General relativity essentials

Metric manifolds. An n-dimensional manifold M is a space which is locally flat, i.e. locally looks like Rⁿ , and furthermore can be constructed from pieces of Rⁿ sewn together smoothly. A Lorentzian or pseudo-Riemannian manifold is a manifold with the notion of 'distance', equivalently 'metric', included. 'Lorentzian' refers to the signature of the metric which in general relativity is taken to be $(-1,+1,+1,+1)$ as opposed to the more familiar/natural 'Euclidean' signature given by $(+1,+1,+1,+1)$ valid for Riemannian manifolds. The metric is usually denoted by ${g}_{\mu \nu }$ while the line element (also called metric in many instances) is written as

$$\begin{eqnarray}&&{{ds}}^{2}={g}_{\mu \nu }{{dx}}^{\mu }{{dx}}^{\nu }.\end{eqnarray} \tag{ 1.12 }$$

For example Minkowski spacetime is given by the flat metric

$$\begin{eqnarray}&&{g}_{\mu \nu }={\eta }_{\mu \nu }=(-1,+1,+1,+1).\end{eqnarray} \tag{ 1.13 }$$

Another extremely important example is Schwarzschild spacetime given by the metric

$$\begin{eqnarray}&&{{ds}}^{2}=-{\left(1-\frac{{R}_{s}}{r}\right){{dt}}^{2}+\left(1-\frac{{R}_{s}}{r}\right)}^{-1}{{dr}}^{2}+{r}^{2}d{{\rm{\Omega }}}^{2}.\end{eqnarray} \tag{ 1.14 }$$

This is quite different from the flat metric ${\eta }_{\mu \nu }$ and as a consequence the curvature of Schwarzschild spacetime is non-zero. Another important curved space is the surface of the two-dimensional sphere on which the metric, which appears as a part of the Schwarzschild metric, is given by

$$\begin{eqnarray}&&{{ds}}^{2}={r}^{2}d{{\rm{\Omega }}}^{2}={r}^{2}(d{\theta }^{2}+{\mathrm{sin}}^{2} \theta d{\phi }^{2}).\end{eqnarray} \tag{ 1.15 }$$

The inverse metric will be denoted by ${g}^{\mu \nu }$, i.e.

$$\begin{eqnarray}&&{g}_{\mu \nu }{g}^{\nu \lambda }={\eta }_{\mu }^{\lambda }.\end{eqnarray} \tag{ 1.16 }$$

Charts. A coordinate system (a chart) on the manifold M is a subset U of M together with a one-to-one map $\phi :U\to {R}^{n}$ such that the image $V=\phi (U)$ is an open set in Rⁿ , i.e. a set in which every point $y\in V$ is the center of an open ball which is inside V. We say that U is an open set in M. Hence we can associate with every point $p\in U$ of the manifold M the local coordinates $({x}^{1},\ldots ,{x}^{n})$ by

$$\begin{eqnarray}&&\phi (p)=({x}^{1},\ldots ,\;{x}^{n}).\end{eqnarray} \tag{ 1.17 }$$

Vectors. A curved manifold is not necessarily a vector space. For example the sphere is not a vector space because we do not know how to add two points on the sphere to get another point on the sphere. The sphere, which is naturally embedded in R³, admits at each point p a tangent plane. The notion of a 'tangent vector space' can be constructed for any manifold which is embedded in Rⁿ . The tangent vector space at a point p of the manifold will be denoted by V_p .

There is a one-to-one correspondence between vectors and directional derivatives in Rⁿ . Indeed, the vector $v=({v}^{1},\ldots ,{v}^{n})$ in Rⁿ defines the directional derivative ${\sum }_{\mu }{v}^{\mu }{\partial }_{\mu }$ which acts on functions on Rⁿ . These derivatives are clearly linear and satisfy the Leibniz rule. We will therefore define tangent vectors at a given point p on a manifold M as directional derivatives which satisfy linearity and the Leibniz rule. These directional derivatives can also be thought of as differential displacements on the spacetime manifold at the point p.

This can be made more precise as follows. First, we define s smooth curve on the manifold M as a smooth map from R into M, namely $\gamma :R\to M$. A tangent vector at a point p can then be thought of as a directional derivative operator along a curve which goes through p. Indeed, a tangent vector T at $p=\gamma (t)\in M$, acting on smooth functions f on the manifold M, can be defined by

$$\begin{eqnarray}&&T( f)=\frac{d}{{dt}}( f \circ \gamma {(t))\bigg | }_{p}.\end{eqnarray} \tag{ 1.18 }$$

In a given chart ϕ the point p will be given by $p={\phi }^{-1}(x)$ where $x\;=({x}^{1},\ldots ,{x}^{n})\in {R}^{n}$. Hence $\gamma (t)={\phi }^{-1}(x)$. In other words, the map γ is mapped into a curve x(t) in Rⁿ . We have immediately

$$\begin{eqnarray}&&T( f)=\frac{d}{{dt}}( f \circ {\phi }^{-1}(x)){\bigg | }_{p}=\sum _{\mu =1}^{n}{X}_{\mu }( f)\frac{{{dx}}^{\mu }}{{dt}}{\bigg | }_{p}.\end{eqnarray} \tag{ 1.19 }$$

The maps ${X}_{\mu }$ act on functions f on the Manifold M as

$$\begin{eqnarray}&&{X}_{\mu }( f)=\frac{\partial }{\partial {x}^{\mu }}( f \circ {\phi }^{-1}(x)).\end{eqnarray} \tag{ 1.20 }$$

These can be checked to satisfy linearity and the Leibniz rule. They are obviously directional derivatives or differential displacements since we may make the identification ${X}_{\mu }={\partial }_{\mu }$. Hence these vectors are tangent vectors to the manifold M at p. The fact that arbitrary tangent vectors can be expressed as linear combinations of the n vectors ${X}_{\mu }$ shows that these vectors are linearly independent, span the vector space V_p and that the dimension of V_p is exactly n. Equation (1.19) can then be rewritten as

$$\begin{eqnarray}&&T=\sum _{\mu =1}^{n}{X}_{\mu }{T}^{\mu }.\end{eqnarray} \tag{ 1.21 }$$

The components ${T}^{\mu }$ of the vector T are therefore given by

$$\begin{eqnarray}&&{T}^{\mu }=\frac{{{dx}}^{\mu }}{{dt}}{\bigg | }_{p}.\end{eqnarray} \tag{ 1.22 }$$

The length l of a smooth curve C with tangent ${T}^{\mu }$ on a manifold M with Riemannian metric ${g}_{\mu \nu }$ is given by

$$\begin{eqnarray}&&l=\int {dt}\sqrt{{g}_{\mu \nu }{T}^{\mu }{T}^{\nu }}.\end{eqnarray} \tag{ 1.23 }$$

The length is parametrization independent. Indeed, we can show that

$$\begin{eqnarray}&&l=\int {dt}\sqrt{{g}_{\mu \nu }{T}^{\mu }{T}^{\nu }}=\int {ds}\sqrt{{g}_{\mu \nu }{S}^{\mu }{S}^{\nu }},\quad {S}^{\mu }={T}^{\mu }\frac{{dt}}{{ds}}.\end{eqnarray} \tag{ 1.24 }$$

In a Lorentzian manifold, the length of a space-like curve is also given by this expression. For a time-like curve for which ${g}_{{ab}}{T}^{a}{T}^{b}\lt 0$ the length is replaced with the proper time τ, which is given by $\tau =\int {dt}\sqrt{-{g}_{{ab}}{T}^{a}{T}^{b}}$. For a light-like (or null) curve for which ${g}_{{ab}}{T}^{a}{T}^{b}=0$ the length is always 0.

We consider the length of a curve C connecting two points $p=C({t}_{0})$ and $q=C({t}_{1})$. In a coordinate basis the length is given explicitly by

$$\begin{eqnarray}&&l={\int }_{{t}_{0}}^{{t}_{1}}{dt}\sqrt{{g}_{\mu \nu }\frac{{{dx}}^{\mu }}{{dt}}\frac{{{dx}}^{\nu }}{{dt}}}.\end{eqnarray} \tag{ 1.25 }$$

The variation in l under an arbitrary smooth deformation of the curve C which keeps the two points p and q fixed is given by

$$\begin{eqnarray}\delta l & = & \frac{1}{2}{\int }_{{t}_{0}}^{{t}_{1}}{dt}{\left({g}_{\mu \nu }\frac{{{dx}}^{\mu }}{{dt}}\frac{{{dx}}^{\nu }}{{dt}}\right)}^{-\frac{1}{2}}\; \left(\frac{1}{2}\delta {g}_{\mu \nu }\frac{{{dx}}^{\mu }}{{dt}}\frac{{{dx}}^{\nu }}{{dt}}+{g}_{\mu \nu }\frac{{{dx}}^{\mu }}{{dt}}\frac{d\delta {x}^{\nu }}{{dt}}\right)\\ & = & \frac{1}{2}{\int }_{{t}_{0}}^{{t}_{1}}{dt}{\left({g}_{\mu \nu }\frac{{{dx}}^{\mu }}{{dt}}\frac{{{dx}}^{\nu }}{{dt}}\right)}^{-\frac{1}{2}}\; \left(\frac{1}{2}\frac{\partial {g}_{\mu \nu }}{\partial {x}^{\sigma }}\delta {x}^{\sigma }\frac{{{dx}}^{\mu }}{{dt}}\frac{{{dx}}^{\nu }}{{dt}}+{g}_{\mu \nu }\frac{{{dx}}^{\mu }}{{dt}}\frac{d\delta {x}^{\nu }}{{dt}}\right)\\ & = & \frac{1}{2}{\int }_{{t}_{0}}^{{t}_{1}}{dt}{\left({g}_{\mu \nu }\frac{{{dx}}^{\mu }}{{dt}}\frac{{{dx}}^{\nu }}{{dt}}\right)}^{-\frac{1}{2}}\\ & & \times \left(\frac{1}{2}\frac{\partial {g}_{\mu \nu }}{\partial {x}^{\sigma }}\delta {x}^{\sigma }\frac{{{dx}}^{\mu }}{{dt}}\frac{{{dx}}^{\nu }}{{dt}}-\frac{d}{{dt}}({g}_{\mu \nu }\frac{{{dx}}^{\mu }}{{dt}})\delta {x}^{\nu }+\frac{d}{{dt}}({g}_{\mu \nu }\frac{{{dx}}^{\mu }}{{dt}}\delta {x}^{\nu })\right).\end{eqnarray} \tag{ 1.26 }$$

We can assume without any loss of generality that the parametrization of the curve C satisfies ${g}_{\mu \nu }({{dx}}^{\mu }/{dt})({{dx}}^{\nu }/{dt})=1$. In other words, we choose dt² to be precisely the line element (interval) and thus ${T}^{\mu }={{dx}}^{\mu }/{dt}$ is the 4-velocity. The last term in the above equation obviously becomes a total derivative, which vanishes by the fact that the considered deformation keeps the two end points p and q fixed. We then obtain

$$\begin{eqnarray}\delta l & = & \frac{1}{2}{\int }_{{t}_{0}}^{{t}_{1}}{dt}\delta {x}^{\sigma }\; \left(\frac{1}{2}\frac{\partial {g}_{\mu \nu }}{\partial {x}^{\sigma }}\frac{{{dx}}^{\mu }}{{dt}}\frac{{{dx}}^{\nu }}{{dt}}-\frac{d}{{dt}}({g}_{\mu \sigma }\frac{{{dx}}^{\mu }}{{dt}})\right)\\ & = & \frac{1}{2}{\int }_{{t}_{0}}^{{t}_{1}}{dt}\delta {x}^{\sigma }\; \left(\frac{1}{2}\frac{\partial {g}_{\mu \nu }}{\partial {x}^{\sigma }}\frac{{{dx}}^{\mu }}{{dt}}\frac{{{dx}}^{\nu }}{{dt}}-\frac{\partial {g}_{\mu \sigma }}{\partial {x}^{\nu }}\frac{{{dx}}^{\nu }}{{dt}}\frac{{{dx}}^{\mu }}{{dt}}-{g}_{\mu \sigma }\frac{{d}^{2}{x}^{\mu }}{{{dt}}^{2}}\right)\\ & = & \frac{1}{2}{\int }_{{t}_{0}}^{{t}_{1}}{dt}\delta {x}^{\sigma }\; \left(\frac{1}{2}\left(\frac{\partial {g}_{\mu \nu }}{\partial {x}^{\sigma }}-\frac{\partial {g}_{\mu \sigma }}{\partial {x}^{\nu }}-\frac{\partial {g}_{\nu \sigma }}{\partial {x}^{\mu }}\right)\frac{{{dx}}^{\mu }}{{dt}}\frac{{{dx}}^{\nu }}{{dt}}-{g}_{\mu \sigma }\frac{{d}^{2}{x}^{\mu }}{{{dt}}^{2}}\right)\\ & = & \frac{1}{2}{\int }_{{t}_{0}}^{{t}_{1}}{dt}\delta {x}_{\rho }\; \left(\frac{1}{2}{g}^{\rho \sigma }\left(\frac{\partial {g}_{\mu \nu }}{\partial {x}^{\sigma }}-\frac{\partial {g}_{\mu \sigma }}{\partial {x}^{\nu }}-\frac{\partial {g}_{\nu \sigma }}{\partial {x}^{\mu }}\right)\frac{{{dx}}^{\mu }}{{dt}}\frac{{{dx}}^{\nu }}{{dt}}-\frac{{d}^{2}{x}^{\rho }}{{{dt}}^{2}}\right).\end{eqnarray} \tag{ 1.27 }$$

By definition geodesics are curves which extremize the length l. The curve C extremizes the length between the two points p and q if and only if $\delta l=0$. This leads immediately to the equation

$$\begin{eqnarray}&&{{\rm{\Gamma }}}_{\mu \nu }^{\rho }\frac{{{dx}}^{\mu }}{{dt}}\frac{{{dx}}^{\nu }}{{dt}}+\frac{{d}^{2}{x}^{\rho }}{{{dt}}^{2}}=0.\end{eqnarray} \tag{ 1.28 }$$

This equation is called the geodesic equation. It is the relativistic generalization of Newton's second law of motion (1.2). The Christoffel symbols are defined by

$$\begin{eqnarray}&&{{\rm{\Gamma }}}_{\mu \nu }^{\rho }=-\frac{1}{2}{g}^{\rho \sigma }\; \left(\frac{\partial {g}_{\mu \nu }}{\partial {x}^{\sigma }}-\frac{\partial {g}_{\mu \sigma }}{\partial {x}^{\nu }}-\frac{\partial {g}_{\nu \sigma }}{\partial {x}^{\mu }}\right).\end{eqnarray} \tag{ 1.29 }$$

In the absence of curvature we will have ${g}_{\mu \nu }={\eta }_{\mu \nu }$ and hence ${\rm{\Gamma }}=0$. In other words, the geodesics are locally straight lines.

Since the length between any two points on a Riemannian manifold (and between any two points which can be connected by a space-like curve on a Lorentzian manifold) can be arbitrarily long, we conclude that the shortest curve connecting the two points must be a geodesic as it is an extremum of length. Hence the shortest curve is the straightest possible curve. The converse is not true: a geodesic connecting two points is not necessarily the shortest path.

Similarly, the proper time between any two points which can be connected by a time-like curve on a Lorentzian manifold can be arbitrarily small and thus the curve with the greatest proper time, if it exists, must be a time-like geodesic as it is an extremum of proper time. On the other hand, a time-like geodesic connecting two points is not necessarily the path with maximum proper time.

Tangent (contravariant) vectors. Tensors are a generalization of vectors. Let us start then by giving a more precise definition of the tangent vector space V_p . Let ${ \mathcal F }$ be the set of all smooth functions f on the manifold M, i.e. $f:M\to R$. We define a tangent vector v at the point $p\in M$ as a map $v:{ \mathcal F }\to R$ which is required to satisfy linearity and the Leibniz rule. In other words,

$$\begin{eqnarray}v({af}+{bg}) & = & {av}( f)+{bv}(g),\\ v({fg}) & = & f(p)v(g)+g(p)v( f),\; a,b\in R, f,g\in { \mathcal F }.\end{eqnarray} \tag{ 1.30 }$$

The vector space V_p is simply the set of all tangents vectors v at p. The action of the vector v on the function f is given explicitly by

$$\begin{eqnarray}&&v( f)=\sum _{\mu =1}^{n}{v}^{\mu }{X}_{\mu }( f),\; {X}_{\mu }( f)\; =\frac{\partial }{\partial {x}^{\mu }}( f \circ {\phi }^{-1}(x)).\end{eqnarray} \tag{ 1.31 }$$

In a different chart $\phi ^{\prime}$ we will have

$$\begin{eqnarray}&&{X}_{\mu }^{^{\prime} }( f)=\frac{\partial }{\partial {x}^{^{\prime} \mu }}( f \circ \phi ^{\prime} -1){\bigg | }_{{x}^{^{\prime} }=\phi ^{\prime} (p)}.\end{eqnarray} \tag{ 1.32 }$$

We compute

$$\begin{eqnarray}{X}_{\mu }( f) & = & \frac{\partial }{\partial {x}^{\mu }}( f\circ {\phi }^{-1}){\bigg | }_{x=\phi (p)}\\ & = & \frac{\partial }{\partial {x}^{\mu }}\;f \circ {\phi }^{^{\prime} -1}(\phi ^{\prime} \circ {\phi }^{-1}){\bigg | }_{x=\phi (p)}\\ & = & \sum _{\nu =1}^{n}\frac{\partial {x}^{^{\prime} \nu }}{\partial {x}^{\mu }}\frac{\partial }{\partial {x}^{^{\prime} \nu }}( f \circ {\phi }^{^{\prime} -1}(x^{\prime} )){\bigg | }_{x^{\prime} =\phi ^{\prime} (p)}\\ & = & \sum _{\nu =1}^{n}\frac{\partial {x}^{^{\prime} \nu }}{\partial {x}^{\mu }}{X}_{\nu }^{^{\prime} }( f).\end{eqnarray} \tag{ 1.33 }$$

This is why the basis elements ${X}_{\mu }$ may be thought of as the partial derivative operators $\partial /\partial {x}^{\mu }$. The tangent vector v can be rewritten as $v={\sum }_{\mu =1}^{n}{v}^{\mu }{X}_{\mu }={\sum }_{\mu =1}^{n}{v}^{^{\prime} \mu }{X}_{\mu }^{^{\prime} }$. We conclude immediately that

$$\begin{eqnarray}&&{v}^{^{\prime} \nu }=\sum _{\nu =1}^{n}\frac{\partial {x}^{^{\prime} \nu }}{\partial {x}^{\mu }}{v}^{\mu }.\end{eqnarray} \tag{ 1.34 }$$

This is the transformation law of tangent vectors under the coordinate transformation ${x}^{\mu }\to {x}^{^{\prime} \mu }$.

Cotangent dual (covariant) vectors or 1-forms. Let ${V}_{p}^{*}$ be the space of all linear maps ${\omega }^{*}$ from V_p into R, namely ${\omega }^{*}:{V}_{p}\to R$. The space ${V}_{p}^{*}$ is the so-called dual vector space to V_p where addition and multiplication by scalars are defined in an obvious way. The elements of ${V}_{p}^{*}$ are called dual vectors. The dual vector space ${V}_{p}^{*}$ is also called the cotangent dual vector space at p and the vector space of one-forms at p. The elements of ${V}_{p}^{*}$ are then called cotangent dual vectors. Another nomenclature is to refer to the elements of ${V}_{p}^{*}$ as covariant vectors, as opposed to the elements of V_p which are referred to as contravariant vectors.

The basis $\{{X}^{\mu *}\}$ of ${V}_{p}^{*}$ is called the dual basis to the basis $\{{X}_{\mu }\}$ of V_p . The basis elements of ${V}_{p}^{*}$ are given by vectors ${X}^{\mu *}$ defined by

$$\begin{eqnarray}&&{X}^{\mu *}({X}_{\nu })={\delta }_{\nu }^{\mu }.\end{eqnarray} \tag{ 1.35 }$$

We have the transformation law

$$\begin{eqnarray}&&{X}^{\mu *}= \sum _{\nu =1}^{n}\frac{\partial {x}^{\mu }}{\partial {x}^{^{\prime} \nu }}{X}^{\nu *^{\prime} }.\end{eqnarray} \tag{ 1.36 }$$

From this result we can think of the basis elements ${X}^{\mu *}$ as the gradients ${{dx}}^{\mu }$, namely

$$\begin{eqnarray}&&{X}^{\mu *}\equiv {{dx}}^{\mu }.\end{eqnarray} \tag{ 1.37 }$$

Let $v={\sum }_{\mu }{v}^{\mu }{X}_{\mu }$ be an arbitrary tangent vector in V_p , then the action of the dual basis elements ${X}^{\mu *}$ on v is given by

$$\begin{eqnarray}&&{X}^{\mu *}(v)={v}^{\mu }.\end{eqnarray} \tag{ 1.38 }$$

The action of a general element ${\omega }^{*}={\sum }_{\mu }{\omega }_{\mu }{X}^{\mu *}$ of ${V}_{p}^{*}$ on v is given by

$$\begin{eqnarray}&&{\omega }^{*}(v)=\sum _{\mu }{\omega }_{\mu }{v}^{\mu }.\end{eqnarray} \tag{ 1.39 }$$

Again we conclude the transformation law

$$\begin{eqnarray}&&{\omega }_{\nu }^{^{\prime} }=\sum _{\nu =1}^{n}\frac{\partial {x}^{\mu }}{\partial {x^{\prime} }^{\nu }}{\omega }_{\mu }.\end{eqnarray} \tag{ 1.40 }$$

Generalization. A tensor T of type (k, l) over the tangent vector space V_p is a multilinear map form $({V}_{p}^{*}\times {V}_{p}^{*}\times \cdots \; \times {V}_{p}^{*})\times ({V}_{p}\times {V}_{p}\times \cdots \; \times {V}_{p})$ into R given by

$$\begin{eqnarray}&&T :{V}_{p}^{*}\times {V}_{p}^{*}\times \cdots \times {V}_{p}^{*}\times {V}_{p}\times {V}_{p}\times \cdots \times {V}_{p}\to R.\end{eqnarray} \tag{ 1.41 }$$

The domain of this map is the direct product of k cotangent dual vector space ${V}_{p}^{*}$ and l tangent vector space V_p . The space ${ \mathcal T }\quad (k,l)$ of all tensors of type (k, l) is a vector space of dimension ${n}^{k}.{n}^{l}$ since $\mathrm{dim}{V}_{p}=\mathrm{dim}{V}_{p}^{*}=n$.

The tangent vectors $v\in {V}_{p}$ are therefore tensors of type $(1,0)$, whereas the cotangent dual vectors $v\in {V}_{p}^{*}$ are tensors of type $(0,1)$. The metric g is a tensor of type $(0,2)$, i.e. a linear map from into R, which is symmetric and non-degenerate.

A covariant derivative is a derivative which transforms covariantly under coordinates transformations $x\to x^{\prime}$. In other words, it is an operator ∇ on the manifold M which takes a differentiable tensor of type (k, l) to a differentiable tensor of type $(k,l+1)$. It must clearly satisfy the obvious properties of linearity and Leibniz rule but also satisfies other important rules such as the torsion free condition given by

$$\begin{eqnarray}&&{{\rm{\nabla }}}_{\mu }{{\rm{\nabla }}}_{\nu } f={{\rm{\nabla }}}_{\nu }{{\rm{\nabla }}}_{\mu }f, f\in { \mathcal F }.\end{eqnarray} \tag{ 1.42 }$$

Furthermore, the covariant derivative acting on scalars must be consistent with tangent vectors being directional derivatives. Indeed, for all $f\in { \mathcal F }$ and ${t}^{\mu }\in {V}_{p}$ we must have

$$\begin{eqnarray}&&{t}^{\mu }{{\rm{\nabla }}}_{\mu } f=t( f)\equiv {t}^{\mu }{\partial }_{\mu }f.\end{eqnarray} \tag{ 1.43 }$$

In other words, If ∇and $\tilde{{\rm{\nabla }}}$ are two covariant derivative operators, then their action on scalar functions must coincide, i.e.

$$\begin{eqnarray}&&{t}^{\mu }{{\rm{\nabla }}}_{\mu } f={t}^{\mu }{\tilde{{\rm{\nabla }}}}_{\mu } f=t( f).\end{eqnarray} \tag{ 1.44 }$$

We now compute the difference ${\tilde{{\rm{\nabla }}}}_{\mu }( f{\omega }_{\nu })-{{\rm{\nabla }}}_{\mu }( f{\omega }_{\nu })$ where ω is some cotangent dual vector. We have

$$\begin{eqnarray}{\tilde{{\rm{\nabla }}}}_{\mu }( f{\omega }_{\nu })-{{\rm{\nabla }}}_{\mu }( f{\omega }_{\nu }) & = & {\tilde{{\rm{\nabla }}}}_{\mu } f.{\omega }_{\nu }+f{\tilde{{\rm{\nabla }}}}_{\mu }{\omega }_{\nu }-{{\rm{\nabla }}}_{\mu } f.{\omega }_{\nu }-f{{\rm{\nabla }}}_{\mu }{\omega }_{\nu }\\ & = & f({\tilde{{\rm{\nabla }}}}_{\mu }{\omega }_{\nu }-{{\rm{\nabla }}}_{\mu }{\omega }_{\nu }).\end{eqnarray} \tag{ 1.45 }$$

We use without proof the following result. Let ${\omega }_{\nu }^{^{\prime} }$ be the value of the cotangent dual vector ${\omega }_{\nu }$ at a nearby point $p^{\prime}$, i.e. ${\omega }_{\nu }^{^{\prime} }-{\omega }_{\nu }$ is zero at p. Since the cotangent dual vector ${\omega }_{\nu }$ is a smooth function on the manifold, then for each $p^{\prime} \in M$, there must exist smooth functions ${f}_{(\alpha )}$ which vanish at the point p and cotangent dual vectors ${\mu }_{\nu }^{(\alpha )}$ such that

$$\begin{eqnarray}&&{\omega }_{\nu }^{^{\prime} }-{\omega }_{\nu }=\sum _{\alpha }{f}_{(\alpha )}{\mu }_{\nu }^{(\alpha )}.\end{eqnarray} \tag{ 1.46 }$$

We compute immediately

$$\begin{eqnarray}&&{\tilde{{\rm{\nabla }}}}_{\mu }({\omega }_{\nu }^{^{\prime} }-{\omega }_{\nu })-{{\rm{\nabla }}}_{\mu }({\omega }_{\nu }^{^{\prime} }-{\omega }_{\nu })=\sum _{\alpha }{f}_{(\alpha )}({\tilde{{\rm{\nabla }}}}_{\mu }{\mu }_{\nu }^{(\alpha )}-{{\rm{\nabla }}}_{\mu }{\mu }_{\nu }^{(\alpha )}).\end{eqnarray} \tag{ 1.47 }$$

This is 0 since by assumption ${f}_{(\alpha )}$ vanishes at p. Hence we obtain the result

$$\begin{eqnarray}&&{\tilde{{\rm{\nabla }}}}_{\mu }{\omega }_{\nu }^{^{\prime} }-{{\rm{\nabla }}}_{\mu }{\omega }_{\nu }^{^{\prime} }={\tilde{{\rm{\nabla }}}}_{\mu }{\omega }_{\nu }-{{\rm{\nabla }}}_{\mu }{\omega }_{\nu }.\end{eqnarray} \tag{ 1.48 }$$

In other words, the difference ${\tilde{{\rm{\nabla }}}}_{\mu }{\omega }_{\nu }-{{\rm{\nabla }}}_{\mu }{\omega }_{\nu }$ depends only on the value of ${\omega }_{\nu }$ at the point p, although both ${\tilde{{\rm{\nabla }}}}_{\mu }{\omega }_{\nu }$ and ${{\rm{\nabla }}}_{\mu }{\omega }_{\nu }$ depend on how ${\omega }_{\nu }$ changes as we go away from point p since they are derivatives. Putting this differently we say that the operator ${\tilde{{\rm{\nabla }}}}_{\mu }-{{\rm{\nabla }}}_{\mu }$ is a linear map which takes cotangent dual vectors at a point p into tensors, of type $(0,2)$, at p and not into tensor fields defined in a neighborhood of p. We write

$$\begin{eqnarray}&&{{\rm{\nabla }}}_{\mu }{\omega }_{\nu }={\tilde{{\rm{\nabla }}}}_{\mu }{\omega }_{\nu }-{C}_{\mu \nu }^{\gamma }{\omega }_{\gamma }.\end{eqnarray} \tag{ 1.49 }$$

The tensor ${C}_{\mu \nu }^{\gamma }$ stands for the map ${\tilde{{\rm{\nabla }}}}_{\mu }-{{\rm{\nabla }}}_{\mu }$ and it is clearly a tensor of type $(1,2)$. By setting ${\omega }_{\mu }={{\rm{\nabla }}}_{\mu }f={\tilde{{\rm{\nabla }}}}_{\mu }f$ we obtain ${{\rm{\nabla }}}_{\mu }{{\rm{\nabla }}}_{\nu }f={\tilde{{\rm{\nabla }}}}_{\mu }{\tilde{{\rm{\nabla }}}}_{\nu }f-{C}_{\mu \nu }^{\gamma }{{\rm{\nabla }}}_{\gamma }f$. By employing now the torsion free condition (1.42) we immediately obtain

$$\begin{eqnarray}&&{C}_{\mu \nu }^{\gamma }={C}_{\nu \mu }^{\gamma }.\end{eqnarray} \tag{ 1.50 }$$

Let us consider now the difference ${\tilde{{\rm{\nabla }}}}_{\mu }({\omega }_{\nu }{t}^{\nu })-{{\rm{\nabla }}}_{\mu }({\omega }_{\nu }{t}^{\nu })$ where ${t}^{\nu }$ is a tangent vector. Since ${\omega }_{\nu }{t}^{\nu }$ is a function we have

$$\begin{eqnarray}&&{\tilde{{\rm{\nabla }}}}_{\mu }({\omega }_{\nu }{t}^{\nu })-{{\rm{\nabla }}}_{\mu }({\omega }_{\nu }{t}^{\nu })=0.\end{eqnarray} \tag{ 1.51 }$$

On the other hand, we compute

$$\begin{eqnarray}&&{\tilde{{\rm{\nabla }}}}_{\mu }({\omega }_{\nu }{t}^{\nu })-{{\rm{\nabla }}}_{\mu }({\omega }_{\nu }{t}^{\nu })={\omega }_{\nu }({\tilde{{\rm{\nabla }}}}_{\mu }{t}^{\nu }-{{\rm{\nabla }}}_{\mu }{t}^{\nu }+{C}_{\mu \gamma }^{\nu }{t}^{\gamma }).\end{eqnarray} \tag{ 1.52 }$$

Hence, we must have

$$\begin{eqnarray}&&{{\rm{\nabla }}}_{\mu }{t}^{\nu }={\tilde{{\rm{\nabla }}}}_{\mu }{t}^{\nu }+{C}_{\mu \gamma }^{\nu }{t}^{\gamma }.\end{eqnarray} \tag{ 1.53 }$$

For a general tensor ${T}^{{\mu }_{1}\ldots {\mu }_{k}}{}_{{\nu }_{1}\ldots {\nu }_{l}}$ of type (k, l) the action of the covariant derivative operator will be given by the expression

$$\begin{eqnarray}{{\rm{\nabla }}}_{\gamma }{T}^{{\mu }_{1}\ldots {\mu }_{k}}{}_{{\nu }_{1}\ldots {\nu }_{l}} & = & {\tilde{{\rm{\nabla }}}}_{\gamma }{T}^{{\mu }_{1}\ldots {\mu }_{k}}{}_{{\nu }_{1}\ldots {\nu }_{l}}+\sum _{i}{C}^{{\mu }_{i}}{}_{\gamma d}{T}^{{\mu }_{1}\ldots d\ldots {\mu }_{k}}{}_{{\nu }_{1}\ldots {\nu }_{l}}\\ & & - \sum _{j}{C}^{d}{}_{\gamma {\nu }_{j}}{T}^{{\mu }_{1}\ldots {\mu }_{k}}{}_{{\nu }_{1}\ldots d\ldots {\nu }_{l}}.\end{eqnarray} \tag{ 1.54 }$$

Let C be a curve with a tangent vector ${t}^{\mu }$. Let ${v}^{\mu }$ be some tangent vector defined at each point on the curve. The vector ${v}^{\mu }$ is parallel transported along the curve C if and only if

$$\begin{eqnarray}&&{t}^{\mu }{{\rm{\nabla }}}_{\mu }{v}^{\nu }{\bigg | }_{\mathrm{curve}}=0.\end{eqnarray} \tag{ 1.55 }$$

If t is the parameter along the curve C then ${t}^{\mu }={{dx}}^{\mu }/{dt}$ are the components of the vector ${t}^{\mu }$ in the coordinate basis. The parallel transport condition reads explicitly

$$\begin{eqnarray}&&\frac{{{dv}}^{\nu }}{{dt}}+{{\rm{\Gamma }}}^{\nu }{}_{\mu \lambda }{t}^{\mu }{v}^{\lambda }=0.\end{eqnarray} \tag{ 1.56 }$$

By demanding that the inner product of two vectors ${v}^{\mu }$ and ${w}^{\mu }$ is invariant under parallel transport we obtain, for all curves and all vectors, the condition

$$\begin{eqnarray}&&{t}^{\mu }{{\rm{\nabla }}}_{\mu }({g}_{\alpha \beta }{v}^{\alpha }{w}^{\beta })=0\Rightarrow {{\rm{\nabla }}}_{\mu }{g}_{\alpha \beta }=0.\end{eqnarray} \tag{ 1.57 }$$

Thus given a metric ${g}_{\mu \nu }$ on a manifold M the most natural covariant derivative operator is the one under which the metric is covariantly constant.

There exists a unique covariant derivative operator ${{\rm{\nabla }}}_{\mu }$ which satisfies ${{\rm{\nabla }}}_{\mu }{g}_{\alpha \beta }=0$. The proof goes as follows. We know that ${{\rm{\nabla }}}_{\mu }{g}_{\alpha \beta }$ is given by

$$\begin{eqnarray}&&{{\rm{\nabla }}}_{\mu }{g}_{\alpha \beta }={\tilde{{\rm{\nabla }}}}_{\mu }{g}_{\alpha \beta }-{C}_{\mu \alpha }^{\gamma }{g}_{\gamma \beta }-{C}_{\mu \beta }^{\gamma }{g}_{\alpha \gamma }.\end{eqnarray} \tag{ 1.58 }$$

By imposing ${{\rm{\nabla }}}_{\mu }{g}_{\alpha \beta }=0$ we obtain

$$\begin{eqnarray}&&{\tilde{{\rm{\nabla }}}}_{\mu }{g}_{\alpha \beta }={C}_{\mu \alpha }^{\gamma }{g}_{\gamma \beta }+{C}_{\mu \beta }^{\gamma }{g}_{\alpha \gamma }.\end{eqnarray} \tag{ 1.59 }$$

Equivalently

$$\begin{eqnarray}&&{\tilde{{\rm{\nabla }}}}_{\alpha }{g}_{\mu \beta }={C}_{\alpha \mu }^{\gamma }{g}_{\gamma \beta }+{C}_{\alpha \beta }^{\gamma }{g}_{\mu \gamma },\end{eqnarray} \tag{ 1.60 }$$

$$\begin{eqnarray}&&{\tilde{{\rm{\nabla }}}}_{\beta }{g}_{\mu \alpha }={C}_{\mu \beta }^{\gamma }{g}_{\gamma \alpha }+{C}_{\alpha \beta }^{\gamma }{g}_{\mu \gamma }.\end{eqnarray} \tag{ 1.61 }$$

Immediately, we conclude that

$$\begin{eqnarray}&&{\tilde{{\rm{\nabla }}}}_{\mu }{g}_{\alpha \beta }+{\tilde{{\rm{\nabla }}}}_{\alpha }{g}_{\mu \beta }-{\tilde{{\rm{\nabla }}}}_{\beta }{g}_{\mu \alpha }=2{C}_{\mu \alpha }^{\gamma }{g}_{\gamma \beta }.\end{eqnarray} \tag{ 1.62 }$$

In other words,

$$\begin{eqnarray}&&{C}_{\mu \alpha }^{\gamma }=\frac{1}{2}{g}^{\gamma \beta }({\tilde{{\rm{\nabla }}}}_{\mu }{g}_{\alpha \beta }+{\tilde{{\rm{\nabla }}}}_{\alpha }{g}_{\mu \beta }-{\tilde{{\rm{\nabla }}}}_{\beta }{g}_{\mu \alpha }).\end{eqnarray} \tag{ 1.63 }$$

This choice of ${C}_{\mu \alpha }^{\gamma }$ which solves ${{\rm{\nabla }}}_{\mu }{g}_{\alpha \beta }=0$ is unique. In other words, the corresponding covariant derivative operator is unique. The most important case corresponds to the choice ${\tilde{{\rm{\nabla }}}}_{a}={\partial }_{a}$ for which case ${C}_{ab}^{c}$ is denoted ${{\rm{\Gamma }}}_{ab}^{c}$ and is called the Christoffel symbol.

Equation (1.56) is almost the geodesic equation. Recall that geodesics are the straightest possible lines on a curved manifold. Alternatively, a geodesic can be defined as a curve whose tangent vector ${t}^{\mu }$ is parallel transported along itself, i.e. ${t}^{\mu }{{\rm{\nabla }}}_{\mu }{t}^{\nu }=0$, This reads in a coordinate basis as

$$\begin{eqnarray}&&\frac{{d}^{2}{x}^{\nu }}{{{dt}}^{2}}+{{\rm{\Gamma }}}_{\mu \lambda }^{\nu }\frac{{{dx}}^{\mu }}{{dt}}\frac{{{dx}}^{\lambda }}{{dt}}=0.\end{eqnarray} \tag{ 1.64 }$$

This is precisely equation (1.28). This is a set of n-coupled second order ordinary differential equations with n unknown ${x}^{\mu }(t)$. We know, given appropriate initial conditions ${x}^{\mu }({t}_{0})$ and ${{dx}}^{\mu }/{dt}{| }_{t={t}_{0}}$, that there exists a unique solution. Conversely, given a tangent vector ${t}^{\mu }$ at a point p of a manifold M there exists a unique geodesic which goes through p and is tangent to ${t}^{\mu }$.

Definition. The parallel transport of a vector from point p to point q on the manifold M is actually path-dependent. This path dependence is directly measured by the so-called Riemann curvature tensor. The Riemann curvature tensor can be defined in terms of the failure of successive operations of differentiation to commute. Let us start with an arbitrary tangent dual vector ${\omega }_{a}$ and an arbitrary function f. We want to calculate $({{\rm{\nabla }}}_{a}{{\rm{\nabla }}}_{b}-{{\rm{\nabla }}}_{b}{{\rm{\nabla }}}_{a}){\omega }_{c}$. First we have

$$\begin{eqnarray}&&{{\rm{\nabla }}}_{a}{{\rm{\nabla }}}_{b}( f{\omega }_{c})={{\rm{\nabla }}}_{a}{{\rm{\nabla }}}_{b}f.{\omega }_{c}+{{\rm{\nabla }}}_{b}f{{\rm{\nabla }}}_{a}{\omega }_{c}+{{\rm{\nabla }}}_{a}f{{\rm{\nabla }}}_{b}{\omega }_{c}+f{{\rm{\nabla }}}_{a}{{\rm{\nabla }}}_{b}{\omega }_{c}.\end{eqnarray} \tag{ 1.65 }$$

Similarly

$$\begin{eqnarray}&&{{\rm{\nabla }}}_{b}{{\rm{\nabla }}}_{a}( f{\omega }_{c})={{\rm{\nabla }}}_{b}{{\rm{\nabla }}}_{a}f.{\omega }_{c}+{{\rm{\nabla }}}_{a}f{{\rm{\nabla }}}_{b}{\omega }_{c}+{{\rm{\nabla }}}_{b}f{{\rm{\nabla }}}_{a}{\omega }_{c}+f{{\rm{\nabla }}}_{b}{{\rm{\nabla }}}_{a}{\omega }_{c}.\end{eqnarray} \tag{ 1.66 }$$

Thus

$$\begin{eqnarray}&&({{\rm{\nabla }}}_{a}{{\rm{\nabla }}}_{b}-{{\rm{\nabla }}}_{b}{{\rm{\nabla }}}_{a})( f{\omega }_{c})=f({{\rm{\nabla }}}_{a}{{\rm{\nabla }}}_{b}-{{\rm{\nabla }}}_{b}{{\rm{\nabla }}}_{a}){\omega }_{c}.\end{eqnarray} \tag{ 1.67 }$$

We can follow the same set of arguments which led from equations ()–() to conclude that the tensor $({{\rm{\nabla }}}_{a}{{\rm{\nabla }}}_{b}-{{\rm{\nabla }}}_{b}{{\rm{\nabla }}}_{a}){\omega }_{c}$ depends only on the value of ${\omega }_{c}$ at the point p. In other words ${\text{}}{{\rm{\nabla }}}_{a}{{\rm{\nabla }}}_{b}-{\text{}}{{\rm{\nabla }}}_{b}{{\rm{\nabla }}}_{a}$ is a linear map which takes tangent dual vectors into tensors of type (0,3). Equivalently we can say that the action of ${\text{}}{{\rm{\nabla }}}_{a}{{\rm{\nabla }}}_{b}-{\text{}}{{\rm{\nabla }}}_{b}{{\rm{\nabla }}}_{a}$ on tangent dual vectors is equivalent to the action of a tensor of type (1,3). Thus we can write

$$\begin{eqnarray}&&({{\rm{\nabla }}}_{a}{{\rm{\nabla }}}_{b}-{{\rm{\nabla }}}_{b}{{\rm{\nabla }}}_{a}){\omega }_{c}={R}_{{abc}}^{d}{\omega }_{d}.\end{eqnarray} \tag{ 1.68 }$$

The tensor ${R}_{{abc}}^{d}$ is precisely the Riemann curvature tensor. We compute explicitly

$$\begin{eqnarray}{{\rm{\nabla }}}_{a}{{\rm{\nabla }}}_{b}{\omega }_{c} & = & {{\rm{\nabla }}}_{a}\left({\partial }_{b}{\omega }_{c}-{{\rm{\Gamma }}}_{{bc}}^{d}{\omega }_{d}\right)\\ & = & {\partial }_{a}\left({\partial }_{b}{\omega }_{c}-{{\rm{\Gamma }}}_{{bc}}^{d}{\omega }_{d}\right)-\;{{\rm{\Gamma }}}_{{ab}}^{e}\left({\partial }_{e}{\omega }_{c}-{{\rm{\Gamma }}}_{{ec}}^{d}{\omega }_{d}\right)-{{\rm{\Gamma }}}_{{ac}}^{e}\left({\partial }_{b}{\omega }_{e}-{{\rm{\Gamma }}}_{{be}}^{d}{\omega }_{d}\right)\\ & = & {\partial }_{a}{\partial }_{b}{\omega }_{c}-{\partial }_{a}{{\rm{\Gamma }}}_{{bc}}^{d}.{\omega }_{d}-{{\rm{\Gamma }}}_{{bc}}^{d}{\partial }_{a}{\omega }_{d}-{{\rm{\Gamma }}}_{{ab}}^{e}{\partial }_{e}{\omega }_{c}+{{\rm{\Gamma }}}_{{ab}}^{e}{{\rm{\Gamma }}}_{{ec}}^{d}{\omega }_{d}-{{\rm{\Gamma }}}_{{ac}}^{e}{\partial }_{b}{\omega }_{e}\\ & & +\;{{\rm{\Gamma }}}_{{ac}}^{e}{{\rm{\Gamma }}}_{{be}}^{d}{\omega }_{d}.\\ \end{eqnarray} \tag{ 1.69 }$$

Thus

$$\begin{eqnarray}&&({{\rm{\nabla }}}_{a}{{\rm{\nabla }}}_{b}-{{\rm{\nabla }}}_{b}{{\rm{\nabla }}}_{a}){\omega }_{c}=\left({\partial }_{b}{{\rm{\Gamma }}}_{{ac}}^{d}-{\partial }_{a}{{\rm{\Gamma }}}_{{bc}}^{d}+{{\rm{\Gamma }}}_{{ac}}^{e}{{\rm{\Gamma }}}_{{be}}^{d}-{{\rm{\Gamma }}}_{{bc}}^{a}{{\rm{\Gamma }}}_{{ae}}^{d}\right){\omega }_{d}.\end{eqnarray} \tag{ 1.70 }$$

We obtain then the components

$$\begin{eqnarray}&&{R}_{{abc}}^{d}={\partial }_{b}{{\rm{\Gamma }}}_{{ac}}^{d}-{\partial }_{a}{{\rm{\Gamma }}}_{{bc}}^{d}+{{\rm{\Gamma }}}_{{ac}}^{e}{{\rm{\Gamma }}}_{{be}}^{d}-{{\rm{\Gamma }}}_{{bc}}^{e}{{\rm{\Gamma }}}_{{ae}}^{d}.\end{eqnarray} \tag{ 1.71 }$$

The action on tangent vectors can be found as follows. Let t^a be an arbitrary tangent vector. The scalar product ${t}^{a}{\omega }_{a}$ is a function on the manifold and thus

$$\begin{eqnarray}&&({{\rm{\nabla }}}_{a}{{\rm{\nabla }}}_{b}-{{\rm{\nabla }}}_{b}{{\rm{\nabla }}}_{a})({t}^{c}{\omega }_{c})=0.\end{eqnarray} \tag{ 1.72 }$$

This leads immediately to

$$\begin{eqnarray}&&({{\rm{\nabla }}}_{a}{{\rm{\nabla }}}_{b}-{{\rm{\nabla }}}_{b}{{\rm{\nabla }}}_{a}){t}^{d}=-{R}_{{abc}}^{d}{t}^{c}.\end{eqnarray} \tag{ 1.73 }$$

Generalization of this result and the previous one to higher order tensors is given by the following equation

$$\begin{eqnarray}({{\rm{\nabla }}}_{a}{{\rm{\nabla }}}_{b}-{{\rm{\nabla }}}_{b}{{\rm{\nabla }}}_{a}){T}^{{d}_{1}\ldots {d}_{k}}{}_{{c}_{1}\ldots {c}_{l}} & = & - \sum _{i=1}^{k}{R}_{{abe}}^{{d}_{i}}{T}^{{d}_{1}\ldots e\ldots {d}_{k}}{}_{{c}_{1}\ldots {c}_{l}}\\ & & +\; \sum _{i=1}^{l}{R}_{{ab}{c}_{i}}^{c}{}^{e}{T}^{{d}_{1}\ldots {d}_{k}}{}_{{c}_{1}\ldots e\ldots {c}_{l}}. \end{eqnarray} \tag{ 1.74 }$$

Properties. We state without proof the following properties of the curvature tensor:

• •  
  Anti-symmetry in the first two indices:

  $$\begin{eqnarray}&&{R}_{{abc}}^{d}=-{R}_{{bac}}^{d}.\end{eqnarray} \tag{ 1.75 }$$
• •  
  Anti-symmetrization of the first three indices yields 0:

  $$\begin{eqnarray}&&{R}_{[{abc}]}^{d}=0,{R}_{[{abc}]}^{d}=\frac{1}{3}\left({R}_{{abc}}^{d}+{R}_{{cab}}^{d}+{R}_{{bca}}^{d}\right).\end{eqnarray} \tag{ 1.76 }$$
• •  
  Anti-symmetry in the last two indices:

  $$\begin{eqnarray}&&{R}_{{abcd}}=-{R}_{{abdc}},\quad {R}_{{abcd}}={R}_{{abc}}^{e}{g}_{{ed}}.\end{eqnarray} \tag{ 1.77 }$$
• •  
  Symmetry if the pair consisting of the first two indices is exchanged with the pair consisting of the last two indices:

  $$\begin{eqnarray}&&{R}_{{abcd}}={R}_{{cdab}}.\end{eqnarray} \tag{ 1.78 }$$
• •  
  Bianchi identity:

  $$\begin{eqnarray}&&{{\rm{\nabla }}}_{[a}{R}_{{bc}]{d}}^{e}=0,\quad {{\rm{\nabla }}}_{[a}{R}_{{bc}]{d}}^{e}=\frac{1}{3}({{\rm{\nabla }}}_{a}{R}_{{bcd}}^{e}+{{\rm{\nabla }}}_{c}{R}_{{abd}}^{e}+{{\rm{\nabla }}}_{b}{R}_{{cad}}^{e}).\end{eqnarray} \tag{ 1.79 }$$
• •  
  The so-called Ricci tensor R_ac , which is the trace part of the Riemann curvature tensor, is symmetric, namely

  $$\begin{eqnarray}&&{R}_{{ac}}={R}_{{ca}},\quad {R}_{{ac}}={R}_{{abc}}^{b}.\end{eqnarray} \tag{ 1.80 }$$
• •  
  The Einstein tensor can be constructed as follows. By contracting the Bianchi identity and using ${{\rm{\nabla }}}_{a}{g}_{{bc}}=0$ we obtain

  $$\begin{eqnarray}{g}_{e}^{c}({{\rm{\nabla }}}_{a}{R}_{{bcd}}^{e}+{{\rm{\nabla }}}_{c}{R}_{{abd}}^{e}+{{\rm{\nabla }}}_{b}{R}_{{cad}}^{e}) & = & 0\Rightarrow {{\rm{\nabla }}}_{a}{R}_{{bd}}+{{\rm{\nabla }}}_{e}{R}_{{abd}}^{e}-{{\rm{\nabla }}}_{b}{R}_{{ad}}\\ & = & 0.\end{eqnarray} \tag{ 1.81 }$$

  By contracting now the two indices b and d we obtain

  $$\begin{eqnarray}&&{g}^{{bd}}({{\rm{\nabla }}}_{a}{R}_{{bd}}+{{\rm{\nabla }}}_{e}{R}_{{abd}}^{e}-{{\rm{\nabla }}}_{b}{R}_{{ad}})=0\Rightarrow {{\rm{\nabla }}}_{a}R-2{{\rm{\nabla }}}_{b}{R}_{a}^{b}=0.\end{eqnarray} \tag{ 1.82 }$$

  This can be put in the form

  $$\begin{eqnarray}&&{{\rm{\nabla }}}^{a}{G}_{{ab}}=0.\end{eqnarray} \tag{ 1.83 }$$

  The tensor G_ab is called Einstein tensor and is given by

  $$\begin{eqnarray}&&{G}_{{ab}}={R}_{{ab}}-\frac{1}{2}{g}_{{ab}}R.\end{eqnarray} \tag{ 1.84 }$$

  The so-called scalar curvature R is defined by

  $$\begin{eqnarray}&&R={R}_{a}{}^{a}.\end{eqnarray} \tag{ 1.85 }$$

We will mostly be interested in continuous matter distributions which are extended macroscopic systems composed of a large number of individual particles. We will think of such systems as fluids. The energy, momentum, and pressure of fluids are encoded in the stress–energy–momentum tensor ${T}^{\mu \nu }$, which is a symmetric tensor of type $(2,0)$. The component ${T}^{\mu \nu }$ of the stress–energy–momentum tensor is defined as the flux of the component ${p}^{\mu }$ of the 4-vector energy–momentum across a surface of constant ${x}^{\nu }$.

Let us consider an infinitesimal element of the fluid in its rest frame. The spatial diagonal component Tⁱⁱ is the flux of the momentum pⁱ across a surface of constant xⁱ , i.e. it is the amount of momentum pⁱ per unit time per unit area traversing the surface of constant xⁱ . Thus Tⁱⁱ is the normal stress, which we also call pressure when it is independent of direction. We write ${T}^{{ii}}={P}_{i}$. The spatial off-diagonal component T^ij is the flux of the momentum pⁱ across a surface of constant x^j , i.e. it is the amount of momentum pⁱ per unit time per unit area traversing the surface of constant x^j , which means that T^ij is the shear stress.

The component T⁰⁰ is the flux of the energy p⁰ through the surface of constant x⁰, namely it is the amount of energy per unit volume at a fixed instant of time. Thus T⁰⁰ is the energy density, i.e. ${T}^{00}=\rho {c}^{2}$, where ρ is the rest-mass density. Similarly, Tⁱ⁰ is the flux of the momentum pⁱ through the surface of constant x⁰, i.e. it is the i momentum density times c. The T⁰ⁱ is the energy flux through the surface of constant xⁱ divided by c. They are equal by virtue of the symmetry of the stress–energy–momentum tensor, ${T}^{0i}={T}^{i0}$.

We begin with the case of 'dust', which is a collection of a large number of particles in spacetime at rest with respect to each other. The particles are assumed to have the same rest mass m. The pressure of the dust is obviously 0 in any direction since there is no motion of the particles, i.e. the dust is a pressureless fluid. The 4-vector velocity of the dust is the constant 4-vector velocity ${U}^{\mu }$ of the individual particles. Let n be the number density of the particles, i.e. the number of particles per unit volume as measured in the rest frame. Clearly ${N}^{i}={{nU}}^{i}=n(\gamma {u}_{i})$ is the flux of the particles, i.e. the number of particles per unit area per unit time in the xⁱ -direction. The 4-vector number-flux of the dust is defined by

$$\begin{eqnarray}&&{N}^{\mu }={{nU}}^{\mu }.\end{eqnarray} \tag{ 1.86 }$$

The rest-mass density of the dust in the rest frame is clearly given by $\rho ={nm}$. This rest-mass density times c² is the $\mu =0$, $\nu =0$ component of the stress–energy–momentum tensor ${T}^{\mu \nu }$ in the rest frame. We remark that $\rho {c}^{2}={{nmc}}^{2}$ is also the $\mu =0$, $\nu =0$ component of the tensor ${N}^{\mu }{p}^{\nu }$, where ${N}^{\mu }$ is the 4-vector number-flux and ${p}^{\mu }$ is the 4-vector energy–momentum of the dust. We define therefore the stress–energy–momentum tensor of the dust by

$$\begin{eqnarray}&&{T}^{\mu \nu }={N}^{\mu }{p}^{\nu }=({nm}){U}^{\mu }{U}^{\nu }=\rho {U}^{\mu }{U}^{\nu }.\end{eqnarray} \tag{ 1.87 }$$

The next fluid of paramount importance is the so-called perfect fluid. This is a fluid determined completely by its energy density ρ and its isotropic pressure P in the rest frame. Hence ${T}^{00}=\rho {c}^{2}$ and ${T}^{{ii}}=P$. The shear stresses T^ij ($i\ne j$) are absent for a perfect fluid in its rest frame. It is not difficult to convince ourselves that stress–energy–momentum tensor ${T}^{\mu \nu }$ is given in this case in the rest frame by

$$\begin{eqnarray}&&{T}^{\mu \nu }=\rho {U}^{\mu }{U}^{\nu }+\frac{P}{{c}^{2}}({c}^{2}{\eta }^{\mu \nu }+{U}^{\mu }{U}^{\nu })=\left(\rho +\frac{P}{{c}^{2}}\right){U}^{\mu }{U}^{\nu }+P{\eta }^{\mu \nu }.\end{eqnarray} \tag{ 1.88 }$$

This is a covariant equation and thus it must also hold, by the principle of minimal coupling (see below), in any other global inertial reference frame. We give the following examples:

• •  
  Dust: P = 0.
• •  
  Gas of photons: $P=\rho {c}^{2}/3$.
• •  
  Vacuum energy: $P=-\rho {c}^{2}\iff {T}^{{ab}}=-\rho {c}^{2}{\eta }^{{ab}}$.

The stress-energy-momentum tensor ${T}^{\mu \nu }$ is symmetric, i.e. ${T}^{\mu \nu }={T}^{\nu \mu }$. It must also be conserved:

$$\begin{eqnarray}&&{\partial }_{\mu }{T}^{\mu \nu }=0.\end{eqnarray} \tag{ 1.89 }$$

This should be thought of as the equation of motion of the perfect fluid. Explicitly this equation reads

$$\begin{eqnarray}&&{\partial }_{\mu }{T}^{\mu \nu }={\partial }_{\mu }\left(\rho +\frac{P}{{c}^{2}}\right)\cdot {U}^{\mu }{U}^{\nu }+\left(\rho +\frac{P}{{c}^{2}}\right)({\partial }_{\mu }{U}^{\mu }\cdot {U}^{\nu }+{U}^{\mu }{\partial }_{\mu }{U}^{\nu })+{\partial }^{\nu }P=0.\end{eqnarray} \tag{ 1.90 }$$

We project this equation along the 4-vector velocity by contracting it with ${U}_{\nu }$. We obtain (using ${U}_{\nu }{\partial }_{\mu }{U}^{\nu }=0$)

$$\begin{eqnarray}&&{\partial }_{\mu }(\rho {U}^{\mu })+\frac{P}{{c}^{2}}{\partial }_{\mu }{U}^{\mu }=0.\end{eqnarray} \tag{ 1.91 }$$

We project the above equation along a direction orthogonal to the 4-vector velocity by contracting it with ${P}_{\nu }^{\mu }$ given by

$$\begin{eqnarray}&&{P}_{\nu }^{\mu }={\delta }_{\nu }^{\mu }+\frac{{U}^{\mu }{U}_{\nu }}{{c}^{2}}.\end{eqnarray} \tag{ 1.92 }$$

Indeed, we can check that ${P}_{\nu }^{\mu }{P}_{\lambda }^{\nu }={P}_{\lambda }^{\mu }$ and ${P}_{\nu }^{\mu }{U}^{\nu }=0$. By contracting equation (1.90) with ${P}_{\nu }^{\lambda }$ we obtain

$$\begin{eqnarray}&&\left(\rho +\frac{P}{{c}^{2}}\right){U}^{\mu }{\partial }_{\mu }{U}_{\lambda }+\left({\eta }_{\nu \lambda }+\frac{{U}_{\nu }{U}_{\lambda }}{{c}^{2}}\right){\partial }^{\nu }P=0.\end{eqnarray} \tag{ 1.93 }$$

We consider now the non-relativistic limit defined by

$$\begin{eqnarray}&&{U}^{\mu }=(c,{u}_{i}),\; | {u}_{i}| \ll 1,\; P\ll \rho {c}^{2}.\end{eqnarray} \tag{ 1.94 }$$

The parallel equation (1.91) becomes the continuity equation given by

$$\begin{eqnarray}&&{\partial }_{t}\rho +\vec{{\rm{\nabla }}}(\rho \vec{u})=0.\end{eqnarray} \tag{ 1.95 }$$

The orthogonal equation (1.93) becomes Euler's equation of fluid mechanics given by

$$\begin{eqnarray}&&\rho ({\partial }_{t}\vec{u}+(\vec{u}\vec{{\rm{\nabla }}})\vec{u})=-\vec{{\rm{\nabla }}}P.\end{eqnarray} \tag{ 1.96 }$$

The laws of physics in general relativity can be derived from the laws of physics in special relativity by means of the so-called principle of minimal coupling. This consists in writing the laws of physics in special relativity in tensor form and then replacing the flat metric ${\eta }_{\mu \nu }$ with the curved metric ${g}_{\mu \nu }$ and the derivative operator ${\partial }_{\mu }$ with the covariant derivative operator ${{\rm{\nabla }}}_{\mu }$. This recipe works in most cases.

For example take the geodesic equation describing a free particle in special relativity given by ${U}^{\mu }{\partial }_{\mu }{U}^{\nu }=0$. Geodesic motion in general relativity is given by ${U}^{\mu }{{\rm{\nabla }}}_{\mu }{U}^{\nu }=0$. These are the geodesics of the curved metric ${g}_{\mu \nu }$ and they describe freely falling bodies in the corresponding gravitational field.

The second example is the equation of motion of a perfect fluid in special relativity which is given by the conservation law ${\partial }^{\nu }{T}_{\nu \lambda }=0$. In general relativity this conservation law becomes

$$\begin{eqnarray}&&{{\rm{\nabla }}}^{\nu }{T}_{\nu \lambda }=0.\end{eqnarray} \tag{ 1.97 }$$

Also, by applying the principle of minimal coupling, the stress–energy–momentum tensor ${T}_{\mu \nu }$ of a perfect fluid in general relativity is given by equation (1.88) with the replacement $\eta \to g$, i.e.

$$\begin{eqnarray}&&{T}_{\mu \nu }=\left(\rho +\frac{P}{{c}^{2}}\right){U}_{\mu }{U}_{\nu }+{{Pg}}_{\mu \nu }.\end{eqnarray} \tag{ 1.98 }$$

Let us first start by describing tidal gravitational forces in Newtonian physics. The force of gravity exerted by an object of mass M on a particle of mass m a distance r away is $\vec{F}=-\hat{r}{GMm}/{r}^{2}$, where $\hat{r}$ is the unit vector pointing from M to m and r is the distance between the center of M and m. The corresponding acceleration is $\vec{a}=-\hat{r}{GM}/{r}^{2}=-\vec{{\rm{\nabla }}}{\rm{\Phi }}$, ${\rm{\Phi }}=-{GM}/r$. We assume now that the mass m is spherical of radius ${\rm{\Delta }}r$. The distance between the center of M and the center of m is r. The force of gravity exerted by the mass M on a particle of mass dm a distance $r\pm {\rm{\Delta }}r$ away on the line joining the centers of M and m is given by $\vec{F}=-\hat{r}{GMdm}/{(r\pm {\rm{\Delta }}r)}^{2}$. The corresponding acceleration is

$$\begin{eqnarray}&&\vec{a}=-\hat{r}{GM}\frac{1}{{(r+{\rm{\Delta }}r)}^{2}}=-\hat{r}{GM}\frac{1}{{r}^{2}}+\hat{r}{GM}\frac{2{\rm{\Delta }}r}{{r}^{3}}+\cdots \end{eqnarray} \tag{ 1.99 }$$

The first term is precisely the acceleration experienced at the center of the body m due to M. This term does not affect the observed acceleration of particles on the surface of m. In other words, since m and everything on its surface are in a state of free fall with respect to M, the acceleration of dm with respect to m is precisely the so-called tidal acceleration, and is given by the second term in the above expansion, i.e.

$$\begin{eqnarray}{\vec{a}}_{t} & = & \hat{r}{GM}\frac{2{\rm{\Delta }}r}{{r}^{3}}+\cdots \\ & = & -({\rm{\Delta }}\hat{r}\cdot \vec{{\rm{\nabla }}})(\vec{{\rm{\nabla }}}{\rm{\Phi }}).\end{eqnarray} \tag{ 1.100 }$$

In a flat Euclidean geometry two parallel lines always remain parallel. This is not true in a curved manifold. To see this more carefully we consider a one-parameter family of geodesics ${\gamma }_{s}(t)$ which are initially parallel and see what happens to them as we move along these geodesics when we increase the parameter t. The map $(t,s)\to {\gamma }_{s}(t)$ is smooth, one-to-one, and its inverse is smooth, which means in particular that the geodesics do not cross. These geodesics will then generate a two-dimensional surface on the manifold M. The parameters t and s can therefore be chosen to be the coordinates on this surface. This surface is given by the entirety of the points ${x}^{\mu }(s,t)\in M$. The tangent vector to the geodesics is defined by

$$\begin{eqnarray}&&{T}^{\mu }=\frac{\partial {x}^{\mu }}{\partial t}.\end{eqnarray} \tag{ 1.101 }$$

This satisfies therefore the equation ${T}^{\mu }{{\rm{\nabla }}}_{\mu }{T}^{\nu }=0$. The so-called deviation vector is defined by

$$\begin{eqnarray}&&{S}^{\mu }=\frac{\partial {x}^{\mu }}{\partial s}.\end{eqnarray} \tag{ 1.102 }$$

The product ${S}^{\mu }{ds}$ is the displacement vector between two infinitesimally nearby geodesics. The vectors ${T}^{\mu }$ and ${S}^{\mu }$ commute because they are basis vectors. Hence we must have ${[T,S]}^{\mu }={T}^{\nu }{{\rm{\nabla }}}_{\nu }{S}^{\mu }-{S}^{\nu }{{\rm{\nabla }}}_{\nu }{T}^{\mu }=0$ or equivalently

$$\begin{eqnarray}&&{T}^{\nu }{{\rm{\nabla }}}_{\nu }{S}^{\mu }={S}^{\nu }{{\rm{\nabla }}}_{\nu }{T}^{\mu }.\end{eqnarray} \tag{ 1.103 }$$

This can be checked directly by using the definition of the covariant derivative and the way it acts on tangent vectors and equations (1.101) and (1.102).

The quantity ${V}^{\mu }={T}^{\nu }{{\rm{\nabla }}}_{\nu }{S}^{\mu }$ expresses the rate of change of the deviation vector along a geodesic. We will call ${V}^{\mu }$ the relative velocity of infinitesimally nearby geodesics. Similarly the relative acceleration of infinitesimally nearby geodesics is defined by ${A}^{\mu }={T}^{\nu }{{\rm{\nabla }}}_{\nu }{V}^{\mu }$. We compute

$$\begin{eqnarray}{A}^{\mu } & = & {T}^{\nu }{{\rm{\nabla }}}_{\nu }{V}^{\mu }\\ & = & {T}^{\nu }{{\rm{\nabla }}}_{\nu }({T}^{\lambda }{{\rm{\nabla }}}_{\lambda }{S}^{\mu })\\ & = & {T}^{\nu }{{\rm{\nabla }}}_{\nu }({S}^{\lambda }{{\rm{\nabla }}}_{\lambda }{T}^{\mu })\\ & = & ({T}^{\nu }{{\rm{\nabla }}}_{\nu }{S}^{\lambda })\cdot {{\rm{\nabla }}}_{\lambda }{T}^{\mu }+{T}^{\nu }{S}^{\lambda }{{\rm{\nabla }}}_{\nu }{{\rm{\nabla }}}_{\lambda }{T}^{\mu }\\ & = & ({S}^{\nu }{{\rm{\nabla }}}_{\nu }{T}^{\lambda })\cdot {{\rm{\nabla }}}_{\lambda }{T}^{\mu }+{T}^{\nu }{S}^{\lambda }({{\rm{\nabla }}}_{\lambda }{{\rm{\nabla }}}_{\nu }{T}^{\mu }-{R}_{{\nu }{\lambda }{\sigma }}^{\mu }{T}^{\sigma })\\ & = & {S}^{\lambda }{{\rm{\nabla }}}_{\lambda }({T}^{\nu }{{\rm{\nabla }}}_{\nu }{T}^{\mu })-{R}_{{\nu }{\lambda }{\sigma }}^{\mu }{T}^{\nu }{S}^{\lambda }{T}^{\sigma }\\ & = & -{R}_{{\nu }{\lambda }{\sigma }}^{\mu }{T}^{\nu }{S}^{\lambda }{T}^{\sigma }.\end{eqnarray} \tag{ 1.104 }$$

This is the geodesic deviation equation. The relative acceleration of infinitesimally nearby geodesics is 0 if and only if ${R}_{{\nu }{\lambda }{\sigma }}^{\mu }=0$. Geodesics will accelerate towards, or away from, each other if and only if ${R}_{{\nu }{\lambda }{\sigma }}^{\mu }\ne 0$. Thus initially parallel geodesics with ${V}^{\mu }=0$ will fail generically to remain parallel.

We will assume that, in general relativity, the tidal acceleration of two nearby particles is precisely the relative acceleration of infinitesimally nearby geodesics given by equation (1.104), i.e.

$$\begin{eqnarray}{A}^{\mu } & = & -{R}_{{\nu }{\lambda }{\sigma }}^{\mu }{T}^{\nu }{S}^{\lambda }{T}^{\sigma }\\ & = & -{R}_{{\nu }{\lambda }{\sigma }}^{\mu }{U}^{\nu }{\rm{\Delta }}{x}^{\lambda }{U}^{\sigma }.\end{eqnarray} \tag{ 1.105 }$$

This suggests, by comparing with equation (1.100), we make the following correspondence

$$\begin{eqnarray}&&{R}_{{\nu }{\lambda }{\sigma }}^{\mu }{U}^{\nu }{U}^{\sigma }\leftrightarrow {\partial }_{\lambda }{\partial }^{\mu }{\rm{\Phi }}.\end{eqnarray} \tag{ 1.106 }$$

Thus

$$\begin{eqnarray}&&{R}_{{\nu }{\mu }{\lambda }}^{\mu }{U}^{\nu }{U}^{\lambda }\leftrightarrow {\rm{\Delta }}{\rm{\Phi }}.\end{eqnarray} \tag{ 1.107 }$$

By using the Poisson's equation (1.1) we obtain then the correspondence

$$\begin{eqnarray}&&{R}_{{\nu }{\mu }{\sigma }}^{\mu }{U}^{\nu }{U}^{\sigma }\leftrightarrow 4\pi G\rho .\end{eqnarray} \tag{ 1.108 }$$

From the other hand, the stress–energy–momentum tensor ${T}^{\mu \nu }$ provides the correspondence

$$\begin{eqnarray}&&{T}_{\nu \sigma }{U}^{\nu }{U}^{\sigma }\leftrightarrow \rho {c}^{4}.\end{eqnarray} \tag{ 1.109 }$$

We expect therefore an equation of the form

$$\begin{eqnarray}&&\frac{{R}_{{\nu }{\mu }{\sigma }}^{\mu }}{4\pi G}=\frac{{T}_{\nu \sigma }}{{c}^{4}}\Leftarrow {R}_{\nu \sigma }=\frac{4\pi G}{{c}^{4}}{T}_{\nu \sigma }.\end{eqnarray} \tag{ 1.110 }$$

This is the original equation proposed by Einstein. However, it has the following problem. From the fact that ${{\rm{\nabla }}}^{\nu }{G}_{\nu \sigma }=0$, we immediately obtain ${{\rm{\nabla }}}^{\nu }{R}_{\nu \sigma }={{\rm{\nabla }}}_{\sigma }R/2$, and as a consequence ${{\rm{\nabla }}}^{\nu }{T}_{\nu \sigma }={c}^{4}{{\rm{\nabla }}}_{\sigma }R/8\pi G$. This result is in direct conflict with the requirement of the conservation of the stress–energy–momentum tensor given by ${{\rm{\nabla }}}^{\nu }{T}_{\nu \sigma }=0$. An immediate solution is to consider instead the equation

$$\begin{eqnarray}&&{G}_{\nu \sigma }={R}_{\nu \sigma }-\frac{1}{2}{g}_{\nu \sigma }R=\frac{8\pi G}{{c}^{4}}{T}_{\nu \sigma }.\end{eqnarray} \tag{ 1.111 }$$

The conservation of the stress–energy–momentum tensor is now guaranteed. Furthermore, this equation is still in accord with the correspondence ${R}_{\nu \sigma }{U}^{\nu }{U}^{\sigma }\leftrightarrow 8\pi G\rho$. Indeed, by using the result $R=-4\pi {GT}/{c}^{4}$ we can rewrite the above equation as

$$\begin{eqnarray}&&{R}_{\nu \sigma }=\frac{8\pi G}{{c}^{4}}\left({T}_{\nu \sigma }-\frac{1}{2}{g}_{\nu \sigma }T\right).\end{eqnarray} \tag{ 1.112 }$$

We compute ${R}_{\mu \nu }{U}^{\mu }{U}^{\nu }=(8\pi G/{c}^{4})({T}_{\mu \nu }{U}^{\mu }{U}^{\nu }+{c}^{2}T/2)$. By keeping only the $\mu =0$, $\nu =0$ component of ${T}_{\mu \nu }$ and neglecting the other components, the right-hand side is exactly $4\pi G\rho$ as it should be.

The Newtonian limit of general relativity is defined by the following three requirements:

• 1)  
  The particles are moving slowly compared with the speed of light.
• 2)  
  The gravitational field is weak so that the curved metric can be expanded about the flat metric.
• 3)  
  The gravitational field is static.

Geodesic equation. We begin with the geodesic equation, with the proper time τ as the parameter of the geodesic, which is

$$\begin{eqnarray}&&{{\rm{\Gamma }}}_{{\mu }{\nu }}^{\rho }\frac{{{dx}}^{\mu }}{d\tau }\frac{{{dx}}^{\nu }}{d\tau }+\frac{{d}^{2}{x}^{\rho }}{d{\tau }^{2}}=0.\end{eqnarray} \tag{ 1.113 }$$

The assumption that particles are moving slowly compared to the speed of light means that

$$\begin{eqnarray}&&\left| \frac{d\vec{x}}{d\tau }\right| \ll c\left| \frac{{dt}}{d\tau }\right| .\end{eqnarray} \tag{ 1.114 }$$

The geodesic equation becomes

$$\begin{eqnarray}&&{c}^{2}{{\rm{\Gamma }}}_{00}^{\rho }{\left(\frac{{dt}}{d\tau }\right)}^{2}+\frac{{d}^{2}{x}^{\rho }}{d{\tau }^{2}}=0.\end{eqnarray} \tag{ 1.115 }$$

We recall the Christoffel symbols

$$\begin{eqnarray}&&{{\rm{\Gamma }}}_{{a}{b}}^{d}=\frac{1}{2}{g}^{{dc}}({\partial }_{a}{g}_{{bc}}+{\partial }_{b}{g}_{{ac}}-{\partial }_{c}{g}_{{ab}}).\end{eqnarray} \tag{ 1.116 }$$

Since the gravitational field is static we have

$$\begin{eqnarray}&&{{\rm{\Gamma }}}_{00}^{d}=-\frac{1}{2}{g}^{{dc}}{\partial }_{c}{g}_{00}.\end{eqnarray} \tag{ 1.117 }$$

The second assumption that the gravitational field is weak allows us to decompose the metric as

$$\begin{eqnarray}&&{g}_{{ab}}={\eta }_{{ab}}+{h}_{{ab}},\quad | {h}_{{ab}}| \ll 1.\end{eqnarray} \tag{ 1.118 }$$

Thus

$$\begin{eqnarray}&&{{\rm{\Gamma }}}_{00}^{d}=-\frac{1}{2}{\eta }^{{dc}}{\partial }_{c}{h}_{00}.\end{eqnarray} \tag{ 1.119 }$$

The geodesic equation becomes

$$\begin{eqnarray}&&\frac{{d}^{2}{x}^{\rho }}{d{\tau }^{2}}=\frac{{c}^{2}}{2}{\eta }^{{dc}}{\partial }_{c}{h}_{00}{\left(\frac{{dt}}{d\tau }\right)}^{2}.\end{eqnarray} \tag{ 1.120 }$$

In terms of components this reads

$$\begin{eqnarray}&&\frac{{d}^{2}{x}^{0}}{d{\tau }^{2}}=\frac{{c}^{2}}{2}{\eta }^{00}{\partial }_{0}{h}_{00}{\left(\frac{{dt}}{d\tau }\right)}^{2}=0,\end{eqnarray} \tag{ 1.121 }$$

$$\begin{eqnarray}&&\frac{{d}^{2}{x}^{i}}{d{\tau }^{2}}=\frac{{c}^{2}}{2}{\eta }^{{ii}}{\partial }_{i}{h}_{00}{\left(\frac{{dt}}{d\tau }\right)}^{2}=\frac{{c}^{2}}{2}{\partial }_{i}{h}_{00}{\left(\frac{{dt}}{d\tau }\right)}^{2}.\end{eqnarray} \tag{ 1.122 }$$

The first equation says that ${dt}/d\tau$ is a constant. The second equation reduces to

$$\begin{eqnarray}&&\frac{{d}^{2}{x}^{i}}{{{dt}}^{2}}=\frac{{c}^{2}}{2}{\partial }_{i}{h}_{00}=-{\partial }_{i}{\rm{\Phi }},\; {h}_{00}=-\frac{2{\rm{\Phi }}}{{c}^{2}}.\end{eqnarray} \tag{ 1.123 }$$

Einstein's equations. Now we turn to the Newtonian limit of Einstein's equation ${R}_{\nu \sigma }=8\pi G({T}_{\nu \sigma }-\frac{1}{2}{g}_{\nu \sigma }T)/{c}^{4}$ with the stress–energy–momentum tensor ${T}_{\mu \nu }$ of a perfect fluid as a source. The perfect fluid is describing the Earth or the Sun. The stress–energy–momentum tensor is given by ${T}_{\mu \nu }=(\rho +P/{c}^{2}){U}_{\mu }{U}_{\nu }+{{Pg}}_{\mu \nu }$. In the Newtonian limit this can be approximated by the stress–energy–momentum tensor of dust given by ${T}_{\mu \nu }=\rho {U}_{\mu }{U}_{\nu }$ since in this limit pressure can be neglected as it comes from motion which is assumed to be slow. In the rest frame of the perfect fluid we have ${U}^{\mu }=({U}^{0},0,0,0)$ and since ${g}_{\mu \nu }{U}^{\mu }{U}^{\nu }=-{c}^{2}$ we obtain ${U}^{0}=c(1+{h}_{00}/2)$ and ${U}_{0}=c(-1+{h}_{00}/2)$, and as a consequence

$$\begin{eqnarray}&&{T}^{00}=\rho {c}^{2}(1+{h}_{00}),\; {T}_{00}=\rho {c}^{2}(1-{h}_{00}).\end{eqnarray} \tag{ 1.124 }$$

The inverse metric is obviously given by ${g}^{00}=-1-{h}_{00}$ since ${g}^{\mu \nu }{g}_{\nu \rho }={\delta }_{\rho }^{\mu }$. Hence

$$\begin{eqnarray}&&T=-\rho {c}^{2}.\end{eqnarray} \tag{ 1.125 }$$

The $\mu =0$, $\nu =0$ component of Einstein's equation is therefore

$$\begin{eqnarray}&&{R}_{00}=\frac{4\pi G}{{c}^{2}}\rho (1-{h}_{00}).\end{eqnarray} \tag{ 1.126 }$$

We recall the Riemann curvature tensor and the Ricci tensor

$$\begin{eqnarray}&&{R}_{\mu \nu \sigma }^{\lambda }={\partial }_{\nu }{{\rm{\Gamma }}}_{\mu \sigma }^{\lambda }-{\partial }_{\mu }{{\rm{\Gamma }}}_{\nu \sigma }^{\lambda }+{{\rm{\Gamma }}}_{\mu \sigma }^{\delta }{{\rm{\Gamma }}}_{\nu \delta }^{\lambda }-{{\rm{\Gamma }}}_{\nu \sigma }^{\delta }{{\rm{\Gamma }}}_{\mu \delta }^{\lambda },\end{eqnarray} \tag{ 1.127 }$$

$$\begin{eqnarray}&&{R}_{\mu \sigma }={R}_{\mu \nu \sigma }^{\nu }.\end{eqnarray} \tag{ 1.128 }$$

Thus (using in particular ${R}_{000}^{0}=0$)

$$\begin{eqnarray}&&{R}_{00}={R}_{0i0}^{i}={\partial }_{i}{{\rm{\Gamma }}}_{00}^{i}-{\partial }_{0}{{\rm{\Gamma }}}_{i0}^{i}+{{\rm{\Gamma }}}_{00}^{e}{{\rm{\Gamma }}}_{{ie}}^{i}-{{\rm{\Gamma }}}_{i0}^{e}{{\rm{\Gamma }}}_{0e}^{i}.\end{eqnarray} \tag{ 1.129 }$$

The Christoffel symbols are linear in the metric perturbation and thus one can neglect the third and fourth terms in the above equation. We then obtain

$$\begin{eqnarray}&&{R}_{00}={\partial }_{i}{{\rm{\Gamma }}}_{00}^{i}=-\frac{1}{2}{\rm{\Delta }}{h}_{00}.\end{eqnarray} \tag{ 1.130 }$$

Einstein's equation reduces therefore to Newton's equation, i.e.

$$\begin{eqnarray}&&-\frac{1}{2}{\rm{\Delta }}{h}_{00}=\frac{4\pi G}{{c}^{2}}\rho \Rightarrow {\rm{\Delta }}{\rm{\Phi }}=4\pi G\rho .\end{eqnarray} \tag{ 1.131 }$$