The Final States

Section 4 of 6

Black Holes

Part 4: Black Holes — Where Gravity Wins

Black hole

A region of spacetime bounded by an event horizon, formed when collapse exceeds all known pressure support. It is not a material surface or solid object; its mass curves spacetime so strongly that nothing inside the horizon can send signals out.

Event horizon

The causal boundary of a black hole: inside it, every future-directed path leads inward and no signal can escape to distant observers. It is a property of spacetime geometry, not a material surface. For a non-rotating black hole it sits at the Schwarzschild radius.

Artist rendering of a supermassive black hole surrounded by a bright orange accretion disk, with light paths visibly warped by extreme spacetime curvature near the event horizon. — Figure 5What to notice: the black hole itself is not glowing. The light comes from hot gas in the accretion flow, while the dark central region and bent light paths reveal strongly curved spacetime near the event horizon.NASA/STScI

The Schwarzschild Radius: A Newtonian Preview

We can motivate the event-horizon scale with a Newtonian escape-speed argument. The escape speed from radius $R$ around mass $M$ is $v_{\rm esc} = \sqrt{2GM/R}$ . Setting $v_{\rm esc} = c$ and solving gives $R = 2GM/c^2$ , the Schwarzschild radius:

For a non-rotating, uncharged black hole, $R_s$ is the event-horizon radius.

Schwarzschild radius

The event-horizon radius of a non-rotating, uncharged black hole, $R_s = 2GM/c^2 \approx 3.0\,\mathrm{km}\,(M/M_\odot)$ . Compress a mass inside its Schwarzschild radius and it becomes a black hole; the ratio $R_s/R$ measures how relativistic an object’s gravity is.

Worked Example 1The Sun's Schwarzschild Radius

Problem

Compute the Schwarzschild radius of the Sun. Use $G = 6.67 \times 10^{-8}\,\mathrm{cm^3\,g^{-1}\,s^{-2}}$ , $M_\odot = 2.0 \times 10^{33}\,\mathrm{g}$ , $c = 3.0 \times 10^{10}\,\mathrm{cm\,s^{-1}}$ .

StepNumerator and denominator

$2GM_\odot = 2.67 \times 10^{26}\,\mathrm{cm^3\,s^{-2}}, \qquad c^2 = 9.0 \times 10^{20}\,\mathrm{cm^2\,s^{-2}}$

StepDivide and convert

$R_{s,\odot} = \frac{2.67 \times 10^{26}}{9.0 \times 10^{20}}\,\mathrm{cm} = 2.97 \times 10^5\,\mathrm{cm} = 2.97\,\mathrm{km}$

Dimensional check

$\dfrac{\mathrm{cm^3\,s^{-2}}}{\mathrm{cm^2\,s^{-2}}} = \mathrm{cm}$ ✓ — the units reduce to a length.

Result

$R_{s,\odot} \approx 3.0\,\mathrm{km}.$ To make the Sun a black hole, you would compress one solar mass inside a radius of about $3\,\mathrm{km}$ .

Figure 6What to notice: the Schwarzschild radius scales linearly with mass. Doubling the mass doubles the event-horizon radius, which is why the scaling R_s ~ 3.0 km (M/M_sun) is so powerful.ASTR 201 (generated)

Useful scaling relation

Because $R_s \propto M$ , using the Sun as reference gives $R_s \approx 3.0\,\mathrm{km}\,(M/M_\odot)$ .

Object	Mass	Schwarzschild radius
Sun	$1\,M_\odot$	$3.0\,\mathrm{km}$
$10\,M_\odot$ black hole	$10\,M_\odot$	$30\,\mathrm{km}$
Sgr A*	$4 \times 10^6\,M_\odot$	$1.2 \times 10^7\,\mathrm{km}$

Problem

Use $R_s \approx 3.0\,\mathrm{km}\,(M/M_\odot)$ . (1) $R_s$ for a $5\,M_\odot$ black hole? (2) For a $30\,M_\odot$ black hole? (3) Why didn’t you need $G$ and $c$ again?

Compactness: When does gravity become relativistic?

The Schwarzschild radius is useful even for objects that are not black holes. The compactness $R_s/R$ measures how relativistic an object’s gravity is. For Earth, $R_s/R_\oplus \sim 10^{-9}$ ; for the Sun, $R_s/R_\odot \approx 4 \times 10^{-6}$ . For a neutron star with $M=1.4\,M_\odot$ and $R=10\,\mathrm{km}$ , $R_s = 3.0\,\mathrm{km}(1.4) = 4.2\,\mathrm{km}$ , so

\frac{R_s}{R} = \frac{4.2\,\mathrm{km}}{10\,\mathrm{km}} = 0.42.

A compactness of $0.42$ is enormous: general relativity is not a tiny correction for neutron stars, it is essential. The Newtonian escape-speed estimate gives $v_{\rm esc}/c = \sqrt{R_s/R} = \sqrt{0.42} \approx 0.65$ , so $v_{\rm esc} \approx 0.65c$ .

Compactness

The dimensionless ratio $R_s/R$ of an object’s Schwarzschild radius to its actual radius — a measure of how close it is to being a black hole. Tiny for Earth and the Sun, $\sim 0.4$ for neutron stars, and exactly $1$ at a black-hole event horizon. Large compactness means general relativity is essential.

Figure 7What to notice: compactness R_s/R is tiny for Earth and the Sun, larger for white dwarfs, a large fraction of unity for neutron stars, and exactly 1 at a black-hole event horizon. Compactness, not mass alone, tells us when general relativity matters.ASTR 201 (generated)

Quick check

Rank Earth, Sun, white dwarf, neutron star, and black-hole event horizon by compactness $R_s/R$ , smallest to largest. Which require general relativity?

Quick check

A neutron star and a stellar-mass black hole of the same mass both bend light strongly and accrete gas.

(a) State the key physical difference between the neutron star’s surface and the black hole’s event horizon.
(b) Name one observation that would distinguish “has a material surface” from “has a horizon.”

Why General Relativity?

The Newtonian escape-speed argument gives the correct Schwarzschild radius, but it does not explain what an event horizon really is. Newtonian gravity has no universal speed limit; general relativity does. GR changes the picture three ways: nothing travels faster than light; mass and energy curve spacetime; and inside an event horizon, all future-directed paths lead inward. That last point is the key — the horizon is not a wall, it is where the geometry of spacetime prevents outward escape.

The More You Know: Enrichment: Gravitational Redshift Near a Neutron Star

A photon climbing out of a gravitational well loses energy and its wavelength increases — gravitational redshift. For a non-rotating compact object, light emitted from radius $R$ is shifted by $\lambda_{\rm obs}/\lambda_{\rm emit} = 1/\sqrt{1 - R_s/R}$ . For a neutron star with $R_s/R = 0.42$ ,

\frac{\lambda_{\rm obs}}{\lambda_{\rm emit}} = \frac{1}{\sqrt{1 - 0.42}} = \frac{1}{\sqrt{0.58}} \approx 1.31,

so $500\,\mathrm{nm}$ green light is observed at $\approx 655\,\mathrm{nm}$ — shifted toward red. This is not a Doppler shift from motion; it is gravitational redshift from spacetime curvature, and it grows without bound as the emitting surface approaches $R_s$ . (A static emitter exactly at the horizon is not physically possible; the useful lesson is the limiting behavior.)

Gravitational redshift

The stretching of light to longer wavelengths as it climbs out of a gravitational field, $\lambda_{\rm obs}/\lambda_{\rm emit} = 1/\sqrt{1 - R_s/R}$ for a static compact object. It comes from spacetime curvature, not from motion, and grows without bound as the emitter approaches the event horizon.

Figure 8What to notice: no single observation carries the whole argument. Pulses, X-ray binaries, stellar orbits, gravitational waves, and horizon-scale images converge on neutron-star and black-hole models.ASTR 201 (generated)