General Relativity

Table of Contents

Notation

  • IFs is inertial frames

Newtonian spacetime

Notation

Stuff

Reformulation of Poisson laws:

\begin{equation*} \Delta \phi = 4 \pi G_{N} \rho \end{equation*}

in terms of the Newtonian spacetime curvature as

\begin{equation*} R_{00} = 4 \pi G_{N} \rho \end{equation*}

where $R_{00}$ is the 00-component of the Ricci tensor.

Then Newtonian spacetime is tuple $\big( M, \mathcal{O}, \mathcal{A}, \nabla, t \big)$ where

Absolute space

Absolute space at time $\tau$ is defined

\begin{equation*} S_{\tau} := \left\{ p \in M \mid t(p) = \tau \right\} \end{equation*}

and so (Newtonian) spacetime can be written as the disjoint union of the absolute spaces:

\begin{equation*} M = \bigsqcup_{\tau} S_{\tau} \end{equation*}

One way to visualize absolute time is as follows:

newtonian_spacetime_absolute_space_at_times.png

If absolute time didn't "flow uniformly", then we would allow something like the following, where we can have time standing completely still:

newtonian_spacetime_time_stands_still.png

Motivation

Newton's laws

  • Motion: $t \mapsto \mathbf{x}(t)$ and $m_I \ddot{\mathbf{x}} = \mathbf{F}$ where
    • $m_I$ is the inertial mass

  • Gravity: $\mathbf{F} = m_g \mathbf{g}$ where $\mathbf{g}(t, \mathbf{x}) = - \nabla \phi$, which satisfies the (gravitational) Poisson equation

    \begin{equation*} \nabla^2 \phi = 4 \pi G \rho \quad \implies \quad \phi = - \frac{GM}{r} \end{equation*}

    where

    • $G$ is Newton's gravitational constant
    • $\rho$ is density
    • $M$ is the mass of the "source"

  • Experiment: inertial mass is equivalent to gravitational mass, i.e.

    \begin{equation*} m_I = m_g \end{equation*}

    which is called the weak equivalence principle, thus

    \begin{equation*} \ddot{\mathbf{x}}(t) = \mathbf{g}\big(t, \mathbf{x}(t) \big) \end{equation*}

    Therefore we get the fact that locally, gravity is uniform acceleration

    • Non-uniform $\mathbf{g}$ gives rise to tidal forces. Follows from the deviation equation

      \begin{equation*} \begin{split} \ddot{z}(t) & = \mathbf{g}(t, \mathbf{x} + \mathbf{z}) - \mathbf{g}(t, \mathbf{x}) \\ &\approx \mathbf{z} \cdot \nabla \mathbf{g} \end{split} \end{equation*}

Special relativity

  • Spacetime $\mathbb{R}^4$ where points are events, with inertial frames

    \begin{equation*} \mathbf{x} = \big( ct, \mathbf{x} \big) \end{equation*}

    and we use the Minkowski metric

    \begin{equation*} I(\mathbf{x}, \mathbf{x}') = - c^2 \big( t - t' \big)^2 + \norm{ \mathbf{x} - \mathbf{x}' }^2 \end{equation*}

    (a quadratic form), which is invariant in all IFs. Notice $I \le 0$

  • Motion is described by wordlines $\mathbf{x}(\lambda)$
    • Free motion follow straight lines
  • IFs related by Poincare group (which leaves $I(\mathbf{x}, \mathbf{x}')$ invariant:
    • Rotations
    • Translations
    • Lorentz boosts (rotation between space and time)
  • Maxwell's eqns have PoincarĂ© symmetry, but NOT Galilean symmetry (which is what Newton's gravity follows)

gr-motivation.png

General relativity

  1. Laws of physics take same form in any coordinate system, i.e. invariant under general coord changes
    • Will obtain by formuating laws with tensor fields
  2. There exists lcoal IF with no gravity, where laws are as in special relativity (called the /strong equivalence principle)
    • Spacetime is (differentiable) manifold with Lorentzian metric

Postulates

Spacetime is a four-dimensional Lorentzian manifold $(M, g)$.

More specifically, (relativistic) spacetime is defined by the tuple $\big( M, \mathcal{O}, \mathcal{A}, \nabla, g, T \big)$ where

Let $\big( M, \mathcal{O}, \mathcal{A}^{\uparrow}, g \big)$ be a Lorentzian manifold.

Then a time-orientation is given by a smooth vector field $T$ that

  1. do not vanish anywhere, i.e. $T(f) \ne 0$ for all $f \in C^{\infty}(M)$
  2. $g(T, T) < 0$

Here $\mathcal{A}^{\uparrow}$ denotes an oriented atlas.

  • Comparing with Newtonian spacetime we have gone from a "time" $t$ to a metric $g$ and time-orientation $T$.
  • Reason for introducing a metric $g$ now is that we want some way of enforcing the fact that a particle cannot move faster than the speed of light

  • If we instead considered the metric $\diag(- c^2, 1, 1, 1)$ then we observe that

    \begin{equation*} g(T, T) = -c^2 + \underbrace{T^{\alpha} T_{\alpha}}_{= \norm{T}^2} \end{equation*}

    so if $g(T, T) < 0$, we are basically saying that the flow along $T$ has a "speed" which less than the speed of light (as we want).

    • Observe then that (with our convention), that $g(T, T) > 0$ is "illegal"
  • The above produces the well-known light-cone, where if we make a choice of time-orientation to enforce the fact that a particle cannot move into the past

light-cone.png

Connections and curvature

Notation

  • $\left\{ e_a \right\}$ denotes a basis of a vector fields on $M$
  • $\left\{ f^a \right\}$ denotes the dual basis of covector fields on $M4
  • In GR, the term connection will often be used to refer to the tensor $\nabla$; this is because of how it relates to the connection 1-form

  • Covariant derivative as a (1, 1)-tensor field

    \begin{equation*} \Big( \nabla Y \Big)(\lambda, X) = \left\langle \lambda, \nabla_X Y \right\rangle \end{equation*}

    with components

    \begin{equation*} \tensor{\big( \nabla Y \big)}{^{a}_{b}} =: \nabla_b Y^a = \tensor{Y}{^{a}_{;b}} \end{equation*}

  • Components of connection

    \begin{equation*} \Gamma^c_{ba} e_c := \nabla_a e_b = \nabla_{e_a} e_b \end{equation*}
  • Symmetric summation $T_{(ab)} = \frac{1}{2} (T_{ab} + T_{ba})$ and anti-symmetric $T_{[ab]} = \frac{1}{2} (T_{ab} - T_{ba})$

  • Torsion is denoted $T \in T^1_2$ or in components $\tensor{T}{^{a}_{bc}}$ and is given by

    \begin{equation*} T(X, Y) = \nabla_X Y - \nabla_Y X - \comm{X}{Y} \end{equation*}

    and in components

    \begin{equation*} \tensor{T}{^{a}_{bc}} = - 2 \tensor{\Gamma}{^{a}_{[bc]}} \end{equation*}

Covariant derivative

Let $\left\{ e_a \right\}$ be a basis of vector fields on $M$. The components of the connection $\tensor{\Gamma}{^{a}_{bc}}$ in a basis is defined by

\begin{equation*} \nabla_a e_a := \nabla_{e_a} e_b = \tensor{\Gamma}{^{c}_{ba}} e_c \end{equation*}

Letting

\begin{equation*} X = X^a e_a \quad \text{and} \quad Y = Y^a e_a \end{equation*}

the covariant derivative can be written

\begin{equation*} \begin{split} \nabla_X Y &= \nabla_{X^a e_a} \big( Y^b e_b \big) \\ &= X^a \nabla_{e_a} \big( Y^b e_b \big) \\ &= X^a \Big( e_a(Y^b) e_b + Y^b \nabla_a e_b \Big) \\ &= X^a \Big( e_a (Y^c) + \tensor{\Gamma}{^{c}_{ba}} Y^b \Big) e_c \end{split} \end{equation*}

and so

\begin{equation*} \nablaa Y^c = \left\langle f^c, \nabla_{e_a} Y \right\rangle = e_a \big( Y^c \big) + \tensor{\Gamma}{^{c}_{ba}} Y^b \end{equation*}

where $\left\{ f^a \right\}$ is the dual basis of $\left\{ e_a \right\}$. In a particular coordinate basis $\left\{ e_i = \partial_i := \pdv{}{x^i} \right\}$

\begin{equation*} \nabla_i Y^j = \partial_i Y^j + \tensor{\Gamma}{^{j}_{ki}} Y^k \end{equation*}

Under a basis change $\tilde{e}_a = \tensor{\big( A^{-1} \big)}{^{b}_{a}} e_b$ and $\tilde{f}^a = \tensor{A}{^{a}_{b}} f^b$, we have

\begin{equation*} \tensor{\tilde{\Gamma}}{^{a}_{bc}} = \tensor{A}{^{a}_{d}} \tensor{\big( A^{-1} \big)}{^{e}_{b}} \tensor{\big( A^{-1} \big)}{^{f}_{c}} \tensor{\Gamma}{^{d}_{ef}} + \tensor{A}{^{a}_{d}} \tensor{\big( A^{-1} \big)}{^{f}_{c}} e_f \Big( \tensor{\big( A^{-1} \big)}{^{d}_{b}} \Big) \end{equation*}

whre the basis change coefficients $\tensor{A}{^{a}_{b}} \in C^{\infty}(M)$.

In particular, under a change of coordinate basis, this becomes

\begin{equation*} \tensor{\tilde{\Gamma}}{^{i}_{jk}} = \pdv{\tilde{x}^i}{x^l} \pdv{x^m}{\tilde{x}^j} \pdv{x^n}{\tilde{x}^k} \tensor{\Gamma}{^{l}_{mn}} + \pdv{\tilde{x}^i}{x^l} \pdv[2]{x^l}{\tilde{x}^j}{\tilde{x}^k} \end{equation*}

We use $i, j, k$ to denote fixed indices, and $a, b, c$ for contractible ones.

\begin{equation*} \begin{split} \tensor{\tilde{\Gamma}}{^{i}_{jk}} &= \left\langle \tilde{f}^i, \nabla_{\tilde{e}_k} \tilde{e}_j \right\rangle \\ &= \left\langle \tensor{A}{^{i}_{a}} f^a, \nabla_{\tensor{(A^{-1})}{^{c}_{k}} e_c} \tensor{\big( A^{-1} \big)}{^{b}_{j}} e_b \right\rangle \\ &= \tensor{A}{^{i}_{a}} \tensor{\big( A^{-1} \big)}{^{c}_{k}} \left\langle f^a, \nabla_c \tensor{\big( A^{-1} \big)}{^{b}_{j}} e_b \right\rangle \\ &= \tensor{A}{^{i}_{a}} \tensor{\big( A^{-1} \big)}{^{c}_{k}} \Big[ \tensor{\big( A^{-1} \big)}{^{b}_{j}} \underbrace{\left\langle f^a, \nabla_c e_b \right\rangle}_{=: \tensor{\Gamma}{^{a}_{bc}}} + \left\langle f^a , e_b \right\rangle e_c \Big( \tensor{\big( A^{-1} \big)}{^{b}_{j}} \Big)\Big] \\ &= \tensor{A}{^{i}_{a}} \tensor{\big( A^{-1} \big)}{^{c}_{k}} \tensor{\big( A^{-1} \big)}{^{b}_{j}} \tensor{\Gamma}{^{a}_{bc}} + \tensor{A}{^{i}_{a}} \tensor{\big( A^{-1} \big)}{^{c}_{k}} \tensor{\delta}{^{a}_{b}} e_c \Big( \tensor{\big( A^{-1} \big)}{^{b}_{j}} \Big) \\ &= \tensor{A}{^{i}_{a}} \tensor{\big( A^{-1} \big)}{^{c}_{k}} \tensor{\big( A^{-1} \big)}{^{b}_{j}} \tensor{\Gamma}{^{a}_{bc}} + \tensor{A}{^{i}_{a}} \tensor{\big( A^{-1} \big)}{^{c}_{k}} e_c \Big( \tensor{\big( A^{-1} \big)}{^{a}_{j}} \Big) \end{split} \end{equation*}

If we then let $i, j, k = a, b, c$ and $a, b, c = d, e, f$ we obtain our proof (up to commutation between $C^{\infty}$ functions, so we're fine).

Moreover, in a specific chart $(U, x)$ and changed basis $(U, \tilde{x})$ we simply have

\begin{equation*} \tensor{A}{^{i}_{a}} = \pdv{x^a}{\tilde{x}^i} \quad \text{and} \quad \tensor{\big( A^{-1} \big)}{^{c}_{k}} = \pdv{\tilde{x}^c}{x^k} \end{equation*}

Substituting back into the above, we obtain our result.

Let $\nabla$ be a connection on a smooth manifold $M$.

A vector field $Y$ is said to be parallely transported along a curve with tangent vector field $X$ if

\begin{equation*} \nabla_X Y = 0 \end{equation*}

The parallel transport condition in components reads

\begin{equation*} \begin{split} 0 &= \big( \nabla_X Y \big)^a \\ &= X^b \nabla_b Y^a + \tensor{\Gamma}{^{a}_{cb}} Y^c X^b \\ &= X(Y^a) + \tensor{\Gamma}{^{a}_{cb}} Y^c X^b \end{split} \end{equation*}

Along an integral curve $X$ we have $X(f) = \dv{f}{t}$, and so we can write

\begin{equation*} \dv{Y^a}{t} + \tensor{\Gamma}{^{a}_{cb}} Y^c X^b = 0 \end{equation*}

A geodesic is an integral curve of a vector field $X$ that satisfies

\begin{equation*} \nabla_X X = 0 \end{equation*}

In other words, a geodesic vector field is parallely transported wrt. itself.

In a coordinate basis, the integral curves satisfy $\dv{x^i}{t} = X^i \big( x(t) \big)$, and so the geodesic eqation is

\begin{equation*} \dv[2]{x^i}{t} + \tensor{\Gamma}{^{i}_{jk}} \Big( x(t) \Big) \dv{x^j}{t} \dv{x^k}{t} = 0 \end{equation*}

Consider a geodesic $t \mapsto \gamma(t)$ with tangent field $X$.

Suppose we reparametrize the curve so that $t = t(s)$ with $\dv{t}{s} > 0$. The tangent field of the curve $s \mapsto \gamma \big( t(s) \big)$ is then given by $Y = g X$ where $g = \dv{t}{s}$. Thus,

\begin{equation*} \nabla_Y Y = \nabla_{g X} (g X) = g^2 \underbrace{\nabla_X X}_{= 0} + g \underbrace{X(g)}_{= \dv{g}{t} =: f} X = f \underbrace{g X}_{ = Y} = f Y \end{equation*}

Thus, the equation $\nabla_Y Y = fY$ describes the same geodesic curve in a different parametrization.

An affine parameter is one for which $f = 0$. We deduce that this means that $t(s) = a s + b$ for $a \mathbb{R}^{ + }$ and $b \in \mathbb{R}$.

Let $p \in U \subset M$ open for which $\exp$ is a bijection.

Normal coordinates at $p$, of a point $q \in U$, are given by $\big( X_p^i \big)$ where $X_p^i$ are the components of

\begin{equation*} X_p = \exp^{-1}(q) \in T_p M \end{equation*}

in a basis $\left\{ e_i \right\}$.

Consider the geodesic through $p$ and $q = \exp(X_p)$.

In normal coords. the geodesic takes the form

\begin{equation*} x^i(t) = t X_p^i \end{equation*}

Inserting this into the geodesic equation and evaluating at $t = 0$, we deduce that

\begin{equation*} \tensor{\Gamma}{^{i}_{jk}} \big( p \big) X_p^j X_p^k = 0 \end{equation*}

Hence

\begin{equation*} \tensor{\Gamma}{^{i}_{(jk)}} \big( p \big) X_p^j X_p^k = 0 \end{equation*}

Since this is true for all $X_p$ in an open set $U \ni 0$, we have our proof.

Torsion and curvature

Torsion

Let $M$ be a smooth manifold with a connection $\nabla$.

The torsion $T$ is a $(1, 2)$ tensor defined by

\begin{equation*} T(X, Y) = \nabla_X Y - \nabla_Y X - \comm{X}{Y} \end{equation*}

where $X, Y$ are smooth vector fields.

One can easily check that this is indeed a tensor.

Componentwise,

\begin{equation*} \begin{split} \tensor{T}{^{i}_{jk}} &= \left\langle f^i, T(e_j, e_k) \right\rangle \\ &= \left\langle f^i, \nabla_j e_k - \nabla_k e_j \right\rangle \\ &= \tensor{\Gamma}{^{i}_{kj}} - \tensor{\Gamma}{^{i}_{jk}} \\ &= - 2 \tensor{\Gamma}{^{i}_{[jk]}} \end{split} \end{equation*}

Geodesics are determined by the symmetric part of the connection components $\tensor{\Gamma}{^{a}_{(bc)}}$, thus the torsion does not affect the geodesics!

Let $M$ be a manifold with a connection $\nabla$.

For any function $f$,

\begin{equation*} \nabla_a \nabla_b f - \nabla_b \nabla_a f = - \tensor{T}{^{c}_{ab}} \nabla_c f \end{equation*}

We prove this by working in a basis $\left\{ e_i \right\}$.

Since $\nabla_i f = \partial_i f$, the covariant derivative of the covector $\nabla f$ is

\begin{equation*} \nabla_i \nabla_j f = \partial_i \partial_j f - \tensor{\Gamma}{^{k}_{ji}} \partial_k f \end{equation*}

Therefore, antisymmetrising, we get

\begin{equation*} \nabla_i \nabla_j f - \nabla_j \nabla_i f = 2 \tensor{\Gamma}{^{k}_{[ij]}} \partial_k f = - \tensor{T}{^{k}_{ij}} \nabla_k f \end{equation*}

Since this is a relation between tensors, it follows that it must hold in any arbitrary basis.

Curvature

The Riemann curvature of a connection $\nabla$ is a $(1, 3) \text{-tensor field}$ defined by

\begin{equation*} R(X, Y) Z = \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{[X, Y]} Z \end{equation*}

where $X, Y, Z$ are smooth vector fields.

One can verify that this is indeed a tensor by checking that it's linear in $Y$ and $Z$.

The Riemann tensor is the (1, 3) tensor defined by $\big( \lambda, Z, X, Y \big) \mapsto \left\langle \lambda, R(X, Y) Z \right\rangle$, where the ordering of the arguments is standard notation. The components are then defined

\begin{equation*} \tensor{R}{^{i}_{jkl}} = \partial_k \tensor{\Gamma}{^{i}_{jl}} - \partial_l \tensor{\Gamma}{^{i}_{jk}} + \tensor{\Gamma}{^{m}_{jl}} \tensor{\Gamma}{^{i}_{mk}} - \tensor{\Gamma}{^{m}_{jk}} \tensor{\Gamma}{^{i}_{ml}} \end{equation*}

The component-wise expression for the Riemann tensor can be seen as follows:

\begin{equation*} \tensor{R}{^{i}_{jkl}} = \left\langle f^i, R(e_k, e_l) e_j \right\rangle \end{equation*}

where we first observe that

\begin{equation*} \begin{split} R(e_k, e_l) e_j &= \nabla_k \nabla_l e_j - \nabla_l \nabla_k e_j \\ &= \nabla_k \big( \tensor{\Gamma}{^{m}_{jl}} e_m \big) - \nabla_l \big( \tensor{\Gamma}{^{m}_{jk}} e_m \big) \\ &= \partial_k \big( \tensor{\Gamma}{^{m}_{jl}} \big) e_m + \tensor{\Gamma}{^{m}_{jl}} \tensor{\Gamma}{^{n}_{mk}} e_n - \partial_l \big( \tensor{\Gamma}{^{m}_{jk}} \big) e_m - \tensor{\Gamma}{^{m}_{jk}} \tensor{\Gamma}{^{n}_{ml}} e_n \end{split} \end{equation*}

from which the Riemann tensor $\tensor{R}{^{i}_{jkl}}$ immediately follows.

Let $\nabla$ be a torsionless connection.

For any vector field $Z$,

\begin{equation*} \nabla_c \nabla_d Z^a - \nabla_d \nabla_c Z^a = \tensor{R}{^{a}_{bcd}} Z^b \end{equation*}

This is called the Ricci identity.

First, observe that for a torsionless connection $\nabla$, we have

\begin{equation*} \comm{X}{Y}^a = X^b \nabla_b Y^a - Y^b \nabla_b X^a \end{equation*}

Now

\begin{equation*} \begin{split} \tensor{R}{^{a}_{bcd}} X^c Y^d Z^b &= X^c \nabla_c \Big( Y^d \nabla_d Z^a \Big) - Y^d \nabla_d \Big( X^c \nabla_c Z^a \Big) - \comm{X}{Y}^b \nabla_b Z^a \\ &= X^c Y^d \Big( \nabla_c \nabla_d Z^a - \nabla_d \nabla_c Z^a \Big) + \Big( X^c \nabla_c Y^b - Y^c \nabla_c X^b - \comm{X}{Y}^b \Big) \nabla_b Z^a \\ &= X^c Y^d \Big( \nabla_c \nabla_d Z^a - \nabla_d \nabla_c Z^a \Big) \end{split} \end{equation*}

Since this holds for all $X, Y$, we conclude our proof.

Let $\nabla$ be a torsionless connection.

For any covector field $\lambda$

\begin{equation*} \nabla_c \nabla_d \lambda_a - \nabla_d \nabla_c \lambda_a = - \tensor{R}{^{b}_{acd}} \lambda_b \end{equation*}

Ricci identity (for vectors) says that for some vector field $Z$ we have

\begin{equation*} \nabla_c \nabla_d Z^a - \nabla_d \nabla_c Z^a = \tensor{R}{^{a}_{bcd}} Z^b \end{equation*}

We can write $Z^a = g^{ai} Z_i$, which gives us

\begin{equation*} \nabla_c \nabla_d \big( g^{ia} Z_i \big) - \nabla_d \nabla_c \big( g^{ia} Z_i \big) = \tensor{R}{^{a}_{bcd}} g^{bj} Z_j \end{equation*}

We have

\begin{equation*} \nabla_c \nabla_d \big( g^{ia} Z_i \big) = \big( \nabla_c \nabla_d g^{ai} \big) Z_i + g^{ai} \big( \nabla_c \nabla_d Z_i \big) \end{equation*}

so

\begin{equation*} \begin{split} \big( \nabla_c \nabla_d g^{ai} \big) Z_i + g^{ai} \big( \nabla_c \nabla_d Z_i \big) + \big( \nabla_d \nabla_c g^{ai} \big) Z_i + g^{ai} \big( \nabla_d \nabla_c Z_i \big) &= \tensor{R}{^{a}_{bcd}} g^{bj} Z_j \\ \big( \nabla_c \nabla_d g^{ai} + \nabla_d \nabla_c g^{ai} \big) Z_i + g^{ai} \big( \nabla_c \nabla_d Z_i - \nabla_d \nabla_c Z_i\big) &= \tensor{R}{^{a}_{bcd}} g^{bj} Z_j \end{split} \end{equation*}

Locally

\begin{equation*} \nabla_c \nabla_d g^{ai} + \nabla_d \nabla_c g^{ai} = e_c \big( e_d(g^{ai}) \big) - e_d \big( e_c(g^{ai}) \big) = \comm{e_c}{e_d} (g^{ai}) = 0 \end{equation*}

where we have dropped the resulting $\Gamma$ terms since they commute in the lower indices. Therefore

\begin{equation*} \begin{split} g^{ai} \big( \nabla_c \nabla_d Z_i - \nabla_d \nabla_c Z_i\big) &= \tensor{R}{^{a}_{bcd}} g^{bj} Z_j \\ \nabla_c \nabla_d Z_i - \nabla_d \nabla_c Z_i &= g_{ia} \tensor{R}{^{a}_{bcd}} g^{bj} Z_j \\ \nabla_c \nabla_d Z_i - \nabla_d \nabla_c Z_i &= \tensor{R}{^{}_{\textcolor{green}{ib}cd}} g^{bj} Z_j \\ \nabla_c \nabla_d Z_i - \nabla_d \nabla_c Z_i &= - \tensor{R}{^{}_{\textcolor{green}{bi}cd}} g^{bj} Z_j \\ \nabla_c \nabla_d Z_i - \nabla_d \nabla_c Z_i &= - \tensor{R}{^{j}_{icd}} Z_j \end{split} \end{equation*}

Locally the coefficients of a covector field can be expressed as $Z_i$ for some vector field $Z$, so we simply let

\begin{equation*} \lambda_i := Z_i \end{equation*}

and map indices $i \to a$ and $j \to b$, giving us

\begin{equation*} \nabla_c \nabla_d \lambda_a - \nabla_d \nabla_c \lambda_a = - \tensor{R}{^{b}_{acd}} \lambda_b \end{equation*}

Riemann tensor has an important geometrical interpretation.

It can be shown that $R(X, Y) Z$ is the change in $Z$ upon parallel transport around a small quadrilateral whose opposite sides are integral curves of vector fields $X$ and $Y$. Hence, if $R = 0$ parallel transport is locally path-independent.

The torsion and curvature of a connection $\nabla$ vanish if and only if for any $p \in M$, there exists a chart $(U, x)$ such that $p \in U$ and

\begin{equation*} \tensor{\Gamma}{^{i}_{jk}} = 0 \end{equation*}

In short,

\begin{equation*} \nabla = 0 \quad \iff \quad \exist (U, x): \quad p \in U \quad \text{and} \quad \tensor{\Gamma}{^{i}_{jk}} = 0 \end{equation*}

The Ricci tensor is the $(0, 2)$ tensor defined by contraction of the Riemann tensor

\begin{equation*} R(X, Y) = \left\langle f^a, R(e_a, Y) X \right\rangle \end{equation*}

In components,

\begin{equation*} \tensor{R}{^{}_{ab}} = \tensor{R}{^{d}_{abd}} \end{equation*}

Suppose $\nabla$ is a torsionless connection. Then we have

\begin{equation*} \begin{split} \tensor{R}{^{a}_{[bcd]}} &= 0 \\ \tensor{R}{^{a}_{bcd}} &= \frac{2}{3} \Big( \tensor{R}{^{a}_{(bc)d}} - \tensor{R}{^{a}_{(bd)c}} \Big) \end{split} \end{equation*}

We will be working in normal coordinates.

Recall that in normal coordinates at $p \in M$, we have $\tensor{\Gamma}{^{i}_{(jk)}} \big( p \big) = 0$.

Hence, for a torsionless connection, this reduces to $\tensor{\Gamma}{^{i}_{jk}} \big( p \big) = 0$.

Therefore, the Riemann tensor in normal coordinates at $p$ is simply

\begin{equation*} \tensor{R}{^{i}_{jkl}} = \partial_k \tensor{\Gamma}{^{i}_{jl}} - \partial_l \tensor{\Gamma}{^{i}_{jk}} \end{equation*}

  1. Due to anti-symmetry in the last two indices,

    \begin{equation*} \tensor{R}{^{a}_{[bcd]}} = \frac{1}{3} \big( \tensor{R}{^{a}_{bcd}} + \tensor{R}{^{a}_{cdb}} + \tensor{R}{^{a}_{dbc}} \big) \end{equation*}

    Substituting in the above identity (since we are in normal coordinates),

    \begin{equation*} \tensor{R}{^{a}_{[bcd]}} = \frac{1}{3} \big( \partial_c \tensor{\Gamma}{^{a}_{bd}} - \partial_d \tensor{\Gamma}{^{a}_{bc}} + \partial_d \tensor{\Gamma}{^{a}_{cb}} - \partial_b \tensor{\Gamma}{^{a}_{cd}} + \partial_b \tensor{\Gamma}{^{a}_{dc}} - \partial_c \tensor{\Gamma}{^{a}_{db}} \big) = 0 \end{equation*}

    since $\tensor{\Gamma}{^{a}_{bc}} = \tensor{\Gamma}{^{a}_{cb}}$.

  2. We have

    \begin{equation*} \begin{split} \tensor{R}{^{a}_{bcd}} &= \frac{2}{3} \Big( \tensor{R}{^{a}_{(bc)d}} - \tensor{R}{^{a}_{(bd) c}} \Big) \\ &= \frac{1}{3} \Big( \tensor{R}{^{a}_{bcd}} + \tensor{R}{^{a}_{cbd}} - \tensor{R}{^{a}_{bdc}} - \tensor{R}{^{a}_{dbc}} \Big) \\ &= \frac{1}{3} \Big( \tensor{R}{^{a}_{bcd}} + \tensor{R}{^{a}_{cbd}} + \tensor{R}{^{a}_{bcd}} - \tensor{R}{^{a}_{dbc}} \Big) \\ &= \frac{1}{3} \Big( 2 \tensor{R}{^{a}_{bcd}} + \tensor{R}{^{a}_{cbd}} - \tensor{R}{^{a}_{dbc}} \Big) \end{split} \end{equation*}

    So we need the last term to equal $\tensor{R}{^{a}_{bcd}}$. To see this we write the expressions out

    \begin{equation*} \begin{split} \tensor{R}{^{a}_{cbd}} + \tensor{R}{^{a}_{dcb}} &= \partial_b \tensor{\Gamma}{^{a}_{cd}} - \partial_d \tensor{\Gamma}{^{a}_{cb}} + \partial_c \tensor{\Gamma}{^{a}_{db}} - \partial_b \tensor{\Gamma}{^{a}_{dc}} \\ &= \partial_c \tensor{\Gamma}{^{a}_{db}} - \partial_d \tensor{\Gamma}{^{a}_{cb}} \\ &= \partial_c \tensor{\Gamma}{^{a}_{bd}} - \partial_d \tensor{\Gamma}{^{a}_{bc}} \\ &= \tensor{R}{^{a}_{bcd}} \end{split} \end{equation*}

    as we wanted. Substituting into the expression above,

    \begin{equation*} \frac{2}{3} \Big( \tensor{R}{^{a}_{(bc)d}} - \tensor{R}{^{a}_{(bd)c}} \Big) = \frac{1}{3} \Big( 2 \tensor{R}{^{a}_{bcd}} + \tensor{R}{^{a}_{bcd}} \Big) = \tensor{R}{^{a}_{bcd}} \end{equation*}

    as claimed.

Suppose $\nabla$ is a torsionless connection. Then we have

\begin{equation*} \nabla_{[c} \tensor{R}{^{a}_{|b|de]}} = 0 \end{equation*}

This is called the Bianchi identity.

\begin{equation*} \begin{split} \nabla_{[c} \tensor{R}{^{a}_{|b| de]}} &= \frac{1}{6} \Big( \nabla_c \tensor{R}{^{a}_{bde}} - \nabla_c \tensor{R}{^{a}_{bed}} + \nabla_e \tensor{R}{^{a}_{b cd}} - \nabla_e \tensor{R}{^{a}_{bdc}} + \nabla_d \tensor{R}{^{a}_{b ec}} - \nabla_d \tensor{R}{^{a}_{b ce}} \Big) \\ &= \frac{1}{3} \Big( \nabla_c \tensor{R}{^{a}_{bde}} + \nabla_e \tensor{R}{^{a}_{b cd}} + \nabla_d \tensor{R}{^{a}_{b ec}} \Big) \end{split} \end{equation*}

which follows directly from

\begin{equation*} \tensor{R}{^{a}_{b (cd)}} = 0 \end{equation*}

Then observe that

\begin{equation*} \nabla_c \tensor{R}{^{a}_{bde}} + \nabla_e \tensor{R}{^{a}_{bcd}} + \nabla_d \tensor{R}{^{a}_{bec}} = 0 \end{equation*}

since in normal coordinates we have

\begin{equation*} \nabla_m \tensor{R}{^{i}_{jkl}} = \partial_m \partial_k \tensor{\Gamma}{^{i}_{jl}} - \partial_m \partial_l \tensor{\Gamma}{^{i}_{jk}} \end{equation*}

because $\partial \big( \Gamma \Gamma \big) = 2 \Gamma \partial (\Gamma) = 0$ since $\Gamma = 0$. Finally, substituting this into the above expression:

\begin{equation*} \begin{split} & \nabla_c \tensor{R}{^{a}_{bde}} + \nabla_e \tensor{R}{^{a}_{bcd}} + \nabla_d \tensor{R}{^{a}_{bec}} \\ = \quad & \big( \cancel{\textcolor{green}{\partial_c \partial_d \tensor{\Gamma}{^{a}_{be}}}} - \cancel{\textcolor{red}{\partial_c \partial_e \tensor{\Gamma}{^{a}_{bd}}}} \big) + \big( \cancel{\textcolor{red}{\partial_e \partial_c \tensor{\Gamma}{^{a}_{bd}}}} - \cancel{\textcolor{yellow}{\partial_e \partial_d \tensor{\Gamma}{^{a}_{bc}}}} \big) + \big( \cancel{\textcolor{yellow}{\partial_d \partial_e \tensor{\Gamma}{^{a}_{bc}}}} - \cancel{\textcolor{green}{\partial_d \partial_c \tensor{\Gamma}{^{a}_{be}}}} \big) \\ = \quad & 0 \end{split} \end{equation*}

since partials commute.

Levi-Civita connection

Let $\big( M, g \big)$ be a psuedo-Riemannian manifold. There exist a unique connection $\nabla$ with vanishing torsion and satisfying $\nabla g = 0$.

This choice of connection is often called the Levi-Civita connection.

\begin{equation*} \Gamma^i_{jk} = \frac{1}{2} g^{il} \big( \partial_j g_{lk} + \partial_k g_{jl} - \partial_l g_{kj} \big) \end{equation*}

Curvature of Levi-Civita connection

Let $\nabla$ be the Levi-Civita connection. We define the (0, 4)-tensor

\begin{equation*} R_{abcd} = \tensor{g}{^{}_{ae}} \tensor{R}{^{e}_{bcd}} \end{equation*}
\begin{equation*} \tensor{R}{^{}_{abcd}} = \tensor{R}{^{}_{cdab}} \quad \text{and} \quad \tensor{R}{_{abcd}} = - \tensor{R}{^{}_{bacd}} \end{equation*}

We work in normal coordinates at $p \in M$, so

\begin{equation*} \partial_i g_{jk}(p) = 0, \quad g_{jk}(p) \in \left\{ \delta_{jk}, \eta_{jk} \right\} \end{equation*}
\begin{equation*} \begin{split} R_{ijkl} &= g_{im} \tensor{R}{^{m}_{jkl}} \\ &= \tensor{g}{^{}_{im}} \big( \partial_k \tensor{\Gamma}{^{m}_{jl}} - \partial_l \tensor{\Gamma}{^{m}_{jk}} \big) \end{split} \end{equation*}

and so

\begin{equation*} \tensor{g}{^{}_{im}} \tensor{\Gamma}{^{m}_{jk}} = \frac{1}{2} \big( \partial_i \tensor{g}{^{}_{ki}} + \partial_k \tensor{g}{^{}_{ji}} - \partial_i \tensor{g}{^{}_{jk}} \big) \end{equation*}

(at $p$) differentiate wrt. $\partial_l$ and evaluate at $p$:

\begin{equation*} \tensor{g}{^{}_{im}} \partial_l \tensor{\Gamma}{^{m}_{jk}} = \frac{1}{2} \big( \partial_l \partial_j \tensor{g}{^{}_{ki}} + \partial_l \partial_k \tensor{g}{^{}_{ji}} - \partial_l \partial_i \tensor{g}{^{}_{jk}} \big) \end{equation*}

at $p$.

AND MORE.

\begin{equation*} R_{ab} = R_{ba} \end{equation*}

for Levi-Civita connection.

\begin{equation*} R_{ab} = \tensor{R}{^{c}_{acb}} = \tensor{g}{^{cd}_{}} \tensor{R}{^{}_{dacb}} = \tensor{g}{^{cd}_{}} \tensor{R}{^{}_{cbda}} = \tensor{R}{^{d}_{bda}} = \tensor{R}{^{}_{ba}} \end{equation*}

using Proposition proposition:4.61-lectures.

The Ricci scalar of $\big( M, g, \nabla \big)$ is

\begin{equation*} R = \tensor{g}{^{ab}_{}} \tensor{R}{^{}_{ab}} \end{equation*}

The Einstein tensor

\begin{equation*} G_{ab} = \tensor{R}{^{}_{ab}} - \frac{1}{2} R g_{ab} \end{equation*}
\begin{equation*} \nabla^a \tensor{G}{^{}_{ab}} = \tensor{g}{^{ac}_{}} \nabla_c \tensor{G}{^{}_{ab}} = 0 \end{equation*}

From the Bianchi identity

\begin{equation*} \nabla_d \tensor{R}{^{e}_{abc}} + \nabla_c \tensor{R}{^{e}_{adb}} + \nabla_b \tensor{R}{^{e}_{acd}} = 0 \end{equation*}

We contract by $e = d$

\begin{equation*} \begin{split} \nabla_d \tensor{R}{^{d}_{abc}} + \nabla_c \tensor{R}{^{}_{ad}} - \nabla_b \tensor{R}{^{}_{ac}} = 0 \end{split} \end{equation*}

Contract with $g^{ac}$, noting that $\nabla_b R_{ac} = \nabla_b \big( g^{ac} R_{ac} \big)$ by $\nabla g = 0$

\begin{equation*} \begin{split} g^{ac} \nabla^d \tensor{R}{^{}_{dabc}} + \nabla^a R_{ab} - \nabla_b \big( g^{ac} R_{ac} \big) &= 0 \\ \implies \quad g^{ac} \nabla^d \tensor{R}{^{}_{adbc}} (-1)^2 + \nabla^a R_{ab} - \nabla_b R &= 0 \\ \implies \quad \nabla^d R_{db} + \nabla^a R_{ab} - \nabla_b R &= 0 \\ \implies \quad 2 \nabla^2 R_{ab} - \nabla_b R & \iff \quad \nabla^a \big( R_{ab} - \frac{1}{2} g_{ab} R \big) = 0 \end{split} \end{equation*}

Riemann tensor of $(M, g)$ vanishes if and only if for every $p \in M$, there exists a chart containing $p$ such that

\begin{equation*} g_{ij} = \begin{cases} \delta_{ij} & \quad \text{(Riemannian)} \\ \eta_{ij} & \quad \text{(Lorentzian)} \end{cases} \end{equation*}

Recall if torsion vanishes, then Riemann tensor vanishes if and only if there exists charts where $\tensor{\Gamma}{^{i}_{jk}} = 0$. Then

\begin{equation*} 0 = \nabla_i \tensor{g}{^{}_{jk}} = \partial_i g_{jk} \end{equation*}

in such charts.

This implies that $g_{jk}$ is constant in chart, if we choose basis at $p$ to be orthonormal, which implies

\begin{equation*} g_{ij} = \begin{cases} \delta_{ij} & \quad \text{(Riemannian)} \\ \eta_{ij} & \quad \text{(Lorentzian)} \end{cases} \end{equation*}

Theorem thm:Riemann-tensor-vanish-iff-Euclidean-or-Minkowski-metric shows us that the Riemann tensor sort of measure the deviation from Euclidean or Minkowski metric, locally.

Weyl tensor

  • Ricci tensor and Ricci scalar contain information about "traces" of the Riemann tensor
  • Sometimes useful to consider separately those parts of the Riemann tensor $\tensor{R}{^{}_{abcd}}$ which the Ricci tensor $\tensor{R}{^{}_{ab}}$ does not tell us about

The Weyl tensor is defined (in $n \ge 3$ dimensions)

\begin{equation*} \tensor{C}{^{}_{abcd}} = \tensor{R}{^{}_{abcd}} - \frac{2}{(n - 2)} \big( \tensor{g}{^{}_{a [c}} \tensor{R}{^{}_{d] a}} - \tensor{g}{^{}_{b [c}} \tensor{R}{^{}_{d] a}} \big) + \frac{2}{(n - 1)(n - 2)} R \tensor{g}{^{}_{a[c}} \tensor{g}{^{}_{d]c}} \end{equation*}

The Weyl tensor is basically the Riemann tensor with all of its contractions removed. The above formula is designed so that all possible contractions of $C_{abcd}$ vanish, while it retains the symmetries of the Riemann tensor:

\begin{equation*} \begin{split} C_{abcd} &= C_{[ab][cd]} \\ C_{abcd} &= C_{cd ab} \\ C_{a [bcd]} &= 0 \end{split} \end{equation*}

One of the most important properties of the Weyl tensor is that it's invariant under

Special relativity

Spacetime is just a Lorentizan manifold $\big( \mathbb{R}^4, \eta \big)$ (Minkowski spacetime),

\begin{equation*} \eta = \eta_{\mu \nu} \dd{x^{\mu}} \dd{x^{\nu}} \end{equation*}

where

\begin{equation*} \big( \eta_{\mu \nu} \big) = \diag \big( -1, 1, 1, 1 \big) \end{equation*}

and $(x^{\mu})$ are the inertial coordinates, i.e. $\dot{x}^{\mu} = 0$.

Free motion: are timelike geodesics, ligthrays null geodesics

Physics: described by tensor fields on $\big( \mathbb{R}^4, \eta \big)$ which obey evolution equations.

Examples

Scalar field

Let $\Phi(x)$ be a scalar field, then

\begin{equation*} \eta^{\mu \nu} \partial_{\mu} \partial_{\nu} \Phi = 0 \end{equation*}

which is just the wave-equation:

\begin{equation*} \bigg( - \pdv[2]{}{(x^0)} + \pdv[2]{}{(x^1)} + \dots \bigg) \Phi = 0 \end{equation*}

Energy-momentum tensor

\begin{equation*} T_{\mu \nu} = \partial_{\mu} \Phi \partial_{\nu} \Phi - \frac{1}{2} g_{\mu \nu} \Big( \partial^{\rho} \Phi \partial_{\rho} \Phi \Big) \end{equation*}

satisfies

\begin{equation*} \partial^{\mu} T_{\mu \nu} = 0 \end{equation*}

which is conservation of energy / momentum.

Maxwell's theory of E. M.

  • Electromagnetic field strength $F_{\mu \nu} = - F_{\nu \mu}$ which obeys the Maxwell's equations:

    \begin{equation*} \tensor{\eta}{^{\mu \nu}_{}} \partial_{\mu} \tensor{F}{^{}_{\nu \rho}} = 0 \quad \text{and} \quad \partial_{[\mu} F_{\nu\ rho]} = 0 \end{equation*}

  • The corresponding energy-momentum tensor

    \begin{equation*} T_{\mu \nu} = F_{\mu \rho} \tensor{F}{_{\nu}^{\rho}} - \frac{1}{4} \eta_{\mu \nu} F_{\rho \sigma} F^{\rho \sigma} \quad \text{with} \quad \partial^{\mu} T_{\mu \nu} = 0 \end{equation*}

Any matter distribution is described by energy-momentum tensor

\begin{equation*} T_{\mu \nu} = T_{\nu \mu}, \quad \partial^{\mu} T_{\mu \nu} = 0 \end{equation*}

Fluids

  • Described by vector field $u^{\mu}$, and typically normalized such that $u^{\mu} u_{\mu} = - 1$

  • Perfect fluids:

    \begin{equation*} T_{\mu \nu } = (\rho + p) u_{\mu} u_{\nu} + p \eta_{\mu \nu} \end{equation*}

    where

    • $\rho$ is the energy density
    • $p$ is the pressure
  • Then $\partial^{\mu} T_{\mu \nu} = 0$ is the relativitic eqns. of fluid dynamics

General relativity

Main idea: there exist local freely flowing frames with no gravity.

This is achieved by the following postulates.

Postulates:

  1. Space-time is a 4D Lorentzian manifold $(M, g)$
  2. Free particles follow timelike / null-geodesics wrt. Levi-Civita of $(M, g)$

  3. Energy-momentum distribution of matter fields described by $\big( 0, 2 \big)$ symmetric tensor field $T_{ab}$ which is conserved

    \begin{equation*} \nabla^a T_{ab} = 0 \end{equation*}

  4. The curvature of $\big( M, g \big)$ is related to energy-momentum tensor of matter by the Einstein's equations:

    \begin{equation*} R_{ab} - \frac{1}{2} g_{ab} R = 8 \pi G T_{ab} \end{equation*}

    where $G$ is the Newton gravitational constant.

    • Note that $T_{ab}$ might also depend on $g_{ab}$, so we cannot simply fix $T_{ab}$ and solve

Laws of physics governed by:

  1. General covariance: laws indep. of basis / coord system
  2. Equivalence principle: in any local inertial frame (normal coordinate system) laws reduce to the laws in Minkowski spacetime (Minkowski space)

Do not fix laws uniquely! But suggests simple rule (called minimal coupling):

  • given any equation in $\big( \mathbb{R}^4, \eta \big)$ (Minkowski space), we replace

    \begin{equation*} \begin{split} \eta_{\mu \nu} &\longrightarrow g_{ab} \\ \partial_{\mu} & \longrightarrow \nabla_a \quad \text{(Levi-Civita)} \end{split} \end{equation*}

    to get laws on curved spacetime $(M, g)$

  • Rules ensure general covariance as they output tensor equations.

  • Local intertial frames are normal coords at $p \in M$, i.e.

    \begin{equation*} g_{\mu \nu} (p) = \eta_{\mu \nu} \quad \text{and} \quad \partial_{\nu} g_{\rho \sigma}(p) = 0 \end{equation*}

Examples

Wave equation

Applying these rules to wave-equation of S.R. we get

\begin{equation*} g^{ab} \nabla_a \nabla_b \Phi = 0 \end{equation*}

Consider the wave equation

\begin{equation*} \nabla^a \nabla_a \Phi + \alpha R \nabla = 0 \end{equation*}

also reduces to the wave-equation on spacetime $\big(\mathbb{R}^4, \eta \big)$ since $R(p) = 0$.

In local inertial frame, observe that we have

\begin{equation*} g^{ab} \nabla_a \nabla_b \Phi \big|_{p} = \eta^{\mu \nu} \partial_{\mu} \partial_{\nu} \Phi \big|_{p} \end{equation*}

where the RHS is the wave-equation in $\big( \mathbb{R}^4, \eta \big)$.

Postulate 3: we have

\begin{equation*} T_{ab} = \nabla_a \Phi \nabla_b \Phi - \frac{1}{2} g_ab} \big( \nabla^c \Phi \nabla_c \Phi \big) \end{equation*}

and by wave-equation

\begin{equation*} \nabla^a T_{ab} = 0 \end{equation*}

Maxwell's equations

Applying these rules to Maxwell's equations in spacetime, we get

\begin{equation*} g^{ab} \nabla_a F_{bc} = 0 \quad \text{and} \quad \nabla_{[a} F_{bc]} = 0 \quad \text{and} \quad F_{ab} = - F_{ba} \end{equation*}

Postulate 3: we have

\begin{equation*} T_{ab} = F_{ac} \tensor{F}{_{b}^{c}} - \frac{1}{4} g_{ab} F_{cd} F^{cd} \end{equation*}

combined with Maxwell's equations we get

\begin{equation*} \nabla^a T_{ab} = 0 \end{equation*}

Fluids

  • Described by vector field $u^{\mu}$, and typically normalized such that $u^{\mu} u_{\mu} = - 1$

  • Perfect fluids:

    \begin{equation*} T_{ab} = \big( \rho + p \big) u_a u_b + p g_{ab} \end{equation*}

    where

    • $\rho$ is the energy density
    • $p$ is the pressure

Then

\begin{equation*} T_{ab} u^a u^b = \rho + p - p = \rho \end{equation*}

Motion of fluid given by

\begin{equation*} \begin{split} \nabla^a T_{ab} &= 0 \\ \iff \quad u^a \nabla_a \rho + \big( \rho + p \big) \nabla_a u^a &= 0 \quad \text{(mass conservation)}\\ \iff \quad \big( \rho + p \big) u^b \nabla_b u_a + \big( g_{ab} + u_a u_b \big) \nabla^b p &= 0 \end{split} \end{equation*}

Observe that if the pressure vanish, i.e. $p = 0$, then we're left with

\begin{equation*} u^b \nabla_b u_a = 0 \end{equation*}

which is the equation describing geodesic, fluid moves as "free particles".

Einstein's equations

  • Motivation:
    • Newtonian
      • Graviational field described by scalar potential $\phi$

      • Equation of motion

        \begin{equation*} \ddot{x}^i = - \partial^i \phi \end{equation*}

        where $\partial^i = \partial_i$ in cartesian coordinates

      • Relative acceleration of nearby particles ("tildal force") governed by deviation equation

        \begin{equation*} \ddot{z}^i = - z^j \partial_j \partial^i \phi \end{equation*}

        where $z^i$ is the separation vector of the particles

    • GR:

      • Relative acceleration of two nearby particles following timelike geodesics is given by the geodesic deviation equation

        \begin{equation*} \nabla_V \nabla_V z^a = \tensor{R}{^{a}_{cdb}} V^c V^d z^b =: \tensor{\Phi}{^{a}_{b}} z^b \end{equation*}

        where $V^a$ is the tangent to the geodesics and $z^a$ is now the deviation vector.

    • Comparison suggests

      \begin{equation*} \tensor{\Phi}{^{a}_{b}} = \tensor{R}{^{a}_{cdb}} V^c V^d \longleftrightarrow - \partial^i \partial_j \phi \end{equation*}

Spacetimes

(anti-)de Sitter

de Sitter spacetime in $D + 1$ dimensions with radius of curvature $R$ is locally isometric to the quadratic

\begin{equation*} x_1^2 + x_2^2 + \cdots + x_D^2 + x_{D + 1}^2 - x_{D + 2}^2 = R^2 \quad \text{in } \mathbb{R}^{D + 1, 1} \end{equation*}

Anti de Sitter spacetime in $D + 1$ dimensions with radius of curvature $R$ is locally isometric to the quadrics

\begin{equation*} x_1^2 + x_2^2 + \cdots + x_D^2 + x_{D + 1}^2 - x_{D + 2}^2 = \textcolor{red}{-} R^2 \quad \text{in } \mathbb{R}^{D + 1, 1} \end{equation*}

More precisely, the de Sitter spacetimes are the simply-connected universal covers of these quadrics.

Taking the limit $R \to \infty$ is equivalent to the zero curvature limit in which we recover Minkowski spacetime.

The real affince space $\mathbb{A}^{D + 1}$ with a metric which, when expressed relative to affine coordinates, is given by

\begin{equation*} \dd{x_1^2} + \dd{x_2^2} + \cdots + \dd{x_D^2} - c^2 \dd{x_{D + 1}^2} \end{equation*}

where we have introduced the speed of light $c$.

We may take limits:

  • non-relativistic limit (on the co-metric): $c \to \infty$ gives us galilean spacetime
  • ultra-relativistic limit (on the co-metric): $c \to 0$ gives us carrollian spacetime

These spacetimes are no longer lorentzian: the (co-)metric becomes degenerate in the limit, leading to a galilean and a carrollian structure, respectively.

Conformal GR