# 广义相对论手札（上）

## January 20, 2021 • 铃宕 • 阅读设置

Update 21-02-06: 基础知识篇已经更新完结，主体理论请参见下篇~

## 2. Manifolds & Tensor Fields

### 2.1 Manifolds

Spacetime itself does not (as far as we know) naturally live in a higher dimensional Euclidean space, so an abstract definition is much more natural.
• open ball : $|x - y| = \left[\sum_{\mu=1}^{n} (x^{\mu } - y^{\mu })^{2} \right]^{\frac{1}{2}}$
• open set : any set as a union of open balls     $\mathbb{R^{n}}\to Topological\ \ space$

$Manifold\ M$ is a set together with a collection of subsets $\{O_{\alpha }\}$ satisfying:

1. Each $p \in M$ lies in at least one $O_{\alpha }$, i.e., the $\{O_{\alpha }\}$ over $M$.
2. For each $\alpha$, there is a one-to-one, onto, map $\psi : O_{\alpha } \to U_{\alpha }\ (chart)$, where $U_{\alpha}$ is an open subset of $\mathbb{R^{n}}$
3. $O_{\alpha} \cap O_{\beta } \neq \varnothing$ require the subsets to be open & the map $\psi _\beta \cdot \psi _\alpha ^{-1}$ to be $C^{\infty}$

(lack img)

Diffeomorphic manifolds have identical manifold structure.

### 2.2 Vectors

Define a tangent vector $v$ at point $p \in M$ to be a map $v:\mathscr{F} \to \mathbb{R}$ which is linear and obeys the Leibnitz rule:

1. $v (af+bg)=av(f)+bv(g), for\ all\ f,g \in \mathscr{F};\ a,b \in \mathbb{R}$
2. $v(fg)=f(p)v(g) +g(p)v(f)$

Consider a constant function, i.e., $h(q)=c$, then $v(h)=0$,$\gets v(h^{2}) = 2cv(h)=v(ch)=cv(h)$

$\mathrm{THEOREM\ 2.2.1}\qquad \dim V_{p}=n\qquad\qquad\qquad Tangent\ space$

Show proof

$Proof.$ Construct a basis of $V_{p}$, i.e., find n linearly independent tangent vectors which span $V_{p}$.

Define $X_{\mu }: \mathscr{F} \to \mathbb{R} \qquad X_{\mu }(f)= \frac{\partial }{\partial x^{\mu }}\left(f \circ \psi^{-1}\right)|_{\psi (p)}$

Then exist $C^{\infty}$ functions $H_{\mu }$ such that for all $x \in \mathbb{R^{n}}$ we have

$$F(x)=F(a)+\sum_{\mu=1}^{n}\left(x^{\mu}-a^{\mu}\right) H_{\mu}(x) \qquad\gets H_{\mu}(a)=\left.\frac{\partial F}{\partial x^{\mu}}\right|_{x=a}$$

Letting $F=f \circ \psi^{-1}$ and $a =\psi(p)$. For all $q \in O$

$$f(q)=f(p)+\sum_{\mu=1}^{n}\left[x^{\mu} \circ \psi(q)-x^{\mu} \circ \psi(p)\right] H_{\mu}(\psi(q))$$

We wish to show that $v$ is a linear combination of $X_1,....,X_{n}$

\begin{aligned} v(f)=& v[f(p)]+\sum_{\mu=1}^{n}\left\{\left.\left[x^{\mu} \circ \psi(q)-x^{\mu} \circ \psi(p)\right]\right|_{q=p} v\left(H_{\mu} \circ \psi\right)\right.\\ &+\left.\left.\left(H_{\mu} \circ \psi\right)\right|_{p} v\left[x^{\mu} \circ \psi-x^{\mu} \circ \psi(p)\right]\right\} \\ =& \sum_{\mu=1}^{n}\left[H_{\mu} \circ \psi(p)\right] v\left(x^{\mu} \circ \psi\right) \\ =& \sum_{ \mu =1 }^{ n }X_{\mu }(f)\ v\left(x^{\mu} \circ \psi\right) \\ =& \sum_{ \mu =1 }^{ n } v^{\mu }X_{\mu }(f) \qquad\qquad \gets v^{\mu }=\left(x^{\mu} \circ \psi\right) \end{aligned}

Using the chain rule we can change the coordinate:

$$X_{\mu}=\left.\sum_{\mu=1}^{n} \frac{\partial }{\partial x^{\mu}}\right|_{\psi (p)} =\left.\sum_{\nu=1}^{n} \frac{\partial x^{\prime \nu}}{\partial x^{\mu}}\right|_{\psi(p)} X_{\nu}^{\prime}\qquad \qquad v^{\prime\nu}=\sum_{ \mu =1 }^{ n } v^{ \mu }\frac{\partial x^{\prime \nu }}{\partial x^{\mu }}$$

Also is known as the vector transformation law.

#### Curve

lack img

A smooth $curve \ C$ is simply a $C^{\infty}$ map. $\ C:\mathbb{R} \to M$.
At each point lying on the curve we can associate $C$ with a tangent vector $T$ :$\quad T(f)=\frac{d}{d t}(f \circ C)$

Coordinate basis vector $X_{\mu }$ associated with $x^{\mu }$ as the parameter $t$ of the curve. $\ \gets x^{\mu }$ coordinate line

$$T(f)=\frac{d}{d t}(f \circ C)=\sum_{\mu} \frac{\partial}{\partial x^{\mu}}\left(f \circ \psi^{-1}\right) \frac{d x^{\mu}}{d t}=\sum_{\mu} \frac{d x^{\mu}}{d t} X_{\mu}(f) \qquad\to\ \ T^{\mu}=\frac{d x^{\mu}}{d t} \quad\quad$$

$Commutator:$ a new vector field with two smooth vector fields $v$ and $w$

$$[v,w] (f) = v[w(f)] - w[v(f)]$$

Note that the commutator of any two vector fields $X_{\mu }$ and $X_{\nu }$ occurring in a coordinate basis vanishes. Conversely,given a collection $X_1,...,X_{n}$ of nonvanishing vector fields that commute with each other and are linearly independent at each point,we can always find a chart for which they are the coordinate basis vector fields.

### 2.3 Tensor & the Metric Tensor

$Dual\ Vector \quad v^{\mu^{*}}\left(v_{\nu}\right)=\delta^{\mu}{ }_{\nu} \qquad\qquad V^{*}:V\to R$

$dual\ basis\ \ \{dx^{u} \}\qquad\qquad\qquad\quad\ \ \quad\omega_{\mu^{\prime}}^{\prime}=\sum_{\mu=1}^{n} \omega_{\mu} \frac{\partial x^{\mu}}{\partial x^{\prime \mu^{\prime}}}$

Double dual vector $\ V^{**}:$ naturally isomorphic to original vector space $V$

$Tensor\ T, \ of\ type\left(k,l\right)\ over\ V$ is a multilinear map:

$$T: \underbrace{V^{*} \times \ldots \times V^{*}}_{k} \times \underbrace{V \times \ldots \times V}_{l} \rightarrow \mathbb{R}$$

In other words, given $k$ dual vectors and $l$ ordinary vectors, $T$ product a number.

$Outer\ product\ \otimes$

If ${v_{\mu }}$ is a basis of $V$ and $v^{\nu }$ is its dual basis, then the $n^{k+l}$ tensor $\left\{v_{\mu_{1}} \otimes \cdots \otimes v_{\mu_{k}} \otimes v^{\nu_{1}^{*}} \otimes \cdot \cdots \otimes v^{\nu_{l}^{*}}\right\}$ yield a basis of $\mathscr{T}(k,l)$. Thus, every tensor $T$ can be expressed as :

$$T=\sum_{\mu_{1}, ..., \nu_{l}=1}^{n} T^{\mu_{1} \cdots \mu_{k}}\!\,_{\nu_{1} \cdots \nu_{l}} v_{\mu_{1}} \otimes \cdots \otimes v^{\nu_{l}^{*}}$$

$Contraction\ \ \ C:\ \mathscr{T}(k, l) \rightarrow \mathscr{T}(k-1, l-1)$

$$(C T)^{\mu_{1} \cdots \mu_{k-1}}\!\,_{\nu_{1} \cdots \nu_{l-1}}=\sum_{\sigma=1}^{n} T^{\mu_{1} \cdots \sigma \cdots \mu_{k-1}}\!\, _{\nu_{1} \dots \sigma \dots \nu_{l-1}}$$

$\mathscr{T}(1,1)$ could be viewed as Linear Transformation Not just Linear Map $\ \ \to$ coordinate invariant : Trace

$$C(v\otimes \omega )=v^{\mu }\omega_{\mu }=\omega(v)=v(\omega ) \qquad C_{2}^{1}(T\otimes v)=T(\,\cdot\, ,v)$$

$\textit{Tensor Transformation Law}$

$$v^{\prime \mu^{\prime}}=\sum_{\mu=1}^{n} v^{\mu} \frac{\partial x^{\prime \mu^{\prime}}}{\partial x^{\mu}} \qquad \qquad d\omega_{\mu^{\prime}}^{\prime}=\sum_{\mu=1}^{n} \omega_{\mu} \frac{\partial x^{\mu}}{\partial x^{\prime \mu^{\prime}}}$$

$$T^{\prime \mu_{1}^{\prime} \cdots \mu_{k}^{\prime}}\!\,_{\nu_{1}^{\prime} \cdots \nu_{l}^{\prime}}=\sum_{\mu_{1}, \ldots, \nu_{l}=1}^{n} T^{\mu_{1} \cdots \mu_{k}}\!\,_{\nu_{1} \cdots \nu_{l}} \frac{\partial x^{\prime \mu_{1}^{\prime}}}{\partial x^{\mu_{1}}} \cdot \cdots \frac{\partial x^{\nu_{l}}}{\partial x^{\prime \nu_{l}^{\prime}}}$$

#### Metric

$Metric\ g$  linear map from $V_{p} \times V_{p}\ \to \mathbb{R}\ \ \ \leftrightarrow\ \,$ tensor of type (0,2)

By symmetric :  for all $v_1,v_2 \in V_{p}$ we have $g(v_1,v_2)=g(v_2,v_{2})$

By nondegenerate :  for all $v\in V_{p}$, if$\ g(v,v_{1})=0 \Rightarrow v_1=0$

In other words, a metric is an inner product on the tangent space at each point.

So it give definition of the $length$ or $magnitude$ of vector $v:\ \ \ |v|=\sqrt[]{|g(v,v)| }$

Expand a metric $g$ in a coordinate basis:

$$g=\sum_{\mu, \nu} g_{\mu \nu} d x^{\mu} \otimes d x^{\nu} \qquad or \qquad d s^{2}=\sum_{\mu, \nu} g_{\mu \nu} d x^{\mu} d x^{\nu}$$

$orthonormal\ basis\ v_1,\dots,v_{n} \Rightarrow g\left(v_{\mu}, v_{\nu}\right)=0$, if $\mu \neq \nu$ and $g\left(v_{\mu}, v_{\mu}\right)=\pm 1$

The number of $+$ and $-$ signs occurring is called the $signature$ of the metric.

Positive definite metrics are called $Riemannian\ +\ +\ +\ +$

Metrics with signatures like spacetime are called $Lorentzian\ -\ +\ +\ +$

$$Curve\ length\ :\quad l=\int \sqrt{|g(T,T)|}\ dt\qquad \qquad T =\partial /\partial_{t}\qquad$$

With $dl=\sqrt{|g(T,T)|}\ dt$, we could set $l$ as parameter and we will have unit line length.$\ \gets|g(T,T)|=1$

However, we can also view $g$ as a linear map from $V_p \to V_{p}^{*}$ via $v \to g(\,\cdot\, ,v)$, for the non-degeneracy, this map is one-to-one and onto, so the inverse map exists. Thus we can establish a one-to-one correspondence between vectors and dual vectors.

### 2.4 The Abstract Index Notation

• The Latin indices should be reminders of the number & type of variables the tensor acts on. Any lowercase Latin letters can be placed in any slot, but in any equation, the same letter must be used to represent the same slot on both sides of the equation.
• Greek letters denote the basis component notation.
• Denote the contraction by using the same letter but repeating the index on the contracted slots.
• Raised or lowered indices denote application of the metric or inverse metric on that slot.
Apply the metric to a vector, $v^{a}$, we get the dual vector $g_{ab}v^{b}$, simply as $v_{a}\gets$ isomorphism
The inverse of $g_{ab}\to\ (g^{-1})^{ab}$ could simply donoted as $g^{ab}$
$g^{ab}g_{bc}= \delta^{a}\!\,_{c}$, where $\delta^{a}\!\,_{c}$ (viewed as a map from $V_{p}\to V_{p}$) is the identity map.
##### Symmetry properties

The equation $\ T_{ab}=T_{ba}$ says that the tensor $T_{ab}$ is symmetric. $\quad \gets\,\,$ interchanging the order

The notation for totally symmetric and totally antisymmetric parts:

$$\begin{array}{l} T_{(a b)}=\frac{1}{2}\left(T_{a b}+T_{b a}\right) \\ T_{[a b]}=\frac{1}{2}\left(T_{a b}-T_{b a}\right) \end{array}$$

More generally, for a tensor $T_{a_{1} \cdots a_{l}}$ we define

$$\begin{array}{l} T_{\left(a_{1} \cdots a_{l}\right)}=\frac{1}{l !} \sum_{\pi} T_{a_{\pi(1)} \cdots a_{\pi(l)}} \\ T_{\left[a_{1} \ldots a_{l}\right]}=\frac{1}{l !} \sum \delta_{\pi} T_{a_{\pi(1)} \cdots a_{\pi(l)}} \end{array}$$

where the sum is taken over all permutations, $\pi$ of $1,\dots,l$ and $\delta_{\pi }$ is $+1$ for even and $-1$ for odd

1. $T_{a_{1} \cdots a_{l}}=T_{\left(a_{1} \cdots a_{l}\right)}\ \Rightarrow\ T_{\left[a_{1} \cdots a_{l}\right]}=0\ \Rightarrow\ T_{a_{1} \cdots a_{l}}=T_{a_{\pi(1)} \cdots a_{\pi(l)}}$

$T_{a_{1} \cdots a_{l}}=T_{\left[a_{1} \cdots a_{l}\right]}\ \Rightarrow\ T_{\left(a_{1} \cdots a_{l}\right)}=0\ \Rightarrow\ T_{a_{1} \cdots a_{l}}= \delta_{\pi}T_{a_{\pi(1)} \cdots a_{\pi(l)}}$
2. Brackets can be infectious under contraction.   $T_{(abc)}S^{abc}=T_{(abc)}S^{(abc)}=T_{abc}S^{(abc)}$
3. The same brackets within the parentheses can be added or deleted at will.   $T_{[(ab)c]}=T_{[abc]}=0$
4. The heterogeneous brackets in parentheses get zero.   $T_{[(ab)c]}=0$
5. The heterogeneous brackets are contracted to zero.   $T^{(abc)}S_{[abc]}=0$

$Differential\ \ l\text{-}form:\,$ A totally antisymmetric tensor field $T_{a_{1} \cdots a_{l}}=T_{\left[a_{1} \cdots a_{l}\right]}$

## 3 Curvature

### 3.1 Derivative Operator

$Derivative\ operator\ \nabla:$ a map take smooth tensor field $(k,l)\to (k,l+1)$ and satisfies:

1. Linearity: $\qquad A, B \in \mathscr{T}(k, l)$ and $\ \alpha, \beta \in R$

\begin{aligned} \nabla_{c}\left(\alpha A^{a_{1} \cdots a_{k}}\!\,_{b_{1} \cdots b_{l}}+\right.&\left.\beta B^{a_{1} \cdots a_{k}} \!\,_{b_1 \cdots b_{l}}\right) \\ &=\alpha \nabla_{c} A^{a_{1} \cdots a_{k}}\!\, _{b_1\cdots b_l}+\beta \nabla_{c} B^{a_{1} \cdots a_{k}}\!\,_{b_1\cdots b_{l}} \end{aligned}

1. Leibnitz rule: $\qquad A\in \mathscr{T}(k, l),\ B\in \mathscr{T}(k^\prime, l^{\prime})$

\begin{aligned} \nabla_{e}\left[A^{a_{1} \cdots a_{k}}\!\,_{b_{1} \cdots b_{l}} B^{c_{1} \cdots c_{k^{\prime}}}\!\,_{d_{1} \cdots d_{l^{\prime}}}\right] =&\left[\nabla_{e} A^{a_{1} \cdots a_{k}}\!\,_{b_{1} \cdots b_{l}} \right] B^{c_{1} \cdots c_{k^{\prime}}}\!\,_{ d_{1} \cdots d_{l^{\prime}}} \\ &+ A^{a_{1} \cdots a_{k}}\!\,_{b_{1} \cdots b_{l}}\left[\nabla_{e} B^{c_{1} \cdots c_{k^{\prime}}}\!\,_{ d_{1} \ldots d_{l^{\prime}}}\right] \end{aligned}

1. Commutativity with contraction: $\qquad A\in \mathscr{T}(k, l)$

$$\nabla_{d}\left(A^{a_{1} \cdots c \cdots a_{k}}\!\,_{b_{1} \ldots c \cdots b_{l}}\right)=\nabla_{d} A^{a_{1} \cdots c \cdots a_{k}}\!\,_{b_{1} \cdots c \cdots b_{l}}$$

1. Consistency with tangent vectors as directional derivatives on scalar fields: $\quad f \in \mathscr{F},\ t^{a} \in V_{p}$

$$t(f)=t^{a} \nabla_{a} f$$

1. Torsion free: $\qquad f \in \mathscr{F}$

$$\nabla_{a} \nabla_{b} f=\nabla_{b} \nabla_{a} f$$

Commutator:

\begin{aligned} [v, w](f) &=v\{w(f)\}-w\{v(f)\} \\ &=v^{a} \nabla_{a}\left(w^{b} \nabla_{b} f\right)-w^{a} \nabla_{a}\left(v^{b} \nabla_{b} f\right) \\ &=\left\{v^{a} \nabla_{a} w^{b}-w^{a} \nabla_{a} v^{b}\right\} \nabla_{b} f \\ \to\ \quad[v,w]^{b} &=v^{a} \nabla_{a} w^{b}-w^{a} \nabla_{a} v^{b} \end{aligned}

$Ordinary\ derivative:\ \partial_{c} T^{a_{1} \cdots a_{k}}\!\,_{b_{1} \cdots b_{l}}$ with components $\ \partial\left(T^{\mu_{1} \cdots \mu_{k}}\!\,_{\nu_{1} \ldots \nu_{l}}\right) / \partial x^{\sigma}$

1-5 was satisfied from the standard properties of partial derivatives. Indeed, by the equality of mixed partial derivatives, the 5th condition was held for all tensor fields, not just scalar fields.

Note that a different coordinate system $\psi$ yields a different operator $\partial a$. Thus, a given ordinary derivative operator is coordinate dependent, i.e., It's not naturally associated with the structure of the manifold.

How unique are derivative operators?

\begin{aligned} \tilde{\nabla}_{a}\left(f \omega_{b}\right)-\nabla_{a}\left(f \omega_{b}\right) &=(\tilde{\nabla}_{a} f) \omega_{b}+f \tilde{\nabla}_{a} \omega_{b}-\left(\nabla_{a} f\right) \omega_{b}-f \nabla_{a} \omega_{b} \\ &=f(\tilde{\nabla}_{a} \omega_{b}-\nabla_{a} \omega_{b}) \end{aligned}

$\tilde{\nabla}_{a} -\nabla_{a}$ depends only on the value of $w_{b}$ at point p, not its neighbor. $\quad \gets$ a tensor of type $(1,2)$ at p

$$\tilde{\nabla}_{a} \omega_{b}-\nabla_{a} \omega_{b}=C_{a b}^{c} \omega_{c}$$

A symmetry property follows immediately : $\quad$ let $\omega_{b}=\nabla_{b} f=\tilde{\nabla}_{b} f$

$$\nabla_{a} \nabla_{b} f-\tilde{\nabla}_{a} \tilde{\nabla}_{b} f=C^{c}\!\,_{a b} \nabla_{c} f \qquad$$

$$\Rightarrow\ \ C^{c}\!\,_{ab}=C^{c}\!\,_{ba}$$

The action of $\tilde{\nabla}_{a} -\nabla_{a}$ on vector fields:
$$\left(\tilde{\nabla}_{a}-\nabla_{a}\right)\left(\omega_{b} t^{b}\right)=\left(C^{c}\!\,_{a b} \omega_{c}\right) t^{b}+\omega_{b}\left(\tilde{\nabla}_{a}-\nabla_{a}\right) t^{b}=0$$

Index substituting on contracted indices,
$$\omega_{b}\left[\left(\tilde{\nabla}_{a}-\nabla_{a}\right) t^{b}+C^{b}\!\,_{a c} t^{c}\right]=0$$

for all $w_{b}$. This implies
$$\nabla_{a} t^{b}=\tilde{\nabla}_{a}t^{b}+C^{b}\!\,_{a c} t^{c}$$

The general formula on arbitrary tensor field:

\begin{aligned} \nabla_{a} T^{b_{1} \cdots b_{k_{1}}}\!\,_{c_1 \ldots c_{l}}=& \tilde{\nabla}_{a} T^{b_{1} \cdots b_{k}}\!\,_{c_{1} \cdots c_{l}}+\sum_{i} C^{b_{i}}{ }_{a d} T^{b_{1} \cdots d \cdots b_{k_{1}}}\!\,_{c_1 \ldots c_{l}} \\ &-\sum_{j} C^{d}\!\,_{a c_{j}} T^{b_{1} \cdots b_{l}}\!\,_{c_1 \ldots d \ldots c_{l}} \end{aligned}

$Christoffel\ symbol:\ \,$ The tensor field $C^{c}\!\,_{ac}$ when $\tilde{\nabla}_{a}$ is a $\partial_a$

$$\nabla_{a} t^{b}=\partial_{a} t^{b}+\Gamma^{b}\!\,_{a c} t^{c}$$

$\Gamma^{b}\!\,_{a c}$ was associated with the coordinate system defined $\partial _a$

#### Parallel transport

Given a $\nabla_{a}$ we can define the parallel transport of a vector along a curve $C$ with a tangent $t^{a}$

$$t^{a} \nabla_{a} v^{b}=0$$

with a tensor of arbitrary rank:

$$t^{a} \nabla_{a} T^{b_{1} \cdots b_{k}}\!\,_{ c_1 \ldots c_{l}}=0$$

Choose a coordinate to express:

$$t^{a} \partial_{a} v^{b}+t^{a} \Gamma^{b}\!\,_{a c} v^{c}=0$$

in terms of components:   (The coordinate base $\partial /\partial x_{\mu }$ is omitted here.)

$$\frac{d v^{\nu}}{d t}+\sum_{\mu, \lambda} t^{\mu} \Gamma^{\nu}\!\,_{\mu \lambda} v^{\lambda}=0$$

The equation above always has a unique solution for any given initial value of $v^{a}$.

Actually, This also applies to normal transport. Note why is $\nabla$ not $\partial\ ?$

$$T^{b} \nabla_{b} v^{a}|_{p} = \lim\limits_{\Delta t\to 0 }\frac{1}{\Delta t}(\tilde v^{a}|_{p} - v^{a}|_{p})$$

where $\ \tilde v^{a}|_{p}$ is the pararallel transport result of $v^{a}|_{p}$ along curve $C(t)$.

$Connection:$ The mathematical structure arising from such a

curve-dependent identification of the tangent spaces of different points.

Natural choice $\nabla_{a}$ with a metric $g_{ab}$ on the manifold

Consider two parallel transport vector $v^{a}, w^{b}$, we demand their inner poduct remain unchanged along any curve:

$$t^{a} \nabla_{a}\left(g_{b c} v^{b} w^{c}\right)=0 \qquad \Rightarrow \qquad t^{a} v^{b} w^{c} \nabla_{a} g_{b c}=0$$

Equation hold for all curves and parallelly transported vectors if and only if

$$\nabla_{a} g_{b c}=0$$

$\mathrm{THEOREM\ 3.1.1} \quad \textit{There exist a unique derivative operator } \nabla_{a}\ \textit{satisfying } \nabla_{a}g_{bc}=0$

$Proof.$ we will prove the theorem by showing that a unique solution for $C^{c}\!\,_{ab}$ exists.

$$0=\nabla_{a} g_{b c}=\tilde{\nabla}_{a} g_{b c}-C^{d}\!\,_{ab} g_{d c}-C_{a c}^{d} g_{b d} \quad \Rightarrow \quad C_{c a b}+C_{b a c}=\tilde{\nabla}_{a} g_{b c}$$

By index substitution:

$$\begin{array}{l} C_{c b a}+C_{a b c}=\tilde{\nabla}_{b} g_{a c} \\ C_{b c a}+C_{a c b}=\tilde{\nabla}_{c} g_{a b} \end{array}$$

Using the symmetry property of $C^{c}_{ab}:$

$$2 C_{c a b}=\tilde{\nabla}_{a} g_{b c}+\tilde{\nabla}_{b} g_{a c}-\tilde{\nabla}_{c} g_{a b}$$

that is,

$$C^{c}\!\,_{ab}=\frac{1}{2} g^{c d}\left\{\tilde{\nabla}_{a} g_{b d}+\tilde{\nabla}_{b} g_{a d}-\tilde{\nabla}_{d} g_{a b}\right\}$$

$\tilde{\nabla}$ is any derivate operator, so the choice of $C^{c}\!\,_{ab}$ is manifestly unique.

In terms of Christoffel symbol:

$$\Gamma^{c}\!\,_{ab}=\frac{1}{2} g^{c d}\left\{\partial_{a} g_{b d}+\partial_{b} g_{a d}-\partial_{d} g_{a b}\right\}$$

$$\Gamma^{\rho }\!\,_{\mu \nu}=\frac{1}{2} \sum_{\sigma} g^{\rho \sigma}\left\{\frac{\partial g_{\nu \sigma}}{\partial x^{\mu}}+\frac{\partial g_{\mu \sigma}}{\partial x^{\nu}}-\frac{\partial g_{\mu \nu}}{\partial x^{\sigma}}\right\}$$

### 3.2 Curvature

Define the Riemann curvature tensor in terms of the failure of successive operations of differentiation to commute when applied to a dual vector field.

\begin{aligned} \nabla_{a} \nabla_{b}\left(f \omega_{c}\right) &=\nabla_{a}\left(\omega_{c} \nabla_{b} f+f \nabla_{b} \omega_{c}\right) \\ &=\left(\nabla_{a} \nabla_{b} f\right) \omega_{c}+\nabla_{b} f \nabla_{a} \omega_{c}+\nabla_{a} f \nabla_{b} \omega_{c}+f \nabla_{a} \nabla_{b} \omega_{c} \end{aligned}

Subtract from the tensor $\nabla_{b} \nabla_{a} (fw_{c})$: the first three terms vanished.

$$\left(\nabla_{a} \nabla_{b}-\nabla_{b} \nabla_{a}\right)\left(f \omega_{c}\right)=f\left(\nabla_{a} \nabla_{b}-\nabla_{b} \nabla_{a}\right) \omega_{c}$$

The tensor $(\nabla_{a} \nabla_{b}-\nabla_{b} \nabla_{a})w_{c}$ at point $p$ depends only on the value of $w_{c}$ at $p$.

Consequently, defines a linear map from $w_{c}$ at $p$ to type $(0,3)$ tensors. $\ \gets\,\,$ a tensor of type $(0,4)$

$$\nabla_{a} \nabla_{b} \omega_{c}-\nabla_{b} \nabla_{a} \omega_{c}=R_{a b c}\!\,^{d} \omega_{d}$$

$R_{a b c}\!\,^{d}$ is called the $\textit{Riemann curvature tensor}$.  it is directly related to the failure of a vector to return to its initial value when parallel transported around a small closed curve.

We can conveniently construct a small closed loop at $p \in M$ by choosing a two-dimensional surface $S$ through $p$ and choosing coordinates $t$ and $s$ in the surface [for simplicity, $p$ to be $(0, 0)$]

let $v^{a}$ be a vector parallelly transported around this closed loop.

(lack img)

For small $\Delta t$ the change, $\delta_{1}$, in $v^{a}w_{a}$ on the first leg of the path is

$$\delta_{1}=\left.\Delta t \frac{\partial}{\partial t}\left(v^{a} \omega_{a}\right)\right|_{(\Delta t / 2,0)}$$

where, by evaluating the derivative at the midpoint, we have ensured that this
expression is accurate to second order in the displacement $\Delta t$. Rewrite $\delta_1$ as

\begin{aligned} \delta_{1} &=\left.\Delta t\ T^{b} \nabla_{b}\left(v^{a} \omega_{a}\right)\right|_{(\Delta t / 2,0)} \\ &=\left.\Delta t\ v^{a} T^{b} \nabla_{b} \omega_{a}\right|_{(\Delta t / 2,0)} \end{aligned}

the two " $\Delta t$ variations," $\delta_1$, $\delta_3$ combine to yield

$$\delta_{1}+\delta_{3}=\Delta t\left\{\left.v^{a} T^{b} \nabla_{b} \omega_{a}\right|_{(\Delta t / 2,0)}-\left.v^{a} T^{b} \nabla_{b} \omega_{a}\right|_{(\Delta t / 2, \Delta s)}\right\}$$

Since the term in brackets vanishes as $\Delta S \to 0$, this shows that to first order in $\Delta t$ and $\Delta S$,
the total change in $v^{a} w_{a}$ vanishes. i.e., parallel transport is path-independent to first order in $\Delta t$ and $\Delta S$.

To first order, the term $T^{b} \nabla_{b} \omega_{a}$ at $(\Delta t/2, \Delta S)$ will differ from the parallel transport of that quantity from $(\Delta t/2,0)$ by the amount $\Delta s S^{c} \nabla_{c}\left(T^{b} \nabla_{b} \omega_{a}\right)$. Thus, to second order:

$$\delta_{1}+\delta_{3}=-\Delta t\ \Delta s\ v^{a} S^{c} \nabla_{c}\left(T^{b} \nabla_{b} \omega_{a}\right)$$

Adding the similar contribution for $\delta_2$ and $\delta_{4}$, the total change is:

\begin{aligned} \delta\left(v^{a} \omega_{a}\right) &=\Delta t\ \Delta s\ v^{a}\left\{T^{c} \nabla_{c}\left(S^{b} \nabla_{b} \omega_{a}\right)-S^{c} \nabla_{c}\left(T^{b} \nabla_{b} \omega_{a}\right)\right\} \\ &=\Delta t\ \Delta s\ v^{a} T^{c} S^{b}\left(\nabla_{c} \nabla_{b}-\nabla_{b} \nabla_{c}\right) \omega_{a} \\ &=\Delta t\ \Delta s\ v^{a} T^{c} S^{b} R_{c b a}\!\,^{d} \omega_{d} \end{aligned}

in the second step we used the fact that $[T^{c}, S^{b}]=T^{c} \nabla_{c}S^{b}- S^{b} \nabla_{b} T^{c} =0 \quad\ \gets\partial_{s}\partial_{t} =0$

Since this equation can behold for all $\omega_{a}$ if and only if

$$\delta v^{a}=\Delta t\ \Delta s\ v^{d} T^{c} S^{b} R_{c b d}\!\,^{a}$$

It shows that the Riemann tensor indeed directly measures the path dependence of parallel transport.

Similarly, we will obtain
$$\left(\nabla_{a} \nabla_{b}-\nabla_{b} \nabla_{a}\right) t^{c}=-R_{a b d}\!\,^{c} t^{d}$$

For an arbitrary tensor field $\ T$

\begin{aligned}\left(\nabla_{a} \nabla_{b}-\nabla_{b} \nabla_{a}\right) T^{c_{1} \cdots c_{k}}\!\,_{d_{1} \cdots d_{l}}=&-\sum_{i=1}^{k} R_{a b e}\!\,^{c_{i}} T^{c_{1} \cdots e \cdots c_{k}}\!\,_{d_{1} \cdots d_{l}} \\ &+\sum_{j=1}^{l} R_{a b d_{j}}\!\,^{e} T^{c_{1} \cdots c_{k}}\!\,_{d_{1} \cdots e \cdots d_{l}} \end{aligned}

Four key properties of the Riemann tensor:

• $R_{abc}\!\,^{d}=-R_{bac}\!\,^{d}$
• $R_{[abc]}\!\,^{d}=0$
• For the $\nabla_{a}$ naturally associated with the metric, $\nabla_{a} g_{bc}=0$

$$R_{abcd}=-R_{abdc}$$
$$R_{abcd}=R_{cdab}$$

• The Bianchi identity hold:

$$\nabla_{[a} R_{b c] d}\!\,^{e}=0$$

Show proof

$Proof: \quad$property 2 & 3:

$$0=2 \nabla_{[a} \nabla_{b} \omega_{c]}=\nabla_{[a} \nabla_{b} \omega_{c]}-\nabla_{[b} \nabla_{a} \omega_{c]}=R_{[a b c]}\!\,^{d} \omega_{d}$$

$$0=\left(\nabla_{a} \nabla_{b}-\nabla_{b} \nabla_{a}\right) g_{c d}=R_{a b c}\!\,^{e} g_{e d}+R_{a b d}\!\,^{e} g_{c e}=R_{a b c d}+R_{a b d c}$$

Bianchi identity:

$$\left(\nabla_{a} \nabla_{b}-\nabla_{b} \nabla_{a}\right) \nabla_{c} \omega_{d}=R_{a b c}\!\,^{e} \nabla_{e} \omega_{d}+R_{a b d}\!\,^{f} \nabla_{c} \omega_{f}$$

$$\nabla_{a}\left(\nabla_{b} \nabla_{c} \omega_{d}-\nabla_{c} \nabla_{b} \omega_{d}\right)=\nabla_{a}\left(R_{b c d}\!\, ^{e} \omega_{e}\right)=\omega_{e} \nabla_{a} R_{b c d}\!\, ^{e}+R_{b c d}\!\, ^{e} \nabla_{a} \omega_{e}$$

$$R_{[a b c]}^{e} \nabla_{e} \omega_{d}+R_{[a b|d|}^{f} \nabla_{c]} \omega_{f}=\omega_{e} \nabla_{[a} R_{b c] d}\!\, ^{e}+R_{[b c|d|}\!\, ^{e} \nabla_{a]} \omega_{e}$$

where the vertical bars indicate that we do not antisymmetrize over $d$, the first term on the left-hand side vanishes for the property 2 while the second terms on both
sides cancel each other. Finally, for all $\omega_{e}$
$$\omega_{e} \nabla_{[a} R_{b c] d}\!\,^{e}=0$$

$Ricci\ tensor\ R_{ac}:\quad$ TRACE PART

$$R_{ac} = R_{abc}\!\, ^d$$

$R_{ab}$ satisfies the symmetry property:

$$R_{ac} = R_{ca}$$

The $scalar\ curvature,\ R$ is defined as the trace of the Ricci tensor:
$$R= R_{a}\!\, ^a$$

$W\!eyl \ tensor \ C_{abcd}:\quad$ TRACE FREE PART

Defined for the manifolds of dimension $n\geqslant 3$ by the equation:

$$R_{a b c d}\,=\ C_{a b c d}+\frac{2}{n-2}\left(g_{a[c} R_{d] b}-g_{b[c} R_{d] a}\right)-\frac{2}{(n-1)(n-2)} R\ g_{a[c} g_{d] b}$$

The Weyl tensor satisfies the symmetry properties 1, 2, and 3 of the Riemann tensor as well as being trace free on all its indices . It also behaves in a very simple manner under conformal transformations of the metric and for this reason is sometimes called the $conformal\ tensor$ !!

With Bianchi identity and property 1 $:\ \to\ \nabla_{a}R_{bcd}\!\, ^{e} + \nabla_{b}R_{cad}\!\, ^{e} + \nabla_{c}R_{abd}\!\, ^{e} = 0$

Contract $a$ with $e: \ \nabla_{a} R_{b c d}\!\, ^{a}+\nabla_{b} R_{c d}-\nabla_{c} R_{b d}=0$

Raising the index $d$ with the metric and contracting over $b$ and $d:$

$$\ \nabla_{a} R_{c}\!\, ^{a}+\nabla_{b} R_{c}\!\, ^{b}-\nabla_{c} R=2\nabla_{a}R_{c}\!\, ^a - \nabla_{c}R =0$$

$$\nabla^{a}G_{ab}=0$$
where (with $R_{ab}=R_{ba}$)
$$G_{a b}=R_{a b}-\frac{1}{2} R\ g_{a b}$$

The tensor $G_{ab}$ is called the $Einstein \ tensor$.

### 3.3 Geodesics

$Geodesic:$ a curve whose tangent vector is parallel propagated along itself.

$$T^{a}\nabla_{a}T^{b}=0$$

The intuitive requirement that the curve is "as straight as possible."

Maintain the same direction instead of the same length.

$$T^{a}\nabla_{a}T^{b}=\alpha T^{b}$$

Given a curve satisfies the above, we can always reparameterize it so that it satisfies the geodesic equation.

The parameterization is called as $\textit{affine parameterization}$.

Writing out the components of this equation in a coordinate basis:

$$\frac{d T^{\mu}}{d t}+\sum_{\sigma, \nu} \Gamma^{\mu }\!\, _{\sigma \nu} T^{\sigma} T^{\nu}=0$$

with $T^{\mu }= dx^{\mu}/dt:$

$$\frac{d^{2} x^{\mu}}{d t^{2}}+\sum_{\sigma, \nu} \Gamma^{\mu }\!\, _{\sigma \nu} \frac{d x^{\sigma}}{d t} \frac{d x^{\nu}}{d t}=0$$

This is a coupled system of $n$ second order ordinary differential equations for the $n$ functions $x^{\mu }(t)$.

There always is a unique solution of equation exists for any given initial value of $x^{\mu }$ and $dx^{u}/dt$.

In mechanics: Given an initial position and velocity, a unique solution exists .

$\textit{Given} \ p \in M \, \textit{and any tangent vector}, \ T^{a} \in V_{p}\\ \textit{there always exists a unique geodesic through p with tangent } \,T^{a}$

$Exponential\ map:$ mapping each $T^{a}$ into the point in $M$ lying at unit affine parameter form $p$ along the geodesic through $p$ with tangent $T^{a}$.

• $\textit{Riemannian normal coordinates}:$ use exponential map to get a coordinate.   $\Gamma ^\mu\!\, _{\sigma \mu}$ vanish

This vector $n^{a}$ is said to be $normal$ to $S$. In the case of a Riemannian metric, $n^{a}$ cannot lie in $V_{p}$, in the case of a metric of indefinite signature, $n^{a}$ could be a null vector, $g_{ab}n^{a}n^{b}=0$, in this case $S$ is said to be a $\textit{null hypersurface}$ at point $p$. Or we may normalize $n^{a}$ by the condition $\ g_{ab}n^{a}n^{b}=\pm 1$

$\textit{Imbedded submanifold}: \quad dimS\leqslant dim M \qquad \psi_{*}v^{a}=0 \Rightarrow v^{a}=0$

• $\textit{Gaussian normal coordinates}:$ defined for any non-null $\textit{hypersurface}\ \ (dimS=n-1)$

For each $p \in S$ we construct the unique geodesic through p with tangent $n^{a}$ and coordinate basis fields $X^{a}_{n-1}$

Gaussian normal coordinates satisfy the important property that the geodesics remain orthogonal to all the hypersurfaces $S_{t}$ defined by the equation $t\ =$ constant. By construction for the hypersurface $S_{0} = S$

\begin{aligned} n^{b} \nabla_{b}\left(n_{a} X^{a}\right) &=n_{a} n^{b} \nabla_{b} X^{a} \\ &=n_{a} X^{b} \nabla_{b} n^{a} \quad\gets [n,X]^{b}=n^{a} \nabla_{a}X^{b}-X^{a} \nabla_{a}n^{b}=0 \\ &=\frac{1}{2} X^{b} \nabla_{b}\left(n^{a} n_{a}\right) \\ &=0 \end{aligned}

The normalization condition $n^{a} n_{a}= \pm 1$ on $S$ is preserved by parallel transport so that $n^{a}n_{a}$ will remain $0$ on $M$

$length,\ l\,$ is defined by

$$l=\int\left(g_{a b} T^{a} T^{b}\right)^{1 / 2} d t$$

For a metric of Lorentz signature $-\ +\ +\ +\,$,

• a curve is said to be $timelike\,$ if the norm of its tangent is everywhere negative,$\ g_{ab}T^{a}T^{b}<0$.
• a curve is said to be $null\,$ if $g_{ab}T^{a}T^{b} = 0$.
• a curve is said to be $spacelike\,$ if $g_{ab}T^{a}T^{b} > 0$

For timelike curves, we use the term $\ proper\ time,\tau\,$, rather than length,

$$\boldsymbol{\tau}=\int\left(-g_{a b} T^{a} T^{b}\right)^{1 / 2} d t$$

Note that since a geodesic tangent is parallel transported and thus its norm is constant,

Geodesics in a Lorentz manifold cannot change from timelike to spacelike or null.

The length (or proper time) of a curve does not depend on the way in which the curve is parameterized .

With a new parameterization $s = s (t)$, the new tangent will be $S^{a} = (dt/ds)T^{a}$ and the new length will be

$$l^{\prime}=\int\left[g_{a b} S^{a} S^{b}\right]^{1 / 2} d s=\int\left[g_{a b} T^{a} T^{b}\right]^{1 / 2} \frac{d t}{d s} d s=l$$

We wish now to derive the condition on a curve which makes it extremize the length between its endpoints,

i.e., wish to find those curves whose length does not change to first order under an arbitrary smooth deformation which keeps the endpoint fixed.

$$l=\int_{a}^{b}\left[\sum_{\mu, \nu} g_{\mu \nu} \frac{d x^{\mu}}{d t} \frac{d x^{\nu}}{d t}\right]^{1 / 2} d t$$

The variation in $l$ is

$$\delta l=\int_{a}^{b}\left[\sum_{\mu, \nu} g_{\mu \nu} \frac{d x^{\mu}}{d t} \frac{d x^{\nu}}{d t}\right]^{-1 / 2} \sum_{\alpha, \beta}\left\{g_{\alpha \beta} \frac{d x^{\alpha}}{d t} \frac{d\left(\delta x^{\beta}\right)}{d t}+\frac{1}{2} \sum_{\sigma} \frac{\partial g_{\alpha \beta}}{\partial x^{\sigma}} \delta x^{\sigma} \frac{d x^{\alpha}}{d t} \frac{d x^{\beta}}{d t}\right\} d t$$

With the choice of parameterization:
$$g_{a b} T^{a} T^{b}=1=\sum_{\mu, \nu} g_{\mu \nu} \frac{d x^{\mu}}{d t} \frac{d x^{\nu}}{d t}$$

the extremization condition is

\begin{aligned} 0 &=\int_{a}^{b} \sum_{\alpha, \beta}\left\{g_{\alpha \beta} \frac{d x^{\alpha}}{d t}\frac{d\left(\delta x^{\beta}\right)}{dt}+\frac{1}{2} \sum_{\sigma} \frac{\partial g_{\alpha \beta}}{\partial x^{\sigma}} \frac{d x^{\alpha}}{d t} \frac{d x^{\beta}}{d t} \delta x^{\sigma}\right\} d t \\ &=\int_{a}^{b} \sum_{\alpha, \beta}\left\{-\frac{d}{d t}\left(g_{\alpha \beta} \frac{d x^{\alpha}}{d t}\right)+\frac{1}{2} \sum_{\lambda} \frac{\partial g_{\alpha \lambda}}{\partial x^{\beta}} \frac{d x^{\alpha}}{d t} \frac{d x^{\lambda}}{d t}\right\} \delta x^{\beta} d t \end{aligned}

Equation will hold for arbitrary $\delta x^{\beta }$ if and only if

$$-\sum_{\alpha} g_{\alpha \beta} \frac{d^{2} x^{\alpha}}{d t^{2}}-\sum_{\alpha, \lambda} \frac{\partial g_{\alpha \beta}}{\partial x^{\lambda}} \frac{d x^{\lambda}}{d t} \frac{d x^{\alpha}}{d t}+\frac{1}{2} \sum_{\alpha, \lambda} \frac{\partial g_{\alpha \lambda}}{\partial x^{\beta}} \frac{d x^{\alpha}}{d t} \frac{d x^{\lambda}}{d t}=0$$

Using our formula for $\Gamma ^{\sigma}\!\, _{\alpha \lambda }$, we see the equation is just the geodesic equation.
(Had we not chosen the parameterization $g_{ab}T^{a}T^{b} = 1$ for $C$, we would have obtained the geodesic equation $T^{a} \nabla_{a} T^{b}=\alpha T^{b}$)

However, a given geodesic connecting two points need not be the shortest path between them. For a manifold with a Lorentz metric, given two points that can be connected by a timelike curve, one can always find timelike curves of arbitrarily small proper time connecting the points

Our final task is to derive the geodesic deviation equation, the equation which relates the tendency of geodesics to accelerate toward or away from each other to the curvature of the manifold .

Let $\gamma _s(t)$ denote a smooth one-parameter family of geodesics (see Fig), that is for each $s \in \mathbb{R}$, the curve $\gamma_s(t)$ is a geodesic (parameterized by affine parameter $t$);

We may choose $s$ and $t$ as coordinates, with tangent $T$:

$$T^{a} \nabla_{a} T^{b}=0$$

The vector field $X^{a}=(\partial /\partial s)^{a}$ represents the displacement to an infinitesimally nearby geodesic, and is called the $deviation\ vector$.

There is "Gauge freedom" in $X^{a}$ that $T^{a}$ under a change of the affine parameterization of the geodesics
$$\gamma _s(t):\quad t\to t' = b(s)t+c(s)$$

It's worth noting that in the case the geodesics arise from the derivative operator associated with a metric $g_{ab}$ ,

$X^{a}$ always can be chosen orthogonal to $T^{a}$, Namely, by re-scaling t by an s-dependent factor, we may ensure that $\,g_{ab}T^{a}T^{b}$ (which is auto constant along each geodesic) does not vary with s.

$Proof:$

First, tangent satisfies $T^{a}\nabla_{a} T^{b}=\alpha T^{b}$ can be reparameterized to satisfies $T^{a}\nabla_{a} T^{b}=0$

$\qquad \quad$ with $\qquad \quad \sum_{\sigma} T^{\sigma} \nabla_{\sigma} T^{\mu}=\alpha T^{\mu}$

\begin{aligned} \sum_{\sigma} S^{\sigma} \nabla_{\sigma} S^{\mu} &=\frac{d S^{\mu}}{d s}+\sum_{\sigma, \nu} \Gamma_{\sigma \nu}^{\mu} S^{\sigma} S^{\nu} \\ &=\frac{d}{d s}\left(T^{\mu} \frac{d t}{d s}\right)+\sum_{\sigma, \nu} \Gamma_{\sigma \nu}^{\mu} T^{\sigma} T^{\nu}\left(\frac{d t}{d s}\right)^{2} \\ &=T^{\mu} \frac{d^{2} t}{d s^{2}}+\frac{d T^{\mu}}{d t}\left(\frac{d t}{d s}\right)^{2}+\sum_{\sigma, \nu} \Gamma_{\sigma \nu}^{\mu} T^{\sigma} T^{\nu}\left(\frac{d t}{d s}\right)^{2} \\ &=T^{\mu} \frac{d^{2} t}{d s^{2}}+\alpha T^{\mu}\left(\frac{d t}{d s}\right)^{2} \end{aligned}

This shows that it's sufficient to choose $s$ such that

$$\frac{d^{2} t}{d s^{2}}+\alpha \left(\frac{d t}{d s}\right)^{2}=0 \quad\to \ \quad S^{a}\nabla_{a} S^{b}=0$$

Second, suppose this time that

$$\sum_{\sigma} T^{\sigma} \nabla_{\sigma} T^{\mu}=0 \qquad \quad \sum_{\sigma} S^{\sigma} \nabla_{\sigma} S^{\mu}=0$$

then

$$0=\sum_{\sigma} S^{\sigma} \nabla_{\sigma} S^{\mu}=T^{\mu} \frac{d^{2} t}{d s^{2}}+\left(\frac{d t}{d s}\right)^{2} \cancel{ \sum_{\mathscr{\sigma }} T^{\sigma} \nabla_{\sigma} T^{\mu}}$$

It follows that $\,d^{2}t/ds^{2}=0$, so $t=as+b$ for some constants $a,b$.

Since $X^{a}$and $T^{a}$ are coordinate vector fields, they commute:
$$T^{b} \nabla_{b} X^{a}=X^{b} \nabla_{b} T^{a}$$

The quantity $v^{a} = T^b\nabla_{b} X^{a}$ gives the rate of change along a geodesic of the displacement to an infinitesimally nearby geodesic. Thus, we may interpret $v_{a}$ as the relative velocity of an infinitesimally nearby geodesic . Similarly, we may interpret

$$a^{a}=T^{c} \nabla_{c} v^{a}=T^{c} \nabla_{c}\left(T^{b} \nabla_{b} X^{a}\right)$$

as the relative acceleration of an infinitesimally nearby geodesic in the family.

\begin{aligned} a^{a} &=T^{c} \nabla_{c}\left(T^{b} \nabla_{b} X^{a}\right) \\ &=T^{c} \nabla_{c}\left(X^{b} \nabla_{b} T^{a}\right) \\ &=\left(T^{c} \nabla_{c} X^{b}\right)\left(\nabla_{b} T^{a}\right)+X^{b} T^{c} \nabla_{c} \nabla_{b} T^{a} \\ &=\left(X^{c} \nabla_{c} T^{b}\right)\left(\nabla_{b} T^{a}\right)+X^{b} T^{c} \nabla_{b} \nabla_{c} T^{a}-R_{c b d}\!\, ^{a} X^{b} T^{c} T^{d} \\ &=X^{c} \nabla_{c}\left(T^{b} \nabla_{b} T^{a}\right)-R_{c b d}\!\, ^{a} X^{b} T^{c} T^{d} \\ &=-R_{c b d}\!\, ^{a} X^{b} T^{c} T^{d} \end{aligned}

This is known as the $geodesic\ deviation\ equation$.

We have $a^{a}=0$ for all families of geodesics if and only $R_{c b d}\!\, ^{a}=0$.

Thus, some geodesics will accelerate toward or away from each other if and only if $R_{c b d}\!\, ^{a}\ne0$

(Some initially parallel geodesics---i.e, goedesics with $v_{a}=T^{b} \nabla_{b} X^{a}=0$ initially--- will fail to remain parallel)

### 3.4 Methods for Computing Curvature

#### 3.4a Coordinate Component Method

Fora dual vector field $w_{a}$, we have

$$\nabla_{b} \omega_{c}=\partial_{b} \omega_{c}-\Gamma_{b c}^{d} \omega_{d}$$

thus,

\begin{aligned} \nabla_{a} \nabla_{b} \omega_{c}=& \partial_{a}\left(\partial_{b} \omega_{c}-\Gamma_{b c}^{d} \omega_{d}\right) \\ &-\Gamma_{a b}^{e}\left(\partial_{e} \omega_{c}-\Gamma_{e c}^{d} \omega_{d}\right) \\ &-\Gamma_{a c}^{e}\left(\partial_{b} \omega_{e}-\Gamma_{b e}^{d} \omega_{d}\right) \end{aligned}

Hence, the Riemann eq could be expressed as

$$R_{a b c}{ }^{d} \omega_{d}=\left[-2 \partial_{[a} \Gamma_{b] c}^{d}+2 \Gamma_{c[a}^{e} \Gamma_{b] e}^{d}\right] \omega_{d}$$

since the aboved eq holds for all $w_{d}$. Taking the components
of the tensors appearing in this equation in the coordinate basis associated with our
chart, we obtain the formula

$$R_{\mu \nu \rho}{ }^{\sigma}=\frac{\partial}{\partial x^{\nu}} \Gamma_{\mu \rho}^{\sigma}-\frac{\partial}{\partial x^{\mu}} \Gamma_{\nu \rho}^{\sigma}+\sum_{\alpha}\left(\Gamma_{\mu \rho}^{\alpha} \Gamma_{\alpha \nu}^{\sigma}-\Gamma_{\mu \rho}^{\alpha} \Gamma_{\alpha \mu}^{\sigma}\right)$$

Ricci equation

\begin{aligned} R_{\mu \rho} &=\sum_{\nu} R_{\mu \nu \rho}^{\nu} \\ &=\sum_{\nu} \frac{\partial}{\partial x^{\nu}} \Gamma_{\mu \rho}^{\nu}-\frac{\partial}{\partial x^{\mu}}\left(\sum_{\nu} \Gamma_{\nu \rho}^{\nu}\right)+\sum_{\alpha, \nu}\left(\Gamma_{\mu \rho}^{\alpha} \Gamma_{\alpha \nu}^{\nu}-\Gamma_{\nu \rho}^{\alpha} \Gamma_{\alpha \mu}{ }_{\alpha}\right) \end{aligned}

A simple formula may be derived for the contracted Christoffel symbol $\Gamma^{a}\!_{ab}$

$$\Gamma_{a \mu}^{a}=\sum_{\nu} \Gamma_{\nu \mu}^{\nu}=\frac{1}{2} \sum_{\nu, \alpha} g^{\nu \alpha} \frac{\partial g_{\nu \alpha}}{\partial x^{\mu}}$$

using the formula for the inverse of a matrix $g=g_{\mu \nu}A^{\mu \nu }$ 代数余子式

$$\sum_{\nu, \alpha} g^{\nu \alpha} \frac{\partial g_{\nu \alpha}}{\partial x^{\mu}}=\frac{1}{g} \frac{\partial g}{\partial x^{\mu}}$$

Thus,

$$\Gamma_{a \mu}^{a}=\frac{1}{2} \frac{1}{g} \frac{\partial g}{\partial x^{\mu}}=\frac{\partial}{\partial x^{\mu}} \ln \sqrt{|g|}$$

In terms of its coordinate basis components

\begin{aligned} \nabla_{a} T^{a} &=\partial_{a} T^{a}+\Gamma_{a b}^{a} T^{b} \\ &=\sum_{\mu} \frac{1}{\sqrt{|g|}} \frac{\partial}{\partial x^{\mu}}\left(\sqrt{|g|} T^{\mu}\right) \end{aligned}