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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.

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

$$\int \mathbf{A} \cdot \mathrm{d} \mathbf{s}=\int \nabla \times \mathbf{A} \cdot \mathrm{d} \boldsymbol{\Sigma}$$

$$\omega=a_{1, \ldots, p}(x) \mathbf{d} x^1 \wedge \cdots \wedge \mathbf{d} x^p .$$

$$\mathbf{d} \boldsymbol{\omega}=\frac{\partial a_{i_1 \cdots i_p}}{\partial x^k} \mathbf{d} x^k \wedge \mathbf{d} x^{i_1} \wedge \cdots \wedge \mathbf{d} x^{i_p} .$$

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

$$\partial(\partial c)=\partial^2 c=0$$

$$\partial^2=0$$

$$\int_{\partial c} \omega=\int_c d \omega .$$

$$\int_{\partial c} a \cdot \mathrm{d} l=\int_c(\nabla \times a) \cdot n \mathrm{~d} \Sigma$$

$$\int_{\partial V} b \cdot n \mathrm{~d} \Sigma=\int_V(\nabla \cdot b) \mathrm{d} V .$$

## MATLAB代写

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

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

## avatest™帮您通过考试

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## 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.

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

$$\nabla f=\frac{\partial f}{\partial x} \mathbf{i}+\frac{\partial f}{\partial y} \mathbf{j}+\frac{\partial f}{\partial z} \mathbf{k}$$

\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
$$x=u+v, \quad y=u-v, \quad z=2 u v+w$$

$$u=1 / 2(x+y), \quad v=1 / 2(x-y), \quad w=z-1 / 2\left(x^2-y^2\right) .$$

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

$$\boldsymbol{\theta}^\mu \rightarrow \boldsymbol{\theta}^\mu=\frac{\partial x^{\prime \mu}}{\partial x^v} \boldsymbol{\theta}^v$$

$$\omega_v \rightarrow \omega_v^{\prime}=\frac{\partial x^\lambda}{\partial x^{\prime v}} \omega_\lambda$$

$$\boldsymbol{\omega} \rightarrow \omega_\mu^{\prime} \boldsymbol{\theta}^{\prime \mu}=\frac{\partial x^\lambda}{\partial x^{\prime \mu}} \frac{\partial x^{\prime \mu}}{\partial x^\kappa} \omega_\lambda \boldsymbol{\theta}^\kappa=\omega_\lambda \boldsymbol{\theta}^\lambda=\boldsymbol{\omega},$$

$$\begin{array}{ll} \text { covariant vector: } \quad V^\mu(x) & \rightarrow V^{\prime \mu}(x)=\frac{\partial x^{\prime \mu}}{\partial x^\lambda}(x) V^\lambda(x), \ \text { contravariant vector: } V_v(x) & \rightarrow V_v^{\prime}(x)=\frac{\partial x^\lambda}{\partial x^{\prime \nu}}(x) V_\lambda(x) . \end{array}$$

## MATLAB代写

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

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

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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

$$u^\mu=\frac{\mathrm{d} x^\mu}{\mathrm{d} \tau}=\left(c \frac{\mathrm{d} t}{\mathrm{~d} \tau}, \frac{\mathrm{d} x}{\mathrm{~d} \tau}, \frac{\mathrm{d} y}{\mathrm{~d} \tau}, \frac{\mathrm{d} z}{\mathrm{~d} \tau}\right) .$$

$$\eta_{\mu v} u^\mu u^v=u^\mu u_\mu=-c^2$$

$$\frac{\mathrm{d}}{\mathrm{d} \tau}\left(u^\mu u_\mu\right)=0=2 \dot{u}^\mu u_\mu$$

$$\eta_{\mu v} a^\mu u^v=a^\mu u_\mu=0 .$$

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

$$\mathrm{d} s^2=-c^2 \mathrm{~d} t^2+\mathrm{d} r^2+r^2 \mathrm{~d} \phi^2+\mathrm{d} z^2 .$$
$S^{\prime}$中的坐标与$S$中的坐标相关
$$t^{\prime}=t, \quad r^{\prime}=r, \quad \phi^{\prime}=\phi-\omega t, \quad z^{\prime}=z$$

\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}

$$\mathrm{d} s^2=-\left(c^2-\omega^2 r^2\right) \mathrm{d} t^2+2 \omega r^2 \mathrm{~d} \phi \mathrm{d} t+\mathrm{d} r^2+r^2 \mathrm{~d} \phi^2+\mathrm{d} z^2$$

$$\mathrm{d} s^2=g_{\mu v} \mathrm{~d} x^\mu \mathrm{d} x^v=g_{00}\left(\mathrm{~d} x^0\right)^2+2 \mathrm{~g}{0 i} \mathrm{~d} x^0 \mathrm{~d} x^i+g{i k} \mathrm{~d} x^i \mathrm{~d} x^k,$$

$$g_{\mu v}=\left(\begin{array}{cccc} -\left(1-\frac{\omega^2 r^2}{c^2}\right) & 0 & \frac{\omega r^2}{c} & 0 \ 0 & 1 & 0 & 0 \ \frac{\omega r^2}{c} & 0 & r^2 & 0 \ 0 & 0 & 0 & 1 \end{array}\right) .$$

## MATLAB代写

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

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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
$$\mathbf{F}=m_{\mathrm{g}} \mathbf{g} .$$

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

$$\mathbf{F}=-\frac{M m G}{r^3} \mathbf{r}$$

$$\mathbf{F}(t)=-\frac{M(t) m G}{r^3} \mathbf{r}$$

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

$$\mathbf{F}=m_{\mathrm{i}} \mathbf{a}$$

$$\mathbf{F}=m_{\mathrm{g}} \mathbf{g} .$$

## MATLAB代写

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

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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}

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

$$b / d_{|}=b / d=\tan p \approx p, \quad d_{|}=b / p$$

$$\vec{r}\left(r^{\prime}\right)=r^{\prime} \hat{n}+\vec{d}\left[\left(1-k r^2\right)^{1 / 2}-\left[1-\left(1-k d^2\right)^{1 / 2}\right] \frac{r^{\prime} \hat{n} \cdot \hat{d}}{d}\right] .$$

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

\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}

$$H=\frac{1}{Q} \frac{d Q}{d t},-q=\frac{1}{H^2 Q} \frac{d^2 Q}{d t^2}=1+\frac{1}{H^2} \frac{d H}{d t},$$

\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}

## MATLAB代写

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

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

## avatest™帮您通过考试

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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}

## 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，
$$\frac{\partial}{\partial t} \propto \bar{v} / r$$

$$\Gamma_{\mu \nu}^{\xi}=g^{\xi \chi}\left(g_{\mu \chi}, \nu+g_{\nu \chi}, \mu-g_{\mu \nu}, \chi\right) / 2 .$$

\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

\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}

$$R_{\mu \nu}=R_{\mu \xi \nu}^{\xi}=\Gamma_{\mu \nu}^\chi \Gamma_{\xi \chi}^{\xi}-\Gamma_{\xi \mu}^\chi \Gamma_{\nu \chi}^{\xi}+\Gamma_{\mu \nu, \xi}^{\xi}-\Gamma_{\xi \mu}^{\xi}, \nu$$

\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}

## MATLAB代写

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

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

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

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

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

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

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

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

## MATLAB代写

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