### 物理代写|广义相对论代写General relativity代考|PHYC90012

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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|The Line Element

The distance between two neighboring points $P\left(\vec{r}\left(x^i\right)\right)$ and $F\left(\vec{r}\left(x^i\right)+d \vec{r}\left(x^i\right)\right)\left(x^i\right.$ are the coordinates of the space) in an $n$-dimensional space is given by (see Fig. 2)
$$d s^2=d \vec{r} \cdot d \vec{r}=g_{a b} d x^a d x^b$$

Here,
$$d \vec{r}\left(x^1\right)=\frac{\partial \vec{r}}{\partial x^1} d x^1+\frac{\partial \vec{r}}{\partial x^2} d x^2+\ldots \ldots \ldots+\frac{\partial \vec{r}}{\partial x^n} d x^n=\alpha_1 d x^1+\alpha_2 d x^2+\ldots+\alpha_n d x^n$$
with
$$\alpha_i=\frac{\partial \vec{r}}{\partial x^i} \text { and } g_{a b}=\alpha_a \cdot \alpha_b .$$
The distance between two neighboring points is referred as line element and is given by Eq. (1.10).

Here, $g_{a b}$ are known as metric tensor, which are functions of $x^a$. If $g=\left|g_{a b}\right| \neq 0$ and $d s$ is adopted to be invariant, then the space is called Riemannian space.

In mathematics, Riemannian space is used for a positive-definite metric tensor, whereas in theoretical physics, spacetime is modeled by a pseudo-Riemannian space in which the metric tensor is indefinite.
The metric tensor $g_{a b}$ is also called fundamental tensor (covariant tensor of order two).
In Euclidean space:
$$d s^2=d x^2+d y^2+d z^2 .$$
In Minkowski flat spacetime, the line element
$$d s^2=d x^{0^2}-d x^{1^2}-d x^{2^2}-d x^{3^2} .$$
Since the distance $d s$ between two neighboring points is real, the Eq. (1.10) will be amended to
$$d s^2=e g_{i j} d x^i d x^j,$$
where $e$ is known as the indicator and assumes the value $+1$ or $-1$ in order that $d s^2$ be always positive.

## 物理代写|广义相对论代写General relativity代考|Levi-Civita Tensor or Alternating Tensor

Levi-Civita tensor is a tensor of order three in three dimensions and is denoted by $\epsilon_{a b c}$ and defined as
$$\epsilon_{a b c}=+1,$$
if a,b,c is an even permutation of $1,2,3$, i.e., in cyclic order.
$$=-1,$$
if a,b,c is odd permutation of $1,2,3$, i.e., not in cyclic order.
$$=0$$
if any two indices are equal.

Levi-Civita tensor is a tensor of order four in four dimensions and denoted by $\epsilon^{a b c d}$.
$$\epsilon^{a b c d}=+1,$$
if a,b,c,d is an even permutation of $0,1,2,3$, i.e., in cyclic order.
$$=-1,$$
if a,b,c,d is odd permutation of $0,1,2,3$, i.e., not in cyclic order.
$$=0$$
if any two indices are equal.
The components of $\epsilon_{a b c d}$ can be found from $\epsilon^{a b c d}$ by lowering the indices in a typical way, just multiplying it by $(-g)^{-1}$ :
$$\epsilon_{a b c d}=g_{a \mu} g_{b v} g_{c \gamma} g_{d \sigma}(-g)^{-1} \epsilon^{\mu v \gamma \sigma} .$$
For example,
\begin{aligned} \epsilon_{0123} & =g_{0 \mu} g_{1 v} g_{2 \gamma} g_{3 \sigma}(-g)^{-1} \epsilon^{\mu \gamma \gamma \sigma} \ & =(-g)^{-1} \operatorname{det} g_{\mu v}=-1 \end{aligned}

## 物理代写|广义相对论代写General relativity代考|The Line Element

$$d s^2=d \vec{r} \cdot d \vec{r}=g_{a b} d x^a d x^b$$

$$d \vec{r}\left(x^1\right)=\frac{\partial \vec{r}}{\partial x^1} d x^1+\frac{\partial \vec{r}}{\partial x^2} d x^2+\ldots \ldots \ldots+\frac{\partial \vec{r}}{\partial x^n} d x^n=\alpha_1 d x^1+\alpha_2 d x^2+\ldots+\alpha_n$$

$$\alpha_i=\frac{\partial \vec{r}}{\partial x^i} \text { and } g_{a b}=\alpha_a \cdot \alpha_b .$$

$$d s^2=d x^2+d y^2+d z^2 .$$

$$d s^2=d x^{0^2}-d x^{1^2}-d x^{2^2}-d x^{3^2} .$$

$$d s^2=e g_{i j} d x^i d x^j,$$

## 物理代写|广义相对论代写General relativity代考|Levi-Civita Tensor or Alternating Tensor

Levi-Civita 张量是三维空间中的三阶张量，表示为 $\epsilon_{a b c}$ 并定义为
$$\epsilon_{a b c}=+1,$$

$$=-1,$$

$$=0$$

Levi-Civita 张量是四维四阶张量，表示为 $\epsilon^{a b c d}$.
$$\epsilon^{a b c d}=+1,$$

$$=-1,$$

$$=0$$

$$\epsilon_{a b c d}=g_{a \mu} g_{b v} g_{c \gamma} g_{d \sigma}(-g)^{-1} \epsilon^{\mu v \gamma \sigma} .$$

$$\epsilon_{0123}=g_{0 \mu} g_{1 v} g_{2 \gamma} g_{3 \sigma}(-g)^{-1} \epsilon^{\mu \gamma \gamma \sigma} \quad=(-g)^{-1} \operatorname{det} g_{\mu v}=-1$$

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