### 物理代写|广义相对论代写General relativity代考|PHYS4123

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• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

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

In general relativity we often need the integral of a scalar function, which itself is a scalar. The integral of a vector or tensor will not in general have a well-defined transformation law since it is not a quantity defined at a single point. Our task in this section is to obtain an expression for a volume element to be used when integrating a scalar function over all or part of a space.

The appropriate expression for a volume element in a general Riemann space can be obtained by first considering the special case of a diagonal metric and then generalizing to any metric using invariance arguments. For a diagonal metric in any number of dimensions we may write the line element as
$$\mathrm{d} s^{2}=\sum_{i} g_{i i}\left(\mathrm{~d} x^{i}\right)^{2}, \quad \mathrm{~d} \ell_{i} \equiv \sqrt{g_{i i}} \mathrm{~d} x^{i}=\text { physical distance in } i \text { direction. }$$
That is, as we discussed in Sect. $4.3$, the $\mathrm{d} \ell_{i}$ is a physical distance interval. (We assume for the moment that $g_{i i}$ is positive.) What is particularly nice is that this allows us to define a physically meaningful $n$-volume element in a clear and obvious

way, as the product of the physical distances,
$$\mathrm{d} V_{n} \equiv \mathrm{d} \ell \ldots \mathrm{d} \ell_{n}=\sqrt{g_{11 \ldots g_{n n}}} \mathrm{~d} x^{1} \ldots \mathrm{d} x^{n}=\sqrt{|g|} d^{n} x, \ldots$$
where $|g|$ denotes the determinant of the metric tensor. This last expression turns out to be general, except that one must use the absolute value of the determinant if the signature is negative.

To show that the expression $(4.76)$ is the correct volume element we prove the following theorem.

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

The term manifold occurs in more mathematically oriented work. A manifold is an open collection of points $P$ with useful properties for physics applications; because the collection is open there will be by definition a region around each point that is also in the manifold. The points in a 1 -dimensional manifold are in one to one correspondence with an open set of the reals; an open set of the reals is defined as a union of open intervals. Similarly the points in a 2 -dimensional manifold are in one to one correspondence with a pair of reals, and so forth for any dimension $n$. Thus an $n$-dimensional manifold is an open set of points that can be labeled by an $n$-tuple of reals in an open region, that is by coordinates.

A function of the manifold points $P$ can be defined in terms of a function of the coordinates as $f(P)=f\left(x^{k}\right)$. Continuous and differentiable functions are naturally of particular usefulness.

In general the labeling of the points in a manifold is not unique and several coordinate systems may be used to label a region of the manifold; they are often denoted as unprimed and primed, or unbarred and barred as we have done. If there is a differentiable and invertible transformation $\bar{x}^{k}=\bar{x}^{k}\left(x^{i}\right)$ between any two such coordinate systems we say that the manifold is differentiable. This means that a differentiable function of the points in the manifold corresponds to a continuous function in both coordinate systems.

Thus, in short, a manifold is simply the kind of space physics has used for centuries, defined a bit more carefully with an emphasis on coordinate systems and open sets.

## 物理代写|广义相对论代写General relativity代考|The Signature Theorem in Two Dimensions

The Signature Theorem states that one can find a linear transformation to a coordinate system in which the metric tensor is diagonal and has $1,-1$, or 0 as diagonal elements. We will illustrate the proof in two dimensions with signature $(1,1)$ for simplicity. This is clearly a matrix problem so we will use matrix notation rather than index notation. We need deal only with a single point. The transformation between the original system and a new barred system we write in matrix form as
$$x=D \bar{x},$$
where $D$ is a matrix to be determined. The metric $G$ transforms as a second rank tensor so its transformation in matrix form is
$$\bar{G}=D^{\mathrm{T}} G D, \quad G=\left(\begin{array}{ll} g_{11} & g_{12} \ g_{12} & g_{22} \end{array}\right)$$
We first make the metric diagonal by choosing the transformation matrix $D$ with a single parameter $b$ to be determined,
$$D=\left(\begin{array}{ll} 1 & 0 \ b & 1 \end{array}\right)$$
After the transformation the metric becomes
$$\bar{G}=\left(\begin{array}{cc} g_{11}+2 b g_{12}+b^{2} g_{22} & g_{12}+b g_{22} \ g_{12}+b g_{22} & g_{22} \end{array}\right)$$
We make this diagonal by choosing $b=-g_{12} / g_{22}$. Then we have

$$\bar{G}=\left(\begin{array}{cc} g_{11}-\left(g_{12}\right)^{2} / g_{22} & 0 \ 0 & g_{22} \end{array}\right)=\left(\begin{array}{cc} \bar{g}{11} & 0 \ 0 & \bar{g}{22} \end{array}\right)$$
Notice that the 1,1 element of the matrix $\bar{G}$ is the determinate of $G$ divided by $g_{22}$; we will assume this is positive so that the signature will be $(1,1)$. (The reader should work out the case where it is negative and the signature is (1, -1) as in Exercise 4.7)
We next apply a second linear transformation to stretch the coordinates and make both diagonal elements of the metric equal to 1 . Specifically this is done with the obvious stretching matrix
$$\left(\begin{array}{cc} \sqrt{g}{11} & 0 \ 0 & \sqrt{\bar{g}{22}} \end{array}\right)$$

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

ds2=∑一世G一世一世( dX一世)2, dℓ一世≡G一世一世 dX一世= 物理距离 一世 方向。

d在n≡dℓ…dℓn=G11…Gnn dX1…dXn=|G|dnX,…

## 物理代写|广义相对论代写General relativity代考|The Signature Theorem in Two Dimensions

X=DX¯,

G¯=D吨GD,G=(G11G12 G12G22)

D=(10 b1)

G¯=(G11+2bG12+b2G22G12+bG22 G12+bG22G22)

G¯=(G11−(G12)2/G220 0G22)=(G¯110 0G¯22)

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