[Wald General Relativity] 6. The Schwarzschild Solution

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6.1 Derivation of the Schwarzchild Solution

A spacetime is said to be stationary if there exists a one-parameter group of isometries, $\phi_i$, whose orbits are timelike curves. This groups of isometries expresses the "time translation symmetry" of the spacetime. Equivalently, a stationary spacetime is one which possesses a timelike Killing vector field, $\xi^a$. A spacetime is said to be static if it is stationary and if, in addition, there exists a hypersurface $\Sigma$ which is orthogonal to the orbits of the isometry. By Frobenius's theorem this is equivalent to the timelike Killing vector field $xi^a$ satisfy $$\begin{align}\xi_{[a}\nabla_b\xi_{c]}=0.\end{align}$$

The meaning of this extra condition of hypersurface orthogonality perhaps can be seen best by introducing convenient coordinates for static spacetime as follows. If $\xi^a\ne0$ everywhere on $\Sigma$, then in a neighborhood of $\Sigma$, every point will lie on a unique orbit of $\xi^a$ which passes through $\Sigma$. Assuming $\xi^a\ne0$, we choose arbitrary coordinates $\{x^\mu\}$ on $\Sigma$ and label each point $p$ in this neighborhood of $\Sigma$ by the parameter, $t$, of the orbit which starts from $\Sigma$ and ends at $p$, and the coordinates, $x^1$, $x^2$, $x^3$, of the orbit at $\Sigma$. Since this coordinate system employs the Killing parameter, $t$, as one of the coordinates, the metric components in this coordinate basis will be independent of $t$. Furthermore, since the surface $\Sigma_t$ -defined as the set of points whose "time coordinate" has a value $r$- is the image of $\Sigma$ under the isometry $\phi_i$, it follows that each $\Sigma_t$ is also orthogonal to $\xi^a$. Thus, in these coordinates, the metric components take the form $$\begin{align}ds^2=-V^2(x^1,x^2,x^3)dt^2+\sum^3_{\mu,\nu=1}h_{\mu\nu}(x^1,x^2,x^3)dx^\mu dx^\nu,\end{align}$$ where $V^2=-\xi_a \xi^a$, and the absence of $dt\ dx^\nu$ cross terms expresses the orthogonality of $\xi^a$ with $\Sigma$. A stationary but nonstatic metric unavoidablt must have $dt\ dx^\mu$ cross terms in any coordinate system which uses the Killing parameter as one of the coordinates.

A spacetime is said to be spherically symmetric if its isometry group contains a subgroup isomorphic to the group $SO(3)$, and the orbits of this subgroup are two-dimensional spheres. The $SO(3)$ isometries may then be interpreted physically as rotations, and thus a spherically symmetric spacetime is one whose metric remains invariant under rotations. The spcaetime metric induces a metric on each orbit 2-sphere, which, because of the rotation symmetry, must be multiple of the metric of a unit 2-sphere, and thus can be completely characterized by the total area, $A$, of the 2-sphere. In spherically spmmetric spacetime it is convenient to introduce the function, $r$, defined by $$\begin{align}r=(A/4\pi)^{1/2}.\end{align}$$ Thus, in spherical coordinates $(\theta,\phi)$, the metric on each orbit 2-sphere takes the form $$\begin{align}ds^2=r^2(d\theta^2+\sin^2\theta d\phi^2).\end{align}$$ In flat, three-dimensional Euclidean space, $r$ would also be the value of the radius of the sphere, i.e., the distance from the surface of the sphere to its center. However, in a curved space, a sphere need not have a center (e.g., the manifold structure could be, say, $\mathbb{R}\times S^2$, rather than $\mathbb{R}^3$); and even if it does, $r$ need not bear any relation to the distance to the center. Nevertheless, we shal refer to $r$ as the "radial coordinate" of the sphere.

If a spacetime is both static and spherically symmetric, and if the static Killing field $\xi^2$ is unique, then $\xi^a$ must be orthogonal to the orbit 2-spheres. Namely, if $\xi^a$ is unique, it must be invariant under all the rotational isometries. However, this requires its projection onto any orbit sphere to vanish, since a nonvanishing vector field on a sphere cannot be invariant under all rotations. Thus the orbit spheres lie wholly within the hypersurfaces, $\Sigma_t$, orthogonal to $\xi^a$. Convenient coordinates on the spacetime may be chosen as follows: We select a sphere on $\Sigma=\Sigma_0$ and choose spherical coordinates $(\theta,\phi)$ on it. We "carry" these spherical coordinates to the other spheres of $\Sigma$ by means of geodesics orthogonal to the two-sphere. Provided that $\nabla_ar\ne0$, we choose $(r,\theta,\phi)$ as coordinates in $\Sigma_t$ and, finally, we choose $(t,r,\theta,\phi)$ as coordinates for the spacetime according to the prescription described above equation (2). In these coordinates, the metric of an arbitrary static, spherically symmetric spacetime takes the simple form $$\begin{align}ds^2=-f(r)dt^2+h(t)dr^2+r^2(d\theta^2+\sin^2\theta d\phi^2).\end{align}$$ It should be kept in mind, however, that, in addition to the well understood breakdown of the spherical coordinates at the north and south poles of the spheres, this coordinate system breaks down at points where $\xi^a=0$ or $\nabla_ar=0$ (or, more generally, where $\xi^a$ and $\nabla^ar$ become collinear).

A convenient orthonormal basis for the metric of equation (5) is $$\begin{align}(e_0)_a=f^{1/2}(dt)_a,\\ (e_1)_a=h^{1/2}(dr)_a,\\ (e_r)_a=r(d\theta)_a,\\ (e_3)_a=r\sin \theta (d\phi)_a.\end{align}$$ Using the ordinary derivative, $\partial a$, of our coordinate system, we fine $$\begin{align}\partial_{[a}(e_0)_{b]}=\frac{1}{2}f^{-1/2}f'(dt)_{[a}(dt)_{b]},\\ \partial_{[a}(e_1)_{b]}=0,\\ \partial_{[a}(e_2)_{b]}=(dr)_{[a}(d\theta)_{b]},\\ \partial_{[a}(e_3)_{b]}=\sin \theta (dr)_{[a}(d\phi)_{b]}+r\cos \theta(d\theta)_{[a}(d\phi)_{b]},\end{align}$$ where $f'=df/dr$. Thus, according to equation (3.25), we must solve the following equations for the connection one-forms $\omega_{a\mu\nu}=-\omega_{a\nu\mu}$: $$\begin{align}\frac{1}{2}f^{-1/2}f'(dr)_{[a}(dt)_{b]}=h^{1/2}(dr)_{[a}\omega_{b]01}+r(d\theta)_{[a}\omega_{b]02}+r\sin\theta(d\phi)_{[a}\omega_{b]03},\\ 0=f^{1/2}(dt)_{[a}\omega_{b]01}+r(d\theta)_{[a}\omega_{b]12}+r\sin\theta(d\phi)_{[a}\omega_{b]13},\\ (dr)_{[a}(d\theta)_{b]}=-f^{1/2}(dt)_{[a}\omega_{b]20}+h^{1/2}(dr)_{[a}\omega_{b]21}+r\sin\theta(d\phi)_{[a}\omega_{b]23},\\ \sin\theta(dr)_{[a}(d\phi){b]}+r\cos\theta(d\theta)_{[a}(d\phi)_{b]}=-f^{1/2}(dt)_{[a}\omega_{b]30}+h^{1/2}(dr)_{[a}\omega_{b]31}+r(d\theta)_{[a}\omega_{b]32}.\end{align}$$ A plausible guess at solving equation (11) is $$\begin{align}\omega_{b02}=\omega_{b03}=0,\\ \omega_{b01}=\frac{1}{2}\frac{f'}{(fh)^{1/2}}(dt)_b+\alpha_1(dr)_b,\end{align}$$ where the function $\alpha_1$ is undetermined by equation (11). Substitution of this trial solution in equation (12) then requires $\alpha_1=0$. From equation (12), we might also guess $\omega_{b12}=-\omega_{b13}=0$, but this leads to inconsistency later, so we merely conclude that $$\begin{align}\omega_{b12}=\alpha_2(d\theta)_b+\alpha_3(d\phi)_b,\\ \omega_{b13}=\alpha_4(d\phi)_b+\frac{\alpha_3}{\sin\theta}(d\theta)_b.\end{align}$$ Substituting our trial solution in equation (13), we find $$\begin{align}\alpha_2=-h^{-1/2},\\ \omega_{b23}=-\frac{h^{1/2}}{r\sin\theta}\alpha_3(dr)_b+\alpha_5(d\phi)_b.\end{align}$$ Finally, substitution into equation (14) yields $$\begin{align}\alpha_3=0,\\ \alpha_4=-h^{-1/2}\sin\theta,\\ \alpha_5=-\cos\theta.\end{align}$$ Since we have found no inconsistency, this means that our trial solution (generated by the initial guess $\omega_{b02}=\omega_{b03}=0$) is, in fact, the solution. Thus, we have found $$\begin{align}\omega_{b02}=\omega_{b03}=0,\\ \omega_{b01}=\frac{f'}{2(fh)^{1/2}}(dt)_b,\\ \omega_{b12}=-h^{-1/2}(d\theta)_b,\\ \omega_{b13}=-h^{-1/2}\sin\theta(d\phi)_b,\\ \omega_{b23}=-\cos\theta(d\phi)_b.\end{align}$$

From equation (3.4.20) we obtain the Riemann tensor with remarkably little total computation compared with other approaches $$\begin{align}R_{ab01}=-R_{ab10}=\frac{d}{dr}[(fh)^{-1/2}f'](dr)_{[a}(dt){b]},\\ R_{ab02}=-R_{ab20}=f^{-1/2}h^{-1}f'(d\theta)_{[a}(dt)_{b]},\\ R_{ab03}=-R_{ab30}=f^{-1/2}h^{-1}f'\sin\theta(d\phi)_{[a}(dt)_{b]},\\ R_{ab12}=-R_{ab21}=h^{-3/2}h'(dr)_{[a}(d\theta)_{b]},\\ R_{ab13}=-R_{ab31}=\sin\theta h^{3/2}h'(dr)_{[a}(d\phi)_{b]},\\ R_{ab23}=-R_{ab32}=2(1-h^{-1})\sin\theta(d\theta)_{[a}(d\phi)_{b]}.\end{align}$$

The Ricci tensor is easily computed from the Riemann tensor via equation (2.4.22). Equating it to zero, we obtain the vacuum Einstein equation for a static, spherically symmetric spacetime, $$\begin{align}0=R_{00}=R_{010}^1+R_{020}^2+R_{030}^3=\frac{1}{2}(fh)^{-1/2}\frac{d}{dr}[(fh)^{-1/2}f']+(rfh)^{-1}f',\\ 0=R_{11}=-\frac{1}{2}(fh)^{-1/2}\frac{d}{dr}[(fh)^{-1/2}f']+(rh^2)^{-1}h',\\ 0=R_{22}=-\frac{1}{2}(rfh)^{-1}f'+\frac{1}{2}(rh^2)^{-1}h'+r^{-2}(1-h^{-1}),\end{align}$$ where $R_{\mu\nu}\equiv R_{ab}(e_\mu)^a(e_\nu)^b$. We also find that $R_{33}=R_{22}$ and that the off-diagonal components of $R_{\mu\nu}$ vanish identically, as could be predicted from symmetry arguments similar to those used in section 5.2.

Adding equations (35) and (36), we obtain $$\begin{align}f'/f+h'/h=0,\end{align}$$ which implies $$\begin{align}f=Kh^{-1},\end{align}$$ where $K$ is a constant. By re-scaling the time coordinate, $t\rightarrow K^{1/2}t$, we may set $K=1$. Equation (37) now yields $$\begin{align}-f'+\frac{1-f}{r}=0,\end{align}$$ i.e., $$\begin{align}\frac{d}{dr}(rf)=1,\end{align}$$ which implies $$\begin{align}f=1+C/r,\end{align}$$ where $C$ is a constant. Equation (42) together with equation (39) (with $K=1$) solves equations (35)-(37), and thus we have found the general solution, first discovered by Schwarzschild, to the vacuum Einstein equation for static, spherically symmetric spacetimes, $$\begin{align}ds^2=-\left( 1+\frac{C}{r}\right) dt^2+\left( 1+\frac{C}{r}\right)^{-1}dr^2+r^2d\Omega^2,\end{align}$$ where $d\Omega^2$ is shorthand for $(d\theta^2+\sin^2\theta d\phi^2)$. With $M=-C/2$, $$\begin{align}ds^2=-\left(1-\frac{2M}{r}\right) dt^2+\left( 1-\frac{2M}{r}\right)^{-1}dr^2+r^2d\Omega^2.\end{align}$$

6.2 Interior Solutions

We turn out attention, now, to the static, spherically symmetric solutions of Einstein's equation with a perfect fluid stress-energy tensor, $$\begin{align}T_{ab}=\rho u_au_b+P(g_{ab}+u_au_b).\end{align}$$ In order to be compatible with the static symmetry of the spacetime the fluid 4-velocity, $u^a$, must point in the same direction as the static Killing vector field, $\xi^a$, i.e., $$\begin{align}u^a=-(e_0)^a=-f^{1/2}(dt)^a,\end{align}$$ where $f$ is the function appearing the general metric form equation (5).

Einstein's equation with a fluid present is obtained simply by adding the appropriate stress-energy terms to equations (35)-(37). We shall take the three independent equations in the form $$\begin{align}8\pi T_{00}=8\pi\rho=G_{00}=R_{00}+\frac{1}{2}(R_0^0+R_1^1+R_2^2+R_3^3)=(rh^2)^{-1}h'+r^{-2}(1-h^{-1}),\\ 8\pi T_{11}=8\pi P=G_{11}=R_{11}-\frac{1}{2}(R^0_0+R_1^1+R_2^2+R_3^3)=(rfh)^{-1}f'-r^{-2}(1-h^{-1}),\\ 8\pi T_{22}=8\pi P=G_{22}=\frac{1}{2}(fh)^{-1/2}\frac{d}{dr}[(fh)^{-1/2}f']+\frac{1}{2}(rfh)^{-1}f'-\frac{1}{2}(rh^2)^{-1}h'.\end{align}$$

Equation (3) involves only $h$. It can be rewritten in the form $$\begin{align}\frac{1}{r}\frac{d}{dr}[r(1-h^{-1})]=8\pi\rho,\end{align}$$ from which it follows immediately that the solution for $h$ is $$\begin{align}h(r)=\left[ 1-\frac{2m(r)}{r}\right],\end{align}$$ where $$\begin{align}m(r)=4\pi \int^r_0\rho(r')r'^2dr'+a,\end{align}$$ where $a$ is a constant. Smoothness of the metric on $\Sigma$ at $r=0$ requires that as $r\rightarrow 0$ the area of spheres approach $4\pi$ times the square of their proper radius, i.e., that $h(r)\rightarrow 1$ as $r\rightarrow 0$. Thus, in order to avoid a "conical singularity" in the metric at $r=0$, we must set $a=0$. Since $\Sigma$ must be spacelike for a static configuration, we see that a necessary condition for statictiy is $h\ge 0$, i.e., $$\begin{align}r\ge 2m(r).\end{align}$$ 

If $\rho=0$ for $r>R$, our solution for $h$, equation (7), joins on to the vacuum Schwarzschild solution with total mass $$\begin{align}M=m(R)=4\pi \int^R_0\rho(r)r^2dr.\end{align}$$ Equation (10) is formally identical to the expression for total mass in Newtonian gravity. Note, however, that this formal analogy is misleading because the proper volume element of $\Sigma$ is $\sqrt{^{(3)}g}d^3x=h^{1/2}r^2\sin\theta drd\theta d\phi$ so that the total proper mass is $$\begin{align}M_p=4\pi \int^R_0\rho(r)r^2\left[ 1-\frac{2m(r)}{r}\right]^{-1/2}dr.\end{align}$$ The difference between $M$ and $M_\rho$ can be interpreted as the gravitational binding energy, $E_B$, of the configuration, $$\begin{align}E_B=M_\rho-M,\end{align}$$ which is always positive since $M_\rho>M$.

If we write $$\begin{align}f=e^{2\phi},\end{align}$$ equation (4) becomes $$\begin{align}\frac{d\phi}{dr}=\frac{m(r)+4\pi r^3P}{r[r-2m(r)]}.\end{align}$$ In the Newtonian limit, we have $r^3P\ll m(r)$ and $m(r)\ll r$, so equation (14) reduces to $$\begin{align}\frac{d\phi}{dr}\approx \frac{m(r)}{r^2},\end{align}$$ which is simply the spherically symmetric version of Poisson's equation for the Newtonian gravitational potential. Thus, in the static spherically symmetric case we may view $\phi=\frac{1}{2}\ln f$ as the general relativistic analog of the Newtonian potential.

If we substitute our results, equations (7) and (14), into our final equation (5), it is apparent that we will obtain an equation for $dP/dr$. However, the rather messy algebra required for doing this can be circumvented by noting that $$\begin{align}(e_\mu)_b8\pi \nabla_aT^{ab}=(e_\mu)_b\nabla_a(8\pi T^{ab}-G^{ab})=\nabla_a[(e_\mu)_b(8\pi T^{ab}-G^{ab})]-\omega_{ab\mu}(8\pi T^{ab}-G^{ab}),\end{align}$$ where $\omega_{ab\mu}=\nabla_a(e_\mu)_b$. Setting $\mu=1$, we find the first term on the right-hand side of equation (16) vanishes if equation (4) is satisfied. Thus, since $\omega_{221}\ne 0$, we see that the vanishing of $(8\pi T^{22}-G^{22})$ is equivalent to $$\begin{align}(e_1)_b\nabla_a T^{ab}=0,\end{align}$$ and hence we may replace equation (5) by equation (17). We have already calculated this component of $\nabla_a T^{ab}$ for a perfect fluid in equation (4.3.8), and thus, without further work, we obtain [using $(e_1)_b=h^{1/2}(dr)_b=h^{-1/2}(\partial/\partial r)_b$], $$\begin{align}h^{-1/2}\frac{dP}{dr}=-(P+\rho)(e_1)_bu^a\nabla_a u^b=-h^{-1/2}(P+\rho)\frac{d\phi}{dr},\end{align}$$ where equations (2) and (13) were used for the last equality. Using equation (14), we may eliminate $d\phi/dr$ to obtain our final result, $$\begin{align}\frac{dP}{dr}=-(P+\rho)\frac{m(r)+4\pi r^3P}{r[r-2m(r)]}.\end{align}$$ Equation (19) is known as the Tolmann-Oppenheimer-Volkoff euqation of hydro-static equilibrium. In the Newtonian limit ($P\ll \rho$, $m(r)\ll r$) it reduces to the Newtonian hydrostatic equilibrium equation. $$\begin{align}\frac{dP}{dr}\approx -\frac{\rho m(r)}{r^2}.\end{align}$$

In summary, we have found that the spacetime geometry inside a static, spherical fluid star is $$\begin{align}ds^2=-e^{2\phi}dt^2+\left( 1-\frac{2m(r)}{r}\right)^{-1}dr^2+r^2d\Omega^2,\end{align}$$ where $$\begin{align}m(r)=4\pi \int^r_0\rho(r')r'^2dr'\end{align}$$ and $\phi$ is determined from equation (14). The necessary and sufficient condition for equilibrium is that equation (19) be satisfied.

Consider the uniform density configurations, corresponding to an incompressible fluid of density $\rho_0$, $$\begin{align}\rho(r)=\begin{cases}\rho_0& (r\le R)\\ 0& (r>R)\end{cases},\end{align}$$ and hence (in both general relativity and Newtonian theory) $$\begin{align} m(r)=\frac{4}{3}\pi r^3\rho_0\quad (r\le R).\end{align}$$ The Newtonian equation of hydrostatic equilibrium (20) is easily integrated to yield (for $r\le R$), $$\begin{align} P(r)=\frac{2}{3}\pi \rho^2_0(R^2-r^2),\end{align}$$ where the boundary condition $P(R)=0$ has been imposed. Thus, the central pressure of a Newtonian uniform density star is $$\begin{align}P_c=\frac{2}{3}\pi \rho^2_0R^2=\left( \frac{\pi}{6}\right)^{1/3}M^{2/3}\rho^{4/3}_0,\end{align}$$ which is finite for all values of $\rho_0$ and $R$, i.e., equilibrium can be achieved with  sufficiently large pressures for all $\rho_0$ and $R$. On the other hand, the general relativistic equation of hydrostatic equilibrium (10) also can be integrated exactly, as was first done by Schwarzschild (1916b), yielding $$\begin{align}P(r)=\rho_0\left[ \frac{(1-2M/R)^{1/2}-(1-2Mr^2/R^3)^{1/2}}{(1-2Mr^2/R^3)^{1/2}-3(1-2M/R)^{1/2}}\right].\end{align}$$ Thus, the central pressure required for equilibrium of a uniform density star in general relativity is $$\begin{align}P_e=\rho\left[\frac{1-(1-2M/R)^{1/2}}{3(1-2M/R)^{1/2}-1}\right].\end{align}$$ For $R\gg M$, equation (28) reduces to the Newtonian value, equation (26). However, now $P_c$ become infinite when $$\begin{align}3(1-2M/R)^{1/2}=1,\end{align}$$ i.e., when $$\begin{align}R=\frac{9}{4}M.\end{align}$$, Thus, in general relativity, uniform density stars with $M>4R/9$ simply cannot exists. 

6.3 Geodesics of Schwarzschild: Gravitational Redshift, Perihelion Precession, Bending of Light, and Time Delay

It would involve a fair amount of labor to solve directly the geodesics equation in the form (5). Fortunately, almost all of this labor can be avoided by taking advantage of the symmetries of the Schwarzschild solution using proposition C.3.1: The inner product, $u^a\xi_a$, of a Killing filed $\xi^a$ with a geodesic tangent $u^a$ is constant along the geodesic. As we shall see below, these constants of the motion enable us to reduce the problem of finding the geodesics to the problem of one-dimensional motion of a particle in a effective potential.

Proposition C.3.1 immediately allows us to derive a formula for the change between emitted and observed frequency of light signals sent between two static observers, i.e. for the gravitational redshift. Consider two static observers (i.e., observers whose 4-velocity is tangent to the static Killing field $\xi^a$) $O_1$ and $O_2$ whose 4-velocities are $u_1^a$ and $u_2^a$. Suppose $O_1$ emits a light signal at event $P_1$ which is received by $O_2$ at event $P_2$, as illustrated in Figure 6.2. As discussed in section 4.3., in the geometrical optics approximation this light signal travels on a null geodesic, whose tangent we denote by $k^a$. The frequency of emission is $$\begin{align}\omega_1=-(k_au^a_1)|_{P_1},\end{align}$$ while the frequency measured by the observer receiving the signal is $$\begin{align}\omega_2=-(k_au^a_2)|_{P_2}.\end{align}$$ However, since $u_1^a$ and $u_2^a$ both are unit vectors which point in the direction of the timelike Killing field $\xi^a$, we have $$\begin{align}u_1^a=[\xi^a/((-\xi^b\xi_b)^{1/2}]|_{P_1},\\ u_2^a=[\xi^a/((-\xi^b\xi_b)^{1/2}]|_{P_2}.\end{align}$$ By proposition C.3.1 we have $(k_a\xi^a)|_{P_1}=(k_a\xi^a)|_{P_2}$, so we obtain $$\begin{align}\frac{\omega_1}{\omega_2}=\frac{(-\xi^b\xi_b)^{1/2}|_{P_2}}{(-\xi^b\xi_b)^{1/2}|_{P_1}}=\frac{(1-2M/r_2)^{1/2}}{(1-2M/r_1)^{1/2}},\end{align}$$ where the explicit form of $\xi^b\xi_b=g_{tt}=-(1-2M/r)$ for Schwarzschild spacetime has been used and $r_1$ and $r_2$ are, the radial coordinates of $O_1$ and $O_2$. We can also see $$\begin{align}\frac{\Delta \omega}{\omega}\approx -\frac{GM}{c^2r_1}+\frac{GM}{c^2r_2},\end{align}$$ 

The coordinate basis components of the tangent $u^a$ to a curve parametrized by $\tau$ are $$\begin{align}u^\mu=\frac{dx^\mu}{d\tau}\equiv \dot{x}^\mu.\end{align}$$ For timelike geodesics, we choose $\tau$ to be the proper time; and for null geodesics, we choose $\tau$ to be an affine parameter. Thus for these cases we have (recalling that $\theta=\pi/2$ by parity reflection symmetry $\theta\rightarrow \pi-\theta$) $$\begin{align}-\kappa=g_{ab}u^au^b=-(1-2M/r)\dot{t}^2+(1-2M/r)^{-1}\dot{r}^2+r^2\dot{\phi}^2,\end{align}$$ where $$\begin{align}\kappa=\begin{cases}1& (\mbox{timelike geodesics})\\ 0& (\mbox{null geodesics})\end{cases}.\end{align}$$

In the derivation of the gravitational redshift, we already used the fact that the quantity $$\begin{align}E=-g_{ab}\xi^au^b=(1-2M/r)\dot{t}\end{align}$$ is a constant of the motion, where $\xi^a=(\partial/\partial t)^a$ denotes the static Killing field. In the case of timelike geodesics, at large distance from the center of attraction ($r\gg M$) where the metric becomes flat and the norm of $\xi^a$ becomes $-1$, $E$ reduces to the special relativistic formula for the total energy per unit rest mass of a particle as measured by a static observer.

The rotational Killing field $\psi^a=(\partial/\partial \phi)^a$ also yields a constant of the motion, $L$, via proposition C.3.1, $$\begin{align}L=g_{ab}\psi^au^b=r^2\dot{\phi}.\end{align}$$ in the Newtonian limit, equation(13) simply express Kepler's second law.

Substituting equations (12) and (13) in equation (10) and rearranging terms, we obtain our final equation for geodesics with remarkable little labor, $$\begin{align}\frac{1}{2}\dot{r}^2+\frac{1}{2}\left( 1-\frac{2M}{r}\right)\left( \frac{L^2}{r^2}+\kappa\right)=\frac{1}{2}E^2.\end{align}$$ This equation shows that the radial motion of a geodesic is the same as that of a unit mass particle of energy $E^2/$ in ordinary one-dimensional, nonrelativistic mechanics moving in the effective potential, $$\begin{align}V=\frac{1}{2}\kappa-\kappa\frac{M}{r}+\frac{L^2}{2r^2}-\frac{ML^2}{r^3}.\end{align}$$ We called $-\kappa M/r$ as "Newtonian term", $L^2/2r^2$ as "centrifugal barrier term", $-ML^2/r^3$ "new term..?".

6.4 The Kruskal Extension

In the case of the Schwarzschild metric, we already pointed out that the Schwarzschild coordinates will be badly behaved where the timelike Killing field $\xi^a$ becomes collinear with $\nabla^ar$. We shall see below that this occurs at $r=2M$, and that the "singularity" at $r=2M$ is merely a coordinate singularity. On the other hand, the singularity at $r=0$ is a true singularity in the spacetime geometry, as can be demonstrated by calculating via equations (29)`(3) the curvature invariant $R_{abcd}R^{abcd}$.

The Schwarzschild spacetime is, of course, four-dimensional, but because of the spherical symmetry, only the two-dimensional $r-t$ part of the metric is of importance for analyzing the nature of the singularity at $r=2M$. Hence, we shall study the two-dimensional metric $$\begin{align}ds^2=-(1-2M/r)dt^2+(1-2M/r)^{-1}dr^2.\end{align}$$ The null geodesics are easily found from the null condition, $$\begin{align}0=g_{ab}k^ak^b=-(1-2M/r)\dot{t}^2+(1-2M/r)^{-1}\dot{r}^2,\end{align}$$ which implies $$\begin{align}\left( \frac{dt}{dr}\right)^2=\left( \frac{r}{r-2M}\right)^2.\end{align}$$ Thus, the radial null geodesics of Schwarzschild satisfy $$\begin{align}t=\pm r_*+\mbox{constant},\end{align}$$ where the "Regge-Wheeler tortoise coordinate" $r_*$ is defined by $$\begin{align}r_*=r+2M\ln (r/2M-1)\end{align}$$ so that $dr_*/dr=(1-2M/r)^{-1}$. We define the null coordinates $u,v$ by $$\begin{align}u=t-r_*,\\ v=t+r_*.\end{align}$$ In these coordinates, the metric (16) takes the form $$\begin{align}ds^2=-(1-2M/r)du\ dv,\end{align}$$ where $r$ is not viewed as a function of $u$ and $v$, defined implicitly by $$\begin{align}r+2M\ln \left( \frac{r}{2M}-1\right)=r_*=(v-u)/2.\end{align}$$ Using equation (24), we can rewrite equation (23) as $$\begin{align}ds^2=-\frac{2Me^{-r/2M}}{r}e^{(v-u)/4M}du\ dv,\end{align}$$ where we have factored the metric components into a piece, $e^{-r/2M}/r$, which is nonsingular as $r\rightarrow 2M$ (i.e., as $u\rightarrow \infty$ or $v\rightarrow \infty$) times a piece with simple $u$ and $v$ dependence. Comparison with the Rindler case, equation (9) suggests that we define new coordinates $U$ and $V$ by $$\begin{align}U=-e^{-u/4M},\\ V=e^{v/4M},\end{align}$$ in terms of which the metric becomes $$\begin{align}ds^2=-\frac{32M^3e^{-r/2M}}{r}dU\ dV.\end{align}$$ There is now no longer a singularity at $r=2M$ (i.e., at $U=0$ or $V=0$), and thus we can extend the Schwarzschild solution by allowing $U$ and $V$ to take on all values compatible with $r>0$.

If we make the final transformation $T=(U+V)/2$, $X=(V-U)/2$, the full Schwarzschild metric takes the final form given by Kruskal (1960), $$\begin{align}ds^2=\frac{32M^3e^{-r/2M}}{r}(-dT^2+dX^2)+(r^2(d\theta^2+\sin^2\theta d\phi^2).\end{align}$$ The relation between the old coordinates $(t,r)$ and the new coordinate $(T,X)$ is given by $$\begin{align}\left( \frac{r}{2M}-1\right) e^{r/2M}=X^2-T^2,\\ \frac{t}{2M}=\ln \left( \frac{T+X}{X-T}\right)=2\tanh^{-1}(T/X),\end{align}$$ and in equation (29), $r$ is to be viewed as the function of $X$ and $T$ defined by equation (30). The allowed range of the coordinates $X$ and $T$ is given by the condition $r>0$, which yields $$\begin{align}X^2-T^2>-1.\end{align}$$