18.4 Pseudo-Riemannian manifolds

中文

A tensor field $g$ of type (0, 2) on a manifold $M$ is said to be non-singular if $g_p\in T^{(0,2)}$ is a non-singular tensor at every point $p\in M$ . A pseudo-Riemannian manifold $(M,g)$ consists of a differentiable manifold $M$ together with a symmetric non-singular tensor field $g$ of type (0, 2), called a metric tensor. This is equivalent to defining an inner produc $X_p\cdot Y_p=g_p(X_p,Y_p)$ on the tangent space $T_p(M)$ at every point $p\in M$ (see Chapters 5 and 7). We will assume $g$ is a differentiable tensor field, so that for every pair of smooth vector fields $X$, $Y\in \mathcal{T}(M)$ the inner product is a differentiable function,

$$g(X,Y)=g(Y,X)\in \mathcal{F}(M).$$

In any coordinate chart $(U;x^i)$ we can write

$$g=g_{ij}\mathrm{d}{x}^i\otimes \mathrm{d}{x}^j\text{ where }{g}_{ij}=g_{ji}=g({\partial }_{x^i},{\partial }_{x^j}),$$

and $\mathsf{G}=[g_{ij}]$ is a non-singular matrix at each point $p\in M.$ . As in Example 7.7 there exists a smooth inverse metric tensor $g^{-1}$ on $M$, a symmetric tensor field oftype (2, 0), such tha in any coordinate chart $(U;x^i)$ 1

$$g^{-1}=g^{ij}\frac{\partial }{\partial x^i}\otimes \frac{\partial }{\partial x^j}\text{ where }{g}^{ik}{g}_{kj}={\delta }_j^i.$$

It is always possible to find a set of orthonormal vector fields $e_1,\dots ,e_n$ on a neighbourhood of any given point $p,$ , spanning the tangent space at each point of the neighbourhood, such that

$$g(e_i,e_j)={\eta }_{ij}=\{\begin{array}{ll}{\eta }_i& \text{ if }i=j\\ 0& \text{ if }i\ne j\end{array}$$

where ${\eta }_i=\pm 1.\mathrm{At}p$ one can set up coordinates such that $e_i(p)={\left({\partial }_{x^i}\right)}_p$ , so that

$$g_p={\eta }_{ij}(\mathrm{d}{x}^i{)}_p\otimes (\mathrm{d}{x}^j{)}_p,$$

but in general it is not possible to achieve that $e_i={\partial }_{x^i}$ over an entire coordinate chart unless all Lie brackets of the orthonormal fields vanish $[e_i,e_j]=0$ . We say $(M,g)$ is a Riemannian manifold if the metric tensor is everywhere positive definite,

$$g_p(X_p,X_p)>0\text{ for all }{X}_p\ne 0\in T_p(M),$$

or equivalently, $g(X,X)\ge 0$ for all vector fields $X\in \mathcal{T}(M)$ . In this case all ${\eta }_i=1$ in the above expansion. The word Riemannian is also applied to the negative definite case, all ${\eta }_i=-1$ . If the inner product defined by $g_p$ on every tangent space is Minkowskian, as defined in Section 5.1, we say $(M,g)$ is a Minkowskian or hyperbolic manifold. In this case there exists a local orthonormal set of vector fields $e_i$ such that the associated coefficients are ${\eta }_1=ϵ,{\eta }_2=\cdot \cdot \cdot ={\eta }_n=-ϵ$ where $ϵ=\pm 1$

If $\gamma :[a,b]\to M$ is a parametrized curve on a Riemannian manifold, its length between $t_0$ and $t$ is defined to be

$$s={\int }_{t_0}^t\sqrt{g(\dot{\gamma }(u),\dot{\gamma }(u))}du.\text{\hspace{1em}(18.33)}$$

If the curve is contained in a coordinate chart $(U;x^i)$ and is written $x^i=x^i(t)$ we have

$$s={\int }_{t_0}^t\sqrt{g_{ij}\frac{\mathrm{d}{x}^i}{\mathrm{d}u}\frac{\mathrm{d}{x}^j}{u}}\mathrm{d}u.$$

Exercise

Verify that the length of the curve is independent of parametrization; i.e., $s$ is unaltered under a change of parameter $u^{\prime }=f(u)$ in the integral on the right-hand side of Eq. (18.33).

Let the value of $t_0$ in Eq. (18.33) be fixed. Then $\mathrm{d}s$ $/\mathrm{d}t=\sqrt{g(\dot{\gamma }(t),\dot{\gamma }(t))}$ , and

$${\left(\frac{\mathrm{d}s}{\mathrm{d}t}\right)}^2=g_{ij}\frac{\mathrm{d}{x}^i}{\mathrm{d}t}\frac{\mathrm{d}{x}^j}{\mathrm{d}t}.\text{\hspace{1em}(18.34)}$$

If the parameter along the curve is set to be the distance parameter $s$, the tangent vector is a unit vector along the curve,

$$g(\dot{\gamma }(s),\dot{\gamma }(s))=g_{ij}\frac{\mathrm{d}{x}^i}{\mathrm{d}s}\frac{\mathrm{d}{x}^j}{\mathrm{d}s}=1.\text{\hspace{1em}(18.35)}$$

Sometimes Eq. (18.34) is written symbolically in the form

$$\mathrm{d}{s}^2=g_{ij}\mathrm{d}{x}^i\mathrm{d}{x}^j,\text{\hspace{1em}(18.36)}$$

commonly called the metric of the space. It is to be thought of as a symbolic expression for displaying the components of the metric tensor and replaces the more correct $g=g_{ij}:\mathrm{d}{x}^i:\otimes :\mathrm{d}{x}^j$ . This may be done even in the case of an indefinite metric where, strictly speaking, we can have $\mathrm{d}{s}^2<0$

The Riemannian space $R^n$ with metric

$$\mathrm{d}{s}^2={\left(\mathrm{d}{x}^1\right)}^2+{\left(\mathrm{d}{x}^2\right)}^2+\cdots +{\left(\mathrm{d}{x}^n\right)}^2$$

is called Euclidean space and is denoted by the symbol ${\mathbb{E}}^n$ . Of course other coordinates such as polar coordinates may be used, but when we use the symbol ${\mathbb{E}}^n$ we shall usually assume that the rectilinear system is being adopted unless otherwise specified.

Example 18.1

If $(M,\phi :M\to {\mathbb{E}}^n)$ is any submanifold of ${\mathbb{E}}^n$ , it has a naturally induced metric tensor

$$g={\phi }^{\ast }\left(\mathrm{d}{x}^1\otimes \mathrm{d}{x}^1+\mathrm{d}{x}^2\otimes \mathrm{d}{x}^2+\cdots +\mathrm{d}{x}^n\otimes \mathrm{d}{x}^n\right).$$

Let $M$ be the 2-sphere of radius $a,x^2+y^2+z^2=a^2$ , and adopt polar coordinates

$$x=a\sin \theta \cos \varphi ,y=a\sin \theta \sin \varphi ,z=a\cos \theta .$$

It is straightforward to evaluate the induced metric tensor on $S^2$

$$\begin{array}{l}g={\phi }^{\ast }(\mathrm{d}x\otimes \mathrm{d}x+\mathrm{d}y\otimes \mathrm{d}y+\mathrm{d}z\otimes \mathrm{d}z)\\ =(\cos \theta \cos \varphi \mathrm{d}\theta -\sin \theta \sin \varphi \mathrm{d}\varphi )\otimes (\cos \theta \cos \varphi \mathrm{d}\theta -\sin \theta \sin \varphi \mathrm{d}\varphi )+\dots \\ =a^2(\mathrm{d}\theta \otimes \mathrm{d}\theta +{\sin }^2\mathrm{d}\varphi \otimes \mathrm{d}\varphi ).\end{array}$$

Alternatively, let $\theta =\theta (t),\varphi =\varphi (t)$ be any curve lying in $M$. The components ofits tangen vector in ${\mathbb{E}}^3$ are

$$\frac{\mathrm{d}x}{\mathrm{d}t}=\cos \theta (t)\cos \varphi (t)\frac{\mathrm{d}\theta }{\mathrm{d}t}-\sin \theta (t)\sin \varphi (t)\frac{\mathrm{d}\varphi }{\mathrm{d}t},\text{ etc. }$$

and the length of the curve in ${\mathbb{E}}^3$ is

$${\left(\frac{\mathrm{d}s}{\mathrm{d}t}\right)}^2={\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)}^2+{\left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)}^2+{\left(\frac{\mathrm{d}z}{\mathrm{d}t}\right)}^2=a^2\left({\left(\frac{\mathrm{d}\theta }{\mathrm{d}t}\right)}^2+{\sin }^2\theta {\left(\frac{\mathrm{d}\varphi }{\mathrm{d}t}\right)}^2\right).$$

The metric induced on $M$ from ${\mathbb{E}}^3$ may thus be written

$$\mathrm{d}{s}^2=a^2\mathrm{d}{\theta }^2+a^2{\cos }^2\theta \mathrm{d}{\varphi }^2.$$

Riemannian connection

A pseudo-Riemannian manifold $(M,g)$ has a natural connection $D$ defined on it that is subject to the following two requirements:

(i) $D$ is torsion-free.

(ii) The covariant derivative of the metric tensor field vanishes, $Dg=0$

This connection is called the Riemannian connection defined by the metric tensor $g$. An interesting example of a physical theory that does not impose condition (i) is the Einstein– Cartan theory where torsion represents spin [9]. Condition (ii) has the following consequence. Let $\gamma$ be a curve with tangent vector $X(t)=\dot{\gamma }(t)$ , and let $Y$ and $Z$ be vector fields parallel transported along ${\gamma }_{:}$ , so that $D_{X}Y=D_{X}Z=0$ . By Eq. (18.18) it follows that their inner product $g(Y,Z)$ is constant along the curve:

$$\begin{array}{rl}\frac{\mathrm{d}}{\mathrm{d}t}g(Y,Z)& =\frac{D}{\mathrm{d}t}g(Y,Z)=D_X(g(Y,Z))\\ & =(D_{X}g)(Y,Z)+g(D_{X}Y,Z)+g(Y,D_{X}Z)\\ & =Dg(Y,Z,X)=0.\end{array}$$

In particular every vector field $Y$ parallel transported along $\gamma$ has constant magnitude $g(Y,Y)$ along the curve, a condition that is in fact necessary and sufficient for condition (ii) to hold.

Exercise

Prove the last statement.

Conditions (i) and (ii) define a unique connection, for let $(U;x^i)$ be any local coordinate chart, and ${\Gamma }_{jk}^i={\Gamma }_{kj}^i$ the components of the connection with respect to this chart. Condition (ii) can be written using Eq. (18.14)

$$g_{ij;k}=g_{ij,k}-{\Gamma }_{ik}^m{g}_{mj}-{\Gamma }_{jk}^m{g}_{im}=0.\text{\hspace{1em}(18.37)}$$

Interchanging pairs of indices $i$, $k$ and $j$, $k$ results in

$$g_{kj;i}=g_{kj,i}-{\Gamma }_{ki}^m{g}_{mj}-{\Gamma }_{ji}^m{g}_{km}=0,\text{\hspace{1em}(18.38)}$$

$$g_{ik;j}=g_{ik,j}-{\Gamma }_{ij}^m{g}_{mk}-{\Gamma }_{kj}^m{g}_{im}=0.\text{\hspace{1em}(18.39)}$$

The combination Eq. (18.37) Eq. (18.38) Eq. (18.39) gives, on using the symmetry of $g_{ij}$ and ${\Gamma }_{ij}^m$ ,

$$g_{ij,k}+g_{kj,i}-g_{ik,j}=2g_{mj}{\Gamma }_{ik}^m.$$

Multiply through by $g^{jl}$ and, after a change of indices, we have

$${\Gamma }_{jk}^i=\frac{1}{2}{g}^{im}(g_{mj,k}+g_{mk,j}-g_{jk,m}).\text{\hspace{1em}(18.40)}$$

These expressions are called Christoffel symbols; they are the explicit expression for the components of the Riemannian connection in any coordinate system.

Exercise

Show that

$$g_{;k}^{ij}=0.\text{\hspace{1em}(18.41)}$$

Let $\gamma :\mathbb{R}M$ be a geodesic with affine parameter $t$. As the tangent vector $X(t)=\dot{\gamma }$ is parallel propagated along the curve,

$$\frac{DX}{\mathrm{d}t}=D_{X}X=0,$$

it has constant magnitude,

$$g(X,X)=g_{ij}\frac{\mathrm{d}{x}^i}{\mathrm{d}t}\frac{\mathrm{d}{x}^j}{\mathrm{d}t}=\text{ const. }$$

A scaling transformation can be applied to the affine parameter such that

$$g(X,X)=g_{ij}\frac{\mathrm{d}{x}^i}{\mathrm{d}t}\frac{\mathrm{d}{x}^j}{\mathrm{d}t}=\pm 1\text{or}0.$$

In Minkowskian manifolds, the latter case is called a null geodesic. $\mathrm{If}(M,g)$ is a Riemannian space and $p=\gamma (0)\mathrm{then}g(X,X)=1$ and the affine parameter $t$ is identical with the distance parameter along the geodesic.

Exercise

Show directly from the geodesic equation Eq. (18.9) and the Christoffel symbols Eq. (18.40) tha

$$\frac{\mathrm{d}}{\mathrm{d}t}\left(g_{ij}\frac{\mathrm{d}{x}^i}{\mathrm{d}t}\frac{\mathrm{d}{x}^j}{\mathrm{d}t}\right)=0.$$

Example 18.2

In a pseudo-Riemannian manifold the geodesic equations may be derived from a variation principle (see Section 16.5). Geodesics can be thought of as curves of stationary length,

$$\delta s=\delta {\int }_{t_1}^{t_2}\sqrt{|g(\dot{\gamma }(t),\dot{\gamma }(t))|}dt=0.$$

Let $\gamma :[t_1,t_2]\times [-a,a]M$ be a variation of the given curve $\gamma :[t_1,t_2]M_{:}$ , such that $\gamma (t,0)=\gamma (t)$ and the end points of all members of the variation are fixed, $\gamma (t_1,\lambda )=\gamma (t_1),\gamma (t_1,\lambda )=\gamma (t_1)$ for all $\lambda \in [-a,a]$ . Set the Lagrangian $L:TM\to \mathbb{R}$ to be $L(X_p)=\sqrt{|g(X_p,X_p)|}$ , and we follow the argument leading to the Euler–Lagrange equations (16.25):

$$\begin{array}{l}0=\delta s={\int }_{t_1}^{t_2}\delta L(\dot{\gamma }(t))\mathrm{d}t\\ ={\int }_{t_1}^{t_2}\frac{\mathrm{d}}{\mathrm{d}\lambda }\sqrt{|g(\dot{\gamma }(t,\lambda ),\dot{\gamma }(t,\lambda ))|}{|}_{\lambda =0}\mathrm{d}t\\ =\pm {\int }_{t_1}^{t_2}\frac{1}{2L}\left(\delta g_{ij}{\dot{x}}^i{\dot{x}}^j+2g_{ij}{\dot{x}}^i\delta {\dot{x}}^j\right)\mathrm{d}t\\ =\pm {\int }_{t_1}^{t_2}\left\{\frac{1}{2L}\delta g_{ij,k}{\dot{x}}^i{\dot{x}}^j\delta x^k-\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{g_{ik}{\dot{x}}^i}{L}\right)\delta x^k\right\}\mathrm{d}t+{\left[\frac{g_{ij}{\dot{x}}^i}{L}\delta {\dot{x}}^j\right]}_{t_1}^{t_2}\\ =\pm {\int }_{t_1}^{t_2}\left\{\frac{1}{2L}\delta g_{ij,k}{\dot{x}}^i{\dot{x}}^j-\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{g_{ik}{\dot{x}}^i}{L}\right)\right\}\delta x^k\mathrm{d}t\end{array}$$

since $\delta x^k=0$ at the end points $t=t_1$ and $t=t_2$ . Since $\delta x^k$ is arbitrary,

$$\frac{1}{2L}\delta g_{ij,k}{\dot{x}}^i{\dot{x}}^j-\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{g_{ik}{\dot{x}}^i}{L}\right)=0$$

and expanding the second term on the left and multiplying the resulting equation by $L{g}^{km}$ we find

$$\frac{{\mathrm{d}}^2{x}^m}{\mathrm{d}{t}^2}+{\Gamma }_{ij}^m\frac{\mathrm{d}{x}^i}{\mathrm{d}t}\frac{\mathrm{d}{x}^j}{\mathrm{d}t}=\frac{1}{L}\frac{\mathrm{d}L}{\mathrm{d}t}\frac{\mathrm{d}{x}^m}{\mathrm{d}t},\text{\hspace{1em}(18.42)}$$

where ${\Gamma }_{ij}^m$ are the Christoffel symbols given by Eq. (18.40). If we set $t$ to be the distance parameter $t=s$ , then $L=1$ so that $\mathrm{d}L/\mathrm{d}s=0$ and Eq. (18.42) reduces to the standard geodesic equation with affine parameter Eq. (18.9).

While we might think ofthis as telling us that geodesics are curves of‘shortest distance’ connecting any pair of points, this is by no means true in general. More usually there is a critical point along any geodesic emanating from a given point, past which the geodesic is ‘point of inflection’ with respect to distance along neighbouring curves. In pseudo-Riemannian manifolds some geodesics may even be curves of‘longest length’. For timelike geodesics in Minkowski space this is essentially the time dilatation effect – a clock carried on an arbitrary path between two events will indicate less elapsed time than an inertial clock between the two events.

Geodesic coordinates

In cartesian coordinates for Euclidean space we have $g_{ij}={\delta }_{ij}$ and by Eq. (18.40) al components of the Riemannian connection vanish, ${\Gamma }_{jk}^i=0.$ . It therefore follows from Eq. (18.25) that all components of the curvature tensor $R$ vanish. Conversely, if all components of the connection vanish in a coordinate chart $(U;x^i)$ , we have $g_{ij,k}=0$ by Eq. (18.37) and the metric tensor components $g_{ij}$ are constant through the coordinate region $U$.

In Section 18.7 we will show that a necessary and sufficent condition for ${\Gamma }_{jk}^i=0$ in a coordinate chart $(V;y^i)$ is that the curvature tensor vanish throughout an open region of the manifold. However, as long as the torsion tensor vanishes, it is always possible to find coordinates such that ${\Gamma }_{jk}^i(p)=0$ at any given point $p\in M$ . For simplicity assume that $p$ has coordinates $x^i(p)=0$ . We attempt a local coordinate transformation of the form

$$x^i=B_{i^{\prime }}^i{y}^{i^{\prime }}+A_{j^{\prime }{k}^{\prime }}^i{y}^{j^{\prime }}{y}^{k^{\prime }}$$

where $B=[B_{i^{\prime }}^i]$ and $A_{j^{\prime }{k}^{\prime }}^i=A_{k^{\prime }{j}^{\prime }}^i$ are constant coefficients. Since

$${\frac{\partial x^i}{\partial y^{i^{\prime }}}|}_p=B_{i^{\prime }}^i$$

the transformation is invertible in a neighbourhood of $p$ only if $B=[B_{i^{\prime }}^i]$ is a non-singula matrix. The new coordinates of $p$ are again zero, $y^{i^{\prime }}(p)=0$ , and using the transformation formula Eq. (18.11), we have

$${\Gamma }_{j^{\prime }{k}^{\prime }}^{\prime i^{\prime }}(p)=B_{j^{\prime }}^j{B}_{k^{\prime }}^k(B^{-1}{)}_i^{i^{\prime }}{\Gamma }_{jk}^i(p)+2A_{j^{\prime }{k}^{\prime }}^i(B^{-1}{)}_i^{i^{\prime }}=0$$

if we set

$$A_{j^{\prime }{k}^{\prime }}^i=-\frac{1}{2}{B}_{j^{\prime }}^j{B}_{k^{\prime }}^k{\Gamma }_{jk}^i(p).$$

Any such coordinates $(V;y^{j^{\prime }})$ are called geodesic coordinates, or normal coordinates, at $p$ . Their effect is to make geodesics appear locally ‘straight’ in a vanishingly smal neighbourhood of $p$ .

Exercise

Why does this procedure fail if the connection is not torsion free?

In the case of a pseudo-Riemannian manifold all derivatives of the metric tensor vanish in geodesic coordinates at $p,g_{ij,k}(p)=0$ . The constant coefficients $B_{j^{\prime }}^j$ in the above may be chosen to send the metric tensor into standard diagonal form at $p,$ , such that $g^{\prime }{}_{i^{\prime }{j}^{\prime }}(p)=g_{ij}(p)B_{i^{\prime }}^i{B}_{j^{\prime }}^j={\eta }_{i^{\prime }{j}^{\prime }}$ has values $\pm 1$ along the diagonal. Higher than first derivatives of $g_{ij}$ will not in general vanish at $p.$ . For example, in normal coordinates at $p$ the components of the curvature tensor can be expressed, using Eq. (18.40) and Eq. (18.25), in terms of the second derivatives $g_{ij,kl}(p)$

$$\begin{array}{rl}& R_{jkl}^i(p)={\Gamma }_{jl,k}^i(p)-{\Gamma }_{jk,l}^i(p)\\ & =\frac{1}{2}{\eta }^{im}(g_{ml,jk}+g_{jk,ml}-g_{mk,jl}-g_{jl,mk}){|}_p.\end{array}\text{\hspace{1em}(18.43)}$$

Problems

Problem 18.12

(a) Show that in a pseudo-Riemannian space the action principle

$$\delta {\int }_{t_1}^{t_2}L\mathrm{d}t=0$$

where $L=g_{\mu \nu }{\dot{x}}^{\mu }{\dot{x}}^{\nu }$ gives rise to geodesic equations with affine parameter $t$.

(b) For the sphere of radius $a$ in polar coordinates,

$$\mathrm{d}{s}^2=a^2(\mathrm{d}{\theta }^2+{\sin }^2\theta \mathrm{d}{\varphi }^2),$$

use this variation principle to write out the equations of geodesics, and read off from them the Christoffel symbols ${\Gamma }_{\nu \rho }^{\mu }$

(c) Verify by direct substitution in the geodesic equations that $L={\dot{\theta }}^2+{\sin }^2\theta {\dot{\varphi }}^2$ is a constant along the geodesics and use this to show that the general solution of the geodesic equations is given by

$$b\cot \theta =-\cos (\varphi -{\varphi }_0)\text{ where }b,{\varphi }_0=\text{ const }.$$

(d) Show that these curves are great circles on the sphere.

Problem 18.13

Show directly from the tensor transformation laws of $g_{ij}$ and $g^{ij}$ that the Christoffe symbols

$${\Gamma }_{jk}^i=\frac{1}{2}{g}^{ia}(g_{aj,k}+g_{ak,j}-g_{jk,a})$$

transform as components of an affine connection.