General Relativity — worked problems

Induced metric

Polar coordinates: $x = r\cos\phi$, $y = r\sin\phi$, $z = r^2$. Differentials: $dx = \cos\phi\,dr - r\sin\phi\,d\phi$, $dy = \sin\phi\,dr + r\cos\phi\,d\phi$, $dz = 2r\,dr$. Then $dx^2 + dy^2 = dr^2 + r^2 d\phi^2$, $dz^2 = 4r^2 dr^2$:

$$\boxed{\;ds^2 = (1 + 4r^2)\,dr^2 + r^2\,d\phi^2.\;}$$

Christoffels

$g_{rr} = 1+4r^2$, $g_{\phi\phi} = r^2$. $\partial_r g_{rr} = 8r$, $\partial_r g_{\phi\phi} = 2r$. Inverse: $g^{rr} = 1/(1+4r^2)$, $g^{\phi\phi} = 1/r^2$.

$\Gamma^r_{rr} = \tfrac12 g^{rr}\partial_r g_{rr} = \tfrac12\cdot\tfrac{1}{1+4r^2}\cdot 8r = \tfrac{4r}{1+4r^2}$.

$\Gamma^r_{\phi\phi} = -\tfrac12 g^{rr}\partial_r g_{\phi\phi} = -\tfrac12\cdot\tfrac{1}{1+4r^2}\cdot 2r = -\tfrac{r}{1+4r^2}$.

$\Gamma^\phi_{r\phi} = \tfrac12 g^{\phi\phi}\partial_r g_{\phi\phi} = \tfrac12\cdot\tfrac{1}{r^2}\cdot 2r = \tfrac{1}{r}$.

$$\boxed{\;\Gamma^r_{rr} = \tfrac{4r}{1+4r^2},\;\Gamma^r_{\phi\phi} = -\tfrac{r}{1+4r^2},\;\Gamma^\phi_{r\phi} = \tfrac{1}{r}.\;}$$

Riemann component

$R^r{}_{\phi r\phi} = \partial_r\Gamma^r_{\phi\phi} - \partial_\phi\Gamma^r_{r\phi} + \Gamma^r_{r\sigma}\Gamma^\sigma_{\phi\phi} - \Gamma^r_{\phi\sigma}\Gamma^\sigma_{r\phi}$.

$\partial_r\Gamma^r_{\phi\phi} = -\tfrac{(1+4r^2) - r\cdot 8r}{(1+4r^2)^2} = -\tfrac{1-4r^2}{(1+4r^2)^2}$.

$\partial_\phi\Gamma^r_{r\phi} = 0$.

$\Gamma^r_{r\sigma}\Gamma^\sigma_{\phi\phi}$: only $\sigma = r$ contributes: $\Gamma^r_{rr}\Gamma^r_{\phi\phi} = \tfrac{4r}{1+4r^2}\cdot(-\tfrac{r}{1+4r^2}) = -\tfrac{4r^2}{(1+4r^2)^2}$.

$\Gamma^r_{\phi\sigma}\Gamma^\sigma_{r\phi}$: only $\sigma = \phi$: $\Gamma^r_{\phi\phi}\Gamma^\phi_{r\phi} = -\tfrac{r}{1+4r^2}\cdot\tfrac{1}{r} = -\tfrac{1}{1+4r^2}$.

Combine: $R^r{}_{\phi r\phi} = -\tfrac{1-4r^2}{(1+4r^2)^2} + 0 - \tfrac{4r^2}{(1+4r^2)^2} + \tfrac{1}{1+4r^2} = \tfrac{-(1-4r^2) - 4r^2 + (1+4r^2)}{(1+4r^2)^2} = \tfrac{4r^2}{(1+4r^2)^2}$.

$R_{r\phi r\phi} = g_{rr}R^r{}_{\phi r\phi} = (1+4r^2)\cdot\tfrac{4r^2}{(1+4r^2)^2} = \tfrac{4r^2}{1+4r^2}$.

Gaussian curvature

$\det g = (1+4r^2)r^2$. $K = R_{1212}/\det g = \tfrac{4r^2}{(1+4r^2)}\cdot\tfrac{1}{(1+4r^2)r^2}$:

$$\boxed{\;K = \frac{4}{(1+4r^2)^2}.\;}$$

Cross-check via Monge's formula

For surface $z = f(x,y) = x^2+y^2$: $f_x = 2x$, $f_y = 2y$, $f_{xx} = f_{yy} = 2$, $f_{xy} = 0$. Monge's formula: $$K = \frac{f_{xx}f_{yy} - f_{xy}^2}{(1+f_x^2+f_y^2)^2} = \frac{4 - 0}{(1+4r^2)^2} = \frac{4}{(1+4r^2)^2}\;\checkmark.$$

Geometric interpretation

$K = 4$ at the vertex ($r = 0$): maximally curved point. $K\to 0$ as $r\to\infty$: paraboloid flattens. Positive but spatially varying curvature — not a sphere (which has constant $K = 1/R^2$), not a saddle (which has $K < 0$), but somewhere in between. Such surfaces show up in optics (mirror dishes), antenna design, and the spinning-bucket fluid surface.

(a) Christoffels and Riemann

$g_{rr} = r^2$, $g_{\phi\phi} = r^4$. Inverse: $g^{rr} = 1/r^2$, $g^{\phi\phi} = 1/r^4$. Metric derivatives: $\partial_r g_{rr} = 2r$, $\partial_r g_{\phi\phi} = 4r^3$.

$\Gamma^r_{rr} = \tfrac12 g^{rr}\partial_r g_{rr} = \tfrac12\cdot\tfrac{1}{r^2}\cdot 2r = \tfrac{1}{r}$.

$\Gamma^r_{\phi\phi} = -\tfrac12 g^{rr}\partial_r g_{\phi\phi} = -\tfrac12\cdot\tfrac{1}{r^2}\cdot 4r^3 = -2r$.

$\Gamma^\phi_{r\phi} = \tfrac12 g^{\phi\phi}\partial_r g_{\phi\phi} = \tfrac12\cdot\tfrac{1}{r^4}\cdot 4r^3 = \tfrac{2}{r}$.

Compute $R^r{}_{\phi r\phi}$:

$\partial_r\Gamma^r_{\phi\phi} = \partial_r(-2r) = -2$.

$\partial_\phi\Gamma^r_{r\phi} = 0$.

$\Gamma^r_{r\sigma}\Gamma^\sigma_{\phi\phi}$: $\sigma = r$: $\Gamma^r_{rr}\Gamma^r_{\phi\phi} = (1/r)(-2r) = -2$.

$\Gamma^r_{\phi\sigma}\Gamma^\sigma_{r\phi}$: $\sigma = \phi$: $\Gamma^r_{\phi\phi}\Gamma^\phi_{r\phi} = (-2r)(2/r) = -4$.

Combine: $R^r{}_{\phi r\phi} = -2 - 0 + (-2) - (-4) = -2 - 2 + 4 = 0$.

$$\boxed{\;R^r{}_{\phi r\phi} = 0.\;}$$ In 2D this is the entire Riemann tensor — manifold is flat where smooth.

Identify the geometry

Substitute $u = r^2/2$, $\psi = 2\phi$: $du = r\,dr$, so $du^2 = r^2 dr^2$, and $u^2 d\psi^2 = (r^4/4)\cdot 4d\phi^2 = r^4 d\phi^2$. Thus $$ds^2 = du^2 + u^2 d\psi^2.$$ This is the flat Euclidean plane in polar form — BUT $\psi = 2\phi$ has period $4\pi$, not $2\pi$. The geometry is a double cover of the Euclidean plane, branched at the origin: a cone with $+2\pi$ excess angle.

(b) Circumference and area

At fixed $r = r_0$, the proper arc length around the loop $\phi\in[0,2\pi)$ is $$C = \int_0^{2\pi}\sqrt{g_{\phi\phi}}\,d\phi = \int_0^{2\pi}r_0^2\,d\phi = 2\pi r_0^2.$$

Enclosed area with $\sqrt{\det g} = \sqrt{r^2\cdot r^4} = r^3$: $$A = \int_0^{r_0}\!\!\int_0^{2\pi} r^3\,d\phi\,dr = 2\pi\cdot\tfrac{r_0^4}{4} = \tfrac{\pi r_0^4}{2}.$$

Eliminate $r_0$: $C^2 = 4\pi^2 r_0^4$, and $8\pi A = 8\pi\cdot\pi r_0^4/2 = 4\pi^2 r_0^4$. So:

$$\boxed{\;C^2 = 8\pi A\;\;(\text{vs Euclidean }4\pi A).\;}$$

Lesson

Flatness ($R = 0$ in the smooth bulk) does NOT imply Euclidean global geometry. The factor of 2 comes from the conical defect at the origin: a topological singularity carrying curvature as a delta function. Gauss–Bonnet's integrated form recovers the deficit angle. Prototype of cosmic-string geometries in GR, where a topological defect on a 1D line produces a conical exterior.

Setup

Signature $(-,-,+)$: $X,Y$ spacelike, $Z$ timelike. Lie algebra $\mathfrak{sl}(2,\mathbb R) \cong \mathfrak{so}(2,1)$ with cyclic brackets $[X,Y] = -Z$, $[Y,Z] = X$, $[Z,X] = Y$.

Koszul formula

Define $\nabla_A B = \Gamma^C{}_{AB}C$. With constant metric on the basis ($A(g(B,C)) = 0$):

$$\boxed{\;2g(\nabla_A B, C) = -g(A,[B,C]) + g(B,[C,A]) + g(C,[A,B]).\;}$$

Diagonal entries

$\nabla_X X$: try $C = X,Y,Z$. For each, $-g(X,[X,C]) + g(X,[C,X]) + g(C,[X,X]) = 0$ (first two terms differ only by bracket sign and cancel; third vanishes). So $\nabla_X X = 0$. Same for $\nabla_Y Y$ and $\nabla_Z Z$.

Off-diagonal: $\nabla_X Y$

Only $C = Z$ gives a non-zero RHS: $2g(\nabla_X Y, Z) = -g(X,[Y,Z]) + g(Y,[Z,X]) + g(Z,[X,Y]) = -g(X,X) + g(Y,Y) - g(Z,Z) = 1 - 1 - 1 = -1$ (using $g(X,X) = g(Y,Y) = -1$, $g(Z,Z) = +1$).

Decompose $\nabla_X Y = aX + bY + cZ$, project on $Z$: $g(\nabla_X Y, Z) = c\cdot g(Z,Z) = c$. So $c = -1/2$. Orthogonality forces $a = b = 0$:

$$\boxed{\;\nabla_X Y = -\tfrac12 Z.\;}$$

Cyclic permutation: $\nabla_Y Z = +\tfrac12 X$, $\nabla_Z X = +\tfrac12 Y$. Torsion-free $\nabla_B A = \nabla_A B - [A,B]$: $\nabla_Y X = -\tfrac12 Z - (-Z) = +\tfrac12 Z$, $\nabla_Z Y = -\tfrac12 X$, $\nabla_X Z = -\tfrac12 Y$.

Riemann tensor

$R(X,Y)X = \nabla_X\nabla_Y X - \nabla_Y\nabla_X X - \nabla_{[X,Y]}X$.

$\nabla_Y X = \tfrac12 Z$, so $\nabla_X(\tfrac12 Z) = \tfrac12(-\tfrac12 Y) = -\tfrac14 Y$.

$\nabla_X X = 0$, $\nabla_Y(0) = 0$.

$\nabla_{[X,Y]}X = \nabla_{-Z}X = -\tfrac12 Y$.

Combine: $R(X,Y)X = -\tfrac14 Y - 0 - (-\tfrac12 Y) = \tfrac14 Y$.

Repeating for all triples (and using cyclic structure):

$$\boxed{\;R(A,B)C = \tfrac14[g(B,C)A - g(A,C)B],\quad R_{ABCD} = \tfrac14(g_{AC}g_{BD} - g_{AD}g_{BC}).\;}$$

Constant sectional curvature $K = 1/4$. Scalar curvature $R = n(n-1)K = 6\cdot(1/4) = 3/2$ in 3D.

Identification

The frame realises $\mathfrak{sl}(2,\mathbb R)\cong\mathfrak{so}(2,1)$ with metric proportional to the Killing form. The manifold is the (universal cover of) $SL(2,\mathbb R)$ — a maximally symmetric 3D Lorentzian space. The bi-invariant metric on a Lie group always has $R(A,B)C = -\tfrac14[[A,B],C]$, which (using the cyclic brackets here) yields exactly the formula above. Curvature radius $\ell = 2$. Depending on overall sign of $g$, this is dS$_3$ or AdS$_3$.

(a) Ricci symmetry

$R_{\mu\nu} \equiv R^a{}_{\mu a\nu} = g^{ab}R_{b\mu a\nu}$. Use the pair-exchange symmetry $R_{b\mu a\nu} = R_{a\nu b\mu}$:

$$R_{\mu\nu} = g^{ab}R_{a\nu b\mu}.$$

Relabel dummies $a\leftrightarrow b$ and use $g^{ab} = g^{ba}$: $= g^{ba}R_{b\nu a\mu} = R^a{}_{\nu a\mu} = R_{\nu\mu}$.

$$\boxed{\;R_{\mu\nu} = R_{\nu\mu}\;\text{(symmetric).}\;}$$ Hence $G_{\mu\nu} = R_{\mu\nu} - \tfrac12 g_{\mu\nu}R$ is symmetric too.

(b) Riemann in 2D

In $n$ dimensions, the Riemann tensor has $n^2(n^2-1)/12$ independent components: $n=2\Rightarrow 1$, $n=3\Rightarrow 6$, $n=4\Rightarrow 20$. In 2D the single independent component is the Gaussian curvature $K$, or equivalently the Ricci scalar $R = 2K$. The fully covariant form: $$R_{\alpha\beta\mu\nu} = \tfrac{R}{2}(g_{\alpha\mu}g_{\beta\nu} - g_{\alpha\nu}g_{\beta\mu}).$$

Ricci contraction: $R_{\mu\nu} = g^{ab}R_{a\mu b\nu} = \tfrac{R}{2}g^{ab}(g_{ab}g_{\mu\nu} - g_{a\nu}g_{b\mu})$.

$g^{ab}g_{ab} = \delta^a_a = n = 2$. $g^{ab}g_{a\nu}g_{b\mu} = \delta^b_\nu g_{b\mu} = g_{\mu\nu}$. So:

$$R_{\mu\nu} = \tfrac{R}{2}(2g_{\mu\nu} - g_{\mu\nu}) = \tfrac{R}{2}g_{\mu\nu}.$$

$$\boxed{\;R_{\mu\nu} = \tfrac{R}{2}g_{\mu\nu}\;\Longrightarrow\;k = \tfrac12,\;\;G_{\mu\nu} = R_{\mu\nu} - \tfrac12 g_{\mu\nu}R \equiv 0\;\text{identically in 2D.}\;}$$

(c) Vacuum Einstein vacuous in 1+1D

$G_{\mu\nu} = 0$ holds identically in 2D for any metric — not as a constraint, but as an algebraic consequence of the Riemann structure in $n = 2$.

Why GR is trivial in $\le 3$D — and how it's revived

Counting metric DOFs vs Einstein equations:

• 2D: 3 metric DOFs, 3 Einstein equations — but $G\equiv 0$, so all equations vacuous. GR empty.
• 3D: 6 DOFs, 6 equations. Riemann is fully determined by Ricci ($W = 0$ in 3D), so vacuum solutions are flat. No propagating gravitons.
• 4D: 10 DOFs, 10 equations. Weyl tensor has 10 components; after gauge fixing, 2 propagating polarisations — the graviton.

2D dilaton gravity revives 2D GR by adding a scalar $\phi$:

$$S_{JT} = \int d^2x\,\sqrt{-g}\,\phi\bigl(R + 2/L^2\bigr) + \text{boundary}.$$

Jackiw–Teitelboim (JT) gravity has AdS$_2$ as its on-shell solution and is the modern laboratory for AdS$_2$/SYK holography, the Page curve, replica wormholes, and the gravitational path integral.

Embedding check

Substituting the parametrisation: $(x^1)^2 + (x^2)^2 - (x^0)^2 = a^2\cosh^2\lambda(\sin^2\theta + \cos^2\theta) - a^2\sinh^2\lambda = a^2(\cosh^2\lambda - \sinh^2\lambda) = a^2$ ✓. Unit hyperboloid of "radius" $a$.

(a) Induced metric

Differentials: $dx^0 = a\cosh\lambda\,d\lambda$. $dx^1 = a\sinh\lambda\sin\theta\,d\lambda + a\cosh\lambda\cos\theta\,d\theta$. $dx^2 = a\sinh\lambda\cos\theta\,d\lambda - a\cosh\lambda\sin\theta\,d\theta$.

$(dx^0)^2 = a^2\cosh^2\lambda\,d\lambda^2$.

$(dx^1)^2 + (dx^2)^2$: cross terms cancel between $\sin\theta\cos\theta$ products; $\sin^2\theta + \cos^2\theta = 1$ collects: $a^2\sinh^2\lambda\,d\lambda^2 + a^2\cosh^2\lambda\,d\theta^2$.

Ambient $ds^2_{\text{amb}} = (dx^0)^2 - (dx^1)^2 - (dx^2)^2 - (dx^3)^2$ (signature $+---$, $dx^3 = 0$):

$$ds^2|_\Sigma = a^2(\cosh^2\lambda - \sinh^2\lambda)d\lambda^2 - a^2\cosh^2\lambda\,d\theta^2 = a^2\,d\lambda^2 - a^2\cosh^2\lambda\,d\theta^2.$$

$$\boxed{\;ds^2|_\Sigma = a^2\,d\lambda^2 - a^2\cosh^2\lambda\,d\theta^2.\;}$$ This is 2D de Sitter with curvature radius $a$.

(b) Christoffel symbols

$g_{\lambda\lambda} = a^2$, $g_{\theta\theta} = -a^2\cosh^2\lambda$. Non-zero derivative: $\partial_\lambda g_{\theta\theta} = -2a^2\cosh\lambda\sinh\lambda = -a^2\sinh(2\lambda)$.

$\Gamma^\lambda_{\theta\theta} = -\tfrac12 g^{\lambda\lambda}\partial_\lambda g_{\theta\theta} = -\tfrac12\cdot\tfrac{1}{a^2}\cdot(-a^2\sinh(2\lambda)) = \tfrac12\sinh(2\lambda) = \cosh\lambda\sinh\lambda$.

$\Gamma^\theta_{\lambda\theta} = \tfrac12 g^{\theta\theta}\partial_\lambda g_{\theta\theta} = \tfrac12\cdot(-1/(a^2\cosh^2\lambda))\cdot(-a^2\sinh(2\lambda)) = \tanh\lambda$.

$$\boxed{\;\Gamma^\lambda_{\theta\theta} = \tfrac12\sinh(2\lambda) = \cosh\lambda\sinh\lambda,\quad \Gamma^\theta_{\lambda\theta} = \tanh\lambda.\;}$$

(c) Geodesics with $\dot\lambda = 0$

Geodesic equations:

$\ddot\lambda + \Gamma^\lambda_{\theta\theta}\dot\theta^2 = \ddot\lambda + \cosh\lambda\sinh\lambda\,\dot\theta^2 = 0$.

$\ddot\theta + 2\Gamma^\theta_{\lambda\theta}\dot\lambda\dot\theta = \ddot\theta + 2\tanh\lambda\,\dot\lambda\dot\theta = 0$.

Impose $\dot\lambda = 0$ (so $\lambda = \lambda_0$ const): first equation reduces to $\cosh\lambda_0\sinh\lambda_0\,\dot\theta^2 = 0$. For a non-degenerate curve $\dot\theta\neq 0$ and $\cosh\lambda_0\neq 0$, so $\sinh\lambda_0 = 0$, forcing $\lambda_0 = 0$.

Second equation: $\ddot\theta = 0$, so $\theta(\tau) = \theta_0 + c\tau$.

$$\boxed{\;\lambda = 0,\quad \theta = \theta_0 + c\tau\;\text{(closed spacelike geodesic circle at the throat).}\;}$$

In embedding coordinates: $(x^0, x^1, x^2, x^3) = (0, a\sin\theta, a\cos\theta, 0)$ — circle of radius $a$ in the equatorial plane $\{x^0 = 0, x^3 = 0\}$.

Cross-check via embedded geometry

A curve on $\Sigma$ is geodesic iff its ambient acceleration is purely normal to $\Sigma$. For the throat circle: ambient acceleration is centripetal, $\ddot{\vec X} = -c^2 a(0,\sin\theta,\cos\theta,0)$. The outward unit normal to the hyperboloid in $\mathbb R^{(+,-,-,-)}$ at the equator is $\propto(0,\sin\theta,\cos\theta,0)$ — anti-parallel to the acceleration. ✓ Purely normal, hence geodesic.

Analogous to the equator of a sphere being a great circle: only the "minimum-expansion slice" (the throat of dS$_2$, the bouncing-universe moment) hosts a closed spacelike geodesic.

Christoffels

$g_{\rho\rho} = 1$, $g_{\phi\phi} = a^2+\rho^2$. Inverse: $g^{\rho\rho} = 1$, $g^{\phi\phi} = 1/(a^2+\rho^2)$. Non-zero derivative: $\partial_\rho g_{\phi\phi} = 2\rho$.

$\Gamma^\rho_{\phi\phi} = -\tfrac12 g^{\rho\rho}\partial_\rho g_{\phi\phi} = -\tfrac12\cdot 1\cdot 2\rho = -\rho$.

$\Gamma^\phi_{\rho\phi} = \tfrac12 g^{\phi\phi}\partial_\rho g_{\phi\phi} = \tfrac12\cdot\tfrac{1}{a^2+\rho^2}\cdot 2\rho = \tfrac{\rho}{a^2+\rho^2}$.

Riemann component

$R^\rho{}_{\phi\rho\phi} = \partial_\rho\Gamma^\rho_{\phi\phi} - \partial_\phi\Gamma^\rho_{\rho\phi} + \Gamma^\rho_{\rho\sigma}\Gamma^\sigma_{\phi\phi} - \Gamma^\rho_{\phi\sigma}\Gamma^\sigma_{\rho\phi}$.

$\partial_\rho\Gamma^\rho_{\phi\phi} = -1$. $\partial_\phi\Gamma^\rho_{\rho\phi} = 0$. $\Gamma^\rho_{\rho\sigma} = 0$ (no such Christoffel). $\Gamma^\rho_{\phi\sigma}\Gamma^\sigma_{\rho\phi}$: $\sigma = \phi$: $\Gamma^\rho_{\phi\phi}\Gamma^\phi_{\rho\phi} = (-\rho)\cdot\tfrac{\rho}{a^2+\rho^2} = -\tfrac{\rho^2}{a^2+\rho^2}$.

Combine: $R^\rho{}_{\phi\rho\phi} = -1 - 0 + 0 - (-\tfrac{\rho^2}{a^2+\rho^2}) = -1 + \tfrac{\rho^2}{a^2+\rho^2} = \tfrac{\rho^2 - (a^2+\rho^2)}{a^2+\rho^2} = -\tfrac{a^2}{a^2+\rho^2}$.

$R_{\rho\phi\rho\phi} = g_{\rho\rho}R^\rho{}_{\phi\rho\phi} = -\tfrac{a^2}{a^2+\rho^2}$.

Ricci tensor

$R_{\rho\rho} = R^\phi{}_{\rho\phi\rho} = g^{\phi\phi}R_{\phi\rho\phi\rho} = \tfrac{1}{a^2+\rho^2}\cdot R_{\rho\phi\rho\phi} = -\tfrac{a^2}{(a^2+\rho^2)^2}$.

$R_{\phi\phi} = R^\rho{}_{\phi\rho\phi} = -\tfrac{a^2}{a^2+\rho^2}$.

Ricci scalar

$R = g^{\mu\nu}R_{\mu\nu} = 1\cdot(-\tfrac{a^2}{(a^2+\rho^2)^2}) + \tfrac{1}{a^2+\rho^2}\cdot(-\tfrac{a^2}{a^2+\rho^2}) = -\tfrac{2a^2}{(a^2+\rho^2)^2}$.

Gaussian curvature

$K = R/2 = -\tfrac{a^2}{(a^2+\rho^2)^2}$.

$$\boxed{\;K = -\frac{a^2}{(a^2+\rho^2)^2},\qquad R = -\frac{2a^2}{(a^2+\rho^2)^2}.\;}$$

Geometry

Curvature is negative everywhere: saddle-like surface. Maximum $|K| = 1/a^2$ at the "waist" $\rho = 0$; asymptotically flat as $|\rho|\to\infty$. Smooth realisation of bounded-curvature hyperbolic-like geometry — the analog of a catenoid in $\mathbb R^3$. Useful as a regularisation of the pseudosphere (Beltrami's tractricoid), which has constant negative curvature but a singular edge.

Christoffels (from Problem 2.28)

$\Gamma^\lambda_{\lambda a} = \Gamma^\lambda_{a\lambda} = 1/a$, $\Gamma^a_{\lambda\lambda} = a$, all others zero.

Compute $R^\lambda{}_{a\lambda a}$

$R^\lambda{}_{a\lambda a} = \partial_\lambda\Gamma^\lambda_{aa} - \partial_a\Gamma^\lambda_{\lambda a} + \Gamma^\lambda_{\lambda\sigma}\Gamma^\sigma_{aa} - \Gamma^\lambda_{a\sigma}\Gamma^\sigma_{\lambda a}$.

$\partial_\lambda\Gamma^\lambda_{aa} = 0$ ($\Gamma^\lambda_{aa} = 0$).

$\partial_a\Gamma^\lambda_{\lambda a} = \partial_a(1/a) = -1/a^2$.

$\Gamma^\lambda_{\lambda\sigma}\Gamma^\sigma_{aa}$: both $\Gamma^\lambda_{\lambda\lambda} = 0$ and $\Gamma^\sigma_{aa} = 0$ (no such Christoffel non-zero), so = 0.

$\Gamma^\lambda_{a\sigma}\Gamma^\sigma_{\lambda a}$: $\sigma = \lambda$ contributes: $\Gamma^\lambda_{a\lambda}\Gamma^\lambda_{\lambda a} = (1/a)(1/a) = 1/a^2$.

Combine: $R^\lambda{}_{a\lambda a} = 0 - (-1/a^2) + 0 - 1/a^2 = 1/a^2 - 1/a^2 = 0$.

$$\boxed{\;R^\mu{}_{\nu\rho\sigma} = 0\;\text{everywhere on the Rindler wedge.}\;}$$

Why this is expected

Rindler coordinates are just Minkowski spacetime in accelerated coordinates ($t = a\sinh\lambda$, $x = a\cosh\lambda$). Minkowski is flat, and curvature is a tensor — vanishing in one coordinate system implies vanishing in all. The non-zero Christoffels merely reflect the non-Cartesian chart.

The horizon

The Rindler wedge covers only $x > |t|$. The boundary $a = 0$ corresponds to $x = \pm t$ — the past and future Rindler horizons. At $a = 0$ the metric component $g_{\lambda\lambda} = a^2$ vanishes, but Riemann is still zero. This is a coordinate singularity, not a physical one.

Prototype of the analogous distinction in Schwarzschild: the metric coefficient $g_{tt} = 1 - 2M/r$ vanishes at $r = 2M$ (event horizon), but Riemann is regular there. The genuine curvature singularity is at $r = 0$, where $R_{abcd}R^{abcd} \sim 1/r^6$ diverges. The Rindler-vs-Schwarzschild parallel was made precise by the Unruh effect — accelerated observers see thermal radiation from their horizon, analogous to Hawking radiation from a black-hole horizon.

Parametrisation

$t = \sinh\lambda$, $x = \cosh\lambda\cos\varphi$, $y = \cosh\lambda\sin\varphi$. Check: $t^2 - x^2 - y^2 = \sinh^2\lambda - \cosh^2\lambda = -1$ ✓.

Induced metric (cf. Problem 2.38 with $a = 1$)

$$ds^2 = d\lambda^2 - \cosh^2\lambda\,d\varphi^2.$$

Christoffels

$\Gamma^\lambda_{\varphi\varphi} = \cosh\lambda\sinh\lambda = \tfrac12\sinh(2\lambda)$, $\Gamma^\varphi_{\lambda\varphi} = \tanh\lambda$.

Riemann component

$R^\lambda{}_{\varphi\lambda\varphi} = \partial_\lambda\Gamma^\lambda_{\varphi\varphi} - 0 + 0 - \Gamma^\lambda_{\varphi\sigma}\Gamma^\sigma_{\lambda\varphi}$.

$\partial_\lambda\Gamma^\lambda_{\varphi\varphi} = \cosh^2\lambda + \sinh^2\lambda = \cosh(2\lambda)$.

$\Gamma^\lambda_{\varphi\sigma}\Gamma^\sigma_{\lambda\varphi}$: $\sigma = \varphi$: $\Gamma^\lambda_{\varphi\varphi}\Gamma^\varphi_{\lambda\varphi} = \cosh\lambda\sinh\lambda\cdot\tanh\lambda = \sinh^2\lambda$.

So $R^\lambda{}_{\varphi\lambda\varphi} = \cosh(2\lambda) - \sinh^2\lambda = (\cosh^2\lambda + \sinh^2\lambda) - \sinh^2\lambda = \cosh^2\lambda$.

Ricci tensor

$R_{\lambda\lambda} = R^\varphi{}_{\lambda\varphi\lambda} = g^{\varphi\varphi}R_{\varphi\lambda\varphi\lambda}$. Compute $R_{\varphi\lambda\varphi\lambda} = g_{\varphi\varphi}R^\varphi{}_{\lambda\varphi\lambda}$ where $R^\varphi{}_{\lambda\varphi\lambda} = ?$ — by pair symmetry $R^\lambda{}_{\varphi\lambda\varphi}$ gives the relevant info. Going through carefully: $R_{\lambda\lambda} = -1$, $R_{\varphi\varphi} = \cosh^2\lambda$.

Check: $R_{\lambda\lambda}/g_{\lambda\lambda} = -1/1 = -1$. $R_{\varphi\varphi}/g_{\varphi\varphi} = \cosh^2\lambda/(-\cosh^2\lambda) = -1$. Same constant: $R_{\mu\nu} = -1\cdot g_{\mu\nu}$.

$$\boxed{\;R_{\mu\nu} = -\Lambda g_{\mu\nu},\quad \Lambda = 1.\;}$$

Identification: 2D de Sitter

Maximally symmetric 2D Lorentzian space with positive sectional curvature $K = 1$. $R = g^{\mu\nu}R_{\mu\nu} = -\Lambda\cdot n = -2$ in 2D (with this signature convention). The Einstein tensor $G_{\mu\nu} = R_{\mu\nu} - \tfrac12 g_{\mu\nu}R = -g_{\mu\nu} + g_{\mu\nu} = 0$ identically (Problem 2.44), so dS$_2$ also trivially satisfies vacuum Einstein. The cosmological-constant content is hidden in the choice of metric, not in the dynamics.

Cosmological prototype: dS$_4$ in 3+1D is the model for inflation ($w = -1$ dark-energy equation of state). The 2D version captures the essential geometry without the extra 3-sphere spatial directions.

Setup (from Problem 2.27)

Metric $ds^2 = v[(r^2-1)dt^2 - dr^2/(r^2-1)]$. Christoffels $\Gamma^0_{01} = r/(r^2-1)$, $\Gamma^1_{00} = r(r^2-1)$, $\Gamma^1_{11} = -r/(r^2-1)$.

Compute Riemann

$R^0{}_{101} = \partial_0\Gamma^0_{11} - \partial_1\Gamma^0_{01} + \Gamma^0_{0\sigma}\Gamma^\sigma_{11} - \Gamma^0_{1\sigma}\Gamma^\sigma_{01}$.

$\partial_0\Gamma^0_{11} = 0$, $\partial_1\Gamma^0_{01} = \partial_r[r/(r^2-1)] = -(r^2+1)/(r^2-1)^2$ (quotient rule).

$\Gamma^0_{0\sigma}\Gamma^\sigma_{11}$: $\sigma = 1$: $\Gamma^0_{01}\Gamma^1_{11} = [r/(r^2-1)]\cdot[-r/(r^2-1)] = -r^2/(r^2-1)^2$.

$\Gamma^0_{1\sigma}\Gamma^\sigma_{01}$: $\sigma = 0$: $\Gamma^0_{10}\Gamma^0_{01} = r^2/(r^2-1)^2$.

Combine: $R^0{}_{101} = 0 + (r^2+1)/(r^2-1)^2 + (-r^2/(r^2-1)^2) - r^2/(r^2-1)^2 = (r^2+1 - 2r^2)/(r^2-1)^2 = (1-r^2)/(r^2-1)^2 = -1/(r^2-1)$.

Ricci components

By symmetry of AdS$_2$ (maximally symmetric, constant negative curvature):

$$\boxed{\;R_{\mu\nu} = -\frac{1}{v}g_{\mu\nu},\quad R = -\frac{2}{v}.\;}$$

Explicitly: $R_{00} = -(r^2-1)$, $R_{11} = 1/(r^2-1)$, $R_{01} = 0$.

Physical significance

AdS$_2$ is a foundational geometry in modern theoretical physics:

• Near-horizon of extremal Reissner–Nordström black holes: the throat geometry is $\text{AdS}_2\times S^2$. Entropy counting in this geometry is exactly tractable, contributing to the Strominger–Vafa programme.
• Holographic dual of 1D conformal quantum mechanics: the simplest setting of AdS/CFT.
• Gravity side of the SYK model: at strong coupling, the SYK quantum-mechanical model develops emergent conformal symmetry whose gravity dual is AdS$_2$ Jackiw–Teitelboim gravity (Problem 2.44).
• Replica wormholes and Page curve: AdS$_2$ JT gravity is the setting where these were first computed exactly, contributing to the resolution of the black-hole information paradox.

Parametrisation

Let $x = \alpha\sinh\chi$, $t = \alpha\cosh\chi\cos\theta$, $u = \alpha\cosh\chi\sin\theta$. Check: $t^2 + u^2 - x^2 = \alpha^2\cosh^2\chi(\cos^2\theta+\sin^2\theta) - \alpha^2\sinh^2\chi = \alpha^2$ ✓.

Induced metric

Differentials: $dx = \alpha\cosh\chi\,d\chi$; $dt = \alpha\sinh\chi\cos\theta\,d\chi - \alpha\cosh\chi\sin\theta\,d\theta$; $du = \alpha\sinh\chi\sin\theta\,d\chi + \alpha\cosh\chi\cos\theta\,d\theta$.

$dt^2 + du^2 = \alpha^2\sinh^2\chi\,d\chi^2 + \alpha^2\cosh^2\chi\,d\theta^2$ (cross terms cancel).

$dx^2 = \alpha^2\cosh^2\chi\,d\chi^2$.

Ambient metric $ds^2_{\text{amb}} = dt^2 + du^2 - dx^2$:

$$ds^2 = \alpha^2\sinh^2\chi\,d\chi^2 + \alpha^2\cosh^2\chi\,d\theta^2 - \alpha^2\cosh^2\chi\,d\chi^2 = -\alpha^2 d\chi^2 + \alpha^2\cosh^2\chi\,d\theta^2.$$

$$\boxed{\;ds^2 = -\alpha^2 d\chi^2 + \alpha^2\cosh^2\chi\,d\theta^2.\;}$$

Signature $(-,+)$ — Lorentzian, $\chi$ timelike, $\theta$ spacelike.

Christoffels

$g_{\chi\chi} = -\alpha^2$, $g_{\theta\theta} = \alpha^2\cosh^2\chi$. $\partial_\chi g_{\theta\theta} = \alpha^2\sinh(2\chi)$. Inverse: $g^{\chi\chi} = -1/\alpha^2$, $g^{\theta\theta} = 1/(\alpha^2\cosh^2\chi)$.

$\Gamma^\chi_{\theta\theta} = -\tfrac12 g^{\chi\chi}\partial_\chi g_{\theta\theta} = -\tfrac12\cdot(-1/\alpha^2)\cdot\alpha^2\sinh(2\chi) = \tfrac12\sinh(2\chi) = \sinh\chi\cosh\chi$.

$\Gamma^\theta_{\chi\theta} = \tfrac12 g^{\theta\theta}\partial_\chi g_{\theta\theta} = \tfrac12\cdot\tfrac{1}{\alpha^2\cosh^2\chi}\cdot\alpha^2\sinh(2\chi) = \tanh\chi$.

$$\boxed{\;\Gamma^\chi_{\theta\theta} = \sinh\chi\cosh\chi,\quad \Gamma^\theta_{\chi\theta} = \tanh\chi.\;}$$

Ricci scalar

Following the same computation as Problem 2.48 (with the curvature radius $\alpha$ in place of 1):

$$\boxed{\;R = \frac{2}{\alpha^2}\;\text{(constant positive curvature).}\;}$$

This is 2D de Sitter of curvature radius $\alpha$, embedded in $\mathbb R^{2,1}$ with the alternative signature $(+,+,-)$. Same intrinsic geometry as Problem 2.48 with the timelike-spacelike roles swapped between $\chi$ and $\theta$. Constant Gaussian curvature $K = 1/\alpha^2$ — the simplest cosmological-constant model in 2D.