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

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 物理代写|广义相对论代写General relativity代考|Physics Consequences

First, consider simultaneity. $\mathrm{O}$ and $\mathrm{O}^{\prime}$ are coincident. $\mathrm{O}^{\prime}$ says two events occur at the same time so $d t^{\prime}=0$, but at positions from the origin $\pm d z^{\prime}$. $\mathrm{O}$ says that the time differences from zero of the two events are,
$$d t_{\pm}=\pm \gamma V d z^{\prime}, \quad d t_{+}-d t_{-} \neq 0 .$$
In general, observers in relative motion do not agree on simultaneity. Only if events are spatially coincident will observers so agree. Thus, to compare times, clocks at the same spatial position have to be compared. Also note, the present position of $\mathrm{O}^{\prime}$, when all clocks are synchronized, is connected with all points in the past, present, and future of $\mathrm{O}$. In spacetime, all space and time points are available. Time isn’t a special quantity, it’s just a one-dimensional projection of spacetime.

Next consider causality. In the unprimed frame, observer $\mathrm{O}$ fires a bullet at $t=0$ from the origin. It hits a target at time $d t$ and position $d z=v_{b} d t$, where $v_{b}$ is the speed of the bullet. $\mathrm{O}^{\prime}$ also says the bullet was fired from the origin at $t^{\prime}=0$, as that’s where the clocks were synchronized. $\mathrm{O}^{\prime}$ says the target was hit at
$$d t^{\prime}=\gamma\left(d t-V\left[v_{b} d t\right]\right)=d t \gamma\left(1-V v_{b}\right)$$
According to $\mathrm{O}^{\prime}$, if $d t^{\prime}<0$, the target is hit before the bullet is fired. That would violate causality, and can happen only if $v_{b}>1$. Thus, $c=1$ is the limiting speed.

## 物理代写|广义相对论代写General relativity代考|Spacetime Diagrams

Many GR texts stress the concept of spacetime diagrams. Analytic calculations are favored by this author. However, for completeness, the former are briefly discussed in this section. One draws on a flat sheet, and must be concerned with four coordinates. Thus, one-dimensional motion, in the $z$-direction, is considered.

Draw a set of axes for frame $\mathrm{O}$ as shown on the top of Fig. 1.5. The upward vertical axis represents increasing time $t$, while the rightward horizontal axis represents increasing $z$. At some time $t_{1}$, particle $\mathrm{A}$ is at $z_{1}$. If particle A has rest mass, it can remain at rest with respect to O. Its world line is just a vertical line upward from $\left(z_{1}, t_{1}\right)$. On the clock attached to $\mathrm{A}$, the coordinate time duration $d t$ is just the proper-time duration $d \tau$. Photons $\mathrm{B}$ and $\mathrm{C}$, starting from the same position and time, must move with unit speed $\left|\frac{d z}{d t}\right|=1$. Events outside of the triangle, defined by world lines $\mathrm{B}$ and $\mathrm{C}$, are inaccessible starting from $\left(z_{1}, t_{1}\right)$. This triangle is called the future triangle. If motion in two spatial directions is considered, the triangle becomes a cone.

A massive particle $\mathrm{D}$, that moves with constant speed $V<1$ relative to $\mathrm{O}$, has a world line with $\left|\frac{d z}{d t}\right|<1$. Its proper-time duration, measured relative to $t_{1}$, is $d \tau=d t / \gamma$. Here, $d t$ is the time on a clock at rest with respect to $\mathrm{A}$ that, according to $\mathrm{D}$, is coincident with D’s position. Let twins $\mathrm{O}$ and $\mathrm{O}^{\prime \prime}$ start at this spacetime point. O notes that her twin, moving with accelerated motion, always stays within the future triangle, at first moving away from $\mathrm{O}$, but later returning. The gist of the twin “paradox” is that it is only $\mathrm{O}^{\prime \prime}$ that experiences acceleration. If not created at $\left(z_{1}, t_{1}\right)$, then $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}$, and $\mathrm{O}^{\prime \prime}$ can get there only from the past triangle, obtained by extending $\mathrm{B}$ and $\mathrm{C}$ backwards in time.

## 物理代写|广义相对论代写General relativity代考|Physics Consequences

$$d t_{\pm}=\pm \gamma V d z^{\prime}, \quad d t_{+}-d t_{-} \neq 0 .$$

$$d t^{\prime}=\gamma\left(d t-V\left[v_{b} d t\right]\right)=d t \gamma\left(1-V v_{b}\right)$$

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## MATLAB代写

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