Derivation of the Schwarzschild Solution

The Schwarzschild solution is the first exact solution of Einstein's field equations, written down by Karl Schwarzschild in December 1915 — within weeks of Einstein's final paper, and from a calculation he did while serving on the Russian front. It describes the geometry of empty spacetime outside any static, spherically symmetric body: a star, a planet, or a non-rotating black hole. Deriving it is the canonical rite of passage in every graduate course, because it is where the abstract machinery of curvature — metrics, Christoffel symbols, the Ricci tensor — first collapses into a single, concrete, physically readable spacetime. The whole derivation is a study in how much work symmetry can do. We demand that the spacetime be static (a timelike Killing vector, hypersurface-orthogonal, so no time dependence and no $\mathrm{d}t\,\mathrm{d}r$ cross term) and spherically symmetric (invariant under the rotation group $SO(3)$ , so the angular part is a round 2-sphere). These two demands collapse the ten independent components of a general metric down to just two unknown functions of the radial coordinate. Computing the Ricci tensor for that ansatz and imposing the vacuum equations $R_{\mu\nu}=0$ then locks the two functions together and integrates to a one-parameter family. Only a single constant survives the geometry; everything else is symmetry and gauge. That last constant is where physics re-enters. The correspondence principle — GR must reduce to Newton far away, where gravity is weak — identifies it as $2GM$ , the geometric mass. And the punchline, proved by Birkhoff in 1923, is that staticity was never needed: any spherically symmetric vacuum is automatically static and equals Schwarzschild, so a pulsating spherical star radiates nothing. We build the static solution in full, then note where Birkhoff makes it inescapable.

Key takeaways

The Schwarzschild solution is the unique static, spherically symmetric vacuum solution of Einstein's equations, obtained by solving $R_{\mu\nu}=0$ for a metric with two unknown radial functions $\alpha(r),\beta(r)$ .
Symmetry reduces the general ten-component metric to $\mathrm{d}s^2=-e^{2\alpha}\mathrm{d}t^2+e^{2\beta}\mathrm{d}r^2+r^2\mathrm{d}\Omega^2$ ; the vacuum equations force $\alpha+\beta=\text{const}$ and $e^{2\alpha}=1-C/r$ , leaving one constant.
Matching $g_{tt}$ to the Newtonian potential $\Phi=-GM/r$ in the weak field fixes $C=2GM$ , defining the geometric mass — the source's only imprint on the exterior.
The result is $\mathrm{d}s^2=-(1-\tfrac{2GM}{r})\mathrm{d}t^2+(1-\tfrac{2GM}{r})^{-1}\mathrm{d}r^2+r^2\mathrm{d}\Omega^2$ , with a coordinate singularity at $r=2GM$ and a true curvature singularity at $r=0$ .
The coordinate at $r=2GM$ is harmless: the Kretschmann scalar $R_{\mu\nu\rho\sigma}R^{\mu\nu\rho\sigma}=48G^2M^2/r^6$ stays finite there and only diverges at $r=0$ .
By Birkhoff's theorem the assumption of staticity was unnecessary — spherical symmetry alone forces this metric, so a radially pulsating star emits no gravitational waves.
Key idea: symmetry does almost all the work — staticity and spherical symmetry reduce the metric to two radial functions, the vacuum equations $R_{\mu\nu}=0$ pin both down up to one integration constant, and the Newtonian limit fixes that constant to be $2GM$ .

The core insight

Symmetry does almost all the work: staticity and spherical symmetry reduce the metric to two unknown radial functions, the vacuum equations $R_{\mu\nu}=0$ pin both down up to a single integration constant, and the Newtonian limit fixes that constant to be $2GM$ — the geometric mass.

Where this lives today

The metric derived here is the working backdrop of modern strong-gravity physics. The Event Horizon Telescope's images of M87 (2019) and Sagittarius A (2022) are, in effect, photographs of a vacuum solution of $R_{\mu\nu}=0$ at the horizon scale — the bright ring traces photon orbits set by exactly this geometry (generalized to spin by Kerr). LIGO, Virgo and KAGRA extract black-hole masses from merger waveforms whose late "ringdown" is Schwarzschild (or Kerr) quasinormal modes, and the first event, GW150914, settled to a single Schwarzschild-like remnant. On the ground, GPS satellites must correct for the Schwarzschild time dilation encoded in $g_{tt}=-(1-2GM/r)$ : clocks deeper in Earth's potential run slow, and the net relativistic offset of tens of microseconds per day would render the system useless within hours if ignored. At the galactic centre the GRAVITY instrument tracked the star S2 through its 2018 perihelion and measured the gravitational redshift predicted by this same $g_{tt}$ . And in stellar astrophysics Birkhoff's theorem makes Schwarzschild the mandatory exterior boundary condition for the interior Tolman-Oppenheimer-Volkoff equations, fixing the structure and maximum mass of neutron stars.

The physical picture

Think of the derivation as building the metric by elimination. The most general metric on four-dimensional spacetime has ten independent functions of four coordinates — an intractable mess. Spherical symmetry insists the geometry look identical in every direction, which forces the angular sector to be a round 2-sphere $r^2\,\mathrm{d}\Omega^2$ and forbids any preferred angular direction. Staticity insists nothing changes in time and there is no swirling — no $\mathrm{d}t\,\mathrm{d}r$ term. What survives is a bare two functions of $r$ : the redshift factor $e^{2\alpha}$ in front of $\mathrm{d}t^2$ , and the radial stretch $e^{2\beta}$ in front of $\mathrm{d}r^2$ . Ten functions have become two, before any physics. The vacuum equations then act almost like a conservation law on those two functions. One particular combination of the Ricci components instantly reveals that $\alpha(r)+\beta(r)$ is constant — the time-warp and the radial-stretch are rigidly locked. A second component is a first-order ordinary differential equation that integrates to $e^{2\alpha}=1-C/r$ . The structure is so tight that no functional freedom remains; the entire physical content of the exterior field is one number, $C$ , the mass felt at infinity.

Figure

Symmetry collapses the ten metric functions to two radial functions α(r) and β(r), the vacuum equations Rμν = 0 lock them and integrate to one constant C, and the Newtonian limit fixes C = 2GM — the geometric mass is the source's only imprint on the exterior

The Newtonian limit is the bridge back to the laboratory. Drop a slow test particle far from the mass; its geodesic must reproduce $\ddot{\vec x}=-\nabla\Phi$ , and that only works if $g_{tt}\approx-(1+2\Phi)$ — the warping of time is the Newtonian potential. Equating $-(1-C/r)$ with $-(1-2GM/r)$ delivers $C=2GM$ with no further input. The metric functions that result are worth seeing plotted: the lapse dips smoothly to zero at $r=2GM$ while the radial component blows up there, yet the geometry itself, as we will see, is perfectly regular.

Figure

The Schwarzschild metric functions versus r/rₛ: the lapse −gtt = 1 − rₛ/r climbs from 0 at the horizon to 1 far away, while grr = 1/(1 − rₛ/r) blows up at r = rₛ — both are smooth for r beyond rₛ, and for the Sun rₛ = 2GM/c² = 2.95 km

The deepest lesson is that we never had to say what the source is. The exterior of a neutron star, a white dwarf, the Sun, or a black hole of the same mass is one and the same Schwarzschild geometry. The mass is the only handle — an early foreshadowing of the no-hair theorems, and the reason the entire zoo of stellar objects shares a single external field.

The core equations

The starting point is the most general metric compatible with the two symmetries. Spherical symmetry fixes the angular part to a round sphere of areal radius $r$ ; staticity kills all time dependence and any $\mathrm{d}t\,\mathrm{d}r$ cross term; writing the surviving functions as exponentials guarantees the correct signature for all real $\alpha,\beta$ .

\mathrm{d}s^2 = -e^{2\alpha(r)}\,\mathrm{d}t^2 + e^{2\beta(r)}\,\mathrm{d}r^2 + r^2\left(\mathrm{d}\theta^2 + \sin^2\theta\,\mathrm{d}\phi^2\right) \qquad (1)

where: $\alpha(r)$ — radial function fixing the time-warp (redshift) factor $e^{2\alpha}$ ; $\beta(r)$ — radial function fixing the radial proper-length stretch $e^{2\beta}$ ; $r$ — areal radial coordinate, defined so a 2-sphere at $r$ has area $4\pi r^2$ ; $\mathrm{d}\Omega^2=\mathrm{d}\theta^2+\sin^2\theta\,\mathrm{d}\phi^2$ — line element of the unit 2-sphere. Two functions of one variable have replaced ten of four. Outside the source there is no matter, so the stress-energy tensor vanishes and the field equations reduce to the Ricci-flat condition. Tracing $G_{\mu\nu}=R_{\mu\nu}-\tfrac12 g_{\mu\nu}R=0$ gives $R=0$ , and back-substitution collapses the Einstein equations to the compact vacuum statement below — the single hypothesis that drives the entire calculation.

R_{\mu\nu} = 0 \qquad (2)

where: $R_{\mu\nu}$ — Ricci tensor of the metric (1). Ricci-flat is far weaker than flat: the full Riemann tensor need not vanish, which is exactly why curvature — tides — survives in the vacuum around a mass. Feeding the ansatz (1) into the Ricci tensor produces only three independent nonzero components (the $\phi\phi$ equation duplicates the $\theta\theta$ one). The angular component is the workhorse of the derivation; it alone, once the two functions are locked, integrates to the answer.

R_{\theta\theta} = e^{-2\beta}\left[\,r\,(\partial_r\beta - \partial_r\alpha) - 1\,\right] + 1 \qquad (3)

where: $\partial_r$ — ordinary derivative with respect to $r$ ; $R_{\phi\phi}=\sin^2\theta\,R_{\theta\theta}$ carries no new information. Setting this to zero is a first-order equation for the metric functions once $\beta$ is expressed through $\alpha$ . Combining the time-time and radial-radial vacuum equations cancels every second derivative and leaves a first-order statement whose entire content is that the two functions sum to a constant. That constant is pure gauge — a rescaling of $t$ absorbs it — so we may set $\beta=-\alpha$ .

\partial_r(\alpha+\beta)=0 \quad\Longrightarrow\quad \alpha(r)+\beta(r)=\text{const}\ \xrightarrow{\ t\text{-rescale}\ }\ \beta=-\alpha \qquad (4)

where: the constant multiplies $g_{tt}$ by an overall factor $e^{2\,\text{const}}$ , undone by $t\to e^{-\text{const}}t$ ; choosing it zero normalizes the metric to Minkowski at infinity. With $\beta=-\alpha$ the angular equation becomes a total derivative and integrates in one step. A single constant of integration $C$ appears — the whole physical content of the exterior solution.

e^{2\alpha} = 1 - \frac{C}{r},\qquad e^{2\beta} = \left(1 - \frac{C}{r}\right)^{-1} \qquad (5)

where: $C$ — the lone integration constant, fixed below to $2GM$ . Because $e^{2\beta}=e^{-2\alpha}$ , the radial and time parts are exact reciprocals — a hallmark of the Schwarzschild form. Identifying $C$ by the Newtonian limit gives the Schwarzschild metric. In geometrized units ( $c=1$ ) the constant is $2GM$ ; restoring $c$ makes it $2GM/c^2$ , the Schwarzschild radius $r_s$ .

\mathrm{d}s^2 = -\left(1 - \frac{2GM}{r}\right)\mathrm{d}t^2 + \left(1 - \frac{2GM}{r}\right)^{-1}\mathrm{d}r^2 + r^2\,\mathrm{d}\Omega^2 \qquad (6)

where: $M$ — total mass of the source as measured at infinity; $G$ — Newton's constant; $r_s=2GM$ (or $2GM/c^2$ ) — Schwarzschild radius, where $g_{tt}$ vanishes and $g_{rr}$ diverges. The two special radii, $r=r_s$ and $r=0$ , are of completely different character. The nature of those two radii is settled by a coordinate-independent curvature invariant, the Kretschmann scalar. It stays finite at $r=r_s$ (a coordinate artifact) and diverges only at $r=0$ (a genuine singularity of the geometry).

K \equiv R_{\mu\nu\rho\sigma}R^{\mu\nu\rho\sigma} = \frac{48\,G^2M^2}{r^6} \qquad (7)

where: $K$ — Kretschmann scalar, a scalar built from the full Riemann tensor; finite at $r=r_s$ , divergent as $r\to0$ . Because it is a scalar, no change of coordinates can remove its divergence at the origin, which is what makes $r=0$ physical.

Derivation ALocking the two radial functions together

The goal of this first step is to show that the vacuum condition, applied to the time-time and radial-radial equations, immediately ties the redshift function and the radial-stretch function into the single relation $\alpha(r)+\beta(r)=\text{const}$ , so that a metric which began with two unknown functions is reduced to just one before any integration is attempted. This is where the rigidity of the problem first becomes visible.

Step 1. Compute the two diagonal Ricci components from the ansatz. A direct calculation of the Christoffel symbols and their curvature gives $R_{tt}=e^{2(\alpha-\beta)}\big[\partial_r^2\alpha+(\partial_r\alpha)^2-\partial_r\alpha\,\partial_r\beta+\tfrac{2}{r}\partial_r\alpha\big]$ and $R_{rr}=-\big[\partial_r^2\alpha+(\partial_r\alpha)^2-\partial_r\alpha\,\partial_r\beta-\tfrac{2}{r}\partial_r\beta\big]$ . Why: these two components share an identical block of second-derivative terms, which is exactly what makes their combination simplify so violently.

Step 2. Form the combination $e^{-2(\alpha-\beta)}R_{tt}+R_{rr}$ . The bracketed second-derivative pieces cancel term by term, and what remains is $\tfrac{2}{r}\big(\partial_r\alpha+\partial_r\beta\big)$ . Why: cancelling every second derivative turns a pair of coupled second-order equations into one clean first-order statement — the algebraic miracle at the heart of the derivation.

Step 3. Impose the vacuum condition. Since both $R_{tt}$ and $R_{rr}$ vanish, the combination vanishes, so $\partial_r(\alpha+\beta)=0$ and therefore $\alpha+\beta=\text{const}$ . Why: this says the time-warp and the radial-stretch are not independent — fixing one fixes the other, halving the unknowns.

Step 4. Fix the constant by gauge. A nonzero constant $k$ merely multiplies $g_{tt}$ by $e^{2k}$ , which is undone by rescaling the time coordinate $t\to e^{-k}t$ ; choosing $k=0$ makes the metric Minkowski at infinity and gives $\beta=-\alpha$ . Why: the constant carries no geometry, only a choice of clock, so we are free to set it to zero and carry a single function forward.

Go deeper · Derivation B · 1 minSolving the radial vacuum equation

Derivation BSolving the radial vacuum equation

With the two functions locked as $\beta=-\alpha$ , the goal now is to integrate the remaining angular equation $R_{\theta\theta}=0$ and obtain the explicit form $e^{2\alpha}=1-C/r$ . The key move is recognizing that, after substitution, the equation is not merely first-order but an exact total derivative, so it integrates in a single stroke and produces exactly one constant of integration — the only physical parameter of the solution.

Step 1. Substitute $\beta=-\alpha$ , hence $\partial_r\beta=-\partial_r\alpha$ , into the angular component (3): $R_{\theta\theta}=e^{2\alpha}\big[r(-\partial_r\alpha-\partial_r\alpha)-1\big]+1=e^{2\alpha}\big(-2r\,\partial_r\alpha-1\big)+1$ . Why: with only one function left, the angular equation becomes a self-contained ODE for $e^{2\alpha}$ alone, decoupled from everything else.

Step 2. Set $R_{\theta\theta}=0$ and rearrange to $e^{2\alpha}+2r\,e^{2\alpha}\partial_r\alpha=1$ . Notice the left side is precisely $\tfrac{\mathrm{d}}{\mathrm{d}r}\!\big(r\,e^{2\alpha}\big)$ , since $\tfrac{\mathrm{d}}{\mathrm{d}r}(r e^{2\alpha})=e^{2\alpha}+r\cdot 2e^{2\alpha}\partial_r\alpha$ . Why: spotting the total derivative converts an ODE into a one-line integration and is the cleanest route to the answer — no integrating factor needed.

Step 3. Integrate once: $r\,e^{2\alpha}=r-C$ , so $e^{2\alpha}=1-C/r$ and $e^{2\beta}=e^{-2\alpha}=(1-C/r)^{-1}$ , where $C$ is the constant of integration. Why: the single constant $C$ is the entire residual freedom of the exterior geometry, and it is what the Newtonian limit will interpret physically.

Step 4. Check consistency. Substituting $e^{2\alpha}=1-C/r$ back into $R_{tt}$ and $R_{rr}$ shows both vanish identically, so no vacuum equation is left unsatisfied; the angular equation together with $\alpha+\beta=0$ exhausts the content of $R_{\mu\nu}=0$ . Why: verifying that the remaining components are automatically solved confirms the system was not over-determined — two independent equations for two functions, as the Bianchi identity guarantees.

Go deeper · Derivation C · 1 minFixing the constant from the Newtonian limit

Derivation CFixing the constant from the Newtonian limit

The final step gives the constant $C$ its physical meaning. The strategy is the correspondence principle: in the weak-field, slow-motion regime the geodesic equation of GR must reduce to Newton's second law with the potential $\Phi=-GM/r$ , and demanding this match forces $g_{tt}=-(1+2\Phi)$ and hence $C=2GM$ . Nothing about the strong field is assumed; the constant is read off entirely from behaviour far away, where both theories are known to agree.

Step 1. Take a slowly moving particle in a weak static field, so that spatial velocities are negligible against $\mathrm{d}t/\mathrm{d}\tau$ and $g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}$ with $|h_{\mu\nu}|\ll1$ . The geodesic equation then keeps only the $\Gamma^i_{tt}$ term: $\tfrac{\mathrm{d}^2x^i}{\mathrm{d}\tau^2}+\Gamma^i_{tt}\big(\tfrac{\mathrm{d}t}{\mathrm{d}\tau}\big)^2=0$ . Why: dropping spatial velocities isolates the single Christoffel symbol that plays the role of the Newtonian gravitational acceleration.

Step 2. Evaluate that symbol in the static weak field: $\Gamma^i_{tt}=-\tfrac12 g^{ij}\partial_j g_{tt}\approx-\tfrac12\partial_i h_{tt}$ , since $\partial_t$ terms vanish for a static metric and $g^{ij}\approx\delta^{ij}$ to leading order. Why: staticity removes the time-derivative pieces of the Christoffel symbol, leaving only a gradient — exactly the shape of $-\nabla\Phi$ .

Step 3. Match to Newton. The geodesic reduces to $\tfrac{\mathrm{d}^2x^i}{\mathrm{d}t^2}=\tfrac12\partial_i h_{tt}$ , which must equal $-\partial_i\Phi$ ; therefore $h_{tt}=-2\Phi$ , i.e. $g_{tt}=-(1+2\Phi)$ . Why: identifying the metric perturbation with the Newtonian potential is the bridge that turns a geometric coefficient into a measurable field.

Step 4. Insert the point-mass potential $\Phi=-GM/r$ to get $g_{tt}=-(1-2GM/r)$ , and compare with the exact $g_{tt}=-(1-C/r)$ from (5). The two agree only if $C=2GM$ (restoring $c$ : $C=2GM/c^2$ ). Why: this is the whole payoff — the abstract integration constant is the source's gravitational mass, and the Schwarzschild radius emerges as the scale $2GM/c^2$ .

Limiting cases

Every corner of the Schwarzschild metric reproduces a familiar piece of physics — each a consistency check on the derivation.

Case	Result	Meaning
Weak field / large $r$	$g_{tt}\to-(1-\tfrac{2GM}{r}),\ g_{rr}\to 1+\tfrac{2GM}{r}$	Linearizing for $r\gg 2GM$ recovers the Newtonian potential $\Phi=-GM/r$ in $g_{tt}$ — the matching that fixes $C=2GM$ .
Vanishing mass $M\to0$	$\mathrm{d}s^2\to-\mathrm{d}t^2+\mathrm{d}r^2+r^2\mathrm{d}\Omega^2$	With no source the metric is flat Minkowski spacetime in spherical coordinates, as the equivalence principle demands.
Spatial infinity $r\to\infty$	$g_{\mu\nu}\to\eta_{\mu\nu}$	Asymptotically flat: far from the mass spacetime is Minkowski, which is what lets mass be defined by Newtonian matching at infinity.
Horizon $r\to2GM$	$g_{tt}\to0,\ g_{rr}\to\infty$ , but $K=48G^2M^2/r^6$ finite	A coordinate singularity, not curvature; the divergence is an artifact of these coordinates and is removed by Kruskal-Szekeres.
Origin $r\to0$	$K\to\infty$	A genuine curvature singularity where tidal forces diverge and the classical geometry ends.

Common trapThe blow-up at r = 2GM means spacetime is destroyed at the horizon

It does not. Two distinct traps hide here. First, the divergence of $g_{rr}$ at $r=2GM$ is a coordinate singularity, not a physical one: every scalar curvature invariant, such as the Kretschmann scalar $K=48G^2M^2/r^6$ , stays perfectly finite there. An infalling observer crosses $r=2GM$ in finite proper time and feels nothing special locally; the coordinates $t,r$ simply fail, and regular coordinates (Eddington-Finkelstein, Kruskal-Szekeres) show the geometry is smooth. The truly pathological point is $r=0$ , where $K$ diverges. Second, the radial coordinate $r$ is not the proper distance from the centre — it is the areal radius, defined so that a sphere at $r$ has area $4\pi r^2$ . The proper radial distance is $\int e^{\beta}\,\mathrm{d}r=\int(1-2GM/r)^{-1/2}\,\mathrm{d}r$ , which exceeds $\Delta r$ and stretches without bound as one approaches the horizon along a static slice.

Worked examples

1 · The geometric mass of the Sun and Earth (confidence builder)

The derivation fixes the one constant as $C=r_s=2GM/c^2$ . Compute the Schwarzschild radius of the Sun and of the Earth, and check that the solar surface sits deep in the weak-field regime where the Newtonian matching is valid.

Step 1. For the Sun, $M_\odot=1.989\times10^{30}$ kg, so the gravitational length is $GM_\odot/c^2=(6.674\times10^{-11})(1.989\times10^{30})/(2.998\times10^8)^2=\mathbf{1477\ m}$ . The Schwarzschild radius is twice this: $r_s=2GM_\odot/c^2=\mathbf{2.95\ km}$ .

Step 2. For the Earth, $M_\oplus=5.972\times10^{24}$ kg gives $r_s=2GM_\oplus/c^2=\mathbf{8.9\ mm}$ — the whole planet's mass would have to be squeezed inside a marble to become a black hole. Both bodies are vastly larger than their $r_s$ , so their surfaces lie in ordinary space, not near any horizon.

Step 3. Check the weak field at the solar surface: $r_s/R_\odot=2954/(6.957\times10^8)=\mathbf{4.2\times10^{-6}}$ . Because $2GM/r\ll1$ everywhere outside the Sun, the exact $g_{tt}=-(1-r_s/r)$ is indistinguishable from the Newtonian $-(1+2\Phi)$ to parts in $10^{6}$ — which is precisely why the constant could be read off by Newtonian matching in the first place. The Sun's rₛ is 2.95 km and the Earth's 8.9 mm; with rₛ/R⊙ ≈ 4×10⁻⁶ the weak-field matching that fixed C = 2GM is self-consistent.

2 · The horizon is a coordinate singularity — Kretschmann at r = rₛ (fully worked)

The derivation flags $r=2GM$ as a place where the metric components blow up. Use the invariant $K=48G^2M^2/r^6$ to prove the geometry is nevertheless regular there, by evaluating it at the horizon of a solar-mass black hole and of the supermassive hole Sagittarius A ( $M=4.3\times10^6\,M_\odot$ ), and read off how tidal strength scales with mass.*

Step 1. At the horizon $r=r_s=2GM/c^2$ , the Kretschmann scalar simplifies to $K(r_s)=48G^2M^2/r_s^6=12/r_s^4$ . It is finite for any $r_s>0$ , which is the whole point: the divergence of $g_{rr}$ at $r_s$ is not matched by any divergence of curvature.

Step 2. Solar-mass hole, $r_s=2.95\times10^3$ m: $K(r_s)=12/(2.95\times10^3)^4=\mathbf{1.6\times10^{-13}\ m^{-4}}$ . The associated curvature radius is $K^{-1/4}\approx\mathbf{1.6\ km}$ — comparable to $r_s$ itself, so tidal forces at a stellar-mass horizon are ferocious (spaghettification before you arrive).

Step 3. Sagittarius A*, $M=4.3\times10^6\,M_\odot$ so $r_s=1.27\times10^{10}$ m $\approx0.085$ AU: $K(r_s)=12/(1.27\times10^{10})^4=\mathbf{4.6\times10^{-40}\ m^{-4}}$ , a curvature radius $K^{-1/4}\approx6.8\times10^9$ m $\approx0.045$ AU. Both values are finite — the horizon is smooth in both cases — but the supermassive horizon is $10^{26}$ times gentler.

Step 4. Read off the scaling. Since $K(r_s)=12/r_s^4\propto M^{-4}$ , the tidal curvature $\sqrt{K}\propto M^{-2}$ : doubling the mass makes the horizon four times gentler. K stays finite at r = rₛ (1.6×10⁻¹³ m⁻⁴ for a solar-mass hole, 4.6×10⁻⁴⁰ for Sgr A*), proving the horizon is a coordinate singularity; tidal strength at the horizon scales as 1/M², so an astronaut could cross a supermassive horizon unharmed.

Physics of the portrait

The Schwarzschild solution is a monument built in extraordinary circumstances — a first exact solution of a brand-new theory, computed under artillery fire, by a man who would be dead within months.

Origins · 1 minKarl Schwarzschild and how this was discovered

Karl Schwarzschild

1873–1916 · the first exact solution of general relativity

Karl Schwarzschild was already a distinguished astronomer and director of the Potsdam Observatory when the First World War broke out; a patriot in his forties, he volunteered and was sent to the Russian front to compute artillery trajectories. There, in the winter of 1915, having read Einstein's field-equation papers only weeks old, he solved them exactly for the field outside a spherical mass — the first exact solution anyone found, and one Einstein himself had not expected so soon. He mailed the calculation to Einstein, who presented it to the Prussian Academy on his behalf in January 1916, writing back that he had not imagined the exact solution could be formulated so simply. Schwarzschild's original coordinates differed from the modern form, and he worked with the interior solution too, but the exterior metric $-(1-2GM/r)\mathrm{d}t^2+\cdots$ bears his name in every textbook. He did not live to see its importance: he contracted pemphigus, a painful autoimmune skin disease, at the front, was invalided home, and died in May 1916 at forty-two. The full meaning of the surface $r=2GM$ — that it is a horizon, not a singularity — would take another four decades and the work of Eddington, Lemaître, Finkelstein, Kruskal and Szekeres to untangle. The mass parameter $M$ he introduced is now the defining quantity of every non-rotating black hole, and the radius $r_s=2GM/c^2$ that carries his name marks the boundary of the observable universe's most extreme objects.

History

1915

— Einstein presents the final generally covariant field equations to the Prussian Academy in November

1916

— Schwarzschild derives the exterior vacuum solution from the Russian front and mails it to Einstein

1916

— Schwarzschild dies in May of an illness contracted during the war, months after his paper appears

1923

— Birkhoff proves the solution is the unique spherically symmetric vacuum, whether or not staticity is assumed

1960

— Kruskal and Szekeres publish coordinates regular across r = 2GM, revealing it as a horizon

Going further

Texts. Misner, Thorne & Wheeler, Gravitation (1973), §31.2 · Wald, General Relativity (1984), §6.1 · Carroll, Spacetime and Geometry (2004), §5.2 · Hartle, Gravity (2003), Ch. 9 · Schutz, A First Course in General Relativity, 2nd ed. (2009), Ch. 10 · Schwarzschild, Sitzungsber. Preuss. Akad. Wiss. (1916), 189. Key numbers to keep in mind: $GM_\odot/c^2=1477$ m so $r_s(\odot)=2.95$ km; $r_s(\oplus)=8.9$ mm; $r_s/R_\odot=4.2\times10^{-6}$ ; Kretschmann $K=48G^2M^2/r^6$ with $K(r_s)=12/r_s^4=1.6\times10^{-13}\ \mathrm{m^{-4}}$ (solar-mass horizon) and $4.6\times10^{-40}\ \mathrm{m^{-4}}$ (Sgr A*, $M=4.3\times10^6\,M_\odot$ , $r_s=0.085$ AU). The classical tests give Mercury perihelion $43''$ /century and light deflection $1.75''$ .

Prerequisites. The Einstein Field Equations · Christoffel Symbols and the Ricci Tensor · Killing Vectors and Conserved Quantities in Schwarzschild · The Newtonian Limit of General Relativity. Leads to. The Schwarzschild Metric · Birkhoff's Theorem · The Schwarzschild Radius and Geometric Mass · Coordinate versus Curvature Singularities.

Ask the tutor: Ask me to grind through any single step — the $R_{tt}+R_{rr}$ cancellation that gives $\beta=-\alpha$ , the total-derivative trick that integrates $R_{\theta\theta}=0$ , or the Newtonian match that pins $C=2GM$ — with every index written out.

Why are only two of the metric functions unknown, when a general metric has ten components?

Symmetry annihilates the other eight. Spherical symmetry — invariance under the rotation group $SO(3)$ — forces the angular part to be a round 2-sphere $r^2\mathrm{d}\Omega^2$ with no cross terms mixing angles into $t$ or $r$ , and it forbids any preferred angular direction. Staticity adds a timelike Killing vector that is hypersurface-orthogonal, which removes all $t$ -dependence and kills the $\mathrm{d}t\,\mathrm{d}r$ cross term. What is left is a diagonal metric with an arbitrary coefficient on $\mathrm{d}t^2$ and on $\mathrm{d}r^2$ , each a function of $r$ only — two functions, written $e^{2\alpha}$ and $e^{2\beta}$ . The exponential form is a convenience that keeps the signature correct and tidies the Ricci tensor.

Where exactly did the assumption of staticity get used, and can it be dropped?

Staticity was used only to write down the ansatz: it justified dropping time dependence and the $\mathrm{d}t\,\mathrm{d}r$ cross term at the very start. Remarkably, it turns out to be unnecessary. Birkhoff's theorem (1923) shows that if you allow the functions to depend on both $t$ and $r$ and include a cross term, the vacuum equations $R_{\mu\nu}=0$ still force the metric to be static and equal to Schwarzschild. Spherical symmetry alone suffices. The physical consequence is striking: a radially pulsating or collapsing spherical star produces a static exterior and emits no gravitational waves — the gravitational analogue of a spherically symmetric charge distribution having a static Coulomb field outside.

Is the radial coordinate r the distance from the centre of the mass?

No, and this is one of the most common misreadings of the metric. The coordinate $r$ is the areal radius: it is defined so that the sphere of constant $r$ (and $t$ ) has surface area exactly $4\pi r^2$ . It is not the proper radial distance you would measure with a ruler running inward. That proper distance is $\int(1-2GM/r)^{-1/2}\,\mathrm{d}r$ , which is always larger than the coordinate difference $\Delta r$ because the radial direction is stretched by the factor $e^{\beta}=(1-2GM/r)^{-1/2}$ . Near the horizon the stretch grows without bound, so a modest change in $r$ can correspond to a large physical distance — a fact obscured if one naively reads $r$ as "distance from the middle".

Why does matching to Newtonian gravity fix the constant, rather than some strong-field condition?

Because the constant $C$ is a global label of the solution, not something local to any one radius, it can be read off wherever the physics is already understood — and the one place we independently know the answer is far away, where gravity is weak and Newton's law is verified to exquisite precision. In that regime the geodesic equation for a slow particle must reduce to $\ddot{\vec x}=-\nabla\Phi$ with $\Phi=-GM/r$ , which forces $g_{tt}=-(1+2\Phi)=-(1-2GM/r)$ . Comparing with the exact $g_{tt}=-(1-C/r)$ gives $C=2GM$ . No strong-field assumption is needed or wanted; the beauty is that a number governing the horizon is pinned down entirely by Newtonian behaviour at infinity.

Does Rμν = 0 mean the spacetime around a star is flat?

No — Ricci-flat is far weaker than flat. The Riemann tensor splits into a Ricci part and a trace-free Weyl part; the vacuum equation $R_{\mu\nu}=0$ switches off only the Ricci piece. The Weyl curvature can be large and is precisely what carries tidal forces and gravitational radiation. Schwarzschild spacetime is Ricci-flat everywhere outside the source yet strongly curved near the horizon, as the Kretschmann scalar $K=48G^2M^2/r^6$ shows directly: it is nonzero for every finite $r$ . Genuine flatness would require the entire Riemann tensor to vanish, which happens only in the $M\to0$ limit or asymptotically as $r\to\infty$ .

How can two integration constants appear but only one be physical?

The integration produces two constants, but they play very different roles. The first arises from $\alpha+\beta=\text{const}$ ; it multiplies $g_{tt}$ by an overall factor and is pure gauge — a rescaling of the time coordinate $t\to e^{-k}t$ removes it, and we conventionally set it to zero so the metric is Minkowski at infinity. The second, $C$ , appears when integrating the radial equation and cannot be gauged away: it changes the geometry itself, controlling how strongly spacetime curves. That is the one physical parameter, and the Newtonian limit identifies it as $2GM$ . The lesson generalizes — in any gauge theory some constants are coordinate conventions and some are physics, and telling them apart is half the work.