物理代写| Parallel Transport and the Covariant Derivative 相对论代考
物理代写
To gain insight into the procedure let us first take the simple case of a vector field with contravariant components, $A^{i}\left(x^{k}\right) \equiv\left(A^{r}(r, \theta), A^{\theta}(r, \theta)\right)$ defined on a plane in polar coordinates. Our goal is to parallely transport this vector to the point $(r+d r, \theta+$ $d \theta)$
Note that we are in a plane and on which we have the luxury of a Cartesian system $(x, y)$ which covers the plane. We can use this coordinate system to achieve our end. We must transform the polar components of the vector $\left(A^{r}, A^{\theta}\right)$ to Cartesian components $\left(A^{x}, A^{y}\right) .$ Then use the constancy of Cartesian components $\left(A^{x}, A^{y}\right)$ to transport it to the neighbouring point given by the coordinates $(r+d r, \theta+d \theta)$. We have the following transformation equations between the coordinates:
$x=r \cos \theta, \quad y=r \sin \theta ; \quad r=\sqrt{x^{2}+y^{2}}, \quad \theta=\tan ^{-1} \frac{y}{x}$
Using the transformation law for contravariant components, we have:
$A^{\theta}=\frac{\partial \theta}{\partial x} A^{x}+\frac{\partial \theta}{\partial y} A^{y}=\frac{1}{r}\left(-\sin \theta A^{x}+\cos \theta\right.$
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$4.2$ Parallel Transport and the Covariant Derivative
$$
\begin{aligned}
A^{r} &=\frac{\partial r}{\partial x} A^{x}+\frac{\partial r}{\partial y} A^{y}=\cos \theta A^{x}+\sin \theta A^{y} \
A^{\theta} &=\frac{\partial \theta}{\partial x} A^{x}+\frac{\partial \theta}{\partial y} A^{y}=\frac{1}{r}\left(-\sin \theta A^{x}+\cos \theta A^{y}\right)
\end{aligned}
$$
Now the constancy of Cartesian components implies that under parallel transport $\delta A^{x}=\delta A^{y}=0$-there is no change in the Cartesian components. Using this fact and the above equations we readily obtain:
$\begin{aligned} \delta A^{r} &=\left(-A^{x} \sin \theta d \theta+\cos \theta A^{y} d \theta\right)=r A^{\theta} d \theta \ \delta A^{\theta} &=-\frac{1}{r} A^{\theta} d r-\frac{1}{r} A^{r} d \theta \end{aligned}$
Thus the parallely transported vector at $(r+d r, \theta+d \theta)$ is $\left(A^{r}+\delta A^{r}, A^{\theta}+\delta A^{\theta}\right)$ where the changes in the components are given by the above equations and is parallel to the former vector. It preserves the constancy of Cartesian components, i.e. $A^{x}(r, \theta)=A^{x}(r+d r, \theta+d \theta)$ and $A^{y}(r, \theta)=A^{y}(r+d r, \theta+d \theta) .$
We could parallely transport the vector as above because we could fall back on the Cartesian coordinate system which was available in the plane because it was a flat space. But what about in curved spaces such as a sphere? Here we do not have a Cartesian coordinate system to refer to. The usual spherical coordinate system $(\theta, \phi)$ is not Cartesian, because the metric components are not constants. So then how can we solve the problem? The answer is that we can still parallely transport a vector but only locally, that is for differential displacements. Note that in the example above we could have just as easily parallely transported the vector for finite displacements. We purposely chose only differential displacements in that example, because we foresaw that we would need to port a similar procedure on curved spaces. Observe that the sphere is locally flat and it is possible to choose a local Cartesian coordinate system in a small patch. But what is a local Cartesian system? Note that a Cartesian system is one for which the metric components are constants. Accordingly, we define a local Cartesian system to be one in which the metric components $g_{i j}$ are locally constants. More specifically, in order to have local Cartesian system around a point P say, we demand that $\left(g_{i j, k}\right)_{P}=0$, where the comma denotes partial derivative with respect to the coordinate $x^{k}$. We assert that we can always do this in a Riemannian manifold. This is argued later in Sect. 4.4. We may sometimes refer to the local Cartesian system also as locally flat.
Below we give the general procedure which carries out parallel transport in a Riemannian manifold by taking a vector $A^{i}$ at a point $P\left(x^{k}\right)$ in the manifold and parallely transporting this vector to a nearby point $Q\left(x^{k}+d x^{k}\right)$.
- Choose a locally Cartesian system around $P$ such that $\left(g_{i j, k}\right)_{P}=0$.
- Transform the components of the vector $A^{i}$ from the general coordinate system $x^{i}$ to a local Cartesian system say $x^{\prime i}$ so that we have the quantities $A^{\prime i}$.
- Use the constancy of Cartesian components $A^{\prime i}$ in the local Cartesian system to parallely transport the vector, that is, set $\delta A^{\prime i}=0$.
物理代考
4.1 简介
在曲线坐标系中比较流形的两个不同点的两个向量是非常重要的。在平面空间中，我们习惯于比较两个不同位置的两个向量。为此，我们自然而然地使用笛卡尔坐标系。例如，我们将恒定电场 $\mathbf{E}$ 定义为具有相同分量 $\left(E_{x}, E_{y}, E_{z}\右)$ 在每个位置 $(x, y, z)$。事实上，我们将所有笛卡尔分量的恒定性定义为向量场恒定性的充分必要条件。但是，如果我们转到曲线坐标，我们通常的分量恒定性概念就会变得模棱两可。例如，在平面上的极坐标中，如果向量场 $\mathbf{E}$ 在每个 $ 处具有恒定的径向分量 $E_{r}=E$ 和方位角分量 $E_{\theta}=0$ (r, \theta)$，向量场 $\mathbf{E}$ 按照我们通常的理解并不是一个常数向量场，因为它对于不同的 $\theta$ 值指向不同的方向——这个场是径向的。如果我们选择了其他坐标系（不同于笛卡尔坐标系和极坐标系），我们会得到不同的结果。相反，沿“$x$-axis”$(\theta=0$ 线) 的常数向量场 $\mathbf{E}$ 将在空间中的不同位置有不同的分量 $E_{r}=E \sin \theta$ 和 $E_{\theta}=$ $E \cos \theta$。图 4.1 提供了图形解释。因此，要比较曲线坐标中两个不同位置的两个向量，需要“平行传输”的概念——我们需要在 $Q\left(x^{k}+dx^{k}\right)$ 处定义一个向量比如说，平行于 $P\left(x^{k}\right)$ 处的给定向量 – 对于小位移 $dx^{k}$ 最方便。并行传输$P\left(x^{k}\right)$-对于小位移$d x^{k}$ 最方便。由向量场的笛卡尔分量的恒定性定义的平行输运导致了黎曼平行输运的概念。也可以为各种其他应用定义其他类型的并行传输。在本书中，我们将把自己局限于黎曼并行传输，正如我们将看到的那样，它是通过度量定义的。
定义并行传输的主要原因是以一致的方式定义张量的导数：即我们要求张量的导数也应该是张量。这种需求主要来自物理学，其中我们需要坐标独立量，而张量是坐标独立量。并行传输通过在同一点减去两个张量来实现这一点，
(C) 作者，获得 Springer Nature Switzerland AG 2022 的独家许可
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S. Dhurandhar 和 S. Mitra，广义相对论和引力波，
UNITEXT 物理，https://doi.org/10.1007/978-3-030-92335-8_4
4 弯曲空间的几何与张量演算
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图 4.1 此图
显示一个常数
电场 $\mathbf{E}$ 可以有
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