物理代写| The Schwarzschild Metric 相对论代考代写 - 代考代写：100%准时可靠您的作业代写专家

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1 物理代写| The Schwarzschild Metric 相对论代考

物理代写| The Schwarzschild Metric 相对论代考

物理代写

6.2 The Schwarzschild Metric
The Schwarzschild spacetime is a spherically symmetric solution of Einstein’s vacuum equations-this means the equations are solved in the region in which the energy-momentum tensor $T_{i k}$ vanishes identically $-T_{i k} \equiv 0$. As we will show that the solution is also static-this is not assumed but can be deduced from Einstein’s equations. This is the essence of what is called as the Birkoff’s theorem, which states that any spherically symmetric vacuum solution of Einstein’s equations is a piece of Schwarzschild geometry. Such a spacetime can be realised in vacuum outside any spherically symmetric matter distribution. The matter distribution could also be dynamic, for example, a spherically symmetric collapse of a dust cloud. But as long as spherical symmetry is maintained, the solution in vacuum is still static and a piece
(1) The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 S. Dhurandhar and S. Mitra, General Relativity and Gravitational Waves,
UNITEXT for Physics, https://doi.org/10.1007/978-3-030-92335-8_6 98 6 Schwarzschild Solution and Black Holes of Schwarzschild geometry. It is also possible that the matter has completely collapsed into a singularity, in which case we have the Schwarzschild black hole with the inevitable event horizon.

We will start with a general form of a spherically symmetric metric not necessarily static. Such a metric is given by:
$$
d s^{2}=A(r, t) d r^{2}+B(r, t) c d r d t+C(r, t) c^{2} d t^{2}+D(r, t)\left(d \theta^{2}+\sin ^{2} \theta d \phi^{2}\right)
$$
where $A, B, C, D$ are arbitrary functions of $r$ and $t$ (Weinberg (1972)). Because of spherical symmetry, the coordinates can be chosen so that there is only one cross term in $d r d t$. It is easy to show that by changing to new coordinates $r^{\prime}(r, t)$ and $t^{\prime}(r, t)$, we can eliminate the $d r d t$ cross term and also reduce $D(r, t)=-r^{2} .$ We the metric in the form:
$$
d s^{2}=e^{v} c^{2} d t^{2}-e^{\lambda} d r^{2}-r^{2}\left(d \theta^{2}+\sin ^{2} \theta d \phi^{2}\right)
$$
where $v$ and $\lambda$ are functions of $r$ and $t$. The centre of symmetry is at $r=0$. The metric is usually written in this form so that far away from the centre, the metric confirms to the usual signature of $(+,-,-,-)$ and moreover, we expect that as $r \longrightarrow \infty$, the metric approaches that of SR or flat spacetime so that the gravitational field decays to zero. Such a solution is called asymptotically flat. This is in fact a boundary condition that we will use when deriving the solution.

The general form of the metric is in terms of $v$ and $\lambda$. Our task is to determine these functions from the Einstein’s equations and boundary conditions. For this purpose we need to compute the Einstein tensor and in effect the Ricci tensor which in turn is obtained from the Christoffel symbols. Thus our first job is to compute the Christoffel symbols. One could do this directly from the metric or more efficiently via the geodesic equations obtained from variational principles.

In order to simplify the notation, we write $x^{0}=c t, x^{1}=r, x^{2}=\theta, x^{3}=\phi$ and denote derivatives with respect of $x^{0}$ and $x^{1}$ by dot and prime respectively. The non-zero Christoffel symbols are:
$$
\begin{aligned}
&\Gamma_{00}^{0}=\frac{1}{2} \dot{v}, \quad \Gamma_{01}^{0}=\Gamma_{10}^{0}=\frac{1}{2} v^{\prime}, \quad \Gamma_{11}^{0}=\frac{1}{2} \dot{\lambda} e^{(\lambda-v)} \
&\Gamma_{11}^{1}=\frac{1}{2} \lambda^{\prime}, \quad \Gamma_{01}^{1}=\Gamma_{10}^{1}=\frac{1}{2} \dot{\lambda}, \quad \Gamma_{00}^{1}=\frac{1}{2} v^{\prime} e^{(\nu-\lambda)} \
&\Gamma_{22}^{1}=-r e^{-\lambda}, \quad \Gamma_{33}^{1}=-r \sin ^{2} \theta e^{-\lambda} \
&\Gamma_{12}^{2}=\Gamma_{21}^{2}=\frac{1}{r}, \quad \Gamma_{33}^{2}=-\sin \theta \cos \theta \
&\Gamma_{13}^{3}=\Gamma_{31}^{3}=\frac{1}{r}, \quad \Gamma_{23}^{3}=\Gamma_{32}^{3}=\cot \theta
\end{aligned}
$$
$6.2$ The Schwarzschild Metric
99
If the $T_{i k}=0$, then the vacuum Einstein’s equations are $R_{i k}-\frac{1}{2} R g_{i k}=0$. Taking the trace of this equation, that is multiplying by $g^{i k}$ and contracting over the indices gives the scalar curvature $R=0$. This in turn implies that the vacuum Einstein equations are $R_{i k}=0$. So from the above Christoffel symbols we must compute Ricci tensor components and set them to zero. This procedure will give us differential equations for $v$ and $\lambda$ which we must solve with boundary conditions to obtain the Schwarzschild metric.

For ready reference let us write down the Ricci tensor in terms of the Christoffel symbols:
$$
R_{i k}=\Gamma_{i k, l}^{l}-\Gamma_{l i, k}^{l}+\Gamma_{i k}^{l} \Gamma_{l m}^{m}-\Gamma_{i l}^{m} \Gamma_{k m}^{l} .
$$
From this expression for the Ricci tensor and Eq. (6.2.3) we obtain the expressions for the Ricci tensor components in terms of the functions $v$ and $\lambda$. We start with $R_{01}$ :
$$
R_{01}=\frac{\dot{\lambda}}{r}=0 .
$$
This shows that $\lambda$ is a function of $r$ only and does not depend on $t$. This fact is very useful in simplifying the rest of the equations. Using the equations $R_{00}=R_{11}=0$, we obtain the following two equations:
$$
\begin{aligned}
2 v^{\prime \prime}+v^{\prime 2}-v^{\prime} \lambda^{\prime}+4 \frac{v^{\prime}}{r} &=0, \
-\frac{1}{2} v^{\prime \prime}+\frac{1}{4} v^{\prime} \lambda^{\prime}-\frac{1}{4} v^{\prime 2}+\frac{\lambda^{\prime}}{r} &=0 .
\end{aligned}
$$
Multiplying the second equation by 4 and adding to the first equation gives the result:
$$
\lambda^{\prime}+v^{\prime}=0 .
$$
Integrating the above equation with respect to $r$ we obtain,
$$
\lambda+v=f(t),
$$
where $f$ is some arbitrary function of $t$. We can use this equation to eliminate $v$. Because, we have in the metric the term:
$$
e^{v} c^{2} d t^{2}=e^{f(t)-\lambda(r)} c^{2} d t^{2}
$$
Defining a new time coordinate by,
$$
t^{\prime}=\int d t e^{\frac{1}{2} f(t)},
$$
we obtain the metric in the form:

物理代考

6.2 史瓦西度量
史瓦西时空是爱因斯坦真空方程的球对称解——这意味着方程在能量-动量张量 $T_{i k}$ 相同地消失 $-T_{ik} \equiv 0$ 的区域内求解。正如我们将展示的那样，解也是静态的——这不是假设的，但可以从爱因斯坦方程中推导出来。这就是所谓的 Birkoff 定理的本质，它指出爱因斯坦方程的任何球对称真空解都是 Schwarzschild 几何的一部分。这样的时空可以在任何球对称物质分布之外的真空中实现。物质分布也可以是动态的，例如，尘埃云的球对称坍塌。但只要保持球对称性，真空中的解仍然是静止的，是一片片的
(1) 作者，获得 Springer Nature Switzerland AG 2022 S. Dhurandhar 和 S. Mitra，广义相对论和引力波的独家许可，
UNITEXT for Physics，https://doi.org/10.1007/978-3-030-92335-8_6 98 6 Schwarzschild 解决方案和 Schwarzschild 几何的黑洞。物质也有可能完全坍缩成一个奇点，在这种情况下，我们就有了史瓦西黑洞和不可避免的事件视界。

我们将从球对称度量的一般形式开始，不一定是静态的。这样的度量由下式给出：
$$
ds^{2}=A(r, t) dr^{2}+B(r, t) cdrd t+C(r, t) c^{2} dt^{2}+D(r, t) \left(d \theta^{2}+\sin ^{2} \theta d \phi^{2}\right)
$$
其中 $A, B, C, D$ 是 $r$ 和 $t$ 的任意函数 (Weinberg (1972))。由于球对称，可以选择坐标使得$d r d t$ 中只有一个交叉项。很容易证明，通过更改新坐标 $r^{\prime}(r, t)$ 和 $t^{\prime}(r, t)$，我们可以消除 $drdt$ 交叉项，并且减少 $D(r, t)=-r^{2} .$ 我们的度量形式为：
$$
ds^{2}=e^{v} c^{2} dt^{2}-e^{\lambda} dr^{2}-r^{2}\left(d \theta^{2}+ \sin ^{2} \theta d \phi^{2}\right)
$$
其中$v$ 和$\lambda$ 是$r$ 和$t$ 的函数。对称中心在$r=0$。度量通常以这种形式编写，以便远离中心，度量确认通常的签名 $(+,-,-,-)$ 而且，我们期望 $r \longrightarrow \infty$,该度量接近SR或平坦时空的度量，因此引力场衰减为零。这种解称为渐近平坦。这实际上是我们在推导解决方案时将使用的边界条件。

度量的一般形式是 $v$ 和 $\lambda$。我们的任务是从爱因斯坦方程和边界条件中确定这些函数。为此，我们需要计算爱因斯坦张量，实际上是里奇张量，而里奇张量又是从 Christoffel 符号中获得的。因此，我们的第一项工作是计算 Christoffel 符号。人们可以直接从度量中做到这一点，或者通过从变分原理获得的测地线方程更有效地做到这一点。

为了简化符号，我们写成 $x^{0}=ct, x^{1}=r, x^{2}=\theta, x^{3}=\phi$ 并表示关于$x^{0}$ 和 $x^{1}$ 分别按点和素数表示。非零 Christoffel 符号为：
$$
\开始{对齐}
&\Gamma_{00}^{0}=\frac{1}{2} \dot{v}, \quad \Gamma_{01}^{0}=\Gamma_{10}^{0}=\frac{ 1}{2} v^{\prime}, \quad \Gamma_{11}^{0}=\frac{1}{2} \dot{\lambda} e^{(\lambda-v)} \
&\Gamma_{11}^{1}=\frac{1}{2} \lambda^{\prime}, \quad \Gamma_{01}^{1}=\Gamma_{10}^{1}=\ frac{1}{2} \dot{\lambda}, \quad \Gamma_{00}^{1}=\frac{1}{2} v^{\prime} e^{(\nu-\lambda) } \
&\Gamma_{22}^{1}=-re^{-\lambda}, \quad \Gamma_{33}^{1}=-r \sin ^{2} \theta e^{-\lambda} \ \
&\Gamma_{12}^{2}=\Gamma_{21}^{2}=\frac{1}{r}, \quad \Gamma_{33}^{2}=-\sin \theta \cos \ θ\
&\Gamma_{13}^{3}=\Gamma_{31}^{3}=\frac{1}{r}, \quad \Gamma_{23}^{3}=\Gamma_{32}^{3 }=\cot \theta
\end{对齐}
$$
$6.2$ 史瓦西公制
99
如果$T_{ik}=0$，则真空爱因斯坦方程为$R_{ik}-\frac{1}{2} R g_{ik}=0$。取这个方程的轨迹，即乘以 $g^{ik}$ 并在索引上收缩，得到标量曲率 $R=0$。这反过来意味着真空爱因斯坦方程是$R_{ik}=0$。因此，从上面的 Christoffel 符号中，我们必须计算 Ricci 张量分量并将它们设置为零。此过程将为我们提供 $v$ 和 $\lambda$ 的微分方程，我们必须使用边界条件求解这些方程以获得 Schwarzschild 度量。

为了方便参考，让我们用 Christoffel 符号写下 Ricci 张量：
$$
R_{ik}=\Gamma_{ik, l}^{l}-\Gamma_{li, k}^{l}+\Gamma_{ik}^{l} \Gamma_{lm}^{m}-\Gamma_ {il}^{m} \Gamma_{km}^{l} 。
$$
从 Ricci 张量和方程式的这个表达式。 (6.2.3) 我们根据函数 $v$ 和 $\lambda$ 获得 Ricci 张量分量的表达式。我们从 $R_{01}$ 开始：
$$
R_{01}=\frac{\dot{\lambda}}{r}