General Relativity — worked problems

Parametrisation

Constraint $(ct)^2 - x^2 - y^2 = -K^2$, with $r^2 = x^2 + y^2$: $(ct)^2 = r^2 - K^2$, so $ct = \sqrt{r^2 - K^2}$ on one sheet (requires $r > K$). Use spatial polar $(r,\phi)$.

Induced metric

Differentiate: $d(ct) = (r/\sqrt{r^2-K^2})\,dr$, so $d(ct)^2 = r^2 dr^2/(r^2-K^2)$. Spatial part: $dx^2 + dy^2 = dr^2 + r^2 d\phi^2$. Ambient metric in signature $(+--)$ is $ds^2 = d(ct)^2 - dx^2 - dy^2$:

$$ds^2 = \frac{r^2}{r^2-K^2}dr^2 - dr^2 - r^2 d\phi^2 = \frac{r^2 - (r^2-K^2)}{r^2-K^2}dr^2 - r^2 d\phi^2.$$

$$\boxed{\;ds^2 = \frac{K^2}{r^2 - K^2}\,dr^2 - r^2\,d\phi^2.\;}$$

Signature analysis

$g_{rr} = K^2/(r^2-K^2) > 0$ for $r > K$; $g_{\phi\phi} = -r^2 < 0$. So the signature is $(+,-)$ — Lorentzian: $r$ is timelike (in this signature convention) and $\phi$ is spacelike. The "time" coordinate on this dS$_2$ surface is the radial coordinate.

Identification

This is 2D de Sitter space $\text{dS}_2$ of curvature radius $K$. From $K = 1/\sqrt\Lambda$ in units where $c = 1$, $\Lambda = 1/K^2$ is the cosmological constant. It is the simplest maximally symmetric Lorentzian space with positive curvature — an inflating universe model with constant Hubble parameter.

Geodesic structure: spacelike geodesics are circles (constant $r$), timelike geodesics correspond to free-fall trajectories ascending and descending in $r$. The spacelike conformal boundary is at $r\to\infty$.

(a) Lorentzian induced metric

Ambient $\mathbb R^{2,2}$ with metric $\eta = \mathrm{diag}(+,+,-,-)$. The constraint surface is the level set of $f(x) = (x^1)^2 + (x^2)^2 - (x^3)^2 - (x^4)^2 - 1 = 0$. The unit outward normal is the gradient: $$n^\mu \propto \eta^{\mu\nu}\partial_\nu f = (x^1, x^2, -x^3, -x^4)/\sqrt{f+1}.$$ On the surface ($f = 0$), $n\cdot n = (x^1)^2 + (x^2)^2 - (x^3)^2 - (x^4)^2 = 1 > 0$ — the normal is spacelike.

The induced metric is the ambient metric restricted to the tangent space $T_p\Sigma = n^\perp$. Removing a spacelike direction from $(+,+,-,-)$ leaves signature

$$(+,+,-,-) - (+) = (+,-,-)$$

— Lorentzian, with 1 timelike (the surviving $+$) and 2 spacelike directions.

(b) Killing vectors of the surface

The ambient $\mathbb R^{2,2}$ has isometry group $O(2,2)$ with six independent Killing vectors:

• 3 rotations/boosts in the $(x^1,x^2)$-block: $L_{12}$ only (the others mix with $(x^3,x^4)$).
• 3 rotations/boosts in the $(x^3,x^4)$-block: $L_{34}$ only (others mix).
• 4 mixed boosts between blocks.

The constraint $f$ is invariant under the block-internal generators $L_{12}$ and $L_{34}$ but generally not under mixed boosts. So on the surface:

$$\boxed{\;L_{12} = x^1\partial_2 - x^2\partial_1,\qquad L_{34} = x^3\partial_4 - x^4\partial_3\;}$$

are tangent to $\Sigma$ and survive as Killing vectors of the induced metric. For geodesics on the surface, the conserved Noether charges are:

$$L_{12}^{\text{cons}} = x^1\dot x^2 - x^2\dot x^1,\qquad L_{34}^{\text{cons}} = x^3\dot x^4 - x^4\dot x^3.$$

Identification: AdS$_3$

This surface is 3D anti-de Sitter space, $\mathrm{AdS}_3$. It is the bulk space of the AdS$_3$/CFT$_2$ correspondence (Brown–Henneaux 1986), with Virasoro symmetry algebra at the boundary. The BTZ black hole is constructed as a quotient of AdS$_3$ by a discrete isometry. The $L_{12}, L_{34}$ Killing vectors correspond to the $SL(2,\mathbb R)\times SL(2,\mathbb R)$ isometry of AdS$_3$.

Setup

Integral curves $\gamma(t)$ of $X$ satisfy $\dot\gamma(t) = X(\gamma(t))$. For $\gamma$ to be a geodesic (with $t$ affine), require $$\nabla_{\dot\gamma}\dot\gamma = 0\quad\Longleftrightarrow\quad \nabla_X X = 0\;\text{along the flow}.$$

Component form

$\nabla_X X = X^j\nabla_j X^i\hat e_i = X^j(\partial_j X^i + \Gamma^i_{jk}X^k)\hat e_i$. Setting each component to zero:

$$\boxed{\;X^j\partial_j X^i + \Gamma^i_{jk}X^j X^k = 0\quad (i = 1,\ldots,n).\;}$$

This is a system of $n$ first-order quasilinear PDEs on the components $X^i(x)$. Geometrically: the vector field is "its own parallel transport" — the directional derivative of $X$ in the direction of $X$ itself, taken covariantly, vanishes.

Examples that satisfy / don't satisfy the condition

Killing vectors are NOT generally geodesic. A Killing vector $K$ satisfies $\nabla_{(\mu}K_{\nu)} = 0$, which does not imply $K^\mu\nabla_\mu K^\nu = 0$. E.g. in Schwarzschild, $K = \partial_t$ is Killing but corresponds to static observers, whose worldlines are non-geodesic (they require thrust to stay at constant $r$).

Comoving FRW observers ARE geodesic. In FRW spacetime, the vector field $X = \partial_t$ in comoving coordinates is the 4-velocity of comoving matter. It satisfies $\nabla_X X = 0$ because the comoving worldlines are geodesics (matter free-falls with the expanding cosmos).

Physical significance

The condition $\nabla_X X = 0$ for a fluid's 4-velocity distinguishes pressureless dust (free-falling, geodesic worldlines) from pressure-supported fluids (where pressure gradients accelerate the fluid off geodesics). This is captured by the relativistic Euler equation: $$(\rho + p)u^\mu\nabla_\mu u^\nu = -(g^{\mu\nu} + u^\mu u^\nu)\nabla_\mu p.$$ Setting $p = 0$ recovers $\nabla_u u = 0$ (dust geodesics); $p \neq 0$ accelerates the fluid off the geodesic congruence.

Setup

$SU(2)$ matrices are $g(x) = \exp(i\sigma\cdot x) = \mathbb 1\cos r + i(\hat x\cdot\sigma)\sin r$, where $r = |x|$ and $\hat x = x/r$. The map $x \mapsto g(x)$ identifies $SU(2)$ with the 3-sphere $S^3$ in $\mathbb R^4 = (a^0, a^1, a^2, a^3)$ via

$$g \leftrightarrow (a^0,\vec a) = (\cos r,\,\hat x\sin r),\qquad (a^0)^2 + |\vec a|^2 = 1.$$

Induced metric on $S^3$

In polar-spherical coordinates $(r,\theta,\phi)$ on $S^3$ (with $\hat x$ parametrised by $\theta,\phi$ on a unit 2-sphere): $$ds^2 = dr^2 + \sin^2 r\,(d\theta^2 + \sin^2\theta\,d\phi^2).$$

1-parameter subgroups

A 1-parameter subgroup with direction $\hat a$ (unit vector) is $g(t) = e^{-t\sigma\cdot\hat a/2}\cdot 2 = $ … more carefully, $g(t) = \exp(it\sigma\cdot\hat a/2)$ traces $r(t) = t/2$ (or just $r(t) = t$ with appropriate normalisation), $(\theta,\phi)$ fixed (the direction $\hat a$ is constant).

Verify geodesic equations

From the metric, the Lagrangian is $\mathcal L = \tfrac12[\dot r^2 + \sin^2 r(\dot\theta^2 + \sin^2\theta\dot\phi^2)]$.

$r$-equation: $\ddot r - \sin r\cos r(\dot\theta^2 + \sin^2\theta\dot\phi^2) = 0$. With $\dot r = 1, \ddot r = 0, \dot\theta = \dot\phi = 0$: $0 - 0 = 0$ ✓.

$\theta$- and $\phi$-equations: both have $\dot\theta = \dot\phi = 0$ identically, so all terms vanish ✓.

$$\boxed{\;\text{1-parameter subgroups are geodesics of }S^3\text{ from the identity.}\;}$$

General theorem

On any Lie group with bi-invariant metric (one for which both left and right translations are isometries), the Riemannian exponential map at the identity equals the Lie-theoretic exponential map: $\exp_e(X) = e^X$ for $X\in\mathfrak g$. This is a beautiful confluence of Lie theory and Riemannian geometry — and the structural reason $SU(2) \cong S^3$ inherits a natural round-sphere metric. On non-bi-invariant Lie groups (most non-compact ones), the two notions of exp differ.

(a) Fundamental theorem derivation

Metric compatibility: $D_k g_{\mu\nu} = \partial_k g_{\mu\nu} - \Gamma^\lambda_{k\mu}g_{\lambda\nu} - \Gamma^\lambda_{k\nu}g_{\mu\lambda} = 0$. Rearranging:

$$\partial_k g_{\mu\nu} = \Gamma_{k\mu\nu} + \Gamma_{k\nu\mu}, \quad\text{where }\Gamma_{k\mu\nu} \equiv g_{\lambda\nu}\Gamma^\lambda_{k\mu}.\qquad(\star)$$

Three cyclic permutations of $(k,\mu,\nu)$ generate three equations. Adding the first two and subtracting the third, then applying the symmetry $\Gamma_{k\mu\nu} = \Gamma_{\mu k\nu}$ (from torsion-free $\Gamma^\lambda_{\mu\nu} = \Gamma^\lambda_{\nu\mu}$):

$$\partial_k g_{\mu\nu} + \partial_\mu g_{\nu k} - \partial_\nu g_{k\mu} = 2\Gamma_{k\mu\nu}.$$

Raise the third index with $g^{\lambda\nu}$:

$$\boxed{\;\Gamma^\lambda_{\mu\nu} = \tfrac12 g^{\lambda\rho}(\partial_\mu g_{\nu\rho} + \partial_\nu g_{\mu\rho} - \partial_\rho g_{\mu\nu}).\;}$$

This is the Fundamental Theorem of Riemannian Geometry: the metric uniquely determines a torsion-free metric-compatible connection (the Levi-Civita connection).

(b) Computation of $T_\mu{}^\nu$ in Cartesian

In flat Minkowski $\Gamma = 0$, so $T_\mu{}^\nu = D_\mu V^\nu = \partial_\mu V^\nu$. With $V^\mu = (x, -t)$ in $(t,x)$ coords:

$T_0{}^0 = \partial_t x = 0$, $T_0{}^1 = \partial_t(-t) = -1$, $T_1{}^0 = \partial_x x = 1$, $T_1{}^1 = \partial_x(-t) = 0$.

$$T_\mu{}^\nu = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\quad(\text{Cartesian}).$$

Transformation to Rindler

Rindler: $t = a\sinh\lambda$, $x = a\cosh\lambda$. Convert $V$ to Rindler components $V'^\mu = (\partial x'^\mu/\partial x^\nu) V^\nu$:

$V'^\lambda = (\partial\lambda/\partial t)V^t + (\partial\lambda/\partial x)V^x = (\cosh\lambda/a)\cdot a\cosh\lambda + (-\sinh\lambda/a)\cdot(-a\sinh\lambda) = \cosh^2\lambda + \sinh^2\lambda = \cosh(2\lambda)$.

$V'^a = (\partial a/\partial t)V^t + (\partial a/\partial x)V^x = (-\sinh\lambda)(a\cosh\lambda) + (\cosh\lambda)(-a\sinh\lambda) = -2a\sinh\lambda\cosh\lambda = -a\sinh(2\lambda)$.

Covariant derivative in Rindler

$T'^\lambda{}_\lambda = D'_\lambda V'^\lambda = \partial_\lambda V'^\lambda + \Gamma'^\lambda_{\lambda b}V'^b$.

$\partial_\lambda V'^\lambda = \partial_\lambda\cosh(2\lambda) = 2\sinh(2\lambda)$.

$\Gamma'^\lambda_{\lambda a} = 1/a$ (Problem 2.28), so the Christoffel correction is $(1/a)\cdot V'^a = (1/a)(-a\sinh(2\lambda)) = -\sinh(2\lambda)$.

$$\boxed{\;T'^\lambda{}_\lambda = 2\sinh(2\lambda) - \sinh(2\lambda) = \sinh(2\lambda).\;}$$

Same intrinsic tensor, different coordinate components — the meaning of "covariant under general coordinate transformations".

(a) General formulas

Tensor transformation for a $(2,0)$ tensor:

$$\boxed{\;S'^{\mu\nu} = \frac{\partial x'^\mu}{\partial x^\alpha}\frac{\partial x'^\nu}{\partial x^\beta}\,S^{\alpha\beta}.\;}$$

Covariant divergence:

$$D_\mu S^{\mu\nu} = \partial_\mu S^{\mu\nu} + \Gamma^\mu_{\mu\lambda}S^{\lambda\nu} + \Gamma^\nu_{\mu\lambda}S^{\mu\lambda}.$$

Using $\Gamma^\mu_{\mu\lambda} = \partial_\lambda\ln\sqrt{|g|}$ (Problem 2.33a):

$$\boxed{\;D_\mu S^{\mu\nu} = \frac{1}{\sqrt{|g|}}\partial_\mu(\sqrt{|g|}\,S^{\mu\nu}) + \Gamma^\nu_{\mu\lambda}S^{\mu\lambda}.\;}$$

(b) Polar transformation of $S^{\mu\nu}$

Cartesian $\to$ polar: $x = r\cos\phi$, $y = r\sin\phi$. Inverse Jacobian: $\partial r/\partial x = \cos\phi$, $\partial r/\partial y = \sin\phi$, $\partial\phi/\partial x = -\sin\phi/r$, $\partial\phi/\partial y = \cos\phi/r$.

Non-zero $S$: $S^{12} = 2xy$, $S^{21} = -2xy$. Compute:

$$S'^{r\phi} = \frac{\partial r}{\partial x^\alpha}\frac{\partial\phi}{\partial x^\beta}S^{\alpha\beta} = \cos\phi\cdot\frac{\cos\phi}{r}\cdot 2xy + \sin\phi\cdot\frac{-\sin\phi}{r}\cdot(-2xy).$$

$= (2xy/r)(\cos^2\phi + \sin^2\phi) = 2xy/r$. Using $2xy = 2r^2\sin\phi\cos\phi = r^2\sin(2\phi)$: $$S'^{r\phi} = r\sin(2\phi).$$

By antisymmetry $S'^{\phi r} = -r\sin(2\phi)$. Diagonal entries vanish by similar computation.

$$\boxed{\;S'^{r\phi} = -S'^{\phi r} = r\sin(2\phi),\quad S'^{rr} = S'^{\phi\phi} = 0.\;}$$

Covariant divergence in Cartesian (flat space)

$\Gamma = 0$, so $D_\mu S^{\mu\nu} = \partial_\mu S^{\mu\nu}$.

$D_\mu S^{\mu 1} = \partial_x S^{x1} + \partial_y S^{y1} = \partial_x 0 + \partial_y(-2xy) = -2x$.

$D_\mu S^{\mu 2} = \partial_x S^{x2} + \partial_y S^{y2} = \partial_x(2xy) + 0 = 2y$.

$$\boxed{\;D_\mu S^{\mu 1} = -2x,\;\; D_\mu S^{\mu 2} = 2y.\;}$$

Antisymmetry preservation

$S'^{r\phi} = -S'^{\phi r}$ exactly — antisymmetry is invariant under coordinate transformations (it's a tensorial property, encoded in the symmetry group of the tensor product space). Equivalently, $S$ is a 2-form, and 2-forms remain 2-forms in every chart.

Key observation

Comparing $S_{00} = a(r^2-1) = (a/v)\cdot v(r^2-1) = (a/v)g_{00}$ and $S_{11} = -a/(r^2-1) = (a/v)\cdot[-v/(r^2-1)] = (a/v)g_{11}$, we see $$S_{\mu\nu} = \frac{a}{v}g_{\mu\nu}.$$ The tensor is proportional to the metric with constant coefficient $a/v$.

(a) Christoffel symbols

Metric: $g_{00} = v(r^2-1)$, $g_{11} = -v/(r^2-1)$. Non-zero metric derivatives: $\partial_r g_{00} = 2vr$, $\partial_r g_{11} = 2vr/(r^2-1)^2$. Inverse: $g^{00} = 1/[v(r^2-1)]$, $g^{11} = -(r^2-1)/v$.

$$\Gamma^0_{01} = \tfrac12 g^{00}\partial_r g_{00} = \frac{1}{2v(r^2-1)}\cdot 2vr = \frac{r}{r^2-1}.$$

$$\Gamma^1_{00} = -\tfrac12 g^{11}\partial_r g_{00} = -\tfrac12\cdot\left(-\frac{r^2-1}{v}\right)\cdot 2vr = r(r^2-1).$$

$$\Gamma^1_{11} = \tfrac12 g^{11}\partial_r g_{11} = \tfrac12\cdot\left(-\frac{r^2-1}{v}\right)\cdot\frac{2vr}{(r^2-1)^2} = -\frac{r}{r^2-1}.$$

(b) Raise indices on $S$; compute divergence

$S^{\mu\nu} = g^{\mu\alpha}g^{\nu\beta}S_{\alpha\beta} = (a/v)g^{\mu\alpha}g^{\nu\beta}g_{\alpha\beta} = (a/v)g^{\mu\nu}$. So $S^{00} = a/[v^2(r^2-1)]$, $S^{11} = -a(r^2-1)/v^2$.

Since $S_{\mu\nu} = (a/v)g_{\mu\nu}$ with $a/v$ constant, and $\nabla g = 0$: $$\boxed{\;D_\mu S^{\mu\nu} = \frac{a}{v}D_\mu g^{\mu\nu} = 0.\;}$$

Covariantly constant tensor — no further computation needed.

(c) Transform to $(\theta,\eta)$

$\theta = at$, $r = \cosh\eta$. Differentials: $dt = d\theta/a$, $dr = \sinh\eta\,d\eta$. Also $r^2 - 1 = \cosh^2\eta - 1 = \sinh^2\eta$.

Metric: $ds^2 = v[\sinh^2\eta\cdot(d\theta/a)^2 - (\sinh^2\eta)^{-1}\sinh^2\eta\,d\eta^2] = \frac{v\sinh^2\eta}{a^2}d\theta^2 - v\,d\eta^2$, so $g'_{\theta\theta} = v\sinh^2\eta/a^2$, $g'_{\eta\eta} = -v$. The horizon at $r = 1$ ($\eta = 0$) is now regular!

Tensor transforms: $S'_{\theta\theta} = (\partial t/\partial\theta)^2 S_{00} = (1/a)^2\cdot a\sinh^2\eta = \sinh^2\eta/a$. $S'_{\eta\eta} = (\partial r/\partial\eta)^2 S_{11} = \sinh^2\eta\cdot[-a/\sinh^2\eta] = -a$. Off-diagonal zero.

$$\boxed{\;S'_{\theta\theta} = \frac{\sinh^2\eta}{a},\quad S'_{\eta\eta} = -a.\;}$$

Ratio check: $S'_{\theta\theta}/g'_{\theta\theta} = (\sinh^2\eta/a)/(v\sinh^2\eta/a^2) = a/v$ ✓. $S'_{\eta\eta}/g'_{\eta\eta} = -a/(-v) = a/v$ ✓. The proportionality $S/g = a/v$ is preserved — as expected since it's a tensor identity. This is the structural reason the cosmological-constant term $\Lambda g_{\mu\nu}$ is automatically conserved in Einstein's equation (Bianchi identity).

(a) Metric and Christoffels

Differentials: $dt = a\cosh\lambda\,d\lambda + \sinh\lambda\,da$, $dx = a\sinh\lambda\,d\lambda + \cosh\lambda\,da$. Then:

$dt^2 - dx^2 = a^2(\cosh^2\lambda - \sinh^2\lambda)d\lambda^2 + (\sinh^2\lambda - \cosh^2\lambda)da^2 + (\text{cross terms cancel}) = a^2 d\lambda^2 - da^2$.

$$\boxed{\;ds^2 = a^2\,d\lambda^2 - da^2.\;}$$

$g_{\lambda\lambda} = a^2$, $g_{aa} = -1$. Only non-zero derivative: $\partial_a g_{\lambda\lambda} = 2a$.

$$\Gamma^\lambda_{\lambda a} = \Gamma^\lambda_{a\lambda} = \tfrac12 g^{\lambda\lambda}\partial_a g_{\lambda\lambda} = \frac{1}{2a^2}\cdot 2a = \frac{1}{a}.$$

$$\Gamma^a_{\lambda\lambda} = -\tfrac12 g^{aa}\partial_a g_{\lambda\lambda} = -\tfrac12\cdot(-1)\cdot 2a = a.$$

$$\boxed{\;\Gamma^\lambda_{\lambda a} = 1/a,\qquad \Gamma^a_{\lambda\lambda} = a.\;}$$

(b) Divergence and Laplacian

$\sqrt{|g|} = a$. Standard formula: $$\boxed{\;\nabla_\mu V^\mu = \frac{1}{a}\partial_\mu(a V^\mu) = \partial_\lambda V^\lambda + \frac{1}{a}\partial_a(aV^a).\;}$$

Scalar Laplacian: $\Box\phi = (1/\sqrt{|g|})\partial_\mu(\sqrt{|g|}g^{\mu\nu}\partial_\nu\phi) = (1/a)\partial_\lambda[(1/a)\partial_\lambda\phi] + (1/a)\partial_a[-a\partial_a\phi]$:

$$\boxed{\;\Box\phi = \frac{1}{a^2}\partial_\lambda^2\phi - \partial_a^2\phi - \frac{1}{a}\partial_a\phi.\;}$$

(c) Transform $T^\mu{}_\nu$ to Rindler

Given $T^0{}_0 = -T^1{}_1 = x^2 - t^2 = a^2$ (on the Rindler wedge). Mixed-tensor transform: $$T'^\mu{}_\nu = \frac{\partial x'^\mu}{\partial x^\rho}\frac{\partial x^\sigma}{\partial x'^\nu}T^\rho{}_\sigma.$$

Jacobian: $\partial\lambda/\partial t = \cosh\lambda/a$, $\partial\lambda/\partial x = -\sinh\lambda/a$, $\partial t/\partial\lambda = a\cosh\lambda$, $\partial x/\partial\lambda = a\sinh\lambda$.

$$T'^\lambda{}_\lambda = \frac{\partial\lambda}{\partial t}\frac{\partial t}{\partial\lambda}T^t{}_t + \frac{\partial\lambda}{\partial x}\frac{\partial x}{\partial\lambda}T^x{}_x = \frac{\cosh\lambda}{a}\cdot a\cosh\lambda\cdot a^2 + \left(-\frac{\sinh\lambda}{a}\right)\cdot a\sinh\lambda\cdot(-a^2)$$

$= a^2\cosh^2\lambda + a^2\sinh^2\lambda = a^2\cosh(2\lambda) = x^2 + t^2$ (using $x^2 + t^2 = a^2(\cosh^2\lambda + \sinh^2\lambda) = a^2\cosh(2\lambda)$).

$$\boxed{\;T'^\lambda{}_\lambda = a^2\cosh(2\lambda) = x^2 + t^2.\;}$$

Trace check

Cartesian: $T^\mu{}_\mu = T^0{}_0 + T^1{}_1 = (x^2-t^2) - (x^2-t^2) = 0$.

Rindler: similar computation gives $T'^a{}_a = -a^2\cosh(2\lambda) = -(x^2+t^2)$. Sum: $T'^\lambda{}_\lambda + T'^a{}_a = 0$ ✓ — the trace is coordinate-invariant.

Setup

Metric $g_{\mu\nu} = \mathrm{diag}(1,-1,-r^2)$ with $(t,r,\phi)$. Inverse $g^{\mu\nu} = \mathrm{diag}(1,-1,-1/r^2)$. $\sqrt{|g|} = r$.

Christoffel symbols

Only $g_{\phi\phi} = -r^2$ depends on coordinates: $\partial_r g_{\phi\phi} = -2r$. From $\Gamma^a_{bc} = \tfrac12 g^{ad}(\partial_b g_{cd}+\partial_c g_{bd}-\partial_d g_{bc})$:

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

$\Gamma^\phi_{r\phi} = \Gamma^\phi_{\phi r} = \tfrac12 g^{\phi\phi}\partial_r g_{\phi\phi} = \tfrac12\cdot(-1/r^2)\cdot(-2r) = 1/r$.

$$\boxed{\;\Gamma^r_{\phi\phi} = -r,\qquad \Gamma^\phi_{r\phi} = \frac{1}{r},\;}$$ all others zero. The sign of $\Gamma^r_{\phi\phi}$ is opposite the Riemannian case because $g_{rr} = -1$ (Lorentzian).

Covariant derivative $\nabla_\mu V^\nu$

$\nabla_\mu V^\nu = \partial_\mu V^\nu + \Gamma^\nu_{\mu\rho}V^\rho$. Component-by-component:

$$\nabla_t V^\nu = \partial_t V^\nu\quad(\text{no }\Gamma^\nu_{t\rho}\text{ non-zero}).$$

$$\nabla_r V^r = \partial_r V^r,\;\;\nabla_r V^\phi = \partial_r V^\phi + \Gamma^\phi_{r\phi}V^\phi = \partial_r V^\phi + V^\phi/r.$$

$$\nabla_\phi V^r = \partial_\phi V^r + \Gamma^r_{\phi\phi}V^\phi = \partial_\phi V^r - rV^\phi.$$

$$\nabla_\phi V^\phi = \partial_\phi V^\phi + \Gamma^\phi_{\phi r}V^r = \partial_\phi V^\phi + V^r/r.$$

Divergence (compact form)

$\nabla_\mu V^\mu = (1/\sqrt{|g|})\partial_\mu(\sqrt{|g|}\,V^\mu) = (1/r)\partial_\mu(r V^\mu)$:

$$\boxed{\;\nabla_\mu V^\mu = \partial_t V^t + \partial_r V^r + \frac{V^r}{r} + \partial_\phi V^\phi.\;}$$

The $V^r/r$ correction is the polar-coordinate geometric piece, identical to the spatial-polar 3D Cartesian-to-polar conversion in classical mechanics. Time component is unaffected since the spatial part is flat.

(a) Geodesic equation from parallel transport of the tangent

Set $V = T = dx/d\lambda$, the tangent to the curve $x(\lambda)$. The parallel-transport condition $T^\mu\nabla_\mu V^\nu = 0$ becomes:

$$T^\mu\nabla_\mu T^\nu = T^\mu\partial_\mu T^\nu + \Gamma^\nu_{\mu\sigma}T^\mu T^\sigma = 0.$$

The first term: $T^\mu\partial_\mu T^\nu = (dx^\mu/d\lambda)\partial_\mu(dx^\nu/d\lambda) = d^2 x^\nu/d\lambda^2 = \ddot x^\nu$ (chain rule along the curve).

$$\boxed{\;\ddot x^\nu + \Gamma^\nu_{\mu\sigma}\dot x^\mu \dot x^\sigma = 0\quad\text{(affine geodesic).}\;}$$

$\lambda$ is then an affine parameter — in the sense that the tangent $T$ is parallel-transported along itself exactly, with no rescaling.

(b) Non-affine parameter and reparametrisation

If the tangent is parallel-transported only "up to rescaling", $T^\mu\nabla_\mu T^\nu = \alpha T^\nu$ for some function $\alpha = \alpha(\lambda)$ along the curve. The geodesic equation becomes:

$$\boxed{\;\ddot x^\nu + \Gamma^\nu_{\mu\sigma}\dot x^\mu\dot x^\sigma = \alpha\dot x^\nu.\;}$$

The shape of the curve is still geodesic; only the parametrisation is non-affine.

Reparametrisation to affine form

Try $\tau = \tau(\lambda)$. Tangent in new parameter: $\tilde T^\mu = dx^\mu/d\tau = (d\lambda/d\tau)T^\mu$. Compute:

$$\tilde T^\mu\nabla_\mu\tilde T^\nu = (d\lambda/d\tau)T^\mu\nabla_\mu[(d\lambda/d\tau)T^\nu] = (d\lambda/d\tau)\left[T^\mu\partial_\mu(d\lambda/d\tau)\cdot T^\nu + (d\lambda/d\tau)T^\mu\nabla_\mu T^\nu\right].$$

$T^\mu\partial_\mu f = df/d\lambda$ along the curve. Using $T^\mu\nabla_\mu T^\nu = \alpha T^\nu$:

$$\tilde T^\mu\nabla_\mu\tilde T^\nu = (d\lambda/d\tau)\left[\frac{d(d\lambda/d\tau)}{d\lambda} + \alpha(d\lambda/d\tau)\right]T^\nu.$$

To make this zero, require the bracket to vanish. Let $u = d\lambda/d\tau$: $du/d\lambda = -\alpha u$, so $u = u_0 e^{-\int\alpha\,d\lambda}$. Hence $$\frac{d\tau}{d\lambda} = \frac{1}{u} = u_0^{-1}e^{\int\alpha\,d\lambda},$$ i.e.

$$\boxed{\;\tau(\lambda) = \int e^{\int\alpha(\lambda')\,d\lambda'}\,d\lambda\;\;\text{gives an affine parameter.}\;}$$

Physical use

This matters for null geodesics (light rays in GR), where proper time vanishes and one must choose an affine parameter (commonly the relativistic frequency-energy variable). Any monotonically related parameter works, but the affine one yields the clean geodesic equation with zero RHS.