Deriving the Schwarzschild Metric

January 4, 2026

general relativity
black holes
differential geometry

In this post I’ll be deriving one of the simplest spacetime metrics: the Schwarzschild Metric. It is named after Karl Schwarzschild who published the solution in 1916, just a little more than a month after the publication of Einstein’s Theory of General Relativity.

It is a metric for a mass that’s:

Spherically symmetric
Non-rotating
Non-charged

and it describes the metric for:

Slowly rotating planets/stars
Non-rotating Black Holes

and it predicts:

Gravitational time dilation around celestial bodies
Gravitational doppler effect
Bending of light due to gravity
Shift in perihelion of orbits
Existence of Black Holes with the event horizon the size of Schwarzschild radius $r_s$

So let’s get into it!

Einstein Field Equations

To find the metric, we first have to solve the Einstein Field Equations.

R_{\mu \nu}-\frac{1}{2}Rg_{\mu \nu}-\Lambda g_{\mu \nu} = \frac{8 \pi G}{c^4} T_{\mu \nu}

We input the energy-momentum tensor and we get out the metric tensor. That is: we give the description of the mass-energy and momentum and we get the description of the geometry of spacetime.

Using the metric, we can find out the connection coefficients:

\Gamma _{\mu \nu}^{\sigma} = \frac{1}{2} g ^{\sigma \alpha}(\partial_{\nu} g_{\alpha \mu} + \partial_{\mu} g_{\alpha \nu} - \partial_{\alpha} g_{\mu \nu})

whence we can derive the geodesic equation which gives us the paths of masses and light:

\frac{d^2 x^{\sigma}}{d \lambda^2}+\Gamma_{\mu \nu}^{\sigma} \frac{dx^{\mu}}{d \lambda}\cdot \frac{dx^{\nu}}{d \lambda} = 0

$T_{\mu \nu} \neq 0$ inside the body, so let’s find the metric outside of our body where $T_{\mu \nu} = 0$ . We can also assume that it’s at non-cosmological scales, i.e. $\Lambda = 0$ . So that gives us:

R_{\mu \nu}-\frac{1}{2}Rg_{\mu \nu}- 0 = \frac{8 \pi G}{c^4} \cdot 0

R_{\mu \nu}-\frac{1}{2}Rg_{\mu \nu}= 0

Multiply both sides by the inverse metric:

R_{\mu \nu}g^{\mu \nu}-\frac{1}{2}Rg_{\mu \nu}g^{\mu \nu}= 0

R_{\mu}^{\mu}-\frac{1}{2}R \delta_{\mu}^{\mu}= 0

R-\frac{1}{2}R \cdot 4= 0

R= 0

R_{\mu \nu}= 0

for vacuum outside the body. This property is called Ricci flat: there’s no immediate volume changes in test particles, there’s only squishing due to curvature of spacetime.

We’re going to assume that as we go farther away from the mass, the effects of gravity are going to be negligible (i.e. described by the Minkowski metric).

We’re also going to assume that the spacetime is static, which means that:

The metric doesn’t depend on time: $\partial_t g_{\mu \nu} = 0$
$t \rightarrow -t$ doesn’t change the metric (this guarantees that the black hole is non-rotating)

We can write this as:

e_t = \frac{\partial}{\partial(ct)} \rightarrow \frac{\partial}{\partial(c(-t))} = -\frac{\partial}{\partial(ct)} = -e_t

g_{ti} = e_t\cdot e_i \rightarrow -e_t \cdot e_i = -g_{ti} \implies g_{ti}=-g_{ti}

e_t\cdot e_i = 0

g_{tt}=e_t\cdot e_t=(-e_t)\cdot(-e_t)=g_{tt}

Putting all this together, we get:

ds^2=-A(\tilde{r})c^2dt^2+B(\tilde{r})d\tilde{r}^2+\tilde{r}^2(d\theta^2+\sin^2\theta d\phi^2)

Since we want spherical symmetry, the $\theta$ and $\phi$ components should use the metric for a sphere of radius $r$ . Hence, to not violate spherical symmetry, the metric becomes:

ds^2=-A(\tilde{r})c^2dt^2+B(\tilde{r})d\tilde{r}^2+\tilde{r}^2(d\theta^2+\sin^2\theta d\phi^2)

where $C(\tilde{r})$ is a function that scales the values and is independent of $\theta$ and $\phi$ .

To make things look cleaner, let’s define:

\sqrt{C(\tilde{r})}\tilde{r} \equiv r

To find $A(r)$ and $B(r)$ , we first have to find the metric tensor $g_{\mu \nu}$ , which we can use to find the connection coefficients $\Gamma_{\mu \nu}^{\sigma}$ , which we can use to find the Ricci tensor $R_{\mu \nu}$ . Then we can force the metric to match Newtonian gravity at weak field and low velocity, hence solve for $A(r)$ and $B(r)$ .

We can calculate the connection coefficients using:

\Gamma _{\mu \nu}^{\sigma} = \frac{1}{2} g ^{\sigma \alpha}(\partial_{\nu} g_{\alpha \mu} + \partial_{\mu} g_{\alpha \nu} - \partial_{\alpha} g_{\mu \nu})

Hence we can find the non-zero connection coefficients to be:

\Gamma_{01}^{0}=\Gamma_{10}^{0}=\frac{1}{2}\frac{1}{A}(\partial_r A), \Gamma_{00}^{1}=\frac{1}{2}\frac{1}{B}(\partial_r A), \Gamma_{11}^{1}=\frac{1}{2}\frac{1}{B}(\partial_r B), \Gamma_{22}^{1}=-\frac{r}{B}, \Gamma_{33}^{1}=-\frac{r \sin^2(\theta)}{B}, \Gamma_{33}^{2}=-\sin(\theta)\cos(\theta), \Gamma_{21}^{2}=\Gamma_{12}^{2}=\frac{1}{r}, \Gamma_{13}^{3}=\Gamma_{31}^{3}=\frac{1}{r}, \Gamma_{23}^{3}=\Gamma_{32}^{3}=\cot(\theta)

Since $R_{\mu \nu}=0$ , that implies $R_{00}=R_{11}=R_{22}=0$ . We can use this to solve for $A$ and $B$ .

The Riemann tensor is:

R_{\sigma \mu \nu}^{\rho} = \partial_{\mu}\Gamma_{\nu \sigma}^{\rho}-\partial_{\nu}\Gamma_{\mu \sigma}^{\rho}+\Gamma_{\nu \sigma}^{\alpha}\Gamma_{\mu \alpha}^{\rho}-\Gamma_{\mu \sigma}^{\beta}\Gamma_{\nu \beta}^{\rho}

Hence, the Ricci tensor is obtained by contraction:

R_{\nu \delta} = R_{\nu \mu \delta}^{\mu}

Calculating $R_{00}$

R_{00}= R_{0 \mu 0}^{\mu} = \underbrace{\partial_{\mu}\Gamma_{00}^{\mu}}_{\neq 0 \text{ only for }\Gamma_{00}^{1}}-\partial_{0}\underbrace{\Gamma_{\mu 0}^{\mu}}_{=0}+\underbrace{\Gamma_{00}^{\alpha}\Gamma_{\mu \alpha}^{\mu}}_{\neq 0 \text{ only for }\alpha = 1}-(\Gamma_{\mu 0}^{\beta}\Gamma_{0 \beta}^{\mu})

=\partial_{1}\Gamma_{00}^{1}-0+\Gamma_{00}^{1}\Gamma_{\mu 1}^{\mu}-(\Gamma_{\mu 0}^{0}\Gamma_{0 0}^{\mu}+\Gamma_{\mu 0}^{1}\Gamma_{01}^{\mu})

=\partial_{1}\Gamma_{00}^{1}+\Gamma_{00}^{1}\Gamma_{0 1}^{0}+\Gamma_{00}^{1}\Gamma_{11}^{1}+\underbrace{\Gamma_{00}^{1}\Gamma_{21}^{2}+\Gamma_{00}^{1}\Gamma_{31}^{3}}_{\Gamma_{21}^{2}=\Gamma_{31}^{3}}-\Gamma_{10}^{0}\Gamma_{00}^{1}-\Gamma_{00}^{1}\Gamma_{01}^{0}

=\partial_{1}\Gamma_{00}^{1}+\Gamma_{00}^{1}\Gamma_{11}^{1}+2 \Gamma_{00}^{1}\Gamma_{21}^{2}-\Gamma_{10}^{0}\Gamma_{00}^{1}

Solving $R_{\mu\nu}=0$

From above, we are working with the static, spherically symmetric ansatz

ds^2=-A(r)c^2dt^2+B(r)dr^2+r^2(d\theta^2+\sin^2\theta\, d\phi^2)

where $A(r)$ and $B(r)$ are functions of $r$ only, and we want the vacuum equations

R_{\mu\nu}=0

It is standard to write derivatives with respect to $r$ using a prime:

A'=\frac{dA}{dr},\quad B'=\frac{dB}{dr},\quad A''=\frac{d^2A}{dr^2}

The Ricci components

Using the connection coefficients (Christoffels) computed earlier, the nontrivial Ricci components reduce to three independent equations. A convenient set is $R_{00}=0$ , $R_{11}=0$ , and $R_{22}=0$ .

A standard simplification is to work with the mixed structure of these equations and write them in terms of $A,B$ and their derivatives. The results are:

The $R_{00}=0$ equation

R_{00}=0

is equivalent to

\frac{A''}{A}-\frac{A'}{2A}\left(\frac{A'}{A}+\frac{B'}{B}\right)+\frac{2A'}{rA}=0

The $R_{11}=0$ equation

R_{11}=0

is equivalent to

-\frac{A''}{A}+\frac{A'}{2A}\left(\frac{A'}{A}+\frac{B'}{B}\right)+\frac{2B'}{rB}=0

The $R_{22}=0$ equation

R_{22}=0

is equivalent to

1-\frac{1}{B}+\frac{r}{2B}\left(-\frac{A'}{A}+\frac{B'}{B}\right)=0

(And $R_{33}=0$ gives the same condition as $R_{22}=0$ , up to the $\sin^2\theta$ factor.)

First key simplification: $A(r)B(r)=\text{constant}$

Add the $R_{00}=0$ and $R_{11}=0$ equations. The $A''$ terms cancel, and the messy middle term cancels too, leaving a very simple relation:

\frac{2A'}{rA}+\frac{2B'}{rB}=0

Multiply both sides by $r/2$ :

\frac{A'}{A}+\frac{B'}{B}=0

This means

\frac{d}{dr}\ln(AB)=0

A(r)B(r)=C

for some constant $C$ .

Now we impose asymptotic flatness: far away from the mass, we want Minkowski space, so

A(r)\to 1,\quad B(r)\to 1\quad \text{as } r\to\infty

Therefore

C=1

and we get the important relation

B(r)=\frac{1}{A(r)}

So the metric becomes a 1-function ansatz:

ds^2=-A(r)c^2dt^2+\frac{1}{A(r)}dr^2+r^2(d\theta^2+\sin^2\theta\, d\phi^2)

Solve for $A(r)$ using $R_{22}=0$

Take the equation

1-\frac{1}{B}+\frac{r}{2B}\left(-\frac{A'}{A}+\frac{B'}{B}\right)=0

and substitute $B=1/A$ .

First note:

\frac{1}{B}=A

and also

\frac{B'}{B}=\frac{d}{dr}\ln B=\frac{d}{dr}\ln\left(\frac{1}{A}\right)=-\frac{A'}{A}

So the bracket becomes

-\frac{A'}{A}+\frac{B'}{B}=-\frac{A'}{A}-\frac{A'}{A}=-\frac{2A'}{A}

Plugging into $R_{22}=0$ gives

1-A+\frac{r}{2B}\left(-\frac{2A'}{A}\right)=0

But $B=1/A$ , so $\frac{1}{B}=A$ and therefore $\frac{1}{2B}=\frac{A}{2}$ . Hence

1-A+r\left(\frac{A}{2}\right)\left(-\frac{2A'}{A}\right)=0

The $A$ cancels nicely:

1-A-rA'=0

Rewrite it as

rA'+A=1

This is a first-order ODE:

\frac{d}{dr}(rA)=1

Integrate:

rA=r+K

A(r)=1+\frac{K}{r}

Asymptotic flatness already ensured the $1$ is correct. The remaining constant $K$ must be related to the mass.

Match to Newtonian gravity (weak-field limit)

In the weak field, the metric should reduce to Newtonian gravity. In particular, the time-time component behaves like

g_{tt}\approx -(1+2\Phi/c^2)c^2

where $\Phi$ is the Newtonian potential

\Phi(r)=-\frac{GM}{r}

In our metric,

g_{tt}=-A(r)c^2

So for large $r$ ,

A(r)\approx 1+\frac{2\Phi}{c^2}=1-\frac{2GM}{c^2 r}

Comparing with

A(r)=1+\frac{K}{r}

we identify

K=-\frac{2GM}{c^2}

Therefore

A(r)=1-\frac{2GM}{c^2 r}

and since $B=1/A$ ,

B(r)=\left(1-\frac{2GM}{c^2 r}\right)^{-1}

The Schwarzschild metric

Substituting $A(r)$ and $B(r)$ back into the line element, we obtain the Schwarzschild metric:

ds^2 =-\left(1-\frac{2GM}{c^2 r}\right)c^2dt^2 +\left(1-\frac{2GM}{c^2 r}\right)^{-1}dr^2 +r^2(d\theta^2+\sin^2\theta\, d\phi^2)

It is common to define the Schwarzschild radius

r_s=\frac{2GM}{c^2}

so the metric becomes

ds^2 =-\left(1-\frac{r_s}{r}\right)c^2dt^2 +\left(1-\frac{r_s}{r}\right)^{-1}dr^2 +r^2(d\theta^2+\sin^2\theta\, d\phi^2)

The coordinate singularity at $r=r_s$ corresponds to the event horizon of a non-rotating, uncharged black hole.

Beyond Schwarzschild: Other Black Hole Solutions

The Schwarzschild solution describes a very special case: a black hole that is non-rotating and uncharged.
In reality, more general black holes can carry electric charge, angular momentum, or both.

Remarkably, General Relativity admits exact analytic solutions for all these cases.

We will now derive (or at least motivate) the remaining classical black hole metrics.

Charged Black Holes: The Reissner–Nordström Metric

Physical setup

We now allow the black hole to carry an electric charge $Q$ , while still assuming:

spherical symmetry
no rotation
vacuum outside the body (except for the electromagnetic field)

This means the stress–energy tensor is not zero, but instead given by the electromagnetic field.

Einstein–Maxwell equations

The Einstein equations become:

R_{\mu\nu} - \frac{1}{2} R g_{\mu\nu} = \frac{8\pi G}{c^4} T_{\mu\nu}^{\text{(EM)}}

where the electromagnetic stress–energy tensor is

T_{\mu\nu}^{\text{(EM)}} = \frac{1}{\mu_0} \left( F_{\mu\alpha}F_{\nu}^{\ \alpha} - \frac{1}{4} g_{\mu\nu} F_{\alpha\beta}F^{\alpha\beta} \right)

with $F_{\mu\nu}$ the electromagnetic field tensor.

For a static, spherically symmetric charge distribution, the only nonzero component is the radial electric field:

E(r)=\frac{Q}{4\pi \varepsilon_0 r^2}

Metric ansatz

Spherical symmetry again forces the metric to take the form

ds^2 =-A(r)c^2dt^2 +\frac{1}{A(r)}dr^2 +r^2(d\theta^2+\sin^2\theta\, d\phi^2)

Solving the coupled Einstein–Maxwell equations yields

A(r) =1-\frac{2GM}{c^2 r} +\frac{GQ^2}{4\pi \varepsilon_0 c^4 r^2}

Reissner–Nordström metric

The resulting line element is

ds^2 =-\left( 1-\frac{2GM}{c^2 r} +\frac{GQ^2}{4\pi \varepsilon_0 c^4 r^2} \right)c^2dt^2 +\left( 1-\frac{2GM}{c^2 r} +\frac{GQ^2}{4\pi \varepsilon_0 c^4 r^2} \right)^{-1}dr^2 +r^2(d\theta^2+\sin^2\theta\, d\phi^2)

Horizons

The horizons are given by the roots of

A(r)=0

which yields

r_\pm =\frac{GM}{c^2} \pm \sqrt{ \left(\frac{GM}{c^2}\right)^2 -\frac{GQ^2}{4\pi \varepsilon_0 c^4} }

Two horizons: outer (event) and inner (Cauchy)
One horizon: extremal black hole
No horizons: naked singularity (unphysical in classical GR)

Rotating Black Holes: The Kerr Metric

Physical setup

Now we allow the black hole to rotate, but remain uncharged.
Rotation breaks spherical symmetry, leaving only:

axial symmetry
stationarity (time-translation invariance)

This drastically complicates the geometry.

Boyer–Lindquist coordinates

The Kerr metric is most naturally expressed in Boyer–Lindquist coordinates
$(t, r, \theta, \phi)$ .

Define:

a=\frac{J}{Mc}

where $J$ is the angular momentum.

Also define the functions

\Delta=r^2-\frac{2GM}{c^2}r+a^2

\rho^2=r^2+a^2\cos^2\theta

Kerr metric

The Kerr line element is

\begin{aligned} ds^2 =&-\left(1-\frac{2GMr}{c^2\rho^2}\right)c^2dt^2 -\frac{4GMar\sin^2\theta}{c^2\rho^2} \, c\,dt\,d\phi \\ &+\frac{\rho^2}{\Delta}dr^2 +\rho^2 d\theta^2 +\left(r^2+a^2+\frac{2GMa^2 r\sin^2\theta}{c^2\rho^2}\right)\sin^2\theta\, d\phi^2 \end{aligned}

Key features

Frame dragging: spacetime itself rotates
Ergosphere: region where no observer can remain stationary
Ring singularity at $\rho^2=0$
Reduction to Schwarzschild when $a\to 0$

Horizons

Horizons are again given by $\Delta=0$ :

r_\pm =\frac{GM}{c^2} \pm \sqrt{\left(\frac{GM}{c^2}\right)^2-a^2}

Charged and Rotating: The Kerr–Newman Metric

The most general stationary black hole solution in four-dimensional GR is the Kerr–Newman metric.

It describes a black hole with:

mass $M$
charge $Q$
angular momentum $J$

Metric functions

Define

\Delta =r^2-\frac{2GM}{c^2}r+a^2+\frac{GQ^2}{4\pi\varepsilon_0 c^4}

and

\rho^2=r^2+a^2\cos^2\theta

Kerr–Newman metric

The line element has the same form as Kerr, with $\Delta$ modified accordingly:

\begin{aligned} ds^2 =&-\left(1-\frac{2GMr}{c^2\rho^2} +\frac{GQ^2}{4\pi\varepsilon_0 c^4\rho^2}\right)c^2dt^2 \\ &-\frac{4GMar\sin^2\theta}{c^2\rho^2} \, c\,dt\,d\phi +\frac{\rho^2}{\Delta}dr^2 +\rho^2 d\theta^2 \\ &+\left(r^2+a^2+\frac{2GMa^2 r\sin^2\theta}{c^2\rho^2}\right)\sin^2\theta\, d\phi^2 \end{aligned}

Horizons and extremality

Horizons occur where $\Delta=0$ :

r_\pm =\frac{GM}{c^2} \pm \sqrt{ \left(\frac{GM}{c^2}\right)^2 -a^2 -\frac{GQ^2}{4\pi\varepsilon_0 c^4} }

The extremal condition is

a^2+\frac{GQ^2}{4\pi\varepsilon_0 c^4} =\left(\frac{GM}{c^2}\right)^2

Summary: The Black Hole Zoo

All classical black holes in four-dimensional GR fall into this family:

| Metric | Mass | Charge | Rotation | | ------------------ | ---- | ------ | -------- | | Schwarzschild | ✓ | ✗ | ✗ | | Reissner–Nordström | ✓ | ✓ | ✗ | | Kerr | ✓ | ✗ | ✓ | | Kerr–Newman | ✓ | ✓ | ✓ |

This remarkable result is summarized by the no-hair theorem:
black holes are completely characterized by just three parameters.

Einstein Field Equations

Calculating R00R_{00}R00​

Solving Rμν=0R_{\mu\nu}=0Rμν​=0

The Ricci components

The R00=0R_{00}=0R00​=0 equation

The R11=0R_{11}=0R11​=0 equation

The R22=0R_{22}=0R22​=0 equation

First key simplification: A(r)B(r)=constantA(r)B(r)=\text{constant}A(r)B(r)=constant

Solve for A(r)A(r)A(r) using R22=0R_{22}=0R22​=0

Match to Newtonian gravity (weak-field limit)

The Schwarzschild metric

Beyond Schwarzschild: Other Black Hole Solutions

Charged Black Holes: The Reissner–Nordström Metric

Physical setup

Einstein–Maxwell equations

Metric ansatz

Reissner–Nordström metric

Horizons

Rotating Black Holes: The Kerr Metric

Physical setup

Boyer–Lindquist coordinates

Kerr metric

Key features

Horizons

Charged and Rotating: The Kerr–Newman Metric

Metric functions

Kerr–Newman metric

Horizons and extremality

Summary: The Black Hole Zoo

Calculating $R_{00}$

Solving $R_{\mu\nu}=0$

The $R_{00}=0$ equation

The $R_{11}=0$ equation

The $R_{22}=0$ equation

First key simplification: $A(r)B(r)=\text{constant}$

Solve for $A(r)$ using $R_{22}=0$