### 物理代写|广义相对论代写General relativity代考|МАТH7105

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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|Transformation of Coordinates

Let there be two reference systems, $S$ with coordinates $\left(x^1, x^2, \ldots, x^n\right)$ and $\bar{S}$ with coordinates $\left(\bar{x}^1, \bar{x}^2, \ldots, \bar{x}^n\right)$ (Fig. 1). The new system $\bar{S}$ depends on the old system $S$ as
$$\bar{x}^i=\phi^i\left(x^1, x^2, \ldots, x^n\right) ; \quad i=1,2, \ldots, n .$$
Here $\phi^i$ are single-valued continuous differentiable functions of $x^1, x^2, \ldots, x^n$ and further the Jacobian
$$\left|\frac{\partial \phi^i}{\partial x^j}\right|=\left|\begin{array}{lllll} \frac{\partial \phi^1}{\partial x^1} & \frac{\partial \phi^1}{\partial x^2} & \frac{\partial \phi^1}{\partial x^3} & \ldots & \frac{\partial \phi^1}{\partial x^n} \ \frac{\partial \phi^2}{\partial x^1} & \frac{\partial \phi^2}{\partial x^2} & \frac{\partial \phi^2}{\partial x^3} & \ldots & \frac{\partial \phi^2}{\partial x^n} \ \ldots & \ldots & \ldots & \ldots & \ldots \ \frac{\partial \phi^n}{\partial x^1} & \frac{\partial \phi^n}{\partial x^2} & \frac{\partial \phi^n}{\partial x^3} & \ldots & \frac{\partial \phi^n}{\partial x^n} \end{array}\right| \neq 0 .$$
Differentiation of Eq. (1.1) yields
$$d \bar{x}^i=\sum_{r=1}^n \frac{\partial \phi^i}{\partial x^r} d x^r=\sum_{r=1}^n \frac{\partial \bar{x}^i}{\partial x^r} d x^r=\sum_{r=1}^n \bar{a}_r^i d x^r .$$
Now and onward, we use the Einstein summation convention, i.e., omit the summation symbol $\sum$ and write the above equations as
$$d \bar{x}^i=\frac{\partial \bar{x}^i}{\partial x^r} d x^r=\vec{a}_r^i d x^r$$

or
$$d x^i=\frac{\partial x^i}{\partial \bar{x}^m} d \bar{x}^m=a_m^i d \bar{x}^m .$$
The repeated index $r$ or $m$ is known as dummy index. The index $i$ is not dummy and is known as free index.
The transformation matrices are inverse to each other
$$\bar{a}_r^i a_i^m=\delta_r^m .$$
The symbol $\delta_r^m$ is Kronecker delta, is defined as
\begin{aligned} \delta_r^m &=1 \text { if } m=r \ &=0 \text { if } m \neq r \end{aligned}
Obviously vectors in $(\bar{S})$ system are linked with $(S)$ system.

## 物理代写|广义相对论代写General relativity代考|Covariant and Contravariant Vector and Tensor

Usually one can describe the tensors by means of their properties of transformation under coordinate transformation. There are two possible ways of transformations from one coordinate system $\left(x^i\right)$ to the other coordinate system $\left(\bar{x}^i\right)$.

Let us consider a set of $n$ functions $A_i$ of the coordinates $x^i$. The functions $A_i$ are said to be the components of covariant vector if these components transform according to the following rule
$$\bar{A}_i=\frac{\partial x^j}{\partial \bar{x}^i} A_j .$$
Also, one can find by multiplying $\frac{\partial x^j}{\partial x^\alpha}$ and using $\frac{\partial x^j}{\partial x^d} \frac{\partial x^j}{\partial x^i}=\delta_k^j$ and $\delta_k^j A_j=A_k$
$$A_k=\frac{\partial \bar{x}^i}{\partial x^k} \bar{A}_i .$$

Here, $A_i$ is known as the covariant tensor of first order or of the type $(0,1)$.
The functions $A^i$ are said to be the components of the contravariant vector if these components transform according to the following rule
$$\bar{A}^i=\frac{\partial \bar{x}^i}{\partial x^j} A^j$$
Also, one can find by multiplying both sides with $\frac{\partial x^k}{\partial \bar{x}^k}$ and using $\delta_j^k A^j=A^k$
$$A^k=\frac{\partial x^k}{\partial \bar{x}^i} \bar{A}^i .$$
Here, $A^i$ is known as the contravariant tensor of first order or of the type $(1,0)$.

## 物理代写|广义相对论代写General relativity代考|Transformation of Coordinates

$$\bar{x}^i=\phi^i\left(x^1, x^2, \ldots, x^n\right) ; \quad i=1,2, \ldots, n .$$

$$d \bar{x}^i=\sum_{r=1}^n \frac{\partial \phi^i}{\partial x^r} d x^r=\sum_{r=1}^n \frac{\partial \bar{x}^i}{\partial x^r} d x^r=\sum_{r=1}^n \bar{a}_r^i d x^r .$$

$$d \bar{x}^i=\frac{\partial \bar{x}^i}{\partial x^r} d x^r=\vec{a}_r^i d x^r$$

$$d x^i=\frac{\partial x^i}{\partial \bar{x}^m} d \bar{x}^m=a_m^i d \bar{x}^m .$$

$$\bar{a}_r^i a_i^m=\delta_r^m .$$

$$\delta_r^m=1 \text { if } m=r \quad=0 \text { if } m \neq r$$

## 物理代写|广义相对论代写General relativity代考|Covariant and Contravariant Vector and Tensor

$$\bar{A}_i=\frac{\partial x^j}{\partial \bar{x}^i} A_j .$$

$$A_k=\frac{\partial \bar{x}^i}{\partial x^k} \bar{A}_i .$$

$$\bar{A}^i=\frac{\partial \bar{x}^i}{\partial x^j} A^j$$

$$A^k=\frac{\partial x^k}{\partial \bar{x}^i} \bar{A}^i .$$

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