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• 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|Killing Equation

The structure of the metric tensor implies the structure of the spacetime.
Question: Does the metric tensor $g_{\mu v}$ change its value under the infinitesimal coordinate transformation
$$\bar{x}^\mu=x^\mu+\epsilon \xi^\mu(x) ?$$
To search the answer to this question, one has to check whether Lie derivative of $g_{\mu v}$ vanish or not. A mapping of the spacetime onto itself of the form
$$\bar{x}^\mu=x^\mu+\epsilon \xi^\mu,$$
[i.e., infinitesimal transformation] is known as isometric mapping if the Lie derivative of the metric tensor vanishes, i.e.,
\begin{aligned} &L_{\xi} g_{\mu v}=0, \ &\Rightarrow \ &\xi^\rho \nabla_\rho g_{\alpha \beta}+g_{\alpha v} \nabla_\beta \xi^v+g_{\mu \beta} \nabla_\alpha \xi^\mu=0, \ &\Rightarrow \ &\nabla_\beta \xi_\alpha+\nabla_\alpha \xi_\beta=0 \equiv A_{\alpha \beta} . \end{aligned}
The equation
$$L_{\xi} g_{\mu v}=\nabla_\beta \xi_\alpha+\nabla_\alpha \xi_\beta=0 .$$ is known as Killing equation. The solutions $\xi^\mu(x)$ of the Killing equation are termed as Killing vectors (KVs).

KV exist $\Rightarrow \exists$ solution of Killing equations $\Rightarrow$ presence of a definite intrinsic symmetry in that spacetime.

No solution of the Killing equation $\Rightarrow$ does not exist $\mathrm{KV} \Rightarrow$ the spacetime has no symmetry whatsoever.

## 物理代写|广义相对论代写General relativity代考|Stationary and Static Spacetimes

A spacetime is said to be stationary if it asserts a time-like $\mathrm{KV}$ field $\xi^\mu(x)$. Thus, the Killing equation
$$\nabla_\mu \xi_v+\nabla_\nu \xi_\mu(x)=0$$
possesses a solution $\xi_\mu$ such that
$$\xi^2=\xi_\mu \xi^\mu>0 .$$
It is conceivable to build world lines (trajectories) of the vector field $\xi^\mu(x)$ in such a way that only time coordinate $x^n$ changes along these trajectories whereas the spatial coordinates $x^1, x^7, x^3$ are not altered. This is feasible as the vector $\xi^\mu(x)$ is time-like. Thus, directions of these trajectories of $\xi^\mu$ coincide with $x^0$ axis (see Fig. 11).

Hence, in this new coordinate system, the spatial components of $\xi^\alpha$ are zero, i.e., $\xi^k=0, k=$ $1,2,3$. Thus, $\xi^\mu(x)=(1,0,0,0)$ is a nonzero $\mathrm{KV}$, which is time-like. Now from Killing equation,
$$L_{\xi} g_{\mu v} \equiv \xi^\rho \frac{\partial g_{\mu v}}{\partial x^\rho}+g_{\mu \rho} \frac{\partial \xi^\rho}{\partial x^\nu}+g_{\rho v} \frac{\partial \xi^\rho}{\partial x^\mu}=0,$$

we get
$$\frac{\partial g_{\mu v}}{\partial x^0}=0 .$$
This is the required condition for a spacetime to be stationary. However, in relation to black holes, stationary only requires a time-like $\mathrm{KV}$ in an asymptotically flat region.

A typical situation of a stationary spacetime is called static if the trajectories of the KV $\xi^\mu$ are orthogonal to a family of hypersurfaces.
The conditions for static spacetime are
$$\frac{\partial g_{\mu v}}{\partial x^0}=0, \quad g_{0 k}=0 .$$
In other words: A spacetime is said to be static if it admits a hypersurface, which has an orthogonal time-like KV field.

# 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|杀戮方程

.

$$\bar{x}^\mu=x^\mu+\epsilon \xi^\mu(x) ?$$

$$\bar{x}^\mu=x^\mu+\epsilon \xi^\mu,$$
[即。，无穷小变换]称为等距映射，如果度规张量的李导数消失，即
\begin{aligned} &L_{\xi} g_{\mu v}=0, \ &\Rightarrow \ &\xi^\rho \nabla_\rho g_{\alpha \beta}+g_{\alpha v} \nabla_\beta \xi^v+g_{\mu \beta} \nabla_\alpha \xi^\mu=0, \ &\Rightarrow \ &\nabla_\beta \xi_\alpha+\nabla_\alpha \xi_\beta=0 \equiv A_{\alpha \beta} . \end{aligned}

$$L_{\xi} g_{\mu v}=\nabla_\beta \xi_\alpha+\nabla_\alpha \xi_\beta=0 .$$称为kill方程。杀伤方程的解$\xi^\mu(x)$称为杀伤向量(KVs)。

KV存在$\Rightarrow \exists$杀戮方程的解$\Rightarrow$在那个时空中存在一个明确的内在对称性

## 物理代写|广义相对论代写广义相对论代考|静止和静止时空

$$\nabla_\mu \xi_v+\nabla_\nu \xi_\mu(x)=0$$

$$\xi^2=\xi_\mu \xi^\mu>0 .$$

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assignmentutor™您的专属作业导师
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