物理代写| The Newtonian Limit of Einstein’s Equations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 Newtonian Limit of Einstein’s Equations 相对论代考

物理代写| The Newtonian Limit of Einstein’s Equations 相对论代考

物理代写

5.3 The Newtonian Limit of Einstein’s Equations
Here we will show that in the weak field slow velocity limit the Einstein’s equations go over to the Newton’s equation. Weak field means that the spacetime in suitable coordinates, which are nearly Cartesian, can be described by the metric:
$$
g_{i k}=\eta_{i k}+h_{i k}, \quad\left|h_{i k}\right| \ll 1 .
$$
So $h_{i k}$ are quantities which are very small compared with unity. We will see how small they are in typical cases later in the section. Secondly, slow velocity means the relevant velocities are small compared with the speed of light $c$. Here we will take the velocities of the sources as well as those of the test particles small compared to the speed of light.

Given these constraints, the Einstein’s equations reduce to just one non-vacuous equation, namely, the one with the 00 component:
$$
R_{00}-\frac{1}{2} R g_{00}=\kappa T_{00}
$$
where $\kappa$ will be determined from Newton’s theory. It is easy to show that the Riemann tensor to the first order in $h_{i k}$ is given by:
$$
R_{i j k l}=\frac{1}{2}\left(h_{i l, j k}+h_{j k, i l}-h_{i k, j l}-h_{j l, i k}\right),
$$
92
5 Einstein’s Equations
where we have dropped terms of $o\left(h^{2}\right)$ and higher. Note that the $\Gamma$ is of order $o(h)$ so that the product of $\Gamma$ s appearing in the Riemann tensor is of order $o\left(h^{2}\right)$ and hence can be neglected. Comparing with Eq. (4.5.9), we can see that the two forms are similar. The physical situations are however different. Here we have taken the gravitational field to be weak, while in Eq. (4.5.9), the gravitational field need not be weak-we have only chosen a locally flat coordinate system.

In order to obtain the Ricci tensor, we must multiply the above with $g^{i k}$ and contract on $i, k$. But since we keep terms to first order in $h$, the $g^{i k}$ can be replaced by $\eta^{i k}$. Thus,
$$
R_{j l}=\eta^{i k} R_{i j k l},
$$
and,
$$
R_{00}=\frac{1}{2} \eta^{i k}\left(h_{i 0,0 k}+h_{0 k, i 0}-h_{i k, 00}-h_{00, i k}\right)
$$
We now drop all the time derivatives-slow motion-and arrive at,
$$
\begin{aligned}
R_{00} &=-\frac{1}{2} \eta^{i k} h_{00, i k} \
& \simeq \frac{1}{2} \nabla^{2} h_{00},
\end{aligned}
$$
where again we have dropped the time derivatives in D’Alembertian to arrive at the Laplacian $\nabla^{2}$. Contracting the Einstein’s equations $R_{i k}-\frac{1}{2} R g_{i k}=\kappa T_{i k}$ leads to $R=-\kappa T$, where $T=g^{i k} T_{i k}$, the trace of $T_{i k}$. We can then write Einstein’s equations as,
$$
R_{i k}=\kappa\left(T_{i k}-\frac{1}{2} g_{i k} T\right) .
$$

For the Newtonian limit the $T_{i k}$ has just one non-zero
component $T_{00}=\rho c^{2}$. Thus the trace is also $T=\rho
c^{2}$. These considerations lead to:
$$
R_{00}=\frac{1}{2} \kappa \rho c^{2}
$$
From Eqs. (5.3.6) and (5.3.8) we obtain:
$$
\nabla^{2} h_{00}=\kappa \rho c^{2} .
$$
The next task is to connect the metric with the Newtonian potential $\phi$. This either can be achieved through the geodesic equation in the weak field slow motion approximation or through gravitational redshift. In this limit, choosing nearly Cartesian coordinates, the geodesic equation reduces to:
$$
\frac{d^{2} x^{i}}{d t^{2}}=-c^{2} \Gamma_{00}^{i} \quad i=1,2,3 .
$$
$5.3$ The Newtonian Limit of Einstein’s Equations
93
The justification for the above is that for low velocities $d s \simeq \mathrm{cd} t$ and on the RHS only the 00 term survives as the other terms are of order $v / c$ or $(v / c)^{2}$. In the same approximation we also have,
$$
\Gamma_{00}^{i}=-\frac{1}{2} \eta^{i k} h_{00, k}=-\frac{1}{2} h_{00}^{i} .
$$
Since we are in Cartesian coordinates, we can pull the index $i$ down and we get,
$$
\frac{d^{2} x^{i}}{d t^{2}}=-\frac{1}{2} c^{2} h_{00, i}
$$
This equation can be compared with the Newtonian one,
$$
\frac{d^{2} x^{i}}{d t^{2}}=-\frac{\partial \phi}{\partial x^{i}}
$$
Comparing the two equations we have:
$$
h_{00}=\frac{2 \phi}{c^{2}} .
$$
This correspondence can also be worked out from the gravitational redshift formula. Comparing Eq. (5.3.9), Eq. (5.3.14) and Newton’s equation $\nabla^{2} \phi=4 \pi G \rho$ we determine $\kappa$ :
$$
\kappa=\frac{8 \pi G}{c^{4}} .
$$
We now work out how small are $h_{i k}$, in particular $h_{00}$ compared to unity in typical situations. Consider the Sun’s field at the position of the Earth’s orbit. An order of magnitude calculation requires us to compute, $\phi / c^{2}$ where $\phi \sim G M_{\odot} / R, M_{\odot} \sim$ $2 \times 10^{33} \mathrm{gm}$ is the mass of the Sun and $R \sim 1.5 \times 10^{8} \mathrm{~km}$, the average approximate distance between the Earth and the Sun. The computation gives $h_{00} \sim 10^{-8}$. Since $h_{00}$ is 8 orders of magnitude less than unity, we are justified in considering the field as weak-we can safely drop terms of the order of $o\left(h^{2}\right)$ and higher. Similarly, the $h_{00}$ at the surface of the Earth due to Earth’s gravitational field is of the order of $10^{-9}$.

物理代写| The Newtonian Limit of Einstein’s Equations 相对论代考

物理代考

5.3 爱因斯坦方程的牛顿极限
在这里，我们将证明在弱场慢速极限下，爱因斯坦方程会过渡到牛顿方程。弱场意味着在合适的坐标中，接近笛卡尔的时空可以用度量来描述：
$$
g_{i k}=\eta_{i k}+h_{i k}, \quad\left|h_{i k}\right| \ll 1 。
$$
所以 $h_{i k}$ 是与单位相比非常小的数量。我们将在本节后面的典型案例中看到它们有多小。其次，慢速意味着与光速$c$相比，相关速度很小。在这里，我们将采用与光速相比较小的源速度以及测试粒子的速度。

鉴于这些约束，爱因斯坦方程简化为只有一个非空方程，即具有 00 分量的方程：
$$
R_{00}-\frac{1}{2} R g_{00}=\kappa T_{00}
$$
其中$\kappa$ 将由牛顿理论确定。很容易证明，$h_{ik}$ 中一阶的黎曼张量由下式给出：
$$
R_{i j k l}=\frac{1}{2}\left(h_{i l, j k}+h_{j k, i l}-h_{i k, j l}-h_{j l, i k}\right)，
$$
92
5 爱因斯坦方程
我们删除了 $o\left(h^{2}\right)$ 和更高的术语。注意 $\Gamma$ 的阶数是 $o(h)$，因此出现在黎曼张量中的 $\Gamma$ 的乘积是阶数为 $o\left(h^{2}\right)$ 的，因此可以忽略不计。与等式比较。 (4.5.9)，我们可以看到这两种形式是相似的。然而，物理情况是不同的。在这里，我们认为引力场很弱，而在方程式中。 (4.5.9)，引力场不必很弱——我们只选择了一个局部平面坐标系。

为了获得 Ricci 张量，我们必须将上述乘以 $g^{i k}$ 并在 $i, k$ 上收缩。但是由于我们在 $h$ 中保持一阶条件，所以 $g^{ik}$ 可以替换为 $\eta^{ik}$。因此，
$$
R_{j l}=\eta^{i k} R_{i j k l}，
$$
和，
$$
R_{00}=\frac{1}{2} \eta^{ik}\left(h_{i 0,0 k}+h_{0 k, i 0}-h_{ik, 00}-h_{00 , ik}\右）
$$
我们现在放弃所有时间导数-慢动作-并到达，
$$
\开始{对齐}
R_{00} &=-\frac{1}{2} \eta^{i k} h_{00, i k} \
& \simeq \frac{1}{2} \nabla^{2} h_{00}，
\end{对齐}
$$
在这里，我们再次放弃了 D’Alembertian 的时间导数，得到了 Laplacian $\nabla^{2}$。收缩爱因斯坦方程 $R_{ik}-\frac{1}{2} R g_{ik}=\kappa T_{ik}$ 导致 $R=-\kappa T$，其中 $T=g^{ik } T_{ik}$，$T_{ik}$ 的踪迹。然后我们可以将爱因斯坦方程写成，
$$
R_{i k}=\kappa\left(T_{i k}-\frac{1}{2} g_{i k} T\right) 。
$$

对于牛顿极限，$T_{ik}$ 只有一个非零
组件 $T_{00}=\rho c^{2}$。因此跟踪也是 $T=\rho
c^{2}$。这些考虑导致：
$$
R_{00}=\frac{1}{2} \kappa \rho c^{2}
$$
从方程式。 (5.3.6) 和 (5.3.8) 我们得到：
$$
\nabla^{2} h_{00}=\kappa \rho c^{2} 。
$$
下一个任务是将度量与牛顿势 $\phi$ 联系起来。这可以通过弱场慢动作近似中的测地线方程或通过引力红移来实现。在此限制下，选择接近笛卡尔坐标，测地线方程可简化为：
$$
\frac{d^{2} x^{i}}{d t^{2}}=-c^{2} \Gamma_{00}^{i} \quad i=1,2,3 。
$$
$5.3$ 爱因斯坦方程的牛顿极限
93
上述理由是，对于低速 $ds \simeq \mathrm{cd} t$ 并且在 RHS 上只有 00 项存在，因为其他项的顺序是 $v / c$ 或 $(v / c)^ {2}$。在同样的近似下，我们也有，
$$
\Gamma_{00}^{i}=-\frac{1}{2} \eta^{ik} h_{00, k}=-\frac{1}{2} h_{00}^{i} 。
$$
由于我们在笛卡尔坐标中，我们可以将索引 $i$ 拉下来，我们得到，
$$
\frac{d^{2} x^{i}}{d t^{2}}=-\frac{1}{2} c^{2} h_{00, i}
$$
这个方程可以与牛顿方程相比较，
$$
\frac{d^{2} x^{i}}{d t^{2}}=-\frac{\partial \phi}{\partial x^{i}}
$$
比较我们得到的两个方程：
$$
h_{00}=\frac{2 \phi}{c^{2}} 。
$$
这种对应关系也可以从引力红移公式中得出。比较方程式。 (5.3.9)，等式。 (5.3.14) 和牛顿方程 $\nabla^{2} \phi=4 \pi G \rho$ 我们确定 $\kappa$ ：
$$
\kappa=\frac{8 \pi G}{c^{4}} 。
$$
我们现在计算出 $h_{ik}$ 有多小，特别是与典型情况下的 unity 相比，$h_{00}$。考虑在地球轨道位置的太阳场。一个数量级的计算需要我们计算，$\phi / c^{2}$ 其中 $\phi \sim G M_{\odot} / R, M_{\odot} \sim$ $2 \times 10^{33 } \mathrm{gm}$ 是太阳的质量，$R \sim 1.5 \times 10^{8} \mathrm{~km}$ 是地球和太阳之间的平均近似距离。计算得到 $h_{00} \sim 10^{-8}$。由于 $h_{00}$ 比单位小 8 个数量级，我们有理由认为该场是弱的——我们可以安全地删除 $o\left(h^{2}\right)$ 量级的项和更高。类似地，由于地球引力场，地球表面的 $h_{00}$ 约为 $10^{-9}$。