18.8 General relativity

中文

The principle of equivalence

The Newtonian gravitational force on a particle of mass $m$ is $\mathbf{F}=-m\nabla \varphi$ where the scalar potential $\varphi$ satisfies Poisson’s equation

$${\nabla }^2\varphi =4\pi G\rho .\text{\hspace{1em}(18.77)}$$

Here $G=6.672\times 10^{-8}{\mathrm{g}}^{-1}{\mathrm{cm}}^3{\mathrm{s}}^{-2}$ is Newton’s gravitational constant and $\rho$ is the density of matter present. While it is in principle possible to generalize this theory by postulating a relativistically invariant equation such as

$$\square \varphi ={\nabla }^2\varphi -\frac{{\partial }^2}{\partial t^2}\varphi =4\pi G\rho ,$$

there are a number of problems with this theory, not least of which is that it does not accord with observations.

Key to the formulation of a correct theory is the principle of equivalence that, in its simplest version, states that all particles fall with equal acceleration in a gravitational field, a fact first attributed to Galileo in an improbable tale concerning the leaning tower of Pisa. While the derivation is by simple cancellation of $m$ from both sides of the Newtonian equation of motion

$$m\ddot{x}=mg,$$

the masses appearing on the two sides of the equation could conceivably be different. On the left we really have inertial mass, $m_i.$ , which measures the particle’s resistance to any force, while on the right the mass should be identified as gravitational mass, measuring the particle’s response to gravitational fields – its ‘gravitational charge’ so to speak. The principle of equivalence can be expressed as saying that the ratio of inertial to gravitational mass will be the same for all bodies irrespective of the material from which they are composed. This was tested to one part in $10^8$ for a vast variety of materials in 1890 by Eötvös using a torsion balance, and repeated in the 1960s to one part in $10^{11}$ by Dicke using a solar balancing mechanism (see C. M. Will’s article in [10]).

The principle of equivalence essentially says that it is impossible to distinguish inertia forces such as centrifugal or Coriolis forces from gravitational ones. A nice example of the equivalence of such forces is the Einstein elevator. An observer in an elevator at rest sees objects fall to the ground with acceleration $g$. However if the elevator is set in free fall, objects around the observer will no longer appear to be subject to forces, much as if he were in an inertial frame in outer space. It has in effect been possible to transform away the gravitational field by going to a freely falling laboratory. Conversely as all bodies ‘fall’ to the floor in an accelerated rocket with the same acceleration, an observer will experience an ‘apparent’ gravitational field.

The effect of a non-inertial frame in special relativity should be essentially indistinguish able from the effects of gravity. The metric interval of Minkowski space (see Chapter 9) in a general coordinate system $x^{\prime {\mu }^{\prime }}=x^{\prime {\mu }^{\prime }}(x^1,x^2,x^3,x^4)$ , where $x^{\nu }$ are inertial coordinates, becomes

$$\mathrm{d}{s}^2=g_{\mu \nu }\mathrm{d}{x}^{\mu }\mathrm{d}{x}^{\nu }=g_{{\mu }^{\prime }{\nu }^{\prime }}^{\prime }\mathrm{d}{x}^{\prime {\mu }^{\prime }}\mathrm{d}{x}^{\prime {\nu }^{\prime }},$$

where

$$g_{{\mu }^{\prime }{\nu }^{\prime }}^{\prime }=g_{\mu \nu }\frac{\partial x^{\mu }}{\partial x^{\prime {\mu }^{\prime }}}\frac{\partial x^{\nu }}{\partial x^{\prime {\nu }^{\prime }}}.$$

Expressed in general coordinates, the (geodesic) equations of motion of an inertial particle, ${\mathrm{d}}^2{x}^{\mu }/\mathrm{d}{s}^2=0$ , are

$$\frac{{\mathrm{d}}^2{x}^{\prime {\mu }^{\prime }}}{\mathrm{d}{s}^2}+{\Gamma }_{{\alpha }^{\prime }{\beta }^{\prime }}^{\prime {\mu }^{\prime }}\frac{\mathrm{d}{x}^{\prime {\alpha }^{\prime }}}{\mathrm{d}s}\frac{\mathrm{d}{x}^{\prime {\beta }^{\prime }}}{\mathrm{d}s}=0,$$

where

$${\Gamma }_{{\alpha }^{\prime }{\beta }^{\prime }}^{\prime {\mu }^{\prime }}=-\frac{{\partial }^2{x}^{\prime {\mu }^{\prime }}}{\partial x^{\mu }\partial x^{\nu }}\frac{\partial x^{\mu }}{\partial x^{\prime {\alpha }^{\prime }}}\frac{\partial x^{\nu }}{\partial x^{\prime {\beta }^{\prime }}}.$$

Figure 18.2 Tidal effects in a freely falling laboratory

The principle of equivalence is a purely local idea, and only applies to vanishingly smal laboratories. A real gravitational field such as that due to the Earth cannot be totally transformed away in general. For example, if the freely falling Einstein elevator has significan size compared to the scale on which there is variation in the Earth’s gravitational field, then particles at different positions in the lift will undergo different accelerations. Particles near the floor of the elevator will have larger accelerations than particles released from the ceiling, while particles released from the sides of the elevator will have a small horizonta acceleration relative to the central observer because the direction to the centre of the Earth is not everywhere parallel. These mutual accelerations or tidal forces can be measured in principle by connecting pairs of freely falling particles with springs (see Fig. 18.2).

Postulates of general relativity

The basic proposition of general relativity (Einstein, 1916) is the following: the world is a four-dimensional Minkowskian manifold (pseudo-Riemannian of index +2) called spacetime. Its points are called events. The world-line, or space-time history of a materia particle, is a parametrized curve $\gamma :\mathbb{R}M$ whose tangent $\dot{\gamma }$ is everywhere timelike. The proper time, or time as measured by a clock carried by the particle between parameter values $\lambda ={\lambda }_1$ and ${\lambda }_2$ , is given by

$$\tau =\frac{1}{c}{\int }_{{\lambda }_1}^{{\lambda }_2}\sqrt{-g(\dot{\gamma },\dot{\gamma })}d\lambda =\frac{1}{c}{\int }_{{\lambda }_1}^{{\lambda }_2}\sqrt{-g_{\mu \nu }\frac{d{x}^{\mu }}{d\lambda }\frac{d{x}^{\nu }}{d\lambda }}d\lambda =\frac{1}{c}{\int }_{s_1}^{s_2}ds.\text{\hspace{1em}(18.78)}$$

A test particle is a particle of very small mass compared to the major masses in its neighbourhood and ‘freely falling’ in the sense that it is subject to no external forces. The world-line of a test particle is assumed to be a timelike geodesic. The world-line of a photon of small energy is a null geodesic. Both satisfy equations

$$\frac{{\mathrm{d}}^2{x}^{\mu }}{\mathrm{d}{s}^2}+{\Gamma }_{\nu \rho }^{\mu }\frac{\mathrm{d}{x}^{\nu }}{\mathrm{d}s}\frac{\mathrm{d}{x}^{\rho }}{\mathrm{d}s}=0,$$

where the Christoffel symbols ${\Gamma }_{\nu \rho }^{\mu }$ are given by Eq. (18.40) with Greek indices subsituted. The affine parameter $s$ is determined by

$$g_{\mu \nu }\frac{\mathrm{d}{x}^{\mu }}{\mathrm{d}s}\frac{\mathrm{d}{x}^{\nu }}{\mathrm{d}s}=\{\begin{array}{ll}-1& \text{ for test particles },\\ 0& \text{ for photons }.\end{array}$$

An introduction to the theory of general relativity, together with many of its developments, can be found in [9, 11–16].

The principle of equivalence has a natural place in this postulate, since all particles have the same geodesic motion independent of their mass. Also, at each event $p$, geodesic norma coordinates may be found such that components of the metric tensor $g=g_{\mu \nu }:\mathrm{d}{x}^{\mu }\otimes \mathrm{d}{x}^{\nu }$ have the Minkowski values $g_{\mu \nu }(p)={\eta }_{\mu \nu }$ and the Christoffel symbols vanish, ${\Gamma }_{\nu \rho }^{\mu }(p)=0$ . In such coordinates the space-time appears locally to be Minkowski space, and gravitational forces have been locally transformed away since any geodesic at $p$ reduces to a ‘local rectilinear motion’ ${\mathrm{d}}^2{x}^{\mu }/\mathrm{d}{s}^2=0$ . When it is possible to find coordinates that transform the metric to constant values on an entire chart, the metric is locally flat and all gravitationa fields are ‘fictitious’ since they arise entirely from non-inertial effects. By Theorem 18.1 such coordinate transformations are possible if and onl if the curvature tensor vanishes. Hence it is natural to identify the ‘real’ gravitational field with the curvature tensor $R_{\nu \rho \sigma }^{\mu }$

The equations determining the gravitational field are Einstein’s field equations

$$G_{\mu \nu }=R_{\mu \nu }-\frac{1}{2}R{g}_{\mu \nu }=\kappa T_{\mu \nu },\text{\hspace{1em}(18.79)}$$

where $G_{\mu \nu }$ is the Einstein tensor defined above in Eqs. Eq. (18.61) and Eq. (18.62),

$$R_{\mu \nu }=R_{\mu \alpha \nu }^{\alpha }\text{ and }R=R_{\alpha }^{\alpha }=g^{\mu \nu }{R}_{\mu \nu },$$

and $T_{\mu \nu }$ is the energy–stress tensor of all the matter fields present. The constant $\kappa$ is known as Einstein’s gravitational constant; we shall relate it to Newton’s gravitational constant directly. These equations have the property that for weak fields, $g_{\mu \nu }\approx {\eta }_{\mu \nu }$ , they reduce to the Poisson equation when appropriate identifications are made (see the discussion of the weak field approximation below), and by the contracted Bianchi identity Eq. (18.60) they guarantee a ‘covariant’ version of the conservation identities

$$T_{;\nu }^{\mu \nu }\equiv T_{,\nu }^{\mu \nu }+{\Gamma }_{\alpha \nu }^{\mu }{T}^{\alpha \nu }+{\Gamma }_{\alpha \nu }^{\nu }{T}^{\mu \alpha }=0.$$

Measurement of the curvature tensor

The equation of geodesic deviation Eq. (18.47) can be used to give a physical interpretation of the curvature tensor. Consider a one-parameter family of timelike geodesics $\gamma (s,\lambda )$ with tangent vectors $U=U^{\mu }{\partial }_{x^{\mu }}\mu$ when expressed in a coordinate chart $(A;x^{\mu })$ , where

$$U^{\mu }(s)={\frac{\partial x^{\mu }}{\partial s}|}_{\lambda =0},U^{\mu }{U}_{\mu }=-1,\frac{D{U}^{\mu }}{\partial s}=0.$$

Suppose the connection vector $Y=Y^{\mu }{\partial }_{x^{\mu }}$ , where $Y^{\mu }=\partial x^{\mu }/\partial \lambda$ , is initially orthogonal to $U$ at $s=s_0$ ,

$${Y^{\mu }{U}_{\mu }|}_{s=s_0}=g(U,Y)(s_0)=0.$$

Figure 18.3 Physical measurement of curvature tensor

We then have $g(U,Y)=0$ for all $s,$ , since $g(U,Y)$ is constant along each geodesic (see Section 18.5). Thus if $e_1,e_2,e_3$ are three mutually orthogonal spacelike vectors at $s=s_0$ on the central geodesic $\lambda =0$ that are orthogonal to $U_{:}$ , and they are parallel propagated along this geodesic, $D{e}_i/\partial s=0$ , then they remain orthogonal to each other and $U$ along this geodesic,

$$g(e_i,e_j)={\delta }_{ij},g(e_i,u)=0.$$

In summary, if we set $e_4=U$ then the four vectors $e_1,\dots ,e_4$ are an orthonormal tetrad of vectors along $\lambda ={\lambda }_0$

$$g(e_{\mu },e_{\nu })={\eta }_{\mu \nu }.$$

The situation is depicted in Fig. 18.3. Let $\lambda ={\lambda }_0+\delta \lambda$ be any neighbouring geodesic from the family, then since we are assuming $\delta x^{\mu }$ is orthogonal to $U^{\mu }$ , the equation of geodesic deviation in the form Eq. (18.48) can be written

$$\frac{D^2\delta x^{\mu }}{\mathrm{d}{s}^2}=\delta \lambda \frac{D^2{Y}^{\mu }}{\mathrm{d}{s}^2}=R_{\alpha \rho \nu }^{\mu }{U}^{\alpha }{U}^{\rho }\delta x^{\nu }.\text{\hspace{1em}(18.80)}$$

Expanding $\delta x^{\mu }$ in terms of the basis $e_{\mu }$ we have, adopting a cartesian tensor summation convention,

$$\delta x^{\mu }={\eta }_j{e}_j^{\mu }\equiv \sum _{j=1}^3{\eta }_j{e}_j^{\mu },$$

where ${\eta }_j={\eta }_j(s)$ , so that

$$\frac{D\delta x^{\mu }}{\mathrm{d}s}=\frac{\mathrm{d}{\eta }_j}{\mathrm{d}s}{e}_j^{\mu }+{\eta }_j\frac{e_i^{\mu }}{\mathrm{d}s}=\frac{\mathrm{d}{\eta }_j}{\mathrm{d}s}{e}_j^{\mu },$$ $$\frac{D^2\delta x^{\mu }}{\mathrm{d}{s}^2}=\frac{{\mathrm{d}}^2{\eta }_j}{\mathrm{d}{s}^2}{e}_j^{\mu }+\frac{\mathrm{d}{\eta }_j}{\mathrm{d}s}\frac{e_i^{\mu }}{\mathrm{d}s}=\frac{{\mathrm{d}}^2{\eta }_j}{\mathrm{d}{s}^2}{e}_j^{\mu }.$$

Substituting Eq. (18.80) results in

$$\frac{{\mathrm{d}}^2{\eta }_i}{\mathrm{d}{s}^2}=e_{i\mu }\frac{D^2\delta x^{\mu }}{\mathrm{d}{s}^2}=R_{\alpha \rho \nu }^{\mu }{e}_{i\mu }{e}_4^{\alpha }{e}_4^{\rho }{e}_j^{\nu }{\eta }_j,$$

which reads in any local coordinates at any point on $\lambda ={\lambda }_0$ such that $e_{\alpha }^{\mu }={\delta }_{\alpha }^{\mu }$

$$\frac{{\mathrm{d}}^2{\eta }_i}{\mathrm{d}{s}^2}=-R_{i4j4}{\eta }_j.\text{\hspace{1em}(18.81)}$$

Thus $R_{i4j4}\equiv R_{\mu \alpha \rho \nu }{e}_i^{\mu }{e}_4^{\alpha }{e}_j^{\rho }{e}_4^{\nu }$ measures the relative accelerations between neighbouring freely falling particles in the gravitational field. Essentially these are what are termed tidal forces in Newtonian physics, and could be measured by the strain on a spring connecting the two particles (see Fig. 18.2 and [17]).

The linearized approximation

Consider a one-parameter family of Minkowskian metrics having components $g_{\mu \nu }=g_{\mu \nu }(x^{\alpha },ϵ)$ such that $ϵ=0$ reduces to flat Minkowski space, $g_{\mu \nu }(0)={\eta }_{\mu \nu }$ . Such a fam ily is known as a linearized approximation of general relativity. If we set

$$h_{\mu \nu }=\frac{\partial g_{\mu \nu }}{\partial ϵ}{|}_{ϵ=0}\text{\hspace{1em}(18.82)}$$

then for $|ϵ|\ll 1$ we have ‘weak gravitational fields’ in the sense that the metric is only slightly different from Minkowski space,

$$g_{\mu \nu }\approx {\eta }_{\mu \nu }+ϵh_{\mu \nu }.$$

From $g^{\mu \rho }{g}_{\rho \nu }={\delta }_{\mu }^{\mu }$ it follows by differentiating with respect to $ϵ$ at $ϵ=0$ that

$$\frac{\partial g^{\mu \rho }}{\partial ϵ}{|}_{ϵ=0}{\eta }_{\rho \nu }+{\eta }^{\mu \rho }{h}_{\rho \nu }=0,$$

whence

$${\frac{\partial g^{\mu \nu }}{\partial ϵ}|}_{ϵ=0}=-h^{\mu \nu }\equiv -{\eta }^{\mu \alpha }\eta \nu \beta .\text{\hspace{1em}(18.83)}$$

In this equation and throughout the present discussion indices are raised and lowered with respect to the Minkowski metric, ${\eta }_{\mu \nu },{\eta }^{\mu \nu }$ . For $ϵ\ll 1$ we evidently have $g^{\mu \nu }\approx {\eta }^{\mu \nu }-ϵh^{\mu \nu }$

Assuming that partial derivatives with respect to $x^{\mu }$ and $ϵ$ commute, it is straightforward to compute the linearization of the Christoffel symbols,

$$\begin{array}{rl}& {\Gamma }_{\nu \rho }^{\mu }\equiv \frac{\partial }{\partial ϵ}{\Gamma }_{\nu \rho }^{\mu }{|}_{ϵ=0}=-\frac{1}{2}{h}^{\mu \alpha }({\eta }_{\alpha \nu ,\rho }+{\eta }_{\alpha \rho ,\nu }-{\eta }_{\nu \rho ,\alpha })+\frac{1}{2}{\eta }^{\mu \alpha }(h_{\alpha \nu ,\rho }+h_{\alpha \rho ,\nu }-h_{\nu \rho ,\alpha })\\ & =\frac{1}{2}{\eta }^{\mu \alpha }(h_{\alpha \nu ,\rho }+h_{\alpha \rho ,\nu }-h_{\nu \rho ,\alpha })\end{array}$$

and

$${\Gamma }_{\nu \rho ,\sigma }^{\mu }=\frac{1}{2}{\eta }^{\mu \alpha }(h_{\alpha \nu ,\rho \sigma }+h_{\alpha \rho ,\nu \sigma }-h_{\nu \rho ,\alpha \sigma }).$$

Thus, from the component expansion of the curvature tensor Eq. (18.25), we have

$$r_{\nu \rho \sigma }^{\mu }\equiv \frac{\partial }{\partial ϵ}{R}_{\nu \rho \sigma }^{\mu }{|}_{ϵ=0}={\Gamma }_{\nu \sigma ,\rho }^{\mu }-{\gamma }_{\nu \rho ,\sigma }^{\mu }$$

since

$$\frac{\partial }{\partial ϵ}{\Gamma }_{\nu \sigma }^{\alpha }{\Gamma }_{\alpha \rho }^{\mu }{|}_{ϵ=0}={\Gamma }_{\nu \sigma }^{\alpha }{\Gamma }_{\alpha \rho }^{\mu }{|}_{ϵ=0}+{\Gamma }_{\nu \sigma }^{\alpha }{|}_{ϵ=0}{\Gamma }_{\alpha \rho }^{\mu }=0,etc.$$

since ${\Gamma }_{\nu \sigma }^{\alpha }{|}_{ϵ=0}=0$ . Thus

$$r_{\nu \rho \sigma }^{\mu }=\frac{1}{2}{\eta }^{\mu \alpha }\left(h_{\alpha \nu ,\sigma \rho }+h_{\alpha \sigma ,\nu \rho }-h_{\nu \sigma ,\alpha \rho }-h_{\alpha \nu ,\rho \sigma }-h_{\alpha \rho ,\nu \sigma }+h_{\nu \rho ,\alpha \sigma }\right)$$

and for small values of the parameter $ϵ$ the Riemann curvature tensor is

$$R_{\mu \nu \rho \sigma }\approx ϵr_{\mu \nu \rho \sigma }=\frac{ϵ}{2}(h_{\mu \sigma ,\nu \rho }+h_{\nu \rho ,\mu \sigma }-h_{\nu \sigma ,\mu \rho }-h_{\mu \rho ,\nu \sigma }).\text{\hspace{1em}(18.84)}$$

It is interesting to compare this equation with the expression in geodesic normal coordinates, Eq. (18.43).

The Newtonian tidal equation is derived by considering the motion oftwo neighbouring particles

$${\ddot{x}}_i=-{\varphi }_{,i}\text{ and }{\ddot{x}}_i+{\ddot{\eta }}_i=-{\varphi }_{,i}(\mathbf{x}+\mathbf{\eta }).$$

Since ${\varphi }_{,i}(\mathbf{x}+\eta )={\varphi }_{,i}(\mathbf{x})+{\varphi }_{ij}(\mathbf{x}){\eta }_j$ we have

$${\ddot{\eta }}_i=-{\varphi }_{,ij}(\mathbf{x}){\eta }_j.$$

Compare with the equation of geodesic deviation Eq. (18.81) with $s$ replaced by $c{t}_{:}$ , which is approximately correct for velocities $|\dot{\mathbf{x}}|\ll c,$

$${\ddot{\eta }}_i=-c^2{R}_{i4j4}{\eta }_j,$$

and we should have, by Eq. (18.84),

$$R_{i4j4}=\frac{ϵ}{2}(h_{i4,4j}+h_{4j,i4}-h_{ij,44}-h_{44,ij})=\frac{{\varphi }_{,ij}}{c^2}.$$

This equation can only hold in a general way if

$$ϵh_{44}\approx -\frac{2\varphi }{c^2}\text{ and }{h}_{i4,4j},h_{ij,44}\ll h_{44,ij},$$

and the Newtonian approximation implies that

$$g_{44}\approx -1+ϵh_{44}\approx -1-\frac{2\varphi }{c^2}(\varphi \ll c^2).\text{\hspace{1em}(18.85)}$$

Note that the Newtonian potentia $\varphi$ has the dimensions of a velocity square – the weak field slow motion approximation ofgeneral relativity arises when this velocity is small compared to the velocity of light $c$.

Multiplying Eq. (18.79) through by $g^{\mu \nu }$ we find

$$R-2R=-R=\kappa T\text{ where }T=T_{\mu }^{\mu }=g^{\mu \nu }{T}_{\mu \nu },$$

and Einstein’s field equations can be written in the ‘Ricci tensor form

$$R_{\mu \nu }=\kappa (T_{\mu \nu }-\frac{1}{2}T{g}_{\mu \nu }).\text{\hspace{1em}(18.86)}$$

Hence

$$R_{44}=R_{4i4}^i\approx \frac{1}{c^2}{\nabla }^2\varphi =\kappa (T_{44}+\frac{1}{2}T).$$

If we assume a perfect fluid, Example 9.4, for low velocities compared to $c$ we have

$$T_{\mu \nu }=\left(\rho +\frac{P}{c^2}\right)V_{\mu }{V}_{\nu }+P{g}_{\mu \nu }\text{ where }{V}_{\mu }\approx (v_1,v_2,v_3,-c),$$

so that

$$T=-c^2(\rho +\frac{P}{c^2})+4P=-\rho c^2+3P\approx -\rho c^2,$$

and

$$T_{44}\approx (\rho +\frac{P}{c^2})c^2+P(-1-\frac{\varphi }{c^2}P)\approx \rho c^2.$$

Substituting in the Ricci form of Einstein’s equations we find

$$\frac{1}{c^2}{\nabla }^2\varphi \approx \frac{1}{2}\kappa \rho c^2,$$

which is in agreement with the Newtonian equation Eq. (18.77) provided Einstein’s gravitational constant has the form

$$\kappa =\frac{8\pi G}{c^4}.\text{\hspace{1em}(18.87)}$$

Exercise

Show that the contracted Bianchi identity Eq. (18.60) implies that in geodesic coordinates at any point representing a local freely falling frame, the conservation identities (9.56) hold $T^{\mu }{}_{\nu ,\mu }=0$

Exercise

Show that if we had assumed field equations of the form $R_{\mu \nu }=\lambda T_{\mu \nu }$ , there would have resulted the physically unsavoury result $T=\mathrm{const}$

Consider now the effect of a one-parameter family of coordinate transformations $x^{\mu }=x^{\mu }(y^{\alpha },ϵ)$ on a linearized approximation $g_{\mu \nu }=g_{\mu \nu }(ϵ)$ and set

$${\xi }^{\mu }=\frac{\partial x^{\mu }}{\partial ϵ}{|}_{ϵ=0}.$$

The transformation of components of the metric tensor results in

$$g_{\mu \nu }(\mathbf{x},ϵ)\to g_{\mu \nu }^{\prime }(\mathbf{y},ϵ)=g_{\alpha \beta }(\mathbf{x}(\mathbf{y},ϵ),ϵ)\frac{\partial x^{\alpha }}{\partial y^{\mu }}\frac{\partial x^{\beta }}{\partial y^{\nu }}$$

and taking $\partial /\partial ϵ$ at $ϵ=0$ gives

$$h_{\mu \nu }^{\prime }(\mathbf{y})=h_{\mu \nu }(\mathbf{y})+{\xi }_{\mu ,\nu }+{\xi }_{\nu ,\mu }.\text{\hspace{1em}(18.88)}$$

These may be thought ofas ‘gauge transformations’ for the weak fields $h_{\mu \nu }$ , comparable with the gauge transformations (9.49), ${\mathcal{A}}_{\mu }^{\prime }={\mathcal{A}}_{\mu }+{\psi }_{,\mu }$ , which leave the electromagnetic field $F_{\mu \nu }$ unchanged. In the present case, it is straightforward to verify that the transformations Eq. (18.88) leave the linearized Riemann tensor Eq. (18.84), or real gravitationalfield, invariant.

We define the quantities ${\phi }_{\mu \nu }$ by

$${\phi }_{\mu \nu }=h_{\mu \nu }-h{\eta }_{\mu \nu }\text{ where }h=h_{\alpha }^{\alpha }={\eta }^{\alpha \beta }{h}_{\alpha \beta }.$$

The transformation of ${\phi }_{\mu ,\mathfrak{i}}^{\nu }$ under a gauge transformation is then

$${\phi }_{\mu ,\nu }^{\prime \nu }={\phi }_{\mu ,\nu }^{\nu }+{\xi }_{\mu ,\nu }^{\nu }={\phi }_{\mu ,\nu }^{\nu }+\square {\xi }_{\mu },$$

where indices are raised and lowered with the Minkowski metric, ${\eta }^{\mu \nu },{\eta }_{\mu \nu }$ . Just as done for the Lorentz gauge (9.51), it is possible (after dropping primes) to find $\xi$ such that

$${\phi }_{\mu ,\nu }^{\nu }=0.\text{\hspace{1em}(18.89)}$$

Such a gauge is commonly known as a harmonic gauge. There are still available gauge freedoms ${\xi }_{\mu }$ subject to solutions of the wave equation $\square {\xi }_{\mu }=0$

A computation of the linearized Ricci tensor $r_{\mu \nu }={\eta }^{\rho \sigma }{r}_{\rho \mu \sigma \nu }$ using Eq. (18.84) give

$$r_{\mu \nu }=\frac{1}{2}(-\square h_{\mu \nu }+{\phi }_{\nu ,\rho \mu }^{\rho }+{\phi }_{\mu ,\rho \nu }^{\rho })=-\frac{1}{2}\square h_{\mu \nu }$$

in a harmonic gauge. The Einstein tensor is thus $G_{\mu \nu }\approx -(ϵ/2){\Theta }_{\mu \nu }$ , and the linearized Einstein equation is

$$ϵ\square {\phi }_{\mu \nu }=-\kappa T_{\mu \nu }=-\frac{16\pi G}{c^4}{T}_{\mu \nu },$$

having solution in terms of retarded Green’s functions (12.23)

$$ϵ{\phi }_{\mu \nu }(\mathbf{x},t)=-\frac{4G}{c^4}\iiint \frac{[T_{\mu \nu }({\mathbf{x}}^{\prime },t^{\prime }){]}_{\mathrm{ret}}}{|\mathbf{x}-{\mathbf{x}}^{\prime }|}{\mathrm{d}}^3{x}^{\prime }.$$

In vacuo, $T_{\mu \nu }$ , Einstein’s field equations can be written $R_{\mu \nu }=0$ , so that in the linearized approximation we have $\boxed{}{h}_{\mu \nu }=0$ ; these solutions are known as gravitational waves (see Problem 18.20 for further details).

The Schwarzschild solution

The vacuum Einstein field equations, $R_{\mu \nu }=0$ are a non-linear set of 10 second-order equations for 10 unknowns $g_{\mu \nu }$ that can only be solved in a handful of special cases. The most important is that of spherical symmetry which, as we shall see in the next chapter, implies that the metric has the form in a set of coordinates $x^1=r,x^2=\theta ,x^3=\varphi ,x^4=ct$

$$\mathrm{d}{s}^2={\mathrm{e}}^{\lambda }\mathrm{d}{r}^2+r^2(\mathrm{d}{\theta }^2+{\sin }^2\theta \mathrm{d}{\varphi }^2)-{\mathrm{e}}^{\nu }{c}^2\mathrm{d}{t}^2\text{\hspace{1em}(18.90)}$$

where $\theta$ and $\varphi$ take the normal ranges ofpolar coordinates ($r$ does not necessarily range from 0 to $\infty$), and $\lambda$ and $\nu$ are functions of $r$ and $t$. We will assume for simplicity that the solutions are static so that they are functions of the radial coordinate $r$ alone, $\lambda =\lambda (r),\nu =\nu (r)$ . A remarkable theorem of Birkhoff assures us that all spherically symmetric vacuum solutions are in fact static for an appropriate choice of the coordinate $t$; a proof may be found in Synge [15].

We will perform calculations using Cartan’s formalism. Many books prefer to calculate Christoffel symbols and do all computations in the coordinate system of Eq. (18.90). Let $e_1,\dots ,e_4$ be the orthonormal basis

$$e_1={\mathrm{e}}^{-\frac{1}{2}\lambda }{\partial }_r,e_2=\frac{1}{r}{\partial }_{\theta },e_3=\frac{1}{r\sin \theta }{\partial }_{\varphi },e_4={\mathrm{e}}^{-\frac{1}{2}v}{c}^{-1}{\partial }_t$$

such that

$$g_{\mu \nu }=g(e_{\mu },e_{\nu })=\eta =\mathrm{diag}(1,1,1,-1)$$

and let ${\epsilon }^1,\dots ,{\epsilon }^4$ be the dual basis

$${\epsilon }^1={\mathrm{e}}^{\frac{1}{2}\lambda }\mathrm{d}r,{\epsilon }^2=r\mathrm{d}\theta ,{\epsilon }^3=r\sin \theta \mathrm{d}\varphi ,{\epsilon }^4={\mathrm{e}}^{\frac{1}{2}\nu }c\mathrm{d}t={\mathrm{e}}^{\frac{1}{2}\nu }\mathrm{d}{x}^4.$$

We will write Cartan’s structural relations in terms of the ‘lowered’ connection forms ${\omega }_{\mu \nu }$ since, by Eq. (18.73), we have ${\omega }_{\mu \nu }=-{\omega }_{\nu \mu }$ . Thus Eq. (18.64) can be written

$$\mathrm{d}{\epsilon }^{\mu }=-{\omega }_{\nu }^{\mu }\wedge {\epsilon }^{\nu }=-{\eta }^{\mu \rho }{\omega }_{\rho \nu }{\epsilon }^{\nu }$$

and setting successively $\mu =1,2,3$ , 4 we have, writing derivatives with respect to $r$ by a prime ,

$$\mathrm{d}{\epsilon }^1=\frac{1}{2}{\mathrm{e}}^{\frac{1}{2}\lambda }{\lambda }^{\prime }\mathrm{d}r\wedge \mathrm{d}r=0=-{\omega }_{12}\wedge {\epsilon }^2-{\omega }_{13}\wedge {\epsilon }^3-{\omega }_{14}\wedge {\epsilon }^4,\text{\hspace{1em}(18.91)}$$

$$\mathrm{d}{\epsilon }^2=r^{-1}{\mathrm{e}}^{-\frac{1}{2}\lambda }{\epsilon }^1\wedge {\epsilon }^2={\omega }_{12}\wedge {\epsilon }^1-{\omega }_{23}\wedge {\epsilon }^3-{\omega }_{24}\wedge {\epsilon }^4,\text{\hspace{1em}(18.92)}$$

$$\mathrm{d}{\epsilon }^3=r^{-1}{\mathrm{e}}^{-\frac{1}{2}\lambda }{\epsilon }^1\wedge {\epsilon }^3+r^{-1}\cot \theta {\epsilon }^2\wedge {\epsilon }^3={\omega }_{13}\wedge {\epsilon }^1+{\omega }_{23}\wedge {\epsilon }^2-{\omega }_{34}\wedge {\epsilon }^4,\text{\hspace{1em}(18.93)}$$

$$\mathrm{d}{\epsilon }^4=\frac{1}{2}{\mathrm{e}}^{-\frac{1}{2}\lambda }{\nu }^{\prime }{\epsilon }^1\wedge {\epsilon }^4=-{\omega }_{14}\wedge {\epsilon }^1-{\omega }_{24}\wedge {\epsilon }^2-{\omega }_{34}\wedge {\epsilon }^4.\text{\hspace{1em}(18.94)}$$

From Eq. (18.93) it follows at once that ${\omega }_{34}={\Gamma }_{344}{\epsilon }^4$ , and substituting in Eq. (18.94) we see that ${\Gamma }_{344}=0$ since it is the sole coefficient of the 2-form basis element ${\epsilon }^3\wedge {\epsilon }^4$ . Similarly, from $(18.94),{\omega }_{24}={\Gamma }_{242}{\epsilon }^2$ and

$${\omega }_{14}=\frac{1}{2}{\mathrm{e}}^{-\frac{1}{2}\lambda }{\nu }^{\prime }{\epsilon }^4+{\Gamma }_{141}{\epsilon }^1.$$

Continuing in this way we find the following values for the connection 1-forms:

$$\begin{array}{ll}{\omega }_{12}=-r^{-1}{\mathrm{e}}^{-\frac{1}{2}\lambda }{\epsilon }^2,& {\omega }_{13}=-r^{-1}{\mathrm{e}}^{-\frac{1}{2}\lambda }{\epsilon }^3,\\ {\omega }_{14}=\frac{1}{2}{\mathrm{e}}^{-\frac{1}{2}\lambda }{v}^{\prime }{\epsilon }^4,& {\omega }_{24}=0,\end{array}\begin{array}{l}{\omega }_{23}=-r^{-1}\cot \theta {\epsilon }^3,\\ {\omega }_{34}=0.\end{array}\text{\hspace{1em}(18.95)}$$

To obtain the curvature tensor it is now a simple matter of substituting these forms in the second Cartan structural equation Eq. (18.67), with indices lowered

$${\rho }_{\mu \nu }=-{\rho }_{\nu \mu }=2\mathrm{d}{\omega }_{\mu \nu }+2{\omega }_{\mu \rho }\wedge {\omega }_{\sigma \nu }{\eta }^{\rho \sigma }.\text{\hspace{1em}(18.96)}$$

For example

$$\begin{array}{l}{\rho }_{12}=2(\mathrm{d}{\omega }_{12}+{\omega }_{13}\wedge {\omega }_{32}+{\omega }_{14}\wedge {\omega }_{42})\\ =2\mathrm{d}(-r^{-1}{\mathrm{e}}^{-\frac{1}{2}\lambda }{\epsilon }^2)\text{ since }{\omega }_{13}\propto {\omega }_{23}\text{ and }{\omega }_{42}=0\\ =-2{\mathrm{e}}^{-\frac{1}{2}\lambda }(r^{-1}{\mathrm{e}}^{-\frac{1}{2}\lambda }{)}^{\prime }{\epsilon }^1\wedge {\epsilon }^2-2r^{-1}{\mathrm{e}}^{-\frac{1}{2}\lambda }\mathrm{d}{\epsilon }^2.\end{array}$$

Substituting for $\mathrm{d}{\epsilon }^2$ using Eq. (18.92) we find

$${\rho }_{12}=r^{-1}{\lambda }^{\prime }{\mathrm{e}}^{-\lambda }{\epsilon }^1\wedge {\epsilon }^2.$$

Similarly,

$$\begin{array}{c}{\rho }_{13}=r^{-1}{\lambda }^{\prime }{\mathrm{e}}^{-\lambda }{\epsilon }^1\wedge {\epsilon }^3,{\rho }_{23}=2r^{-2}(1-{\mathrm{e}}^{-\lambda }){\epsilon }^2\wedge {\epsilon }^3,\\ {\rho }_{14}={\mathrm{e}}^{-\lambda }({\nu }^{\prime \prime }-\frac{1}{2}{\lambda }^{\prime }{\nu }^{\prime }+({\nu }^{\prime }{)}^2){\epsilon }^1\wedge {\epsilon }^4,\\ {\rho }_{24}=r^{-1}{\nu }^{\prime }{\mathrm{e}}^{-\lambda }{\epsilon }^2\wedge {\epsilon }^4,{\rho }_{34}=r^{-1}{\nu }^{\prime }{\mathrm{e}}^{-\lambda }{\epsilon }^3\wedge {\epsilon }^4.\end{array}$$

The components of the Riemann tensor in this basis are given by

$$R_{\mu \nu \rho \sigma }=\overline{R}(e_{\mu },e_{\nu },e_{\rho },e_{\sigma })={\rho }_{\mu \nu }(e_{\rho },e_{\sigma }).$$

The non-vanishing components are

$$\begin{array}{l}{R}_{1212}=R_{1313}=\frac{{\lambda }^{\prime }}{2r}{\mathrm{e}}^{-\lambda },\\ R_{2323}=\frac{1-{\mathrm{e}}^{-\lambda }}{r^2},\\ R_{1414}=\frac{1}{4}{\mathrm{e}}^{-\lambda }(2{\nu }^{\prime \prime }-{\lambda }^{\prime }{\nu }^{\prime }+({\nu }^{\prime }{)}^2),\\ R_{2424}=R_{3434}=\frac{{\nu }^{\prime }}{2r}{\mathrm{e}}^{-\lambda }.\end{array}\text{\hspace{1em}(18.97)}$$

The Ricci tensor components

$$R_{\mu \nu }={\eta }^{\rho \sigma }{R}_{\rho \mu \sigma \nu }=\sum _{i=1}^3{R}_{i\mu i\nu }-R_{4\mu 4\nu }$$

are therefore

$$R_{11}={\mathrm{e}}^{-\lambda }(-\frac{{\nu }^{\prime \prime }}{2}+\frac{{\lambda }^{\prime }{\nu }^{\prime }}{4}-\frac{({\nu }^{\prime }{)}^2}{4}+\frac{{\lambda }^{\prime }}{r}),\text{\hspace{1em}(18.98)}$$

$$R_{44}={\mathrm{e}}^{-\lambda }(\frac{{\nu }^{\prime \prime }}{2}-\frac{{\lambda }^{\prime }{\nu }^{\prime }}{4}+\frac{({\nu }^{\prime }{)}^2}{4}+\frac{{\nu }^{\prime }}{r}),\text{\hspace{1em}(18.99)}$$

$$R_{22}=R_{33}={\mathrm{e}}^{-\lambda }\left(\frac{{\lambda }^{\prime }-{\nu }^{\prime }}{2r}-\frac{1}{r^2}\right)+\frac{1}{r^2}.\text{\hspace{1em}(18.100)}$$

To solve Einstein’s vacuum equations $R_{\mu \nu }=0$ , we see by adding Eq. (18.98) and Eq. (18.99) tha ${\lambda }^{\prime }+{\nu }^{\prime }=0$ , whence

$$\lambda =-\nu +C(C=\text{ const. })$$

A rescaling of the time coordinate, $t{t}^{\prime }={\mathsf{e}}^{C/2}t$ , has the effect of making $C=0$ , which we now assume. By Eq. (18.99), $R_{44}=0$ reduces the second-order differential equation to

$${\nu }^{\prime \prime }+({\nu }^{\prime }{)}^2+\frac{2{\nu }^{\prime }}{r}=0$$

and the substitution $\alpha ={\mathrm{e}}^{\nu }$ results in

$${\left(r^2{\alpha }^{\prime }\right)}^{\prime }=0,$$

whence

$$\alpha ={\mathrm{e}}^{\nu }=A-\frac{2m}{r}(m,A=\text{ const. }).$$

If we substitute this into $R_{22}=0$ we have, by $(18.100),(r\alpha {)}^{\prime }=1$ so that $A=1$ . The most general spherically symmetric solution of Einstein’s vacuum equations is therefore

$$\mathrm{d}{s}^2=\frac{1}{1-2m/r}\mathrm{d}{r}^2+r^2(\mathrm{d}{\theta }^2+{\sin }^2\theta \mathrm{d}{\varphi }^2)-\left(1-\frac{2m}{r}\right)c^2\mathrm{d}{t}^2,\text{\hspace{1em}(18.101)}$$

known famously as the Schwarzschild solution. Converting the polar coordinates to equiv alent cartesian coordinates $x$, $y$, $z$ we have, as $r\to \infty$ ,

$$\mathrm{d}{s}^2\approx \mathrm{d}{x}^2+\mathrm{d}{y}^2+\mathrm{d}{z}^2-c^2\mathrm{d}{t}^2+\frac{2m}{r}(c^2\mathrm{d}{t}^2+\dots )$$

and $g_{\mu \nu }\approx {\eta }_{\mu \nu }+h_{\mu \nu }$ where

$$h_{44}=\frac{2m}{r}\approx \frac{-2\varphi }{c^2},$$

assuming the Newtonian approximation with potential $\varphi$ is applicable in this limit. Since the potential of a Newtonian mass $M$ is given by $\varphi =GM/r$ , it is reasonable to make the identification

$$m=\frac{GM}{c^2}.$$

The constant $m$ has dimensions of length and $2m=2GM/c^2$ , where the metric Eq. (18.101) exhibits singular behaviour, is commonly known as the Schwarzschild radius. For a solar mass, $M_{\odot }=2\times 10^{33}$ $g$, its value is about 3 $\mathrm{km}$. However the Sun would need to collapse to approximately this size before strong corrections to Newtonian theory apply.

When paths ofparticles (timelike geodesics) and photons (null geodesics) are calculated in this metric, the following deviations from Newtonian theory are found for the sola system:

1. There is a slowing of clocks at a lower gravitational potential. At the surface of the Earth this amounts to a redshift from a transmitter to a receiver at a height $h$ above it of

$$z=\frac{G{M}_{e}h}{R_e^2{c}^2}.$$

This amounts to a redshift of about $10^{-15}{\mathrm{m}}^{-1}$ and is measurable using the Mössbauer effect.

2. The perihelion of a planet in orbit around the Sun precesses by an amount

$$\delta \phi =\frac{6\pi M_{\odot }G}{c^{2}a(1-{\mathrm{e}}^2)}\text{ per revolution. }$$

For Mercury this comes out to 43 seconds of arc per century.

3. A beam of light passing the Sun at a closest distance $r_0$ is deflected an amount

$$\delta \phi =\frac{4G{M}_{\odot }}{r_0{c}^2}.$$

For a beam grazing the rim of the Sun $r_0=R_{\odot }$ the deflection is 1.75 seconds of arc.

The limi $r2m$ is of particular interest. Although it appears that the metric Eq. (18.101) is singular in this limit, this is really only a feature of the coordinates, not of the space-time as such. A clue that this may be the case is found by calculating the curvature components Eq. (18.97) for the Schwarzschild solution,

$$R_{1212}=R_{1313}=R_{2424}=R_{3434}=-\frac{m}{r^3},R_{2424}=-R_{1414}=\frac{2m}{r^3}$$

all of which approach finite values as $r2m$

Exercise

Verify these expressions for components of the Riemann tensor.

More specifically, let us make the coordinate transformation from $t$ to $\mathbf{v}=\mathbf{c}\mathbf{t}+\mathbf{r}+$ 2$m$ $\ln (r-2m)$ , sometimes referred to as advanced time since it can be shown to be constant on inward directed null geodesics, while leaving the spatial coordinates $r$, $\theta$, $\varphi$ unchanged. In these Eddington–Finkelstein coordinates the metric becomes

$$\mathrm{d}{s}^2=-\left(1-\frac{2m}{r}\right)\mathrm{d}{v}^2+2\mathrm{d}r\mathrm{d}v+r^2(\mathrm{d}{\theta }^2+{\sin }^2\theta \mathrm{d}{\varphi }^2).$$

$\mathbf{As}r2m$ the metric shows no abnormality in these coordinates. Inward directed timelike geodesics in the region $r>2m$ reach $r=2m$ in finite $v$-time (and also in finite proper time). However, after the geodesic particle crosses $r=2m$ no light signals can be sent ou from it into $r>2m$ (see Fig. 18.4). The surface $r$ 2$m$ acts as a one-way membrane for light signals, called an event horizon. Observers with $r>2m$ can never see any events inside $r=2m$ , an effect commonly referred to as a black hole.

Figure 18.4 Schwarzschild solution in Eddington–Finkelstein coordinates

Problems

Problem 18.20

A linearized plane gravitational wave is a solution of the linearized Einstein equations $\boxed{}{h}_{\mu \nu }=0$ of the form $h_{\mu \nu }=h_{\mu \nu }(u)$ where $u=x^3-x^4=z-ct$ . Show that the harmonic gauge condition Eq. (18.89) implies that, up to undefined constants,

$$h_{14}+h_{13}=h_{24}+h_{23}=h_{11}+h_{22}=0,h_{34}=-\frac{1}{2}(h_{33}+h_{44}).$$

Use the remaining gauge freedom ${\xi }_{\mu }={\xi }_{\mu }(u)$ to show that it is possible to transform $h_{\mu \nu }$ to the form

$$[h_{\mu \nu }]=\left(\begin{array}{cc}\mathsf{H}& \mathsf{O}\\ \mathsf{O}& \mathsf{O}\end{array}\right)\text{ where }\mathsf{H}=\left(\begin{array}{cc}{h}_{11}& h_{12}\\ h_{12}& -h_{11}\end{array}\right).$$

Setting $h_{11}=\alpha (u)$ and $h_{12}=\beta (u)$ , show that the equation of geodesic deviation has the form

$${\ddot{\eta }}_1=\frac{ϵ}{2}{c}^2\left({\alpha }^{\prime \prime }{\eta }_1+{\beta }^{\prime \prime }{\eta }_2\right),{\ddot{\eta }}_2=\frac{ϵ}{2}{c}^2\left({\beta }^{\prime \prime }{\eta }_1-{\alpha }^{\prime \prime }{\eta }_2\right)$$

and ${\ddot{\eta }}_3=0$ . Make a sketch of the distribution of neighbouring accelerations of freely falling parti cles about a geodesic observer in the two cases $\beta =0$ and $\alpha =0$ ). These results are central to the observational search for gravity waves.

Problem 18.21

Show that every two-dimensional space-time metric (signature 0) can be expressed locally in conformal coordinates

$$\mathrm{d}{s}^2={\mathrm{e}}^{2\varphi }\left(\mathrm{d}{x}^2-\mathrm{d}{t}^2\right)\text{ where }\varphi =\varphi (x,t).$$

Calculate the Riemann curvature tensor component $R_{1212}$ , and write out the two-dimensional Einstein vacuum equations $R_{ij}=0$ . What is their general solution?

Problem 18.22

(a) For a perfect fluid in general relativity,

$$T_{\mu \nu }=(\rho c^2+P)U_{\mu }{U}_{\nu }+P{g}_{\mu \nu }(U^{\mu }{U}_{\mu }=-1)$$

show that the conservation identities $T^{\mu \nu }{}_{;\nu }=0$ imply

$$\begin{array}{c}{\rho }_{,\nu }{U}^{\nu }+(\rho c^2+P)U_{;\nu }^{\nu },\\ (\rho c^2+P)U_{;\nu }^{\mu }{U}^{\nu }+P_{,\nu }(g^{\mu \nu }+U^{\mu }{U}^{\nu }).\end{array}$$

(b) For a pressure-free fluid show that the streamlines of the fluid (i.e. the curves $x^{\mu }(s)$ satisfying $\mathrm{d}{x}^{\mu }/\mathrm{d}s=U^{\mu })$ are geodesics, and $\rho U^{\mu }$ is a covariant 4-current, $(\rho U^{\mu }{)}_{,\mu }=0$

(c) In the Newtonian approximation where

$$U_{\mu }=\left(\frac{v_i}{c},-1\right)+O({\beta }^2),P=O({\beta }^2)\rho c^2,\left(\beta =\frac{v}{c}\right)$$

where $|\beta |\ll 1$ and $g_{\mu \nu }={\eta }_{\mu \nu }+ϵh_{\mu \nu }$ with $ϵ\ll 1$ , show that

$$h_{44}\approx -\frac{2\varphi }{c^2},h_{ij}\approx -\frac{2\varphi }{c^2}{\delta }_{ij}\text{ where }{\nabla }^2\varphi =4\pi G\rho$$

and $h_{i4}=O(\beta )h_{44}$ . Show in this approximation that the equation $T^{\mu \nu }{}_{;\nu }=0$ approximate to

$$\frac{\partial \rho }{\partial t}+\nabla \cdot (\rho \mathbf{v})=0,\rho \frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t}=-\nabla P-\rho \nabla \varphi .$$

Problem 18.23

(a) Compute the components of the Ricci tensor $R_{\mu \nu }$ for a space-time that has a metric of the form

$$\mathrm{d}{s}^2=\mathrm{d}{x}^2+\mathrm{d}{y}^2-2\mathrm{d}u\mathrm{d}v+2H\mathrm{d}{v}^2(H=H(x,y,u,v)).$$

(b) Show that the space-time is a vacuum if and only if $H=\alpha (x,y,v)+f(v)u$ where $f(v)$ is an arbitrary function and $\alpha$ satisfies the two-dimensional Laplace equation

$$\frac{{\partial }^2\alpha }{\partial x^2}+\frac{{\partial }^2\alpha }{\partial y^2}=0,$$

and show that it is possible to set $f(v)=0$ by a coordinate transformation $u^{\prime }=ug(v),v^{\prime }=h(v)$ (c) Show that $R_{i4j4}=-H_{,ij}$ for $i,j=1,2$

Problem 18.24

Show that a coordinate transformation $r=h(r^{\prime })$ can be found such that the Schwarzschild solution has the form

$$\mathrm{d}{s}^2=-{\mathrm{e}}^{\mu (r^{\prime })}\mathrm{d}{t}^2+{\mathrm{e}}^{\nu (r^{\prime })}(\mathrm{d}{r}^{\prime 2}+r^{\prime 2}(\mathrm{d}{\theta }^2+{\sin }^2\theta \mathrm{d}{\varphi }^2)).$$

Evaluate the functions ${\mathtt{e}}^{\mu }$ and ${\mathsf{e}}^{\nu }$ explicitly.

Problem 18.25

Consider an oscillator at $r=r_0$ emitting a pulse of light (null geodesic) at $t=t_0$ If this is received by an observer at $r=r_1$ at $t=t_1$ , show that

$$t_1=t_0+{\int }_{r_0}^{r_1}\frac{\mathrm{d}r}{c(1-2m/r)}.$$

By considering a signal emitted at $t_0+\Delta t_0,$ , received at $t_1+\Delta t_1$ (assuming the radial positions $r_0$ and $r_1$ to be constant), show that $t_0=t_1$ and the gravitational redshift found by comparing proper times at emission and reception is given by

$$1+z=\frac{\Delta {\tau }_1}{\Delta {\tau }_0}=\sqrt{\frac{1-2m/r_1}{1-2m/r_0}}.$$

Show that for two clocks at different heights $h$ on the Earth’s surface, this reduces to

$$z\approx \frac{2GM}{c^2}\frac{h}{R},$$

where $M$ and $R$ are the mass and radius of the Earth.

Problem 18.26

In the Schwarzschild solution show the only possible closed photon path is a circula orbit at $r=3m$ , and show that it is unstable.

Problem 18.27

(a) A particle falls radially inwards from rest at infinity in a Schwarzschild solution. Show that it will arrive at $r=2m$ in a finite proper time after crossing some fixed reference position $r_0.$ , but that coordinate time $t\infty \mathrm{as}r2m$

(b) On an infalling extended body compute the tidal force in a radial direction, by parallel propagating a tetrad (only the radial spacelike unit vector need be considered) and calculating $R_{1414}$

(c) Estimate the total tidal force on a person of height 1.8 $m$, weighing 70 $\mathrm{kg}$, falling head-first into a solar mass black hole $(M_{\odot }=2\times 10^{30}\mathrm{kg})$ , as he crosses $r=2m$