### 物理代写|广义相对论代写General relativity代考|Physics 617

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (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|Metric Tensor

In Chapter 1 , the contravariant displacement vector $d r^{\bar{\mu}}$ and the invariant proper-time element $d \tau$ were discussed. It was noted that $(d \tau)^{2} \neq$ $\sum_{\mu=0}^{3} d r^{\bar{\mu}} d r^{\bar{\mu}}$, but rather, from Eqs. (1.4) and (1.5), and the discussion following, summation occurs when the same index appears as both covariant and contravariant. Thus,
\begin{aligned} (d \tau)^{2} &=\left(d r^{\overline{0}}\right)^{2}-\left[\left(d r^{\overline{1}}\right)^{2}+\left(d r^{\overline{2}}\right)^{2}+\left(d r^{\overline{3}}\right)^{2}\right] \ &=(d t)^{2}-\left[(d x)^{2}+(d y)^{2}+(d z)^{2}\right] \ & \equiv-d r_{\bar{\mu}} d r^{\bar{\mu}} \equiv-g_{\bar{\mu} \bar{\nu}} d r^{\bar{\nu}} d r^{\bar{\mu}} \end{aligned}
A vector necessarily has covariant and contravariant components. As shall be seen below, since $d \tau$ is an invariant and $d r^{\bar{\mu}}, d r_{\bar{\mu}}$ are vectors,
\begin{aligned} (d \tau)^{2} &=-d r_{\mu} d r^{\mu}=-g_{\mu \nu} d r^{\nu} d r^{\mu} \ &=-g_{\nu \mu} d r^{\mu} d r^{\nu}=-g_{\nu \mu} d r^{\nu} d r^{\mu} \end{aligned}
The quantity $g_{\mu \nu}$ with two covariant indexes is a tensor of rank 2 called the covariant metric tensor. By convention, when summing over an index that appears as both contravariant and covariant, the sum is over $0-3$ for a Greek index and over 1-3 for a Roman index, e.g., $\delta_{\mu}^{\mu}=4$ but $\delta_{i}^{i}=3$. Equations (2.2) and (2.3) yield $g_{\mu \nu}=g_{\nu \mu}$, which is a symmetric tensor. In an inertial frame using rectangular coordinates, the metric tensor is particularly simple, and given a special symbol $g_{\bar{\mu} \bar{\nu}}=\eta_{\mu \nu}$. From Eq. (2.1),

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

So far we have seen that there are quantities that don’t depend on an index and are invariants, for example, $d \tau$. In view of the discussion following Eq. (1.5), quantities that depend on one index, and transform as in Eq. (1.4), between different reference frames or coordinate systems, are contravariant vectors or tensors of rank 1. The transforms of tensors of higher rank are discussed in Section 2.3. Consider observers $\mathrm{O}$ and $\mathrm{O}^{\prime}$, with coordinates $x^{\mu}$ and $x^{\mu^{\prime}}$, the components of contravariant vectors transform like,
$$V^{\mu}=x^{\mu}, \nu^{\prime} V^{\nu^{\prime}} \quad \text { or } \quad g^{\mu \alpha} V_{\alpha}=x^{\mu}, \nu^{\prime} g^{\beta^{\prime} \nu^{\prime}} V_{\beta^{\prime}},$$
and
\begin{aligned} V_{\sigma} &=\delta_{\sigma}^{\alpha} V_{\alpha}=g_{\sigma \mu} g^{\mu \alpha} V_{\alpha} \ &=g_{\sigma \mu} g^{\beta^{\prime} \nu^{\prime}} x^{\mu}, \nu^{\prime} V_{\beta^{\prime}}=x^{\beta^{\prime}},{ }{\sigma} V{\beta^{\prime}} . \end{aligned}

Equation (2.7) is the rule for transforming the components of covariant vectors. These results show that Eq. (2.1) leads to Eq. (2.2),
\begin{aligned} (d \tau)^{2} &=d r_{\bar{\mu}} d r^{\bar{\mu}}=\left(x^{\chi}, \bar{\mu} d r_{\chi}\right)\left(x^{\bar{\mu}},{ }{\nu} d r^{\nu}\right) \ &=x^{\chi}, \bar{\mu}^{\bar{\mu}} x{\nu} d r_{\chi} d r^{\nu}=\delta_{\nu}^{\chi} d r_{\chi} d r^{\nu} \ &=d r_{\nu} d r^{\nu} \end{aligned}

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

$$(d \tau)^{2}=\left(d r^{\overline{0}}\right)^{2}-\left[\left(d r^{\overline{1}}\right)^{2}+\left(d r^{\overline{2}}\right)^{2}+\left(d r^{\overline{3}}\right)^{2}\right] \quad=(d t)^{2}-\left[(d x)^{2}+(d y)^{2}+(d z)^{2}\right] \equiv-d \gamma$$

$$(d \tau)^{2}=-d r_{\mu} d r^{\mu}=-g_{\mu \nu} d r^{\nu} d r^{\mu} \quad=-g_{\nu \mu} d r^{\mu} d r^{\nu}=-g_{\nu \mu} d r^{\nu} d r^{\mu}$$

(2.3) 产生 $g_{\mu \nu}=g_{\nu \mu}$ ，这是一个对称张量。在使用直角坐标的惯性坐标系中，度量张量特别简单，并且给定一 个特殊的符号 $g_{\bar{\mu} \bar{\nu}}=\eta_{\mu \nu}$. 从方程式。(2.1),

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

$\$ \Vbegin { aligned } lend{对齐 } \ \

$$(d \tau)^{2}=d r_{\bar{\mu}} d r^{\bar{\mu}}=\left(x^{\chi}, \bar{\mu} d r_{\chi}\right)\left(x^{\bar{\mu}}, \nu d r^{\nu}\right) \quad=x^{\chi}, \bar{\mu}^{\bar{\mu}} x \nu d r_{\chi} d r^{\nu}=\delta_{\nu}^{\chi} d r_{\chi} d r^{\nu}=d r_{\nu} d r^{\nu}$$

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