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# Bei's Study Notes

## General Relativity

### General Relativity 02 - Manifold and tensor fieldsLast updated: 2017-04-19 21:55:03 PDT.

(This is from Charpter 2.2 of Wald)

### Vectors

In GR, we discuss about spaces that are not exactly flat. If we can split the space into parts and each part is continously corresponsive to a region of $\mathbb R^n$, then it is called a manifold. $n$ is called toe dimension of the manifold.

Normally, a finite dimensional manifold can be embeded in a higher dimensional Euclidean space. Whereas such embedding might not be natural. In GR, the spacetime does not naturally live in a higher dimensional space, so an abstract definition of a manifold is necessary.

Definition (Manifold): An n-dimensional, $C^\infty$, real manifold $M$ is a set together with a collection of subsets $\{O_\alpha\}$ satisfying the following properties:

1. $\{O_\alpha\}$ is a open cover of $M$.
2. For each $\alpha$, there is a one-to-one, onto, map $\psi_\alpha: O_\alpha \to U_\alpha$ where $U_\alpha$ is an open subset of $\mathbb R^n$.
3. For each $O_\alpha$ and $O_\beta$ that $O_\alpha \cap O_\beta \neq \emptyset$, the map $\psi_\beta \circ \psi_\alpha^{-1}$ maps $\psi_\alpha[O_\alpha \cap O_\beta] \subset U_\alpha$ to $\psi_\beta[O_\alpha \cap O_\beta] \subset U_\beta$. Then both sets must be open and the map must be $C^\infty$.

Each map $\psi_\alpha$ is called a "chart" or a "coordinate system". The definition of $C^k$ manifold and complex manifold simply is the same with some natural changes.

Through out this book, the manifold involed are all assumed to be Hausdorff and paracomact.

NOTE By this definition, the charts, by no means, are required to provide a "straight coordinate" to $\mathbb R^n$, by which I mean the tangent vector can change through out each $O_\alpha$.

Example Euclidean space $\mathbb R^n$ with one chart and map to be identity function.

Example 2-sphere (a 2-dimensional spherical surface embedded in $\mathbb R^3$).

In SR, each coordinate system is applicable to the whole universe. but in GR, a coordinate system is only usable inside its corresponding open set.

With the mapping to $\mathbb R^n$, we can easily define differentiability and smoothness of maps between manifolds.

Defintion (Diffeomorphism): Let $M$ and $M'$ to be manifolds and $\{\psi_\alpha\}$ and $\{\psi_\beta\}$ to be chart maps. A map $f:M\to M'$ is said to be $C^\infty$ if for each $\psi'_\beta \circ f \circ \psi_\alpha^{-1}$ is $C^\infty$ in Euclidean spaces. If $f$ is $C^\infty$, one-to-one and onto, then it is called a diffeomorphism, and $M$ and $M'$ are said to be diffeomorphic.

NOTE diffeomorphism requires the manifolds to have the same dimension.

### Tangent vector in a manifold (without embedding in $\mathbb R^n$):

We can define tangent vectors as directional derivatives. In $\mathbb R^n$, the mapping between vectors and directional derivatives is one-to-one. $(v^1,...,v^n)$ defines derivative operator $\sum_\mu v^n(\partial/\partial x^\mu)$. Directional derivatives are characterized by Leibnitz rules.

Let $\mathcal F$ denote the collections of all $C^\infty$ functions from manifold $M$ into $\mathbb R$. We define a tangent vector $v$ at point $p \in M$ to be a map $v: \mathcal F \to \mathbb R$ which is (1) linear and (2) obeys Leibnitz rules: $\forall f,g \in \mathcal F, a, b \in R,$

1. $v(af+bg)=av(f) + bv(g)$
2. $v(fg) = f(p)v(g) + g(p)v(f)$

NOTE: Be very careful that the second rule also applies the function $f$ and $g$ at $p$.

Prop: $h\in\mathcal F,\ \forall q \in M, h(q) = c \text{ (constant) } \implies v(h) = 0$

Proof.

\begin{align*} p \mapsto h(p)h(p) &= p \mapsto ch(p) \implies hh = ch\\ v(hh) &= 2h(p)v(h) = 2cv(h)\quad,\, \text{whereas}\\ v(ch) &= cv(h)\quad,\text{therefore} \\ cv(h) &= 0 \\ \end{align*}

If $c = 0$, $v(h) = 0$ since linearlity, otherwise divide the equation by $c$, and $v(h) = 0$. $\square$

The maps of a tangent vectors of a point $p$ forms a vector space by adding this addition law: $\forall a \in \mathbb R, v_1 + av_2 \equiv h \mapsto v_1(h) + av_2(h)$.

Prop:

Theorem 2.2.1: Let $M$ be an $n$-dimensional manifold. Let $p\in M$ and let $V_p$ denote the tangent space at $p$, then $\mathrm {dim}(V_p) = n$.

Proof. Given a chart $\psi$ of open set $O$ where $p \in O$, If $f \in \mathcal F$, then $f\circ \psi^{-1} \to \mathbb R$ is $C^\infty$. For $\mu = 1, ..., n$ define functional $X_{p,\mu}: \mathcal F \to \mathbb R$ by

\begin{align*} X_{p,\mu}[f] &= \frac{\partial}{\partial x^\mu}\left[f\circ\psi^{-1}\right](\psi(p)) \\ \end{align*}

This means, the $\mu$th component of $X$ is a functional that takes the $\mu$th partial derivative of function $f\circ\psi^{-1}$ in $\mathbb R^n$, and apply the point $\psi(p)$ to it. It is clear that $X_{p,\mu}$ is a derivative from the chain rule. Now we need to prove that $V_p = \mathrm{span}\left\{X_{p,\mu}\right\}$.

For any function $F: \mathbb R^n \to \mathbb R$, if $F$ is $C^\infty$,

$$\forall a\in\mathbb R^n, \exists H_{a,\mu}(x\in \mathbb R^n) \in C^\infty, s.t.\\ F(x) = F(a) + \sum^n_{\mu=1}(x^\mu-a^\mu)H_{a,\mu}(x)\quad,$$

especially, we have

\begin{align*} H_{a,\mu}(a) &= \lim_{x\to a} H_{a,\mu}(x)\\ &= \frac{\partial}{\partial x^\mu}[F](a)\quad. \end{align*}

Let $F= f\circ \psi^{-1}$, and $a=\psi(p)$ we have

\begin{align*} f(q) &= f(p) + \sum^n_{\mu=1}(x^\mu \circ \psi(q)-x^\mu \circ \psi(p)) H_{\psi(p),\mu}(\psi(q))\quad.\\ H_{\psi(p),\mu}(\psi(p)) &= \frac{\partial}{\partial x^\mu}[F](\psi(p)) \\ &= \frac{\partial}{\partial x^\mu}[f \circ \psi^{-1}](\psi(p)) \\ &= X_{p,\mu}[f]\quad, \end{align*}

where $x^\mu \circ \psi$ denotes $(x\mapsto x^\mu) \circ \psi$, that is, picking the $\mu$th element of the result.

For any $v \in V_p$, apply the functional,

\begin{align*} v[f]&=\left.\cancel{v[f(p)]} + \sum^n_{\mu=1}v[x^\mu \circ \psi-\cancel{x^\mu \circ \psi(p)}]\cdot (H_{\psi(p),\mu}\circ\psi)(p) \\ + \sum^n_{\mu=1}\cancel{\left(x^\mu \circ \psi(p)-x^\mu \circ \psi(p)\right)} v[H_{\psi(p),\mu}\circ\psi]\right.\quad, \\ &= \sum^n_{\mu=1}v[x^\mu \circ \psi]X_{p,\mu}[f]\quad. \end{align*}

This means $v[f]$ is a linear combination of $\{X_{p,\mu}[f]\}$. $\square$

The basis $\{X_\mu\}$ is called a coordinate basis, frequently denoted as simply $\partial/\partial x^\mu$. For each different chart $\psi'$ chosen, there is a different coordinate basis $\{X'_\nu\}$, and

\begin{align*} X'_{p,\nu}[f] &= \sum^n_{\nu=1} X'_{p,\nu}[x^\mu \circ \psi] X_{p,\mu}[f] \\ &= \sum^n_{\nu=1} \frac{\partial}{\partial x'^\nu}[x^\mu \circ \psi\circ\psi'^{-1}](\psi'(p)) X_{p,\mu}[f] \\ &= \sum^n_{\nu=1} \frac{\partial x^\mu}{\partial x'^\nu}(\psi'(p)) X_{p,\mu}[f]\quad, \\ X_{p,\mu} &= \sum^n_{\nu=1} \frac{\partial x'^\nu}{\partial x^\mu}(\psi(p)) X'_{p,\nu}\quad. \\ \end{align*}

We can also get the vector transformation law from it:

$$v'^\nu=\sum^n_{\mu=1}v^\mu\frac{\partial x'^\nu}{\partial x^\mu}\quad.$$

A smooth curve, $C$ on a manifold $M$ is a $C^\infty$ map of $\mathbb R$ into $M$, $C: \mathbb R \to M$. At each point $p \in M \cap C$, there is a tangent vector $T \in V_p$ associated with $C$ as follows.

\begin{align*} T[f]&=\frac{d}{dt}[f\circ C] \\ &=\sum_{\mu=1}^n \frac{d x^\mu}{dt}\frac{\partial}{\partial x^\mu}[f\circ \psi^{-1}] \\ &=\sum_{\mu=1}^n\frac {dx^\mu}{dt} X_\mu(f)\quad.\\ \end{align*}

Therefore the components of $T$ is given by

$$T^\mu = \frac{dx^\mu}{dt}\quad.$$

If $p$ and $q$ are on the manifold, there is no way to correlate them in a general manifold. Another construct ("connection", or "parallel transportation") must be introduced to do so. However, if the curvation is nonzero, the identification of $V_p$ with $V_q$ obtained in this manner will depend on the choice of curve.

A tangent field, $v$, on a manifold $M$ is an assignment of a tangent vector, $v\vert_p \in V_p$ for each point $p \in M$. Despite the fact that the tangent spaces $V_p$ and $V_q$ at different points are different vector spaces, there is a natural notion of what it means for $v$ to vary smoothy from point to point.

A one-parameter group of diffeomorphisms $\phi_t$ is a $C^\infty$ map from $\mathbb R \times M \to M$. In particular:

1. $\forall \in\mathbb R, \phi_t$ is a diffeomorphism, and
2. $\forall s, t \in \mathbb R, \phi_{s+t} = \phi_s \circ \phi_t$.

At each point $p$, $\phi_t(p)$ is a curve, called the orbit of $\phi_t$ which passes through $p$ at $t=0$. Define $v|_p$ to be the tangent vector at $t=0$. Thus we can consider the vector field $v$ to be the generator of such a group.

Conversely, we can ask a question, that if given a vector field $v$, can we find a family of curves s.t. for each point $p \in M$, there is one and only one curve that passes through the point with the tangent vector equals to $v\vert_p$. The answer is yes.

Therefore we have a one-to-one mapping between a tangent field and a diffeomorphism. We can thus consider a tangent field to be a mapping of type $M \to M$:

$$v|_p = \left.\frac{\mathrm d\phi_t(p)}{\mathrm dt}\right|_{t=0}$$

Given two smooth vector fields $v$ and $w$, we can define the commutator field $[v, w]$ as follows:

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

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### Topological space (definition dump)

Definition (Topological space, open set): A topological space $(X,\mathcal J)$ is a set $X$ with a collection $\mathcal J$ of subsets of $X$ that satisfies:

1. $X, \emptyset \in \mathcal J$;
2. If $O_1, O_2, ... \subseteq \mathcal J$, then $\bigcup_\alpha\,O_\alpha \in \mathcal J$;
3. If $n \in \mathbb Z^+, O_1, ... O_n \in \mathcal J$, then

$$\bigcap_{i=1}^n O_i \in \mathcal J\,.$$

Sets in $\mathcal J$ are called "open set"s.

An example of a topology on $\mathbb R$ contains all the open intervals in $\mathbb R$. Thus the name "open set"s.

Definition (Induced topology): If $(X,\mathcal J)$ is a topological space and $A$ is a subset of $X$, we may make $A$ into a topological space by defining the topology $\mathcal F = \{U\mid U=A\cap O, O\in\mathcal J\}$, then $(A, \mathcal F)$ forms a topology space. $\mathcal F$ is called induced (or relative) topology.

Definition (Product topology): If $(X_1,\mathcal J_1)$ and $(X_2,\mathcal J_2)$ are both topological spaces, the direct prodct of both naturally forms a topological space $(X_1\times X_2, \mathcal J)$. $\mathcal J$ is called the product topology.

NOTE This lifts the dimension of the topological space.

Open balls on $\mathbb R^n$ naturally form a topology.

Definition (Continuous mapping): If $(X,\mathcal J)$ and $(Y,\mathcal K)$ are topological spaces, a map $f:X\to Y$ is continuous if the inverse image $f^{-1}[O] \equiv \{x\in X \mid f(x) \in O\}$ maps every open set in $Y$ to an open set in $X$.

Definition (Homeomorphism): If $f$ is continuous, one-to-one, onto, and its inverse is continues, then $f$ is called a homeomorphism, and the spaces are said to be "homeomorphic".

NOTE Not to be confused with homomorphism and homomorphic.

Definition (Closed set): The complement of an open set is called a "closed set". Sets in a topology can be open, close, both, or neither.

Definition (Connected): The topology is said to be conneted if the only subsets that are both open and closed are $X$ and $\emptyset$. $\mathbb R^n$ is connected.

Definition (Closure): If $(X, \mathcal J)$ is a topological space, $\forall A \subseteq X$, the closure $\overline A$ is the intersection of all open sets that contains $A$.

Properties:

1. $\overline A$ is closed;
2. $A \subseteq \overline A$;
3. $\overline A = A \iff A$ is closed.

NOTE Meaning "to make a set closed". Closure of a set is unique and is necessarily in the topology.

Definition (Interior, Boundary): Interior of $A$ is defined as the union of all the open sets contained in $A$. The boundy of $A$, denoted $\dot A$ (or $\partial A$), is defined as elements in $\overline A$ but not the interior of $A$, $\equiv \mathrm{int}(A)$.

NOTE alternatively, $\partial A \equiv \overline A \cap \overline {X \setminus A}$.
NOTE alternatively, $\mathrm{int}(A) = X \setminus \overline {X \setminus A}$.
NOTE alternatively, $\partial A \equiv \{ p \mid p \in X. O \in \mathcal J. p \in O \to (\exists \, a, b \in O. a \in A \wedge b \notin A) \}$.

Definition (Hausdorff): A topological space is Hausdorff if any two distinct points can be included in two disjoint open sets.

$\mathbb R^n$ is Hausdorff.

### Compactness

One of the most powerful notions in topology is that of compactness, which is defined as follows.

Definition (Open cover): If $(X, \mathcal J)$ is a topological space and a collection of open sets $C=\{O_\alpha\}$ has $\bigcup_\alpha\,O_\alpha = X$, then $C$ is said to be an open cover of $X$, and $C$ "covers" $X$. Also if $Y$ is a subset of $X$, and $Y \subseteq \bigcup_\alpha\,O_\alpha$, then $C$ is said to be an open cover of $Y$ and $C$ "covers" $Y$. A subcollection of $C$ forms a subcover if it also covers $X$ (or $Y$).

Definition (Compact space): If every open cover can be written as finite subcover, then the topological space is compact.

Alternative definitions of compact space. The following are equivalent:

1. A topological space $X$ is compact.
2. Every open cover a $X$ has a finite subcover.
3. $X$ has a sub-base such that every cover of the space by members of the sub-base has a finite subcover (Alexander's sub-base theorem).
4. Any collection of closed subsets of $X$ with the finite intersection property has nonempty intersection.
5. Every net on X has a convergent subnet (see the article on nets for a proof).
6. Every filter on X has a convergent refinement.
7. Every ultrafilter on X converges to at least one point.
8. Every infinite subset of X has a complete accumulation point.

Definition (Open cover of a set, subcover of a set): If $(X, \mathcal J)$ is a topological space and $A$ is a subset of $X$. A open cover $U$ is a open cover of $A$ if $A \subset U$. A subcover that also covers $A$ is called a subcover of $A$.

Definition (Compact subset): $A$ is said to be compact if every open cover of A has a finite subcover.

The relation ship between compact space and compact subset is given by these two theorems:

1. Compact subset of a Hausdorff space is closed.
2. Closed subset of a compact space is compect.

Heine-Borel Theorem. A closed interval $[a, b]$ of $\mathbb R$ is compact.

Open interval $(0, 1)$ is not compact (since the open cover $O_\alpha = (1/\alpha, 1)$ has no finite subcover).

A subset of $\mathbb R$ is compact iff it is closed and bounded.

NOTE A unbounded set can totally be closed. For example, $\mathbb Z$ is obviously unbounded and closed.

Let $(X, \mathcal J)$ and $(Y, \mathcal K)$ be topological spaces. Suppose $(X, \mathcal J)$ is compact and $f: X \to Y$ is continuous. Then $f[X] \equiv \{y\in Y \mid y = f(x)\}$ is compact.

NOTE This transfers compactness through homeomorphisms.

A continuous function from a compact topological space into $\mathbb R$ is bounded and attains its maximum and minimum values.

Tychonoff theorem: Product of compact topological spaces is compact. Given the axiom of choice, the number of such spaces can be infinite.

An application of these is that $S^n$ is compact, because 1) the sphere in $\mathbb R^{n+1}$ is closed and bounded, therefore compact; 2) there is a continuous function from $\mathbb R^{n+1}$ to $S^n$.

### Convergence of sequences

To extend the normal definition of sequence convergence, a sequence $\{x_n\}$ of points in a topological space $(X, \mathcal J)$ is said to converge to point $x$ if $\forall O \in \mathcal J .( x \in O \to \exists N \in \mathbb Z. \forall n > N. x_n \in O)$. $x$ is called the limit of the sequence.

A point $y \in X$ is said to be a accumulation point of $\{x_n\}$ if every open neiborhood of $y$ contains infinitely many points of the sequence.

NOTE The difference between a limit and an accumulation point is that the former requires a particular set of infinite points in $\{x_n\}$. For example, the alternating sequence $1, -1, 1, -1, ...$ has two accumulation points $1$ and $-1$, but it does not have a limit.

Definition (First countable): For every point $p$ in $X$, if there is a countable collection of open sets $\{O_\alpha\}$ that for every neiborhood $O$ of $p$, $O$ contains at least one element in $\{O_\alpha\}$.

Definition (Second countable): There is a countable collection of open sets that every open set can be written as the union of some of the sets in the collections. The sets in that collection are called basis.

NOTE The basis of a linear space is a collection of vectors, s.t. every vector in the space is a linear combination of the basis. The basis of a topological space is a collection of open sets, s.t. every opens set in the space is a union of the basis.

NOTE $\mathbb R^n$ is second countable. Open balls with rational radii centered on rational coordinates can form a countable collection of open sets.
NOTE Every second countable space is first countable.

The relationship between compactness and convergence of sequences is expressed by Bolzano-Weierstrass theorem:

Bolzano-Weierstrass theorem Let $(X, \mathcal J)$ be a topological space and let $A \subset X$

1. If $A$ is compact, then every infinite sequence $\{x_n\}$ of points in $A$ has a accumulation point lying in $A$;
2. Conversely, if $(X, \mathcal J)$ is second countable and every sequence in $A$ has an accumulation point in $A$, then $A$ is compact.

Thus, in particular, if $(X, \mathcal J)$ is second countable, $A$ is compact iff every sequence in $A$ has a convergent subsequence whose limit lies in $A$.

### Paracompactness

Definition (Neighborhood) Given $p$ in topological space $(X, \mathcal J)$, a neighborhood $V$ of $p$ is a subset of $X$ that includes an open set $U$ containing $p$:

$$V \subseteq X, \exists U \in \mathcal J, U \subseteq V, p \in U$$

NOTE: $V$ may not be open, but it contains a open set $U$ that contains $p$.

Definition (Refinement of an open cover): Open cover $\{V_\beta\}$ of $X$ is said to be a refinement of open cover $\{O_\alpha\}$ of $X$ if $\forall V_\beta. \exists O_\alpha. V_\beta \subseteq O_\alpha$.

NOTE Refinements forms a partially ordered set.
NOTE Subcover is always a refinement of a open cover. A refinement of a open cover is not always a subcover.

Definition (Locally finite): $\{V_\beta\}$ is locally finite if each $x \in X$ has an open neighborhood $W$ such that only finitely many $V_\beta$ satisfy $W \cap V_\beta \neq \emptyset$.

NOTE Compactness requires a finite subcover, locally finiteness only requires a finte refinement. It is a weaker requirement.

Definition (Paracompactness): A space is paracompact if every open cover has a locally finite refinement.

NOTE: Locally finiteness is weaker than finitenes of subcovers. Therefore every compact space is paracompact.

NOTE: (wiki) Every metric space is paracompact. A topological space is metrizable if and only if it is a paracompact and locally metrizable Hausdorff space. Paracompactness has little to do with the notion of compactness, but rather more to do with breaking up topological space entities into manageable pieces.

A paracomact manifold $M$ implies:

1. $M$ admits a Riemannian metric and
2. $M$ is second countable.

The most important implication is that a paracompact manifold $M$ will have a partition of unity.

Definition (Partition of unity): If $(X, \mathcal T)$ is a topological space, and $R$ is a set of continuous functions from $X$ to unit interval $[0, 1]$, such that for every point $x \in X$:

1. there is a neighborhood of $x$ where all but finite number of functions in $R$ are $0$, and
2. the sum of all the functions at $x$ is $1$:

$$\sum_{\rho\in R}\rho(x) = 1$$

This is for the ease of defining integrals on the manifold.

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### Starting to Learn General RelativityLast updated: 2017-04-19 21:55:03 PDT.

I was told that usually, a physics student can only master either GR or QFT, but not both at the same time. It will be an interesting experiment if I can start both since I'm not in a rush right now. So this is it. This is the textbook I will use: General Relativity by Robert M. Wald.

## General Relativity - UCI OCW

### General Relativity 10Last updated: 2016-08-25 18:41:25 PDT.

Reference: UCI OpenCourseWare - GR 11 (Playlist)

## Einstein's field equation

$$g_{00} = -(1+2\phi(x)) \quad(\text{Weak fields})\\$$

$$\Delta\phi = 4\pi G\,\rho\\ \Delta g_{00} = -8\pi G\,\rho\\$$

$$T_{00}=\rho$$

therefore

$$\Delta g_{00} = -8\pi G T_{00}$$

An extension might be:

$$G_{\alpha\beta}=-8\pi G T_{\alpha\beta}$$

where $G_{\alpha\beta}$ is an unknown object that relates to gravitational field.

Promotes to G.C.:

$$G_{\mu\nu}=-8\pi G T_{\mu\nu}$$

To determine $G_{\mu\nu}$, we write down the properties of $G_{\mu\nu}$:

1. $G_{\mu\nu}$ is a tensor.
2. $G_{\mu\nu}$ is made out of at most second derivatives of the metric tensor, e.g. $\left(\frac{\partial g}{\partial x}\right)^2$, or $\frac{\partial^2 g}{\partial x^2}$.
3. $G_{\mu\nu}=G_{\nu\mu}$.
4. $G^{\mu\nu}{}_{;\nu} = 0$
5. On non-relativistic limit, $G_{00}\to\Delta g_{00}$.

These requirements uniquely determines the form of $G_{\mu\nu}$.

$$G_{\mu\nu} = C_1 R_{\mu\nu}+ C_2 g_{\mu\nu} R$$

$$G^{\mu}{}_{\nu;\mu}=\left(\frac{1}{2}C_1+C_2\right)R_{;\nu}$$

therefore, $G_{\mu\nu}$ has the form $C_1(R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R)$. Take the non-relativistic limit, that is, $u^\mu \approx 0$ and the field is weak ($g_{\mu\nu} = \eta_{\mu\nu}+\delta h_{\mu\nu}$), and so is the affine connection and the Riemann tensor. Also, $R_{ij} \approx \frac{1}{2}g_{ij}R$.

$$R\equiv g^{\mu\nu} R_{\mu\nu}\approx R_{kk}-R_{00}=\frac{3}{2}R - R_{00}\\ R \approx 2R_{00}$$

$$G_{00} = C_1\left(R_{00}-\frac{1}{2}\eta_{00}R\right)=2C_1R_{00}\\ R_{\mu\nu}\approx \eta^{\lambda\sigma}R_{\lambda\mu\sigma\nu}\\ R_{00}\approx \eta^{\lambda\sigma}R_{\lambda 0\sigma 0}\\ R_{\lambda\mu\nu\kappa} = \frac{1}{2}\left[ \left(\frac{\partial^2g_{\lambda\nu}}{\partial x^\kappa\partial x^\nu}\right) _{[\lambda,\mu]} \right]_{[\nu,\kappa]} + o(\Gamma^2)\\ R_{0000} \approx 0\\ R_{i0j0} \approx \frac{1}{2}\frac{\partial^2g_{00}}{\partial x^i\partial x^j}\\ G_{00} \approx 2C_1(\frac{1}{2}g_{00}) = C_1 \Delta g_{00} \approx -8\pi G \rho\\ C_1 = 1$$

Conclusion (Einstein's field equation):

$$R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R=-8\pi G T_{\mu\nu}$$

Equivalent form:

\begin{align*} g^{\mu\nu}R_{\mu\nu}-\frac{1}{2}g^{\mu\nu}g_{\mu\nu}R&=-8\pi G T_{\mu\nu} g^{\mu\nu}\\ R-2R&=-8\pi G T^{\nu}{}_{\nu}\\ R&=8\pi G T^{\nu}{}_{\nu}\\ R_{\mu\nu}&=-8\pi G \left(T_{\mu\nu}-\frac{1}{2}g_{\mu\nu}T^\lambda{}_\lambda\right)\\ \end{align*}

In vacuum, $T=0$, therefore $R_{\mu\nu}=0$, this is how gravitational waves are calculated, which does not imply Riemann tensor is $0$. In fact, even when $R_{\mu\nu}$ is $0$, $g_{\mu\nu}$ can still oscillate through spacetime.

(then a discussion about the cosmological constant).

### General Relativity 09Last updated: 2017-04-19 21:55:03 PDT.

Reference: UCI OpenCourseWare - GR 10 (Playlist)

### Some exercises

Covariant derivative:

$$V^\mu{}_{;\lambda}=\frac{\partial V^\mu}{\partial x^\lambda} + \Gamma^\mu_{\lambda\sigma}V^\sigma$$

is a tensor.

With this, we can rewrite the equations in special relativity with this procedure:

1. $\eta_{\alpha\beta} \to g_{\alpha\beta}(x)$
2. $\partial_\alpha \to \nabla_\alpha$

In SR, a free particle with velocity $u^\alpha$ and spin $S^\alpha$:

$$\frac{d u^\alpha}{d\tau}=0\\ \frac{d S^\alpha}{d\tau}=0\\ u_\alpha S^\alpha = 0$$

Promoted to:

1. $$\frac{D u^\mu}{D\tau}=\frac{d u^\mu}{d\tau}+\Gamma^\mu_{\lambda\sigma}u^\mu u^\sigma = 0\\$$

2. $$\frac{D S_\mu}{D\tau}=\frac{d S_\mu}{d\tau}-\Gamma^\lambda_{\mu\sigma}S_\lambda u^\sigma = 0\\$$

3. $$u^\mu S_\mu = 0$$

Other force $f^\alpha$:

$$\frac{Du^\mu}{D\tau}=\frac{f^\alpha}{m} \\$$

For Lorentz force:

$$f^\alpha = e F^\mu{}_\nu u^\nu$$

Or

$$m\frac{d^2 x^\mu}{d\tau^2}=f^\mu-m\Gamma^\mu_{\lambda\sigma}u^\lambda u^\sigma$$

For spin $S_\mu$:

$$\frac{dS^\alpha}{d\tau} = \left(\frac{u^\alpha f^\beta}{m}\right)S_\beta$$

in GR (Fermi-Walker Transport):

$$\frac{DS^\alpha}{D\tau} = \left(\frac{u^\alpha f^\beta}{m}\right)S_\beta$$

#### Promote E.M. to G.C. form

E.M.:

\begin{align*} \frac{\partial}{\partial x^\alpha} F^{\alpha\beta} &= - J^\beta\\ \partial_\alpha F_{\beta\gamma} + \partial_\gamma F_{\alpha\beta} + \partial_\beta F_{\gamma\alpha} &= 0 \end{align*}

Promotes to:

\begin{align*} F^{\mu\nu}{}_{;\nu} &= - J^\mu\\ F_{\mu\nu}{}_{;\lambda}+F_{\lambda\mu}{}_{;\nu}+F_{\nu\lambda}{}_{;\mu}&=0 \end{align*}

The covariance divergence can be written as:

\begin{align*} \frac{\partial}{\partial x^\mu}\sqrt{g}\,F^{\mu\nu} &= - J^\nu\\ F_{\mu\nu}{}_{,\lambda}+F_{\lambda\mu}{}_{,\nu}+F_{\nu\lambda}{}_{,\mu}&=0 \end{align*}

Also, current conservation:

$$\frac{\partial}{\partial x^\mu}(\sqrt{g} J^\mu) = 0$$

the change is very minimal (!!).

#### Promote Energy Momentum Tensor

$$\frac{\partial T^{\alpha\beta}}{\partial x^\alpha} = G^\beta\quad (\text{external force})$$

Promotes to:

$$T^{\mu\nu}{}_{;\nu} = G^\mu\quad (\text{external force})$$

simplifies to:

$$\frac{1}{\sqrt{g}}\frac{\partial}{\partial x^\mu}(\sqrt{g}T^{\mu\nu}) = G^\nu - \Gamma^\nu_{\lambda\sigma} T^{\lambda\sigma}$$

#### Hydrodynamics

Perfect fluid

$$T^{\alpha\beta} = p \eta^{\alpha\beta} + (p + \rho)u^\alpha u^\beta$$

Promotes to:

$$T^{\mu\nu} = p g^{\mu\nu} + (p + \rho)u^\mu u^\nu$$

Divergence of above:

$$T^{\mu\nu}{}_{;\mu}= \frac{\partial p}{\partial x^\nu} g^{\mu\nu} + \frac{1}{\sqrt{g}}\frac{\partial}{\partial x^\nu} \left(\sqrt{g} (p + \rho)u^\mu u^\nu\right) + \Gamma^\mu_{\nu\lambda}(p + \rho)u^\mu u^\nu = 0$$

My note: it looks like that $\Gamma$ tends show up in non-linear parts of the equation.

## Curvature

In ME, curvature is described as:

$$F_{\mu\nu}=\partial_\mu A_\nu - \partial_\nu A_\mu\quad.$$

The derivative of the field gives the curvature of the space. The problem in GR is that

$$g_{\mu\nu;\lambda}=0\quad.$$

This problem is solved by Riemann and Christoffel by taking derivative of the affine connection. But since affine connection is not a tensor, the form is rather complicated. Einstein started to notice this in 1912, when examine the Ehrenfest paradox and realized that the spacetime has to be curved. Grossman pointed this out to Einstein. At around the same time, Einstein realized that the gravitational field should not be described by a single field $\phi(x)$, but a field of the metric tensors $g_{\mu\nu}(x)$, which turned out to be the most fundamental field in GR.

The point of it, is we write down:

$$\Gamma^\lambda{}_{\mu\nu} (\text{not a tensor})\,$$

so if we take the derivative of it, say:

$$\frac{\partial}{\partial x^\sigma}\Gamma^\lambda{}_{\mu\nu}\quad,$$

it will not be a tensor either. The hope was that by twiddling the indices, we can cancel out the non-tensor part of them and obtain a tensor. This is what Riemann discovered:

$$R^\lambda{}_{\mu\nu\kappa}\stackrel{def}= \Gamma^\lambda_{\mu\nu,\kappa} +\Gamma^\eta_{\mu\nu}\Gamma^\lambda_{\kappa\eta} - (\nu \leftrightarrow \kappa) \quad(\text{Riemann tensor})$$

#### Useful contractions of $R^\lambda{}_{\mu\nu\sigma}$

\begin{align} R_{\mu\kappa}&\stackrel{def}=R^\lambda{}_{\mu\lambda\kappa}\quad&(\text{Ricci tensor})\\ R&\stackrel{def}=g^{\mu\nu} R_{\mu\nu}\quad&(\text{Ricci Scalar, or Scalar Curvature}) \end{align}

#### Parallel transport around a (small) closed loop

Use the analog of Stoke's theorem:

$$\oint_C\,d\vec e \cdot \vec A = \int_\Sigma (\vec \nabla \times \vec A)\cdot \vec n\,d\sigma$$

the result is:

$$\Delta S_\mu = \frac{1}{2} R^\sigma{}_{\mu\nu\rho}\cdot S_\sigma\cdot u^{\rho\nu}$$

where $u^{\rho\nu}$ is call the "area bivector":

$$u^{\mu\nu}\stackrel{def}=\oint \,x^\mu\,dx^\nu$$

e.g. around a parallelogram of sides $\delta a$ and $\delta b$:

$$u^{\mu\nu} = \delta a^\mu \delta b^\nu - (\mu \leftrightarrow \nu)\\ u_{\mu\nu}u^{\mu\nu} = -2 A$$

#### Flat spacetime vs. Riemann tensor

N.S condition for spacetime to be flat ($g_{\mu\nu} \to \eta_{\alpha\beta}$), is:

1. $R^\lambda_{\mu\nu\sigma}(x) = 0$
2. $g_{\mu\nu}$ has the same signature as $\eta_{\alpha\beta}$.

### Some identities of Riemann tensor

Recall that $R^\lambda{}_{\mu\nu\kappa}=\left(\partial_\kappa \Gamma^\lambda_{\mu\nu} + \Gamma^\eta_{\mu\nu}\Gamma^\lambda_{\kappa\eta}\right)_{[\nu,\kappa]}$ and $\Gamma^\lambda_{\mu\nu}=-\frac{1}{2}g^{\lambda\sigma}(\partial_\mu g_{\sigma\nu} + \partial_\nu g_{\sigma_mu} - \partial_\sigma g_{\mu\nu})$. Then:

1. (Antisymmetric on both pairs, Skew symmetry)

$$R_{\lambda\mu\nu\kappa}=-R_{\mu\lambda\nu\kappa}=-R_{\lambda\mu\kappa\nu}\,.$$

2. (Swap two pairs, Interchange symmetry)

$$R_{\lambda\mu\nu\kappa}=R_{\nu\kappa\lambda\mu}.$$

3. (First Bianchi identity)

$$R_{\lambda\mu\nu\kappa}+R_{\lambda\nu\kappa\mu}+R_{\lambda\kappa\mu\nu}=0$$

or

$$R_{\lambda[\mu\nu\kappa]}=0$$

4. (Bianchi identity)

$$(R^{\mu\nu}-\frac{1}{2}g^{\mu\nu}R)_{;\mu}=0$$

### General Relativity 08Last updated: 2016-08-21 22:20:50 PDT.

Reference: UCI OpenCourseWare - GR 09 (Playlist)

### Covariant differentiation

We are attempting to find a new definition of derivative that is covariant to coordinate change. Consider a vector:

\begin{align*} V'^\mu &= \frac{\partial x'^\mu}{\partial x^\nu}\, V^\nu\\ \end{align*}

Its partial derivative:

\begin{align*} \frac{\partial V'^\mu}{\partial x'^\lambda} &= \frac{\partial}{\partial x'^\lambda}\left(\frac{\partial x'^\mu}{\partial x^\nu}\, V^\nu\right)\\ &= \frac{\partial x'^\mu}{\partial x^\nu}\, \frac{\partial V^\nu}{\partial x'^\lambda} + \frac{\partial^2 x'^\mu}{\partial x'^\lambda\partial x^\nu}\, V^\nu\\ &= \frac{\partial x'^\mu}{\partial x^\nu}\, \frac{\partial x^\rho}{\partial x'^\lambda}\, \frac{\partial V^\nu}{\partial x^\rho} + \frac{\partial^2 x'^\mu}{\partial x^\nu\partial x^\rho} \,\frac{\partial x^\rho}{\partial x'^\lambda} \, V^\nu\\ \end{align*}

The appearance of the second derivative indicates that this differentiation is not covariant. Goes back to affine connection:

\begin{align*} \Gamma'^\mu_{\lambda\kappa}\,V'^\kappa &=\frac{\partial x'^\mu}{\partial x^\nu}\, \frac{\partial x^\rho}{\partial x'^\lambda}\, \frac{\partial x^\sigma}{\partial x'^\kappa}\, \Gamma^\nu_{\rho\sigma}\, V'^\kappa + \frac{\partial x'^\mu}{\partial x^\rho} \, \frac{\partial^2 x^\rho}{\partial x'^\lambda \partial x'^\kappa}\, V'^\kappa \\ &=\frac{\partial x'^\mu}{\partial x^\nu}\, \frac{\partial x^\rho}{\partial x'^\lambda}\, \Gamma^\nu_{\rho\sigma}\, V^\sigma + \frac{\partial x'^\mu}{\partial x^\rho}\, \frac{\partial}{\partial x'^\kappa} \left(\frac{\partial x^\rho}{\partial x'^\lambda} \right)\, V'^\kappa \\ \end{align*}

\begin{align*} \frac{\partial V'^\mu}{\partial x'^\lambda} + \Gamma'^\mu_{\lambda\kappa}\,V'^\kappa &= \frac{\partial x'^\mu}{\partial x^\nu}\,\frac{\partial x^\rho}{\partial x'^\lambda}\, \left(\frac{\partial V'^\mu}{\partial x'^\lambda} + \Gamma'^\mu_{\lambda\kappa}\,V'^\kappa\right) + V^\nu\cdot\frac{\partial}{\partial x^\nu} \left( \frac{\partial x'^\mu}{\partial x^\rho}\, \frac{\partial x^\rho}{\partial x'^\lambda} \right)\\ &= \frac{\partial x'^\mu}{\partial x^\nu}\,\frac{\partial x^\rho}{\partial x'^\lambda}\, \left(\frac{\partial V'^\mu}{\partial x'^\lambda} + \Gamma'^\mu_{\lambda\kappa}\,V'^\kappa\right) + \cancel{\frac{\partial \delta^\mu{}_\lambda}{\partial x^\nu} V^\nu} \end{align*}

An alternative derivation can be found here: Wikipedia.

So we can define the covariant derivative as:

$$V^\mu{}_{;\lambda} \stackrel{def} = \frac{\partial V^\mu}{\partial x^\lambda} + \Gamma^\mu_{\lambda\kappa}V^\kappa$$

or

$$\nabla_\lambda V^\mu{} \stackrel{def} = \partial_\lambda V^\mu + \Gamma^\mu_{\lambda\kappa}V^\kappa$$

It is a mixed tensor.

Similarly for covariant vector:

$$V_{\mu;\lambda} \stackrel{def}= \frac{\partial V_\mu}{\partial x^\lambda} - \Gamma^\kappa_{\mu\lambda}V_\kappa$$

More generally:

\begin{align*} T^{\alpha_1\alpha_2...\alpha_r}{}_{\beta_1\beta_2...\beta_s;\gamma} \stackrel{def}= T^{\alpha_1\alpha_2...\alpha_r}{}_{\beta_1\beta_2...\beta_s,\gamma} &+\Gamma^{\alpha_1}_{\delta\gamma}T^{\delta\alpha_2...\alpha_r}{}_{\beta_1\beta_2...\beta_s,\gamma} + ... + \Gamma^{\alpha_r}_{\delta\gamma}T^{\alpha_1\alpha_2...\delta}{}_{\beta_1\beta_2...\beta_s,\gamma}\\ &-\Gamma^{\delta}_{\beta_1\gamma}T^{\alpha_1\alpha_2...\alpha_r}{}_{\delta\beta_2...\beta_s,\gamma} -...-\Gamma^{\delta}_{\beta_s\gamma}T^{\alpha_1\alpha_2...\alpha_r}{}_{\beta_1\beta_2...\delta,\gamma} \end{align*}

Apply this to metric:

\begin{align*} g_{\mu\nu;\lambda}&=\frac{\partial g_{\mu\nu}}{\partial x_\lambda} -\Gamma^\rho_{\lambda\mu}g_{\rho\nu} -\Gamma^\rho_{\lambda\nu}g_{\mu\rho}\\ &=0 \end{align*}

Since $g_{\mu\nu}$ is constant in freely falling frame, the covariant derivative has to be $0$ everywhere.

Taking covariant derivative commutes with raising/lowering indices:

$$(g^{\mu\nu}V_\nu)_{;\lambda}=g^{\mu\nu}V_{\nu;\lambda}$$

#### Recipes for equations in GR

1. Take an equations valid in SR;
2. $\eta_{\alpha\beta} \to g_{\alpha\beta}$;
3. $\partial_\mu \to {}_{;\mu}(\text{or }\nabla_\lambda)$.

"Minimal substitution"

$$p_\mu \to p_\mu - \frac{e}{c}A_\mu$$

in $H=\frac{1}{2m}(\vec p - \frac{e}{c} \vec A)^2$ and

$$\partial_\mu \to \partial_\mu - \frac{ie}{\hbar c}A_\mu$$

This substitution implies the Gauge invariance.

#### More properties of Covariant derivative

Covariant divergence:

\begin{align*} V^\mu{}_{;\mu} &= V^\mu{}_{,\mu}+\Gamma^\mu_{\mu\lambda}V^\lambda\\ &=\partial_\mu V^\mu + \frac{1}{2}g^{\mu\rho}\frac{\partial g_{\rho\mu}}{\partial x^\lambda} \\ &=\frac{1}{\sqrt{g}}\partial_\mu(\sqrt{g}V^\mu)\\ \end{align*}

useful for conservation laws:

\begin{align*} \int\,d^4 \sqrt{g}\,V^\mu{}_{;\mu} &= \int\,d^4\,\partial_\mu(\sqrt{g}V^\mu) \\ &= \text{boundary term} \end{align*}

### Covariant differentiation along a curve

Transporting an vector along a curve $x(\tau)$, we get a family of vectors:

$$A'^\mu(\tau) = \left.\frac{\partial x'^\mu}{\partial x^\nu}\right|_{x=x(\tau)}\,A^\nu(\tau)$$

differentiation with respect to $\tau$.

\begin{align*} \frac{dA'^\mu}{d\tau}&=\frac{d}{d\tau}\left(\frac{\partial x'^\mu}{\partial x^\nu}A^\nu\right) \\ &= \frac{\partial x'^\mu}{\partial x^\nu}\frac{dA^\nu}{d\tau}+ \frac{\partial^2 x'^\mu}{\partial x^\nu\partial x^\lambda} \frac{dx^\lambda}{d\tau} A^\nu \end{align*}

$$\frac{DA'^\mu}{D\tau} \stackrel{def}= \frac{dA^\mu}{d\tau} + \Gamma^\mu_{\nu\lambda}\frac{dx^\lambda}{d\tau}A^\nu$$

This leads to the notion of parallel transport.

#### Parallel transport

A path that $A^\mu$ does not change in the frame attached to particle. Which is supposed to be local inertial frame.

In that local inertial frame:

1. $\Gamma = 0$ (F.F frame). (This has not been mentioned before, but can be found easily).
2. $\frac{dA^\mu}{d\tau} = 0$ if $A^\mu$ does not change direction.
3. $\frac{DA^\mu}{D\tau} = 0$ in any frame. This means:

$$\frac{dA^\mu}{d\tau}=-\Gamma^{\mu}_{\nu\lambda}\frac{dx^\lambda}{d\tau}A^\nu$$

The geodesic equation is of this type:

\begin{align*} \frac{d^2x^\mu}{d\tau^2} + \Gamma^\mu_{\lambda\sigma}\frac{dx^\lambda}{d\tau}\frac{dx^\sigma}{d\tau} = 0\\ \frac{D}{D\tau}\left(\frac{dx^\mu}{d\tau}\right)=0 \end{align*}

Geodesic is not only the shortest path, but also the straightest path.

Parallel transporting a vector along a closed curve, the result vector may not coincide with the original one.

#### Integrating equation of parallel transport

$$A^\mu \to S_\mu$$

Spin is a covariant vector, therefore

\begin{align*} \frac{dS_\mu}{d\tau} = \Gamma^\lambda_{\mu\nu}\,\frac{dx^\nu}{d\tau}\,S_\lambda \\ dS_\mu = \Gamma^\lambda_{\mu\nu}\,S_\lambda\,dx^\nu \\ S' = S + dS = (1+\Gamma^\lambda_{\mu\nu}\,dx^\nu)S \\ \end{align*}

This can be integrated in terms of a rotation matrix

\begin{align*} S'^\mu &= R^\mu{}_\nu S^\nu\\ R^\mu{}_\nu=[\mathcal P \exp\left(-\oint dx'^\lambda\, \Gamma^\mu_{\lambda\nu} \right)] \end{align*}

where $\mathcal P$ is the path-ordering meta operator.

Parallel transport can be used to detect curvature.

#### Pursue analogy with EM

1. Local gauge invariance in E.M. with EM field $A_\mu(x)$, and matter field $\psi(x)$. Equations are preserved if:

\left\{\begin{align*} A_\mu(x) &\to A_\mu(x) + \frac{\partial}{\partial^\mu}\theta(x)\\ \psi(x) &\to e^{ie\theta(x)}\cdot \psi(x) \end{align*}\right.

where $\theta(x)$ is called the gauge function. A useful tool to deal with gauge invariance is to introduce a "gauge invariant derivative":

\begin{align*} D_\alpha \psi(x) &\stackrel{def} = \left(\frac{\partial}{\partial x^\alpha}-ie A_\alpha(x)\right)\psi(x)\\ D_\alpha (e^{ie\theta(x)}\psi(x)) &= e^{ie\theta(x)}\left( \frac{\partial}{\partial x^\alpha}\psi(x) - ie A_\alpha(x)\psi(x) \right)\\ &= e^{ie\theta(x)}D_\alpha(\psi(x))\\ \end{align*}

So when we construct the Lagrangian density $\mathcal L$, we can use this derivative as the first derivative and the equation will be invariant modular a rotation factor:

$$\mathcal L \propto |D_\mu\psi(x)|^2 \quad (\text{Higgs field}) \\ \mathcal L \propto \bar\psi(x)\gamma^\mu D_\mu\psi(x) \quad (\text{Dirac field}) \\$$

It is alway the covariant derivative that makes sure the equations is gauge invariant. (Noether's theorem)

The local gauge invariance leads to conserved EM current:

\begin{align*} J_\alpha(x) &= -ie\left[\psi^+D_\alpha \psi - \psi(D_\alpha \psi)^+\right]\\ \frac{\partial}{\partial x^\alpha} J^\alpha(x) &= 0\\ \end{align*}

Similarly, the local invariance of GR give raise to a local conserved current, that is the energy-momentum tensor.

$$\begin{array}{|c|c|c|} \hline \text{Theory}&\text{EM}&\text{GR}\\\hline \text{Gauge field} & A^\mu & \Gamma^\lambda_{\mu\nu}\\\hline \text{Fundamental field} & A^\mu & g_{\mu\nu}\\\hline \text{Field strength} & F_{\mu\nu} & R_{\mu\nu}\\\hline \text{Invariance}&\text{Phase}&\text{Coordinate}\\\hline \text{Conserved quantity}&\text{charge}&\text{energy-momentum tensor}\\\hline \end{array}$$

The local invariance can not be consistent unless the boson is massless.

### General Relativity 07Last updated: 2016-08-21 22:20:50 PDT.

Reference: UCI OpenCourseWare - GR 08 (Playlist)

## Principle of General Covariance

1. Starting from an equation that is valid in an inertial frame.
2. Generalize them using symmetry (invariance).

We used Lorentz invariant equation to describe SR previously. In GR, we also started with a equation in an inertial frame and generalized it to an arbitrary lab frame. The difference, is that the coordinate transformation in SR is global and in GR is local. This is similar to QED. Schroedinger Equation and Dirac Equation are invariant under phase rotation. When we promote that to local symmetry in QED, that requires us to have a electromagnetic field. That's the reason when we couple the particle into the EM field, we have to change the derivative to:

$$\partial_\mu \to \partial_\mu -\frac{ie}{\hbar c}A_\mu$$

(This is a part that I don't understand.)

Similar to the EM field in QED, we have two field $g_{\mu\nu}(x)$ and $\Gamma^\lambda{}_{\mu\nu}(x)$, and they are gauge fields. They ensure that the theory has the local invariance.

We are going to use the principle of general covariance to construct equations that obeys GR local invariance, similar to what we do in QED.

Noether's theorem states that each continuous symmetry corresponds to a conserved quantity. This is also true to general relativity. The local invariance corresponds to energy momentum conservation.

From the previous observation, we can describe the motion of a free particle basing on the metric tensor $g_{\mu\nu}$ and affine connection $\Gamma^\lambda_{\mu\nu}$. The classical limit of metric tensor $g_{00}=-(1+2\phi)$, also we have Gaussian's law of gravity $-\nabla\cdot\phi=4\pi G\rho$ where $\rho$ is the density. This is, of course, not Lorentz invariant. We can guess the form of the equation that relates the metric tensor and energy-momentum tensor:

$$\cancel{\square g_{\mu\nu} \stackrel{?}= 8 \pi G T_{\mu\nu}}\,,$$

which is similar to EM but this doesn't work because $\square g_{\mu\nu}=0$. It turns out that eventually it is the affine connection and Riemann tensor that show up in the equation, which is also non-linear.

(Feynmann) Einstein's equation is the only one that can be written down if massless spin-2 particle is allowed.

### Vectors and Tensors

Consider general coordinate transformation:

$$x^\mu = x'^\mu$$

$$\begin{array}{|r|c|c|} \hline \text{Scalars} & \pi, \tau, ... & \text{(unchanged)} \\ \hline \text{Contravariant vector} & V^\mu(x) & V'_\mu = \frac{\partial x'^\mu}{\partial x^\nu} V^\nu \\ \hline \text{Covariant vector} & U_\mu(x) & U'_\mu = \frac{\partial x^\nu}{\partial x'^\mu} U_\nu \\ \hline \text{Derivative of a scala field} & \frac{\partial\phi}{\partial x'^\mu} & \frac{\partial\phi}{\partial x'^\mu} = \frac{\partial x^\nu}{\partial x'^\mu}\frac{\partial\phi}{\partial x^\nu} \\ \hline \text{Tensor}&T^\mu{}_\nu{}^\lambda& T^\mu{}_\nu{}^\lambda= \frac{\partial x^{\mu'}}{\partial x^{\mu}} \frac{\partial x^\nu}{\partial x^{\nu'}} \frac{\partial x^{\lambda'}}{\partial x^{\lambda}} T'^{\mu'}{}_{\nu'}{}^{\lambda'} \\ \hline \end{array}$$

#### Metric tensor

$$g_{\mu\nu}(x) \stackrel{def} = \eta_{\alpha\beta}\frac{\partial \xi^\alpha}{\partial x^\mu}\frac{\partial \xi^\beta}{\partial x^\nu}\\ g'_{\mu\nu}(x) = g_{\rho\sigma}\frac{\partial x^\rho}{\partial x^\mu}\frac{\partial x^\sigma}{\partial x^\nu}\\$$

#### Inverse of metric tensor

$$g^{\mu\lambda}g_{\nu\mu} = \delta^\lambda{}_\nu\\ g'^{\mu\lambda} = \frac{\partial x'^\mu}{\partial x^\rho} \frac{\partial x'^\lambda}{\partial x^\sigma} g^{\rho\sigma}$$

#### Kronecker symbol

$\delta^\mu{}_\nu$ is mixed.

\begin{align*} \frac{\partial x'^\mu}{\partial x^\rho} \frac{\partial x^\sigma}{\partial x'^\nu} \delta^\rho{}_\sigma &= \frac{\partial x'^\mu}{\partial x^\rho} \frac{\partial x^\rho}{\partial x'^\nu} \\ &= \frac{\partial x'^\mu}{\partial x'^\nu}\\ &= \delta^\mu{}_\nu \end{align*}

#### Write down the covariant equation

Covariant equation only contains:

• Scalar, Vectors, Tensors
• Linear Combination
• Direct product
• Contraction

For example:

$$R^{\nu\rho}{}_\sigma = g^{\mu\nu} S_\mu{^\rho}_{\sigma}$$

If certain equation can be written as equality of tensors, it will look the same in all kinds of coordinate system.

The metric tensor is useful for lowering and raising indices:

$$g^{\rho\sigma}g_{\sigma\mu}T^{\mu\nu} = T^{\rho\nu} \\ g_{\rho\nu}g_{\sigma\mu}T^{\mu\nu} = T_{\sigma\rho}$$

#### Quantities that are almost tensors (Tensor Densities)

$$g(x) = - \det g_{\mu\nu}$$

(Convention is to put minus before metric tensor.) $g$ is not a scalar:

\begin{align*} g'_{\rho\sigma} &= \frac{\partial x'_\rho}{\partial x_\mu} \frac{\partial x'_\sigma}{\partial x_\nu} g_{\mu\nu}\\ \det g'_{\mu\nu} &= \left|\frac{\partial x_\rho}{\partial x'_\mu}\right| \cdot \left|\frac{\partial x_\sigma}{\partial x'_\nu}\right| \cdot g \\ &=\left|\frac {\partial x}{\partial x'}\right|^2\cdot g \end{align*}

The determine of the metric is a scalar density.

Define "tensor density" of weight $w$:

$$\mathcal T'^{\mu}{}_{\nu} = \left|\frac{\partial x'}{\partial x}\right|^w \frac{\partial x'^\mu}{\partial x^\rho} \frac{\partial x^\sigma}{\partial x'^\nu} \mathcal T^\rho{}_\sigma$$

weight of metric density $w=-2$.

(!!) Volume element in 4d space time: $d^4x$ under $x^\mu \to x'^\mu$:

$$d^4 x' = \left|{\frac{\partial x'}{\partial x}}\right|d^4 x$$

$w=1$ Therefore it is not invariant. But if we multiply it with a tensor of weight $-1$, then it will become invariant, for example:

$$\sqrt{-g}\,d^4x \stackrel{def} = dV$$

Tensors so far: $g_{\mu\nu},g^{\mu\nu},\delta^\mu_\nu$, $\frac{1}{\sqrt{g}}\epsilon_{\mu\nu\lambda\sigma}(\text{Levi-Civita density})$.

Affine Connection:

\begin{align*} \Gamma^\lambda_{\mu\nu} &=\frac{\partial x^\lambda}{\partial \xi^\nu}\cdot \frac{\partial^2 \xi^\nu}{\partial x^\mu\partial x^\nu}\\ \frac{\partial}{\partial x}&=\frac{\partial x'}{\partial x}\frac{\partial }{\partial x'}\\ \frac{\partial}{\partial x'}&=\frac{\partial x}{\partial x'}\frac{\partial }{\partial x}\\ \\ \Gamma'^\lambda_{\mu\nu} &= \frac{\partial x'^\lambda}{\partial \xi^\nu} \cdot \frac{\partial^2 \xi^\nu}{\partial x'^\mu\partial x'^\nu}\\ &= \frac{\partial x^\rho}{\partial \xi^\nu}\frac{\partial x'^\lambda}{\partial x^\rho}\, \frac{\partial x^\tau}{\partial x'^\mu}\, \frac{\partial }{\partial x^\tau} \left( \frac{\partial x^\sigma}{\partial x'^\nu} \,\frac{\partial}{\partial x^\sigma} \xi^\nu \right) \\ &= \frac{\partial x^\rho}{\partial \xi^\nu} \frac{\partial x'^\lambda}{\partial x^\rho}\, \frac{\partial x^\tau}{\partial x'^\mu}\, \, \left( \frac{\partial^2 x^\sigma}{\partial x^\tau\partial x'^\nu}\, \frac{\partial}{\partial x^\sigma}\xi^\nu + \frac{\partial x^\sigma}{\partial x'^\nu} \frac{\partial^2}{\partial x^\sigma\partial x^\tau}\xi^\nu \right)\\ &= \frac{\partial x'^\lambda}{\partial x^\rho}\, \frac{\partial x^\tau}{\partial x'^\mu}\, \, \left( \frac{\partial x^\rho}{\partial \xi^\nu} \frac{\partial^2 x^\sigma}{\partial x^\tau\partial x'^\nu}\, \frac{\partial \xi^\nu}{\partial x^\sigma} + \frac{\partial x^\sigma}{\partial x'^\nu} \frac{\partial x^\rho}{\partial \xi^\nu} \frac{\partial^2 \xi^\nu}{\partial x^\sigma\partial x^\tau} \right)\\ &= \frac{\partial x'^\lambda}{\partial x^\rho}\, \frac{\partial x^\tau}{\partial x'^\mu}\, \, \left( \frac{\partial^2 x^\rho}{\partial x^\tau\partial x'^\nu}\, + \frac{\partial x^\sigma}{\partial x'^\nu} \Gamma^\rho_{\sigma\tau} \right)\\ &= \frac{\partial x'^\lambda}{\partial x^\rho}\, \frac{\partial x^\tau}{\partial x'^\mu}\, \frac{\partial x^\sigma}{\partial x'^\nu}\, \Gamma^\rho_{\sigma\tau} + \frac{\partial x'^\lambda}{\partial x^\rho}\, \frac{\partial^2 x^\rho}{\partial x'^\mu\partial x'^\nu}\,\\ \end{align*}

This is means affine connection is not a tensor.

This raises the concern that the equation of free fall:

$$\frac{d^2x^\mu}{d\tau^2}+\Gamma^\mu_{\lambda\sigma}\cdot\frac{dx^\mu}{d\tau}\cdot\frac{dx^\mu}{d\tau}=0\,,$$

may not be invariant. Fortunately, this equation is indeed invariant. Also the derivative of $x$ is tensor, the second derivative of $x$ is not a tensor. This miraculously reconcile the issue of affine connection.

### General Relativity 06Last updated: 2016-08-14 19:21:06 PDT.

Reference: UCI OpenCourseWare - GR 07 (Playlist)

### Some remarks (Mach's Principle)

Starting from 39:00.

My Note: I've never been convinced of Mach's Principle. In fact, there has been many null results in testing Mach's principle. I don't know this has to be a thing in the first place.

### Non-Cartitian coordinate system

Starting from 1:07:55.

Starting from an example:

$$\xi^\alpha = \begin{cases} x &=& r \sin \theta \cos \phi \\ y &=& r \sin \theta \sin \phi \\ z &=& r \cos \theta \end{cases} \\ x^\alpha = \begin{cases} r &=& \sqrt{x^2 + y^2 + z^2} \\ \theta &=& \cos^{-1}(\frac{z}{r})\\ \phi &=& \tan^{-1}(\frac{y}{x}) \end{cases} \\ \eta_{\alpha\beta} =\delta_{\alpha\beta}$$