物理学的突破。

在平坦空间的几何中，测地线是两点之间的最小距离路径，对于匀速运动，或者是使运动时间最小的路径。真空中的光当然符合这种情况，在GR之前，“光沿直线传播”这句话经常被听到。这是因为引力太弱，以至于偏离直线的距离太小而无法观察到。GR知识渊博的观察者知道，在重力作用下，但不是自由落体，没有什么能比光更快地在两点之间移动。因此，“直线”或测地线实际上是弯曲的路径。

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Tag: PHYS503

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物理代写|广义相对论代写General Relativity代考|Exterior derivative operator: generalised Stokes’ theorem

物理代写|广义相对论代写General Relativity代考|Generalised Stokes’ theorem

物理代写|广义相对论代写General Relativity代考|Exterior derivative operator: generalised Stokes’ theorem

物理代写|广义相对论代写General Relativity代考|Generalised Stokes’ theorem

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物理代写|广义相对论代写General Relativity代考|One-forms

物理代写|广义相对论代写General Relativity代考|Transformation rules

物理代写|广义相对论代写General Relativity代考|One-forms

物理代写|广义相对论代写General Relativity代考|Transformation rules

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物理代写|广义相对论代写General Relativity代考|Twin paradox: accelerations

物理代写|广义相对论代写General Relativity代考|Rotating frames: the Sagnac effect

物理代写|广义相对论代写General Relativity代考|Twin paradox: accelerations

物理代写|广义相对论代写General Relativity代考|Rotating frames: the Sagnac effect

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物理代写|广义相对论代写General Relativity代考|The need for a theory of gravity

物理代写|广义相对论代写General Relativity代考|Gravitation and inertia: the Equivalence Principle in mechanics

物理代写|广义相对论代写General Relativity代考|The need for a theory of gravity

物理代写|广义相对论代写General Relativity代考|Gravitation and inertia: the Equivalence Principle in mechanics

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物理代写|广义相对论代写General Relativity代考|Determining Distance

物理代写|广义相对论代写General Relativity代考|Red Shift Versus Distance Relation

物理代写|广义相对论代写General Relativity代考|Determining Distance

物理代写|广义相对论代写General Relativity代考|Red Shift Versus Distance Relation

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物理代写|广义相对论代写General Relativity代考|The C Symbols

物理代写|广义相对论代写General Relativity代考|Ricci Tensor and Einstein Field Equations

物理代写|广义相对论代写General Relativity代考|The C Symbols

物理代写|广义相对论代写General Relativity代考|Ricci Tensor and Einstein Field Equations

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物理代写|广义相对论代写General Relativity代考|Quadrupole Radiation

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

物理代写|广义相对论代写General Relativity代考|Quadrupole Radiation

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

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物理代写|广义相对论代写General Relativity代考|Plane Waves

物理代写|广义相对论代写General Relativity代考|The Graviton

物理代写|广义相对论代写General Relativity代考|Plane Waves

物理代写|广义相对论代写General Relativity代考|The Graviton

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物理代写|广义相对论代写General Relativity代考|Schwarzschild Solution

物理代写|广义相对论代写General Relativity代考|Conserved Quantities: Massive Particles

物理代写|广义相对论代写General Relativity代考|Schwarzschild Solution

物理代写|广义相对论代写General Relativity代考|Conserved Quantities: Massive Particles

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

物理代写|广义相对论代写General Relativity代考|Geodesics

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

物理代写|广义相对论代写General Relativity代考|Geodesics

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