## 物理代写|广义相对论代写General relativity代考|KYA424

2023年2月6日

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## 物理代写|广义相对论代写General relativity代考|Lie Derivatives and Killing’s Equation

The Lie derivative is a significant concept of differential geometry, named after the discovery by Sophus Lie in the late nineteenth century. It estimates the modification of a tensor field (containing scalar function, vector field), along the flow defined by an additional vector field. Lie derivative can be defined on any differentiable manifold as this change is coordinate invariant.

A vector field $X$ is a linear mapping from $C^{\infty}$ function to $C^{\infty}$ function on a manifold, satisfying
$$\begin{gathered} X(f g)=(X f) g+f X(g), \quad \forall f, g \in C^{\infty}(M) \ X(f)=\sum \frac{d x^\mu}{d v} \frac{\partial f}{\partial x^\mu} \end{gathered}$$
such that
$$X=\sum a^\mu \frac{\partial}{\partial x^\mu} .$$
Suppose $X^\mu(x)$ is a vector field defined over a manifold $M$. Trajectory of $X^\mu$ is obtained by solving
$$\frac{d x^\mu}{d v}=X^\mu(x(v)) .$$
Let us consider a coordinate transformation
$$\bar{x}^\mu=\bar{x}^\mu\left(\epsilon, x^r\right),$$
where $\epsilon$ is a parameter. This is known as one parameter set of transformation. This transformation designates a mapping of the spacetime onto itself. If the transformation takes the form
$$\bar{x}^\mu=x^\mu+\epsilon \xi^\mu(x),$$
then it is called infinitesimal one parameter transformation or infinitesimal mapping.
Here, $\xi^\mu(x)$ is a contravariant vector field defined by
$$\xi^\mu(x)=\left.\frac{\partial \bar{x}^\mu}{\partial \epsilon}\right|_{\epsilon=0} .$$

## 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.

# 广义相对论代考

## 物理代写|广义相对论代写General relativity代考|Lie Derivatives and Killing’s Equation

$$X(f g)=(X f) g+f X(g), \quad \forall f, g \in C^{\infty}(M) X(f)$$

$$X=\sum a^\mu \frac{\partial}{\partial x^\mu} .$$

$$\frac{d x^\mu}{d v}=X^\mu(x(v)) .$$

$$\bar{x}^\mu=\bar{x}^\mu\left(\epsilon, x^r\right),$$

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

$$\xi^\mu(x)=\left.\frac{\partial \bar{x}^\mu}{\partial \epsilon}\right|_{\epsilon=0}$$

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

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

$$\bar{x}^\mu=x^\mu+\epsilon \xi^\mu,$$

$$L_{\xi} g_{\mu v}=0, \quad \Rightarrow \xi^\rho \nabla_\rho g_{\alpha \beta}+g_{\alpha v} \nabla_\beta \xi^v+g_{\mu \beta} \nabla_\alpha \xi^\mu$$

$$L_{\xi} g_{\mu v}=\nabla_\beta \xi_\alpha+\nabla_\alpha \xi_\beta=0$$

$\mathrm{KV}$ 存在 $\Rightarrow \exists$ 杀死方程的解 $\Rightarrow$ 在该时空中存在确定的内 在对称性。

Killing 方程无解 $\Rightarrow$ 不存在 $\mathrm{KV} \Rightarrow$ 时空没有任何对称性。

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