物理代写|广义相对论代写General relativity代考|MATH7105

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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 $g_{\mu \nu}$ vanish or not. A mapping of the spacetime onto itself of the form
$$\bar{x}^\mu=x^\mu+e \xi^\mu$$
[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 \nu}=0 \ & \Rightarrow \ & \xi^\rho \nabla_\rho g_{a \beta}+g_{a v} \nabla_\beta \xi^\nu+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
$$L_{\xi} g_{\mu \nu}=\nabla_\beta \xi_\alpha+\nabla_\alpha \xi_\beta=0$$ is known as Killing equation. The solutions $\xi^\mu(x)$ of the Killing equation are termed as Killing vectors (KVs).
$\mathrm{KV}$ 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 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 $x^0$ changes along these trajectories whereas the spatial coordinates $x^1, x^2, x^3$ 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
$$\frac{\partial g_{\mu \nu}}{\partial x^0}=0$$
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
$$\frac{\partial g_{\mu \nu}}{\partial x^0}=0, \quad g_{0 k}=0 .$$
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General relativity代考|Killing Equation

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

$$\bar{x}^\mu=x^\mu+e \xi^\mu$$

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

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

$\mathrm{KV}$ 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。

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

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

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

$$L_{\xi} g_{\mu v} \equiv \xi^\rho \frac{\partial g_{\mu v}}{\partial x^\rho}+g_{\mu \rho} \frac{\partial \xi^\rho}{\partial x^\nu}+g_{\rho v} \frac{\partial \xi^\rho}{\partial x^\mu}=0$$

$$\frac{\partial g_{\mu \nu}}{\partial x^0}=0$$

$$\frac{\partial g_{\mu \nu}}{\partial x^0}=0, \quad g_{0 k}=0$$

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