Black holes: horizons and orbits

Imagine sending a light pulse outward as you fall toward a spherical object. At which events can that pulse escape to observers arbitrarily far away? We will answer by solving the field equation outside the object, following light through the resulting geometry, and identifying the boundary of escape. That boundary is the black-hole horizon in the spacetime we construct.

17.1 Solving the spherical vacuum equation#

Set $\Lambda=0$. Outside a static spherical source, use the areal radius $r$: a symmetry sphere has area $4\pi r^2$. This is a geometrically meaningful definition, not a promise that radial proper distance equals $r$.

For this subsection the chart is $(t,r,\theta,\phi)$, with $t$ in seconds. Write

Spherical symmetry forbids a preferred angular direction. Staticity permits a diagonal time-radial form in the exterior region. We have reduced ten metric components to two unknown functions, without assuming their values.

A few connection coefficients show the mechanism:

A prime means $d/dr$. The last expression remembers that spheres change size as $r$ changes. Such angular terms are essential even though the unknown functions depend only on radius.

For example, the Ricci calculation gives

The second derivative comes from differentiating the connection; the products come from its $\Gamma\Gamma$ terms; the $2/r$ accounts for the two angular dimensions. Curvature is assembled from precisely the operations learned earlier.

Define $f(r)=e^{-2\beta(r)}$. Two convenient mixed Einstein components are

Vacuum requires both to vanish. The first equation can be reorganized as

A derivative is zero, so the bracket is a constant: $r(1-f)=C$. Thus $f=1-C/r$. This is the central integration, and it explains the inverse-radius form instead of merely announcing it.

Subtracting the two vacuum equations gives

In the static exterior $f\ne0$, hence $\alpha+\beta$ is constant. A constant rescaling of $t$ removes that constant. Therefore $e^{2\alpha}=e^{-2\beta}=f$. The remaining angular vacuum equation is satisfied by these functions; the contracted Bianchi identity explains why all the apparent equations are not independent.

At large $r$, match $g_{tt}=-c^2(1-C/r)$ to the Newtonian clock coefficient $-c^2(1-2G_NM/(rc^2))$. This fixes $C=2G_NM/c^2$. Define the length $m=G_NM/c^2$. The solution is

We assumed staticity to make the derivation accessible. Birkhoff’s theorem says something stronger: a spherically symmetric vacuum region with $\Lambda=0$ is locally Schwarzschild even if the spherical matter boundary moves. A perfectly spherical pulsating star does not broadcast tensor gravitational waves into its vacuum exterior. The theorem does not describe a region filled with an outgoing matter or radiation flux, which is not vacuum. The time-independence step can be checked in the calculation below. David Tong’s black-hole lecture notes discuss the coordinate construction and theorem.

Further calculation: where spherical time dependence goes

In a region where the gradient of the areal radius is spacelike, choose the same diagonal time-radius chart but initially allow $\alpha(t,r)$ and $\beta(t,r)$. The off-diagonal Ricci calculation now gives

$$R_{tr}=\frac{2}{r}\partial_t\beta.$$

For example, in $R_{tr}=\partial_\lambda\Gamma^\lambda{}_{rt}-\partial_r\Gamma^\lambda{}_{\lambda t}+\Gamma^\lambda{}_{\lambda\sigma}\Gamma^\sigma{}_{rt}-\Gamma^\lambda{}_{r\sigma}\Gamma^\sigma{}_{\lambda t}$, the differentiated terms cancel. The remaining terms involving $\partial_t\alpha$ and radial derivatives cancel in pairs. The angular trace $\Gamma^\theta{}_{\theta r}+\Gamma^\phi{}_{\phi r}=2/r$ multiplies $\Gamma^r{}_{rt}=\partial_t\beta$ and remains.

Vacuum therefore requires $\partial_t\beta=0$. The difference of the two diagonal equations still gives $\partial_r(\alpha+\beta)=0$, so $\alpha(t,r)=-\beta(r)+q(t)$. The metric’s only apparent time dependence is the factor $e^{2q(t)}dt^2$. Define $t'=\int e^{q(t)}dt$ to remove it. The radial integration then gives the same Schwarzschild function $f$ as above.

This proves the staticity step in this exterior chart. It does not use this chart through a null gradient of $r$; Section 17.3 supplies a regular extension through the horizon. A moving spherical matter boundary changes which region is vacuum, but does not add a freely varying time function to its vacuum metric.

17.2 Testing the horizon with curvature#

The metric’s radial component diverges at $r=2m$. Its formula also fails as $r$ approaches zero. To distinguish a coordinate failure from a divergent tidal field, calculate curvature before drawing a conclusion.

In a static orthonormal frame outside $r=2m$, the six curvature entries have the pattern calculated in Chapter 9’s vacuum example: $(-2q,q,q,-q,-q,2q)$ with $q=m/r^3$. Their Ricci contractions cancel. Their full squared contraction is $4(4+1+1+1+1+4)q^2=48q^2$, giving the Kretschmann scalar:

It is finite at $r=2m$ and diverges at $r=0$. Finite scalar invariants alone do not prove every conceivable spacetime point is regular, but here an explicit nonsingular chart will establish regularity at the horizon. At $r=0$, the divergent invariant proves the problem cannot be repaired by relabeling coordinates.

Meanwhile $R_{\mu\nu}=0$ and $R=0$ everywhere in the vacuum exterior. The remaining tidal field is Weyl curvature. Vacuum removes the Ricci source in this solution; it leaves these nonzero tidal components.

The horizon radius is

At the horizon, the tidal scale $G_NM/r_s^3$ is proportional to $M^{-2}$. A larger black hole can have gentler horizon tides. Event-horizon status and local violence are different questions.

SPACETIME LAB / 06

The same interval. A longer ruler.

Keep two coordinate radii fixed. Increase the black hole’s mass and measure how much farther apart they become on this spatial slice.

ExploreDiagram

Interactive geometry is loading.

Drag to orbit · pinch to zoom

Mass at a fixed reference radius

Smaller horizonLarger horizon

Proper length of the rose ruler

$$\begin{gathered}r_s=\frac{2GM}{c^2},\quad\Delta r=0.25R\\\Delta\ell=\int_{0.75R}^{R}\frac{dr}{\sqrt{1-r_s/r}}\end{gathered}$$

Read the scene. All displayed lengths use one fixed reference radius $R$. Increasing $r_s/R$ increases mass while the ruler endpoints stay at $0.75R$ and $R$, outside the horizon. The rose path measures proper radial distance; coordinate radius is defined by circumference, not by this ruler. This is Flamm’s embedding of an equatorial, constant Schwarzschild-time slice, with $z=2\sqrt{r_s(r-r_s)}$ up to an additive constant. Height is auxiliary, not time or a force. The outer edge is the chosen viewing radius, not an edge of space.

17.3 Repairing the horizon with Eddington–Finkelstein coordinates#

Define the tortoise coordinate

Then introduce an advanced time $v=t+r_*/c$, still measured in seconds. Since $dt=dv-dr/(cf)$, substitution gives

The troublesome $dr^2/f$ terms cancel exactly. The metric becomes

Its time-radial block has determinant $-c^2$, including at $r=2m$. The time-radius block remains invertible at the future horizon. Away from the usual angular-coordinate poles, the full metric is smooth and nondegenerate there.

For radial light, set $d\Omega=0$ and $ds^2=0$. One family has $dv=0$: ingoing rays. The other satisfies

Outside, these outgoing rays increase their radius. At the horizon they remain on it. Inside, even this outgoing family decreases its areal radius. Future-directed timelike trajectories lie between the two null directions and also move toward smaller $r$.

An outward-directed engine can change the traveler’s timelike direction within the cone. It cannot produce a direction outside it. Local light still travels at $c$; the negative radial rate expresses the shape of the future cone in these coordinates.

The maximal mathematical extension of eternal Schwarzschild has additional regions. A black hole produced by stellar collapse need not contain its white-hole region or second exterior. An exact metric’s maximal extension and the spacetime of a particular formation process are distinct objects.

WORKED EXAMPLE

At a horizon, ask which directions lead into the future

Can a future-directed light ray point outward and still lose areal radius?

Builds onNull curves The Schwarzschild solution Coordinate failure versus curvature

See the idea

At an event, a light cone gives the possible directions of light and the allowed directions of massive travelers. In this radial diagram, left and right measure areal radius $r$: a symmetry sphere has area $4\pi r^2$. Upward means increasing plot time. The horizon is $r=r_s=2G_NM/c^2$. Compare an event outside, on, and inside it; the directions are computed from the metric, not drawn to resemble a funnel.

Work it out

1. 01
   Use a chart that survives horizon crossing

   Set $f=1-r_s/r$. Schwarzschild radial light obeys $c\,dt=\pm dr/f$. Introduce $r_*=r+r_s\ln|r/r_s-1|$, whose derivative is $1/f$, and the advanced time $v=t+r_*/c$ in seconds. Substituting $dt=dv-dr/(cf)$ cancels the divergent radial terms. The time-radial metric block has determinant $-c^2$, including at the horizon.

   $$ds^2=-fc^2dv^2+2c\,dv\,dr+r^2d\Omega^2.$$

   Why this step works The new metric stays finite and nondegenerate at the horizon. The old chart failed there; the geometry did not.

2. 02
   Find both light directions without dividing one away

   Radial light has $d\Omega=0$ and $ds^2=0$. Factoring gives $c\,dv(-fc\,dv+2dr)=0$. One family has $dv=0$: future ingoing light has decreasing $r$. The other family has $dr/dv=cf/2$. Keep the first family before dividing by $dv$, or half the light cone disappears from the calculation.

   $$dv=0\quad\text{or}\quad \frac{dr}{dv}=\frac c2\left(1-\frac{r_s}{r}\right).$$

   Why this step works A product vanishes when either factor vanishes. The two solutions are the two radial null directions.

3. 03
   Translate the directions into the displayed axes

   Use dimensionless radius $\rho=r/r_s$ and dimensionless plot time $\tau_{\rm plot}=cv/r_s-\rho$. This $\tau_{\rm plot}$ is a coordinate, not a traveler’s proper time. For outgoing light, divide $d\rho/d(cv/r_s)=f/2$ by $d\tau_{\rm plot}/d(cv/r_s)=1-f/2$. For ingoing light, $dv=0$ gives $d\tau_{\rm plot}=-d\rho$.

   $$\left.\frac{d\rho}{d\tau_{\rm plot}}\right|_{\rm out}=\frac{\rho-1}{\rho+1},\qquad \left.\frac{d\rho}{d\tau_{\rm plot}}\right|_{\rm in}=-1.$$

   Why this step works Both plotted directions rise toward the future. Their slopes are coordinate rates, not the speed a local observer measures.

4. 04
   Read the permitted futures

   At $\rho=3$, outgoing light has positive slope $1/2$. At $\rho=1$, its slope is zero: it follows the horizon. At $\rho=1/2$, even the outgoing slope is $-1/3$. Future radial timelike directions lie strictly between the null directions. Angular motion narrows the allowed radial projection further. Thus every massive future traveler inside this black-hole region moves toward smaller areal radius.

   $$0<\rho<1\quad\Longrightarrow\quad -1<\frac{d\rho}{d\tau_{\rm plot}}<\frac{\rho-1}{\rho+1}<0\quad\text{(radial timelike)}.$$

   Why this step works Increasing thrust changes which allowed timelike direction is followed. It cannot move that direction outside the local light cone.

Go deeper

This diagram describes the future-horizon extension of the stationary Schwarzschild solution for $r>0$. Its local cone calculation must be distinguished from the global definition of a black hole: in an asymptotically flat spacetime, a black-hole event cannot send any future causal signal to future null infinity, the ideal destination of light that escapes forever. The event horizon is the boundary of that region. It depends on the entire future geometry. The surface $r=r_s$ has this role in the specified Schwarzschild black-hole spacetime; a local “cone tilt” test is not a general horizon detector in arbitrary evolving spacetimes. At $r_s$ the curvature invariant is $12/r_s^4$, finite for $M>0$. At $r=0$ it diverges. A regular horizon and a curvature singularity are different geometric facts.

Further readingDavid Tong · Black holes, §§6.1.3–6.1.5

A STATE YOU CAN CHECK

The two slopes are coordinate directions in the dimensionless radial plot. At each event, both point toward the future. A local observer still measures the speed of light as c.

Ingoing light at every event

$d\rho/d\tau_{\rm plot}=-1$

Outgoing light at ρ = 3

$d\rho/d\tau_{\rm plot}=0.500$

Outgoing light at ρ = 1

$d\rho/d\tau_{\rm plot}=0.000$

Outgoing light at ρ = 0.5

$d\rho/d\tau_{\rm plot}=-0.333$

The worked steps explain these measurements. Interactive controls appear when available.

Open the reference diagram

Test the idea

FIRST, PREDICT

Inside this Schwarzschild black hole, why can an outward-fired light pulse have decreasing $r$?

“Outward” labels the less inward null direction; both future radial null directions have decreasing areal radius.Its locally measured speed has fallen below $c$.An infinite curvature wall pushes it back at $r_s$.

Compare the reasoning

“Outward” labels the less inward null direction; both future radial null directions have decreasing areal radius.

Yes. The two null directions remain distinct. Their radial coordinate rates are both negative inside this horizon.

Its locally measured speed has fallen below $c$.

A local freely falling observer still measures the light pulse’s speed as $c$. The negative quantity in the diagram is a coordinate rate.

An infinite curvature wall pushes it back at $r_s$.

The Schwarzschild horizon is regular. The divergent curvature invariant occurs at $r=0$, not at $r_s$.

A hint

Compare the two null slopes at $\rho=1/2$ and keep local measurement separate from coordinate slope.

NOW CHANGE THE EXAMPLE

At $r=3r_s/4$, calculate the outgoing light slope $d\rho/d\tau_{\rm plot}$.

A hint

Insert $\rho=3/4$ into $(\rho-1)/(\rho+1)$.

Work through the solution

The slope is $(-1/4)/(7/4)=-1/7\simeq-0.142857$. This negative coordinate slope says the outgoing light ray still moves toward smaller areal radius.

The horizon is a boundary of causal escape, not a place where local light slows down.

17.4 A falling clock and a hovering rocket#

For radial timelike motion, time-translation symmetry gives a dimensionless conserved energy per unit rest energy,

Substitute this into $g_{\mu\nu}u^\mu u^\nu=-c^2$:

For an object falling from rest at infinity, $\mathcal E=1$, so

Nothing diverges at the horizon. In this idealized classical trajectory, proper time from horizon to $r=0$ is

This is specific to that radial energy and classical solution, not a universal countdown for every infaller. Schwarzschild coordinate time diverges at horizon crossing because that chart fails there. Signals reaching a distant observer become increasingly delayed and redshifted; the object does not remain as a permanently bright frozen photograph.

Now hold a rocket at constant $r>2m$. Its four-velocity has $u^t=1/\sqrt f$. Its radial four-acceleration is

This component happens to resemble Newton’s acceleration, but the accelerometer measures the invariant magnitude

It diverges on approach to the horizon. The diverging quantity belongs to the family of observers trying to remain static. It does not imply a freely falling observer measures infinite curvature there. At the horizon, being static would require following a null worldline; no massive rocket can do that.

17.5 Circular orbits and their stability#

Spherical symmetry lets a geodesic lie in an equatorial plane. Define $\ell=r^2d\phi/d\tau$. The timelike normalization becomes

Dots here mean proper-time derivatives. Expanding the potential reveals a term proportional to $-1/r^3$, absent from the Newtonian effective potential. It changes the centrifugal barrier near the hole.

A circular orbit requires $V'_{\rm eff}=0$. Solving this algebraic condition gives

A small radial displacement is stable only when $V''_{\rm eff}>0$. At a circular orbit,

Thus circular timelike geodesics exist for $r>3m$ and are stable for $r>6m$. The marginal boundary $r=6m$ is the innermost stable circular orbit, or ISCO. Unstable circular orbits can exist between $3m$ and $6m$; “unstable” does not mean “algebraically nonexistent.”

For null geodesics, the effective potential is proportional to $f/r^2$. Its derivative vanishes at $r=3m$, a maximum. That is the photon sphere. Its circular light orbits are unstable: a small displacement from the potential maximum grows rather than oscillating around it.

The photon sphere is not a material surface. Nor is a black-hole image a direct photograph of the horizon’s coordinate radius: lensing, emission, absorption, and observer geometry intervene.

To recover the orbit equation used in Chapter 16, put $u=1/r$ and use $\dot r=-\ell u'$, where a prime now means $d/d\phi$. The radial energy equation becomes

Differentiate it and collect the common factor $2u'$:

For a noncircular orbit with $\ell\ne0$, divide where $u'\ne0$ and extend the result continuously through isolated turning points. Since $mc^2=G_NM$, this gives $u''+u=G_NM/\ell^2+3mu^2$. Circular orbits satisfy the same equation by the separate condition $V'_{\rm eff}=0$.

17.6 An event horizon knows about the future#

In an asymptotically flat spacetime, the geometry approaches flat spacetime sufficiently far from the isolated system. Future null infinity is the ideal destination of light that escapes indefinitely to larger distances. The black-hole region consists of events that cannot send a future-directed causal signal to that destination. Its boundary is the event horizon. The qualifier “future” means the entire future development matters.

A sufficiently small freely falling laboratory generally cannot determine by purely local experiments whether it has crossed an event horizon. It can measure curvature and tidal forces, but horizon membership is a global causal statement. Chapter 22 introduces another diagnostic by measuring whether both future-directed families of light leaving a closed surface initially decrease its area. That local area calculation answers a different question from whether a signal can escape forever.

WORKED EXAMPLE

Build a global causal map one coordinate change at a time

How can a finite diagram include infinity without distorting who can signal whom?

Builds onHorizon-penetrating coordinates Elementary derivatives

See the idea

An ordinary road map preserves useful routes while changing distances. A causal diagram preserves the possible routes of signals. First choose coordinates in which radial light rays are straight. Then compress each null coordinate with a strictly increasing function. The resulting picture answers causal questions; its ruler does not measure proper distance or elapsed time.

Work it out

1. 01
   Exponentiate the troublesome null coordinates

   Begin in the right Schwarzschild exterior, $\rho=r/r_s>1$. Define $q=ct/r_s$, $\rho_*=\rho+\ln(\rho-1)$, $\bar u=q-\rho_*$ and $\bar v=q+\rho_*$. Constant $\bar u$ and constant $\bar v$ describe radial light. Define $U=-e^{-\bar u/2}$ and $V=e^{\bar v/2}$. Multiplying eliminates $q$ and identifies the radius.

   $$UV=-(\rho-1)e^\rho,\qquad U<0,\ V>0\quad\text{in the right exterior}.$$

   Why this step works These exponentials make the horizon occur at a finite null-coordinate value. Their sign definitions here cover one exterior, not all four regions.

2. 02
   Extend the metric, not an invalid exterior formula

   Differentiating gives $d\bar u=-2dU/U$ and $d\bar v=2dV/V$. Substituting in $ds^2_{\rm radial}=-r_s^2f\,d\bar u\,d\bar v$ cancels the zero of $f$. The result below stays regular at $\rho=1$. It defines an extension to real $U,V$ with $UV<1$, where $\rho>0$ is fixed implicitly by their product. One does not keep $U=-e^{-\bar u/2}$ with real exterior Schwarzschild coordinates inside every region.

   $$ds^2=-\frac{4r_s^2e^{-\rho}}{\rho}\,dU\,dV+r^2d\Omega^2.$$

   Why this step works A smooth, nondegenerate metric is the object being extended. A coordinate expression valid only in the exterior must not be mistaken for a universal definition.

3. 03
   Identify the regions by their signs

   With future chosen so that both $U$ and $V$ increase along radial timelike curves, $(U<0,V>0)$ is the right exterior, $(U>0,V>0)$ the black-hole interior, $(U>0,V<0)$ the left exterior, and $(U<0,V<0)$ the white-hole interior. The null lines $U=0$ and $V=0$ are horizons. The boundary $UV=1$ corresponds to $r=0$: its positive-sign branch is a future singularity, and its negative-sign branch a past singularity. In the axes below, radial light has slope $\pm1$.

   $$T_K=\frac{V+U}{2},\qquad X_K=\frac{V-U}{2},\qquad dU\,dV=dT_K^2-dX_K^2.$$

   Why this step works The singularity is a spacelike boundary, not a stationary object drawn at the center of a spatial funnel.

4. 04
   Compress infinity while keeping null rays straight

   Set $p=\arctan U$ and $q_c=\arctan V$, both in $(-\pi/2,\pi/2)$. Their derivatives are positive, so increasing null coordinates still means the same future direction. Define $T_c=(q_c+p)/2$ and $X_c=(q_c-p)/2$. Curves with constant $U$ or $V$ remain lines at $45$ degrees. Infinite coordinate values become limiting edges; those edges are not nearby physical walls.

   $$dU\,dV=\sec^2p\,\sec^2q_c\,(dT_c^2-dX_c^2).$$

   Why this step works The multiplying factor is positive at interior points. It changes scale but preserves which radial directions are null, timelike, and spacelike.

Go deeper

In the right exterior, future null infinity $\mathscr I^+$ is approached by $V\to+\infty$ with fixed finite $U<0$: escaping outgoing light reaches the boundary $q_c=\pi/2$. Timelike and spacelike infinity are different limiting destinations. Each point of the radial diagram represents a symmetry two-sphere, with the angular directions suppressed. The four-region map describes maximally extended eternal Schwarzschild. It contains a white-hole region and a second exterior. A black hole formed by stellar collapse has a matter interior and a different past causal structure; it need not contain either of those extra regions. No future causal route crosses the eternal bridge from one exterior to the other: from the right exterior, $V$ cannot decrease to the negative values of the left exterior.

Further readingDavid Tong · Black holes, §§6.1.3–6.1.5

Test the idea

FIRST, PREDICT

What survives the compression of a causal diagram?

The numerical proper time between any pair of events.A guarantee that a collapsing star produces all four eternal Schwarzschild regions.The possible causal directions and the order in which a signal can connect events.

Compare the reasoning

The numerical proper time between any pair of events.

Proper time depends on the metric’s scale, which the conformal drawing does not preserve.

A guarantee that a collapsing star produces all four eternal Schwarzschild regions.

The global spacetime must include its formation history. The eternal extension is a different idealized spacetime.

The possible causal directions and the order in which a signal can connect events.

Yes. Monotone changes of the null coordinates and positive interior conformal factors preserve the causal cones.

A hint

The prefactor in the final metric expression changes lengths while leaving the zero of a null interval unchanged.

NOW CHANGE THE EXAMPLE

An event has $\arctan U=-\pi/6$ and $\arctan V=\pi/3$. What is its compressed time $T_c$, expressed as a multiple of $\pi$? Enter the coefficient.

A hint

Use $T_c=(\arctan V+\arctan U)/2$.

Work through the solution

$T_c=[\pi/3-\pi/6]/2=\pi/12$, so the coefficient is $1/12\simeq0.083333$. This is a coordinate value, not a reading in seconds.

A causal map preserves signal routes; its boundaries and extra regions must be interpreted for the spacetime actually specified.

WORKED EXAMPLE

Which initial data can determine this event?

Why does a local light cone not settle every question about predictability?

Builds onCausal maps

See the idea

Suppose you know all the initial measurements only on a finite interval. You can predict an event only if every possible incoming causal influence traces back to that interval. Following one convenient light ray is not enough; an unaccounted path could carry missing information.

Work it out

1. 01
   Separate timelike and causal reachability

   The chronological future $I^+(p)$ consists of events reached from p by future-directed timelike curves. The causal future $J^+(p)$ also allows null curves (and conventionally p itself). In ordinary Minkowski spacetime these are the interior and the closed interior of the future light cone. Time orientation means we have consistently chosen which half-cone is future.

   $$I^+(p)\subseteq J^+(p).$$

   Why this step works A massive traveler follows a timelike path; a light signal can run along the boundary.

2. 02
   Draw the domain of an interval

   Use $c=1$ and an initial segment $S=\{t=0,-L\le x\le L\}$ in 1+1 Minkowski spacetime. An event with $0\le t\le L$ has past-null intersections $x-t$ and $x+t$ with the initial line. Both must remain inside the segment.

   $$D^+(S)=\{(t,x):0\le t\le L,\ |x|\le L-t\}.$$

   Why this step works Every past-inextendible causal curve from a point in this triangle must meet the initial segment.

3. 03
   State the global requirement

   An inextendible causal curve cannot be continued further as a causal curve in the spacetime under discussion. A Cauchy surface meets every inextendible causal curve exactly once. A spacetime admitting such a surface is globally hyperbolic, using the equivalent standard characterization for smooth time-oriented Lorentzian spacetimes. An achronal set contains no two points joined by a timelike curve; achronal does not by itself mean Cauchy.

   $$D(S)=D^+(S)\cup D^-(S).$$

   Why this step works Cauchy data must intercept every possible causal history, not merely form a visually level slice.

Go deeper

The future domain of dependence $D^+(S)$ is defined by every past-inextendible causal curve through the event meeting S. This quantifier is essential. A puncture removed from spacetime can create causal curves that end at the missing point relative to the remaining manifold, reducing the domain. Geodesic completeness is another definition: an affinely parameterized geodesic can be continued to arbitrary parameter values. For timelike geodesics proper time supplies an affine parameter up to a constant scale. Finite affine length of an inextendible geodesic is incompleteness; it need not be diagnosed by a divergent scalar curvature. These are the concepts needed before Chapter 22’s theorem discussion.

Further readingDavid Tong · General relativity

Test the idea

FIRST, PREDICT

A spacelike slice is drawn on a causal diagram. Does its spacelike character prove it is Cauchy?

Only if the coordinate time on it is zero.No. It must meet every inextendible causal curve exactly once.Yes. Spacelike and Cauchy mean the same thing.

Compare the reasoning

Only if the coordinate time on it is zero.

A coordinate label does not decide whether all causal histories cross the slice.

No. It must meet every inextendible causal curve exactly once.

A finite piece of a spacelike slice already fails to intercept histories outside its domain.

Yes. Spacelike and Cauchy mean the same thing.

Spacelike is a local geometric property; the Cauchy condition is global.

A hint

Try a finite segment of the t = 0 line.

NOW CHANGE THE EXAMPLE

For initial data on $[-5,5]$ at $t=0$ with $c=1$, what is the half-width in x of $D^+(S)$ at $t=2$?

A hint

Each inward null boundary moves one unit per unit time.

Work through the solution

$L-t=5-2=3$. The available interval is $[-3,3]$.

Predictability requires control of every possible incoming causal history.

17.7 The geometry of a rotating black hole#

The Kerr solution describes a stationary, isolated rotating vacuum black hole. We will take this exact solution as given and calculate its rotational effects. Obtaining it from the field equation is a separate boundary-value problem; unlike the spherical calculation, we have not derived its metric functions here.

Define

and

In Boyer–Lindquist coordinates, with $t$ in seconds, the metric is

Setting $a_K=0$ recovers Schwarzschild. The new $dt\,d\phi$ term mixes time evolution with angular motion. Because a cross term in $ds^2$ is $2g_{t\phi}dt\,d\phi$, its displayed coefficient is twice the metric component. Read the metric component by dividing that coefficient by two before using it in a momentum or velocity calculation.

An observer with zero conserved axial angular momentum satisfies

Zero angular momentum therefore does not mean zero coordinate angular velocity. This is one operational expression of frame dragging. It is an off-diagonal geometric effect, not viscous friction against an invisible fluid.

The roots of $\Delta=0$ are

For the Kerr black-hole family, $|\chi_{\rm spin}|\le1$. The outer root is the event-horizon radius. The surface where the stationary Killing field becomes null is instead $g_{tt}=0$:

Between this surface and the outer horizon lies the ergoregion. There, remaining at fixed spatial coordinates is impossible for a timelike observer, although escape can still be possible. At the poles the two surfaces meet. Elsewhere they differ: inability to remain stationary is weaker than inability to escape.

“No hair” is not a theorem that every possible gravitating theory has only two black-hole parameters. Kerr uniqueness results apply under substantial assumptions about vacuum Einstein gravity, stationarity, asymptotics, horizon structure, and regularity. For example, a rigorous result establishes Kerr uniqueness within a class of connected, nondegenerate, analytic regular vacuum black holes. Additional fields, different asymptotics, or dynamical settings change the question. See the primary mathematical result, Chruściel and Costa, On uniqueness of stationary vacuum black holes.

The useful physical idea is that an isolated black hole settling into the appropriate stationary vacuum state is described by very few exterior parameters. The assumptions determine when this description applies.

17.8 A relativistic star has an interior#

Imagine a small slab of material inside a star. Gravity pulls it inward. Pressure pushes on both sides; the pressure on its inner face must be greater if the slab is to remain at rest. This is why a supported star needs a pressure that decreases toward its surface. A large pressure with no pressure gradient would push equally from both sides.

Build the star outward from its centre. At each radius keep track of the mass enclosed and the pressure still needed to support the material above. The surface is where that pressure reaches zero. Increasing the central density changes the entire solution, including where its surface lies.

The following experiment uses one specified relation between density and pressure, called an equation of state. Its three curves show pressure falling, density falling, and enclosed mass growing. Each curve is divided by its own reference value so that their shapes can be compared. The horizontal coordinate runs from the centre to the surface. The mass and radius readouts use the model’s chosen scales, rather than solar masses and kilometres; those scales are derived below.

Stellar structure lab

Build a star from its centre

Choose a central density. Follow the pressure outward until it reaches zero: that is the surface of your star.

• Pressure / central pressure
• Density / central density
• Enclosed mass / total mass

Surface radius R0.86584

Gravitating mass M0.15736

Compactness 2M/R0.36349

Change in R when step is halved-3.907e-5

Increase the central density. The radius uses a fixed scale; watch whether a denser star is also a larger star. Colour shows density relative to the centre.

How this is calculated

Dimensionless geometric units G = c = K = 1; rest-mass density ρ₀, pressure p = ρ₀², energy density ε = ρ₀ + p. These are a toy equation of state, not a fitted neutron-star model.

$$\frac{dm}{dr}=4\pi r^2\epsilon$$

$$\frac{dp}{dr}=-\frac{(\epsilon+p)(m+4\pi r^3p)}{r(r-2m)}$$

• Static, spherical, isotropic perfect fluid; zero cosmological constant.
• A regular centre, a zero-pressure surface, and a Schwarzschild exterior.
• RK4 in radius using enthalpy h = ln(1 + 2ρ₀); the surface uses linear interpolation.

A mass–radius point alone does not establish radial stability. Changing the equation of state changes the family. Grid-refinement differences are diagnostics, not certified error bounds.

Oppenheimer and Volkoff · On Massive Neutron Cores ↗

Measurements and notebook

Dimensionless geometric units G = c = K = 1; rest-mass density ρ₀, pressure p = ρ₀², energy density ε = ρ₀ + p. These are a toy equation of state, not a fitted neutron-star model.

r	m	rho	p
0	0	0.2	0.04
0.08001	5.088e-4	0.19663	0.038664
0.16001	0.0039279	0.18678	0.034885
0.24001	0.012493	0.17109	0.02927
0.32001	0.027248	0.15062	0.022686
0.40001	0.047806	0.12677	0.01607
0.48001	0.072427	0.10117	0.010235
0.56001	0.0984	0.075522	0.0057036
0.64001	0.12259	0.051399	0.0026419
0.72001	0.14197	0.03001	9.006e-4
0.80001	0.1541	0.012054	1.453e-4
0.86584	0.15736	0	0

From pressure support to the spacetime equations. Start with a static, spherical perfect fluid, meaning that the local pressure is the same in every spatial direction. Let $\epsilon(r)$ be its rest-frame energy density, including rest energy, and let $p(r)$ be its pressure. Write

Here $r$ is areal radius, $m$ has units of mass, and $\Phi$ is dimensionless. Defining the mass function this way makes the radial metric coefficient a statement about the enclosed spherical gravitational mass; it is not simply the integral of rest-mass density over proper spatial volume.

The time-time Einstein equation and the radial equation give, respectively,

These equations can be checked from the spherical connection in §17.1: replace the constant exterior mass by $m(r)$ before differentiating and retain the nonzero fluid source. The mass equation is the time-time curvature equation; the pressure term in $\Phi'$ is the radial stress source. They are not obtained by assigning a Newtonian potential to a relativistic star.

Conservation supplies the mechanical balance. A static fluid has $u=e^{-\Phi}\partial_t$ and radial covariant acceleration $a_r=c^2\Phi'$. Projecting $\nabla_\mu T^{\mu\nu}=0$ orthogonal to $u$ gives

This is the Tolman–Oppenheimer–Volkoff equation. Three relativistic changes are visible: pressure contributes to inertial energy density, pressure also enters the source of the lapse gradient, and the radial geometry supplies a compactness factor. When $p\ll\epsilon$, $2G_Nm/(rc^2)\ll1$, and $\epsilon\simeq\rho c^2$, it reduces to $p'=-G_N\rho m/r^2$.

The equations need an equation of state, a relation between pressure and energy density supplied by matter physics. Choose central pressure $p_c>0$, impose a regular centre $m(0)=0$, integrate outward, and identify the first zero-pressure surface $R$. Then $M=m(R)$. With no material surface layer, match to an exterior Schwarzschild solution and normalize the clock by $e^{2\Phi(R)}=1-2G_NM/(Rc^2)$. The central lapse is fixed by integrating $\Phi'$ inward from that boundary; it is not an independently adjustable physical clock rate after exterior normalization.

The laboratory above uses an explicit, deliberately simple equation of state. In geometric units $G_N=c=1$, let $p=K\rho_0^2$ and $\epsilon=\rho_0+p$, where $\rho_0$ is rest-mass density in geometric units. Scale lengths and masses by $\sqrt K$ to set $K=1$. Its local sound-speed ratio is $dp/d\epsilon=2\rho_0/(1+2\rho_0)<1$. It is causal as a barotropic toy model, but it is not a fit to nuclear matter. Restoring a chosen $K$ sets the physical mass and radius scales; a plot without that choice is not a neutron-star prediction in solar masses and kilometres.

The numerical integration uses enthalpy $h=\int_0^p dp'/(\epsilon+p')=\ln(1+2\rho_0)$, for which $h'=-(m+4\pi r^3p)/[r(r-2m)]$. Near the centre,

These expansions start the calculation away from the apparent $0/0$ at the origin. The displayed step-refinement difference measures numerical sensitivity. It is not a statement about uncertainty in the equation of state. The zero of enthalpy is located by linear interpolation, so the surface calculation can dominate the error even though the interior integrator is fourth order.

An independent limiting check. In the weak-gravity, low-density limit, this $K=1$ equation of state has the Newtonian solution $\rho_0(r)=\rho_c\sin(\sqrt{2\pi}r)/(\sqrt{2\pi}r)$. Its first zero is $R=\sqrt{\pi/2}$ and its mass is $M=\sqrt{2\pi}\rho_c$. Derive this by eliminating $m$ between $2\rho_0'=-m/r^2$ and $m'=4\pi r^2\rho_0$. The automated model check compares against this independently solved limit.

A different analytic benchmark. A constant-energy-density star, in $G_N=c=1$ units, has $m(r)=Mr^3/R^3$ and

Substitute it into the mass and pressure equations and check $p(R)=0$. Its central pressure diverges as $2M/R\to8/9$. This incompressible model has unphysical infinite sound speed and is a mathematical benchmark, not a viable matter model. The broader Buchdahl bound requires its own assumptions, including static spherical equilibrium, isotropic pressure, regularity, and a nonincreasing density profile; it is not a universal bound on every object called a star.

A turning point on a one-parameter equilibrium mass–radius family can signal a change of radial stability under appropriate assumptions. Establishing stability requires perturbing the equilibrium and checking the resulting mode problem. A visually impressive mass–radius curve alone has not done that calculation.

17.9 Separate the photon from the observers#

A light signal can climb outward while its receiver moves inward to meet it. The climb tends to lower the received frequency; motion toward the incoming light tends to raise it. Compare the two effects by holding the emission and reception events fixed and changing the observers’ velocities at those events.

In this experiment, the frequency ratio is the receiver’s reading divided by the emitter’s reading. One means equal readings; a value above one means a blueshift. Radii are multiples of the Schwarzschild radius. Both events stay outside it, where a hovering observer can provide a local reference for velocity. Each point on the graph describes a possible receiving observer, rather than successive positions of one moving receiver.

Light and observers lab

Follow the photon. Ask each observer.

Send light between two observers. Change their heights and motion, and compare the frequencies they measure.

• Specified receiving observers
• Both observers static

Received / emitted frequency0.75593

Gravitational factor0.75593

Motion factor1

Locally measured light speed / c1

Each wave strip shows the same amount of local time. More crests mean a higher measured frequency. The markers below locate the two measurement events.

How this is calculated

Radii are multiples of rₛ = 2GM/c²; velocities are measured by local static observers in units of c, positive outward. Frequency is normalized to the emitter’s measurement.

$$\frac{\nu_r}{\nu_e}=\sqrt{\frac{f_e}{f_r}}\,\frac{\gamma_r(1-n\beta_r)}{\gamma_e(1-n\beta_e)},\qquad f=1-\frac{r_s}r$$

$$E_{\rm obs}=-p_\mu u^\mu,\qquad n=\pm1$$

• A radial photon in the exterior of a Schwarzschild black hole, r > rₛ.
• Specified local four-velocities at emission and reception; their full worldlines are not evolved.
• The plot samples a family of possible receiving events and observers, not the history of one moving detector.

Static reference frames do not extend through the horizon. The coordinate light rate is not a locally measured speed. Use the regular-coordinate horizon lesson to study crossing.

Tong · Black holes ↗

Measurements and notebook

Radii are multiples of rₛ = 2GM/c²; velocities are measured by local static observers in units of c, positive outward. Frequency is normalized to the emitter’s measurement.

radius	frequencyRatio	gravity	motion	receiverCoordinateSpeed	localLightSpeed
2	1	1	1	0.5	1
2.6	0.90139	0.90139	1	0.61538	1
3.2	0.8528	0.8528	1	0.6875	1
3.8	0.82375	0.82375	1	0.73684	1
4.4	0.8044	0.8044	1	0.77273	1
5	0.79057	0.79057	1	0.8	1
5.6	0.78019	0.78019	1	0.82143	1
6.2	0.77211	0.77211	1	0.83871	1
6.8	0.76564	0.76564	1	0.85294	1
7.4	0.76035	0.76035	1	0.86486	1
8	0.75593	0.75593	1	0.875	1

Now derive the comparison. Let $f(r)=1-r_s/r$. A static observer measures photon energy $E_{\rm static}=E_\infty/\sqrt f$, where $E_\infty$ is the conserved energy associated with the stationary Killing vector normalized at infinity. Thus static source and receiver measure $\nu_r/\nu_e=\sqrt{f_e/f_r}$.

At either event, a radial observer with local velocity $\beta c$ measures a further Doppler factor $\gamma(1-n\beta)$, where $n=+1$ for outward light and $n=-1$ for inward light. Dividing the receiver factor by the emitter factor gives the laboratory’s combined formula. This is an instantaneous comparison of specified four-velocities; it does not assume the moving observer remains at a fixed radius.

For an outward radial null ray, $c\,dt/dr=1/f$. Integrating between exterior radii gives

This is Schwarzschild coordinate time. A static local observer uses $d\tau=\sqrt f\,dt$ and radial proper length $d\ell=dr/\sqrt f$, obtaining $d\ell/d\tau=c$. A changing coordinate slope has not changed the locally measured light speed. Static reference observers require infinite support at the horizon and do not exist inside it; the regular-coordinate lessons handle that different domain.