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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|Lie Derivative of Covariant Vector

The transformation of covariant vector $V_\alpha$ is
$$\bar{V}\alpha(\bar{x})=\frac{\partial x^\mu}{\partial \bar{x}^\alpha} V\mu(x) .$$
From,
$$\bar{x}^\mu=x^\mu+\epsilon \xi^\mu,$$

we have,
$$\delta_\alpha^\mu=\frac{\partial x^\mu}{\partial \bar{x}^\alpha}+\epsilon \frac{\partial \xi^\mu}{\partial \bar{x}^\alpha},$$
Therefore,
$$\frac{\partial x^\mu}{\partial \bar{x}^\alpha}=\delta_\alpha^\mu-\epsilon \frac{\partial \xi^\mu}{\partial \bar{x}^\alpha} .$$
Also,
$$V_\alpha(\bar{x})=V_\alpha(x+\epsilon \xi)=V_\alpha(x)+\epsilon \xi^\mu \frac{\partial V_\alpha(x)}{\partial x^\mu} .$$
Now,
\begin{aligned} L_{\xi} V_\alpha &=\lim {\epsilon \rightarrow 0} \frac{V\alpha(\bar{x})-\bar{V}\alpha(\bar{x})}{\epsilon} \ &=\xi^\mu \frac{\partial V\alpha}{\partial x^\mu}+V_\mu \frac{\partial \xi^\mu}{\partial x^\alpha} . \end{aligned}
$\left[\epsilon \rightarrow 0, \bar{x}\alpha \rightarrow x\alpha\right]$
Similar to the scalar function case, one can replace the partial derivatives by covariant derivatives.
Thus,
\begin{aligned} &L_{\xi} V^\alpha=\xi^\beta \nabla_\beta V^\alpha-\nabla_\beta \xi^\alpha V^\beta, \ &L_{\xi} V_\alpha=\xi^\mu \nabla_\mu V_\alpha+V_\mu \nabla_\alpha \xi^\mu . \end{aligned}

物理代写|广义相对论代写General relativity代考|Lie Derivative of Covariant and Contravariant Tensors

The transformation of $T_{\alpha \beta}$ is
\begin{aligned} \bar{T}{\alpha \beta}(\bar{x}) &=\frac{\partial x^\mu}{\partial \bar{x}^\alpha} \frac{\partial x^v}{\partial \bar{x}^\beta} T{\mu \nu}(x), \ \bar{x}^\mu &=x^\mu+c \xi^\mu \ & \Rightarrow \ \delta_\alpha^\mu &=\frac{\partial x^\mu}{\partial \bar{x}^\alpha}+\epsilon \frac{\partial \xi^\mu}{\partial \bar{x}^\alpha} \end{aligned}
Also, we can write
$$\delta_\beta^v=\frac{\partial x^v}{\partial \bar{x}^\beta}+e \frac{\partial \xi^v}{\partial \bar{x}^\beta} .$$

Hence
\begin{aligned} \bar{T}{\alpha \beta} &=\left(\delta\alpha^\mu-\epsilon \frac{\partial \xi^\mu}{\partial \bar{x}^\alpha}\right)\left(\delta_\beta^v-\epsilon \frac{\partial \xi^v}{\partial \bar{x}^\beta}\right) T_{\mu v}(x) \ &=\delta_\alpha^\mu \delta_\beta^v T_{\mu v}(x)-\epsilon \delta_\alpha^\mu \frac{\partial \xi^v}{\partial \bar{x}^\beta} T_{\mu v}(x)-\epsilon \frac{\partial \xi^\mu}{\partial \bar{x}^\alpha} \delta_\beta^v T_{\mu v}(x)+\epsilon^2 \frac{\partial \xi^\mu}{\partial \bar{x}^\alpha} \frac{\partial \xi^v}{\partial \bar{x}^\beta} T_{\mu v}(x) \end{aligned}
[put first term $\mu=\alpha$ and $v=\beta$, and neglecting the last term]
$$=T_{\alpha \beta}-\epsilon\left(T_{\mu \beta} \frac{\partial \xi^\mu}{\partial \bar{x}^\alpha}+T_{\alpha v} \frac{\partial \xi^v}{\partial \bar{x}^\beta}\right) .$$
Also
$$T_{\alpha \beta}(\bar{x})=T_{\alpha \beta}(x+\epsilon \xi)=T_{\alpha \beta}(x)+\epsilon \xi^\rho\left(\frac{\partial T_{\alpha \beta}}{\partial x^\rho}\right) .$$
Hence
$$L_{\xi} T_{\alpha \beta}=\lim {\epsilon \rightarrow 0} \frac{T{\alpha \beta}(\bar{x})-\bar{T}{\alpha \beta}(\bar{x})}{\epsilon},$$ as $\epsilon \rightarrow 0, \bar{x}\alpha \rightarrow x_\alpha$
$$=\xi^\rho \frac{\partial T_{\alpha \beta}}{\partial x^\rho}+T_{\alpha v} \frac{\partial \xi^v}{\partial x^\beta}+T_{\mu \beta} \frac{\partial \xi^\mu}{\partial x^\alpha}$$
Similarly,
$$L_{\xi} T^{\alpha \beta}=\xi^\rho \frac{\partial T^{\alpha \beta}}{\partial x^\rho}-T^{\alpha v} \frac{\partial \xi^\beta}{\partial x^\nu}-T^{\mu \beta} \frac{\partial \xi^\alpha}{\partial x^\mu}$$

广义相对论代考

物理代写|广义相对论代写广义相对论代考|协变向量的Lie导数

$$\bar{V}\alpha(\bar{x})=\frac{\partial x^\mu}{\partial \bar{x}^\alpha} V\mu(x) .$$
From，
$$\bar{x}^\mu=x^\mu+\epsilon \xi^\mu,$$

$$\delta_\alpha^\mu=\frac{\partial x^\mu}{\partial \bar{x}^\alpha}+\epsilon \frac{\partial \xi^\mu}{\partial \bar{x}^\alpha},$$

$$\frac{\partial x^\mu}{\partial \bar{x}^\alpha}=\delta_\alpha^\mu-\epsilon \frac{\partial \xi^\mu}{\partial \bar{x}^\alpha} .$$

$$V_\alpha(\bar{x})=V_\alpha(x+\epsilon \xi)=V_\alpha(x)+\epsilon \xi^\mu \frac{\partial V_\alpha(x)}{\partial x^\mu} .$$

\begin{aligned} L_{\xi} V_\alpha &=\lim {\epsilon \rightarrow 0} \frac{V\alpha(\bar{x})-\bar{V}\alpha(\bar{x})}{\epsilon} \ &=\xi^\mu \frac{\partial V\alpha}{\partial x^\mu}+V_\mu \frac{\partial \xi^\mu}{\partial x^\alpha} . \end{aligned}
$\left[\epsilon \rightarrow 0, \bar{x}\alpha \rightarrow x\alpha\right]$

\begin{aligned} &L_{\xi} V^\alpha=\xi^\beta \nabla_\beta V^\alpha-\nabla_\beta \xi^\alpha V^\beta, \ &L_{\xi} V_\alpha=\xi^\mu \nabla_\mu V_\alpha+V_\mu \nabla_\alpha \xi^\mu . \end{aligned}

物理代写|广义相对论代写广义相对论代考|协变和逆变张量的Lie导数

.

$T_{\alpha \beta}$的转换是
\begin{aligned} \bar{T}{\alpha \beta}(\bar{x}) &=\frac{\partial x^\mu}{\partial \bar{x}^\alpha} \frac{\partial x^v}{\partial \bar{x}^\beta} T{\mu \nu}(x), \ \bar{x}^\mu &=x^\mu+c \xi^\mu \ & \Rightarrow \ \delta_\alpha^\mu &=\frac{\partial x^\mu}{\partial \bar{x}^\alpha}+\epsilon \frac{\partial \xi^\mu}{\partial \bar{x}^\alpha} \end{aligned}

$$\delta_\beta^v=\frac{\partial x^v}{\partial \bar{x}^\beta}+e \frac{\partial \xi^v}{\partial \bar{x}^\beta} .$$

\begin{aligned} \bar{T}{\alpha \beta} &=\left(\delta\alpha^\mu-\epsilon \frac{\partial \xi^\mu}{\partial \bar{x}^\alpha}\right)\left(\delta_\beta^v-\epsilon \frac{\partial \xi^v}{\partial \bar{x}^\beta}\right) T_{\mu v}(x) \ &=\delta_\alpha^\mu \delta_\beta^v T_{\mu v}(x)-\epsilon \delta_\alpha^\mu \frac{\partial \xi^v}{\partial \bar{x}^\beta} T_{\mu v}(x)-\epsilon \frac{\partial \xi^\mu}{\partial \bar{x}^\alpha} \delta_\beta^v T_{\mu v}(x)+\epsilon^2 \frac{\partial \xi^\mu}{\partial \bar{x}^\alpha} \frac{\partial \xi^v}{\partial \bar{x}^\beta} T_{\mu v}(x) \end{aligned}
[把第一个项$\mu=\alpha$和$v=\beta$，忽略最后一个项]
$$=T_{\alpha \beta}-\epsilon\left(T_{\mu \beta} \frac{\partial \xi^\mu}{\partial \bar{x}^\alpha}+T_{\alpha v} \frac{\partial \xi^v}{\partial \bar{x}^\beta}\right) .$$

$$T_{\alpha \beta}(\bar{x})=T_{\alpha \beta}(x+\epsilon \xi)=T_{\alpha \beta}(x)+\epsilon \xi^\rho\left(\frac{\partial T_{\alpha \beta}}{\partial x^\rho}\right) .$$

$$L_{\xi} T_{\alpha \beta}=\lim {\epsilon \rightarrow 0} \frac{T{\alpha \beta}(\bar{x})-\bar{T}{\alpha \beta}(\bar{x})}{\epsilon},$$ as $\epsilon \rightarrow 0, \bar{x}\alpha \rightarrow x_\alpha$
$$=\xi^\rho \frac{\partial T_{\alpha \beta}}{\partial x^\rho}+T_{\alpha v} \frac{\partial \xi^v}{\partial x^\beta}+T_{\mu \beta} \frac{\partial \xi^\mu}{\partial x^\alpha}$$

$$L_{\xi} T^{\alpha \beta}=\xi^\rho \frac{\partial T^{\alpha \beta}}{\partial x^\rho}-T^{\alpha v} \frac{\partial \xi^\beta}{\partial x^\nu}-T^{\mu \beta} \frac{\partial \xi^\alpha}{\partial x^\mu}$$

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assignmentutor™您的专属作业导师
assignmentutor™您的专属作业导师