General Relativity — worked problems

Setup

Ansatz $ds^2 = e^{\nu(t,r)}c^2 dt^2 - e^{\rho(t,r)}dr^2 - r^2 d\Omega^2$ with $d\Omega^2 = d\theta^2 + \sin^2\theta\,d\phi^2$.

Step 1: Birkhoff's theorem forces static

From Problem 2.52, $G^t{}_r = 0$ implies $\dot\rho = 0$, so $\rho = \rho(r)$. Then $G^r{}_r = 0$ gives $\nu' = \nu'(r)$. After absorbing the pure-$t$ piece by reparametrising $t$, we have $\nu = \nu(r)$ as well. Henceforth treat both as functions of $r$ alone.

Step 2: Components of $G^\mu{}_\nu$

For this spherically symmetric static metric (standard computation):

$$G^t{}_t = \frac{1}{r^2}\Bigl[e^{-\rho}(1 - r\rho') - 1\Bigr]\cdot c^2,$$

$$G^r{}_r = \frac{1}{r^2}\Bigl[e^{-\rho}(1 + r\nu') - 1\Bigr]\cdot c^2.$$

Step 3: Solve the vacuum equations

$G^t{}_t = 0$: $e^{-\rho}(1 - r\rho') = 1$, equivalently $1 - r\rho' = e^\rho$, or $(re^{-\rho})' = 1$. Integrating: $re^{-\rho} = r - r_s$ for some constant $r_s$, so $$e^{-\rho} = 1 - \frac{r_s}{r}.$$

$G^r{}_r = 0$: $e^{-\rho}(1 + r\nu') = 1$. Substituting $e^{-\rho} = 1 - r_s/r$: $(1 - r_s/r)(1 + r\nu') = 1$, giving $r\nu' = r_s/(r - r_s)$, so $\nu' = -\rho'$, hence $\nu = -\rho + \text{const}$. Asymptotic flatness ($\nu \to 0$ at $r\to\infty$) fixes the constant to zero:

$$e^\nu = 1 - \frac{r_s}{r}.$$

Step 4: Identify $r_s$

For a slow weak-field test particle to obey Newton's law $g = GM/r^2$, the asymptotic Newtonian limit of the geodesic equation gives $g_{tt}\approx 1 - 2\Phi/c^2$ with $\Phi = -GM/r$. So $r_s/r = 2GM/(rc^2)$, giving the Schwarzschild radius $$r_s = \frac{2GM}{c^2}.$$

Final result

$$\boxed{\;ds^2 = \Bigl(1-\frac{r_s}{r}\Bigr)c^2 dt^2 - \frac{dr^2}{1 - r_s/r} - r^2 d\Omega^2,\qquad r_s = \frac{2GM}{c^2}.\;}$$

Unique spherically symmetric vacuum solution (Birkhoff). Sun's $r_s \approx 3$ km; Earth's $r_s \approx 9$ mm. Foundational geometry for: Mercury's perihelion precession ($43''$/century, GR's first triumph), light bending by the Sun (Eddington 1919), gravitational redshift (Pound–Rebka 1959), the Schwarzschild black hole (event horizon at $r = r_s$, singularity at $r = 0$). Karl Schwarzschild, late 1915, on the Russian front, weeks before his death.

Setup

Most general spherically symmetric ansatz (allowing $t$-dependence): $$ds^2 = e^{\nu(t,r)}c^2 dt^2 - e^{\rho(t,r)}dr^2 - r^2 d\Omega^2.$$ Goal: show vacuum Einstein equations $G_{\mu\nu} = 0$ force $\dot\nu = \dot\rho = 0$.

Step 1: $G^t{}_r = 0$ forces $\rho = \rho(r)$

Compute the off-diagonal $(t,r)$ component of the Einstein tensor for this ansatz:

$$G^t{}_r = \frac{\dot\rho}{r}.$$

Setting $G^t{}_r = 0$: $\dot\rho = 0$, i.e. $\rho$ is independent of $t$. $\rho = \rho(r)$.

Step 2: $G^r{}_r = 0$ forces $\nu'$ time-independent

With $\rho = \rho(r)$, $G^r{}_r$ depends on $\rho(r), \nu, \nu'$. Setting $G^r{}_r = 0$ and using $\rho = \rho(r)$, the equation can be solved for $\nu'$ in terms of $\rho(r)$ — hence $\nu' = \nu'(r)$ depends only on $r$.

Therefore $\nu(t,r) = f(r) + h(t)$ for some functions $f, h$.

Step 3: Absorb $h(t)$ by coordinate change

Define new time coordinate $t'$ by $$dt' = e^{h(t)/2}dt,\quad\text{i.e.}\quad t' = \int e^{h(t)/2}dt.$$ Then $e^\nu c^2 dt^2 = e^{f(r)+h(t)}c^2 dt^2 = e^{f(r)}c^2(e^{h(t)/2}dt)^2 = e^{f(r)}c^2 dt'^2$. So in $(t',r,\theta,\phi)$ coords, $\nu$ effectively reduces to $f(r)$ — manifestly static.

$$\boxed{\;\text{Any spherically symmetric vacuum solution is static (modulo coordinate transformations).}\;}$$

Implications

(i) Uniqueness of Schwarzschild. Combined with asymptotic flatness, Birkhoff's theorem makes the Schwarzschild solution the unique spherically symmetric vacuum geometry.

(ii) Newton's shell theorem generalised. No monopole gravitational radiation: a radially pulsating star generates a static exterior field, just as in Newton's gravity where the exterior depends only on the total mass.

(iii) Why LIGO sees binaries, not symmetric supernovae. Gravitational waves require quadrupole or higher asymmetry. A spherically symmetric collapsing or pulsating star, no matter how violent, emits NO gravitational waves — the exterior remains Schwarzschild throughout. Asymmetric mergers (inspiralling binaries, neutron-star coalescence) and asymmetric supernovae are the GW sources detected since 2015.

(iv) Black-hole no-hair refinement. A spherically symmetric perturbation of Schwarzschild cannot change the exterior. Part of the no-hair theorem for stationary black holes.

Strategy

Apply $\nabla_\mu$ to each piece of $T^{\mu\nu}$, use the inhomogeneous Maxwell equation $\nabla_\mu F^{\mu\nu} = J^\nu$ (Problem 2.58), and use the Bianchi half $\nabla_{[\alpha}F_{\beta\gamma]} = 0$ (Problem 2.54) to simplify the remaining bilinears.

First term: $\nabla_\mu(F^\mu{}_\lambda F^{\nu\lambda})$

Product rule:

$$\nabla_\mu(F^\mu{}_\lambda F^{\nu\lambda}) = (\nabla_\mu F^\mu{}_\lambda)F^{\nu\lambda} + F^\mu{}_\lambda\nabla_\mu F^{\nu\lambda}.$$

By Maxwell, $\nabla_\mu F^\mu{}_\lambda = J_\lambda$. The first piece gives $J_\lambda F^{\nu\lambda} = F^{\nu\lambda}J_\lambda$.

The second piece $F^\mu{}_\lambda\nabla_\mu F^{\nu\lambda}$ is bilinear in $F$ and $\nabla F$; combine with the trace term below.

Second term: $\tfrac14\nabla^\nu(F_{\rho\sigma}F^{\rho\sigma})$

$$\tfrac14\nabla^\nu(F_{\rho\sigma}F^{\rho\sigma}) = \tfrac12 F^{\rho\sigma}\nabla^\nu F_{\rho\sigma}.$$

Combine bilinears using Bianchi

Need to show: $F^\mu{}_\lambda\nabla_\mu F^{\nu\lambda} + \tfrac12 F^{\rho\sigma}\nabla^\nu F_{\rho\sigma} = 0$.

The Bianchi identity $\nabla_\alpha F_{\beta\gamma} + \nabla_\beta F_{\gamma\alpha} + \nabla_\gamma F_{\alpha\beta} = 0$ permits substituting $\nabla_\mu F_{\nu\lambda} = -\nabla_\nu F_{\lambda\mu} - \nabla_\lambda F_{\mu\nu}$ (raising/lowering as needed). Substituting and using antisymmetry of $F$, the bilinear terms cancel exactly (standard textbook computation, see e.g. Jackson or Weinberg).

Result

$$\boxed{\;\nabla_\mu T^{\mu\nu}_{\text{EM}} = \epsilon_0 F^{\nu\lambda}J_\lambda = \epsilon_0 F^\nu{}_\lambda J^\lambda\;\text{(Lorentz-force density 4-vector).}\;}$$

Physical interpretation

The EM stress-energy is NOT separately conserved: it exchanges 4-momentum with charged matter via the Lorentz force. The 4-vector $F^\nu{}_\lambda J^\lambda$ encodes:

• $\nu = 0$ (energy): $F^0{}_\lambda J^\lambda = \vec E\cdot\vec J$ — Joule heating rate.
• $\nu = i$ (momentum): $F^i{}_0 J^0 + F^i{}_j J^j = \rho\vec E + \vec J\times\vec B$ — Lorentz force density.

Total $T^{\mu\nu}_{\text{tot}} = T^{\mu\nu}_{\text{EM}} + T^{\mu\nu}_{\text{matter}}$ is conserved: $\nabla_\mu T^{\mu\nu}_{\text{matter}} = -\epsilon_0 F^\nu{}_\lambda J^\lambda$, balancing the field's flow into matter. In Einstein's equation $G^{\mu\nu} = \kappa T^{\mu\nu}_{\text{tot}}$, only the total $T$ appears; the partition into field+matter is observer-dependent.

Setup

Show $\partial_\alpha F_{\beta\gamma} + \partial_\beta F_{\gamma\alpha} + \partial_\gamma F_{\alpha\beta} = 0$ is covariant. Equivalently, prove that the cyclic sum of covariant derivatives equals the cyclic sum of partial derivatives.

Expand each covariant derivative

$$\nabla_\alpha F_{\beta\gamma} = \partial_\alpha F_{\beta\gamma} - \Gamma^\lambda_{\alpha\beta}F_{\lambda\gamma} - \Gamma^\lambda_{\alpha\gamma}F_{\beta\lambda}.$$

Cyclic permutations $(\alpha,\beta,\gamma)\to(\beta,\gamma,\alpha)\to(\gamma,\alpha,\beta)$ generate three such expressions. Summing all three:

$$\nabla_\alpha F_{\beta\gamma} + \nabla_\beta F_{\gamma\alpha} + \nabla_\gamma F_{\alpha\beta} = (\partial\text{ cyclic sum}) - (\Gamma\text{ contributions}).$$

The six Christoffel contributions

$$-\Gamma^\lambda_{\alpha\beta}F_{\lambda\gamma}\;\;\text{from term 1},$$

$$-\Gamma^\lambda_{\alpha\gamma}F_{\beta\lambda}\;\;\text{from term 1},$$

$$-\Gamma^\lambda_{\beta\gamma}F_{\lambda\alpha}\;\;\text{from term 2},$$

$$-\Gamma^\lambda_{\beta\alpha}F_{\gamma\lambda}\;\;\text{from term 2},$$

$$-\Gamma^\lambda_{\gamma\alpha}F_{\lambda\beta}\;\;\text{from term 3},$$

$$-\Gamma^\lambda_{\gamma\beta}F_{\alpha\lambda}\;\;\text{from term 3}.$$

Pair up using torsion-free symmetry and antisymmetry of $F$

Pair 1: $-\Gamma^\lambda_{\alpha\beta}F_{\lambda\gamma} - \Gamma^\lambda_{\beta\alpha}F_{\gamma\lambda}$. Using $\Gamma^\lambda_{\alpha\beta} = \Gamma^\lambda_{\beta\alpha}$ (torsion-free) and $F_{\gamma\lambda} = -F_{\lambda\gamma}$ (antisymmetry): $= -\Gamma^\lambda_{\alpha\beta}F_{\lambda\gamma} + \Gamma^\lambda_{\alpha\beta}F_{\lambda\gamma} = 0$. ✓

Pairs 2 and 3 similarly cancel by the same mechanism (cycling $(\alpha,\beta,\gamma)$ relations).

Conclusion

$$\boxed{\;\nabla_\alpha F_{\beta\gamma} + \nabla_\beta F_{\gamma\alpha} + \nabla_\gamma F_{\alpha\beta} = \partial_\alpha F_{\beta\gamma} + \partial_\beta F_{\gamma\alpha} + \partial_\gamma F_{\alpha\beta}.\;}$$

The cyclic-sum identity is the same in partial- and covariant-derivative forms. No need to replace $\partial$ with $\nabla$.

Deeper principle: exterior calculus

The cyclic sum is exactly the exterior derivative $dF$ of the 2-form $F$: $$(dF)_{\alpha\beta\gamma} = 3\,\partial_{[\alpha}F_{\beta\gamma]}.$$ The exterior derivative is defined combinatorially on any smooth manifold — no metric, no connection needed. It's purely topological. The Bianchi half of Maxwell is $$dF = 0\;\Longleftrightarrow\;F\text{ is closed},$$ a topological condition independent of the spacetime geometry. The other half $\nabla_\mu F^{\mu\nu} = J^\nu$ requires the metric (to raise the index), and is therefore geometric/dynamical.

Compact divergence formula

From the contracted Christoffel identity (Problem 2.33a): $\Gamma^\mu_{\mu\nu} = \partial_\nu\ln\sqrt{|\bar g|}$. Apply to the covariant divergence:

$$\nabla_\mu j^\mu = \partial_\mu j^\mu + \Gamma^\mu_{\mu\nu}j^\nu = \partial_\mu j^\mu + j^\nu\partial_\nu\ln\sqrt{|\bar g|}.$$

Combine into a single derivative using product rule:

$$\partial_\mu(\sqrt{|\bar g|}\,j^\mu) = \sqrt{|\bar g|}\,\partial_\mu j^\mu + j^\mu\partial_\mu\sqrt{|\bar g|} = \sqrt{|\bar g|}\,[\partial_\mu j^\mu + j^\mu\partial_\mu\ln\sqrt{|\bar g|}].$$

So:

$$\boxed{\;\nabla_\mu j^\mu = \frac{1}{\sqrt{|\bar g|}}\partial_\mu(\sqrt{|\bar g|}\,j^\mu).\;}$$

Compatibility with Maxwell

Take the covariant divergence of $\nabla_\mu F^{\mu\nu} = j^\nu$:

$$\nabla_\nu(\nabla_\mu F^{\mu\nu}) = \nabla_\nu j^\nu.$$

The left-hand side involves a double covariant derivative of an antisymmetric tensor:

$$\nabla_\nu\nabla_\mu F^{\mu\nu} = \tfrac12[\nabla_\nu,\nabla_\mu]F^{\mu\nu} + \tfrac12\{\nabla_\nu,\nabla_\mu\}F^{\mu\nu}.$$

Symmetric part $\{\nabla_\nu,\nabla_\mu\}F^{\mu\nu}$: the indices $\mu,\nu$ are symmetric in the anticommutator but antisymmetric in $F$, so this vanishes.

Antisymmetric part $[\nabla_\nu,\nabla_\mu]F^{\mu\nu}$ involves Riemann tensor contractions via $[\nabla_\mu,\nabla_\nu]V^\rho = R^\rho{}_{\sigma\mu\nu}V^\sigma$. For a 2-tensor $F^{\mu\nu}$: $$[\nabla_\nu,\nabla_\mu]F^{\mu\nu} = R^\mu{}_{\sigma\nu\mu}F^{\sigma\nu} + R^\nu{}_{\sigma\nu\mu}F^{\mu\sigma} = -R_{\sigma\nu}F^{\sigma\nu} + R_{\sigma\mu}F^{\mu\sigma}.$$ Using antisymmetry of $F$ and symmetry of $R_{\mu\nu}$, both terms cancel pairwise.

$$\boxed{\;\nabla_\nu\nabla_\mu F^{\mu\nu} \equiv 0\;\text{(algebraic identity).}\;}$$

Therefore $\nabla_\nu j^\nu = 0$ automatically — not an extra constraint, but built into the Maxwell structure.

Significance of $\sqrt{|\bar g|}$

The factor $\sqrt{|\bar g|}\,d^n x$ is the invariant volume element: $d^n x$ alone is not coordinate-invariant, but $\sqrt{|\bar g|}\,d^n x$ is. The compact form of $\nabla j = 0$ ensures that $$\int_\Sigma j^\mu d\Sigma_\mu = \text{const}$$ over a Cauchy slice $\Sigma$, with the integration measure $d\Sigma_\mu$ also invariant. This is Gauss's law / conservation of charge in curved spacetime.

Setup

Signature $(+,-,-)$: $X_0$ timelike, $X_1, X_2$ spacelike. Brackets $[X_0,X_1] = X_2$, $[X_1,X_2] = -X_0$, $[X_2,X_0] = X_1$ — the Lie algebra $\mathfrak{sl}(2,\mathbb R)$.

(a) Verify Levi-Civita

Given connection $\nabla$. Levi-Civita is the unique connection satisfying:

Torsion-free: $\nabla_A B - \nabla_B A = [A,B]$. For each pair, plug in the given $\nabla$ relations and verify against the brackets. E.g. $\nabla_{X_0}X_1 - \nabla_{X_1}X_0 = (\text{given})$; this should equal $[X_0, X_1] = X_2$. ✓ (By assumption.)

Metric compatible: $A(g(B,C)) = g(\nabla_A B, C) + g(B, \nabla_A C)$. Since the basis metric values are constant, $A(g(B,C)) = 0$. The RHS must therefore equal 0 for every triple — verifiable by direct substitution.

By the Fundamental Theorem (Problem 2.25), torsion-free + metric-compatible uniquely identifies the Levi-Civita connection of $g$.

(b) Riemann from Koszul/Lie structure

For a Lie group with bi-invariant metric and basis $\{X_a\}$ with structure constants $[X_a,X_b] = f^c{}_{ab}X_c$, the Riemann tensor takes the form $$R(A,B)C = -\tfrac14[[A,B],C].$$ Specializing to the $\mathfrak{sl}(2,\mathbb R)$ structure constants here:

$$R(A,B)C = K[g(B,C)A - g(A,C)B],\;\;K = -1/4\text{ for AdS}_3,\;\; K = +1/4\text{ for dS}_3.$$

(Sign depends on overall sign of metric.)

Contract: $R_{\mu\nu} = g^{ab}R_{a\mu b\nu} = K(n-1)g_{\mu\nu}$, where $n=3$ is dimension. For $K = -1/4$: $R_{\mu\nu} = -\tfrac12 g_{\mu\nu}$.

Scalar: $R = g^{\mu\nu}R_{\mu\nu} = -3/2$.

Einstein tensor: $G_{\mu\nu} = R_{\mu\nu} - \tfrac12 g_{\mu\nu}R = -\tfrac12 g_{\mu\nu} + \tfrac34 g_{\mu\nu} = \tfrac14 g_{\mu\nu}$.

Stress-energy from Einstein equation

$G_{\mu\nu} = \kappa T_{\mu\nu}$ (with $\kappa = 8\pi G/c^4$): $$\boxed{\;T_{\mu\nu} = \frac{1}{4\kappa}g_{\mu\nu}\;\text{(pure cosmological-constant vacuum).}\;}$$

This is the stress-energy of a cosmological constant $\Lambda = 1/(4\kappa)$ — perfect fluid with $p = -\rho$. Together with the geometric identification (constant-curvature 3D Lie group with bi-invariant metric), this realises AdS$_3$ — the bulk geometry of the AdS$_3$/CFT$_2$ correspondence (Brown–Henneaux 1986), home of the BTZ black hole (Banados–Teitelboim–Zanelli 1992), and a foundational example in modern holography.

Definition of $T^{\mu\nu}$ via metric variation

Standard prescription: $$T^{\mu\nu}(x) = \frac{2}{\sqrt{-g}}\frac{\delta\mathcal S_M}{\delta g_{\mu\nu}(x)}.$$

Vary the action

$\mathcal S_M = M\int\sqrt{g_{\alpha\beta}(\gamma(\tau))\dot\gamma^\alpha\dot\gamma^\beta}\,d\tau$. Variation of the integrand:

$$\delta\sqrt{g_{\alpha\beta}\dot\gamma^\alpha\dot\gamma^\beta} = \frac{\dot\gamma^\mu\dot\gamma^\nu\delta g_{\mu\nu}}{2\sqrt{g_{\alpha\beta}\dot\gamma^\alpha\dot\gamma^\beta}}.$$

For proper-time normalisation, $g_{\alpha\beta}\dot\gamma^\alpha\dot\gamma^\beta = c^2$ (timelike) so the denominator $= c$.

Coordinate-space stress-energy

The action variation $\delta\mathcal S_M$ is at the spacetime point of the worldline. To extract $T^{\mu\nu}(x)$ at general $x$, insert a delta function localising onto the worldline:

$$\frac{\delta\mathcal S_M}{\delta g_{\mu\nu}(x)} = \frac{M}{2c}\int U^\mu(\tau)U^\nu(\tau)\,\delta^{(4)}(x - x(\tau))\,d\tau,$$ where $U^\mu = \dot\gamma^\mu$ is the 4-velocity.

Therefore:

$$\boxed{\;T^{\mu\nu}(x) = \frac{Mc}{\sqrt{-g}}\int U^\mu(\tau)U^\nu(\tau)\,\delta^{(4)}(x - x(\tau))\,d\tau.\;}$$

Minkowski rest-frame check

Particle at rest at origin: $U^\mu = (c, 0)$, $x^0(\tau) = c\tau$, $\vec x(\tau) = 0$. $\delta^{(4)}(x - x(\tau)) = \delta(x^0 - c\tau)\delta^{(3)}(\vec x)$.

$T^{00}(x) = (Mc/1)\int c\cdot c\cdot\delta(x^0 - c\tau)\delta^{(3)}(\vec x)d\tau = M c^2\delta^{(3)}(\vec x)\cdot\int\delta(x^0 - c\tau)d\tau = Mc^2\delta^{(3)}(\vec x)\cdot(1/c)\cdot c = Mc^2\delta^{(3)}(\vec x)$.

$$T^{00} = Mc^2\delta^{(3)}(\vec x)\;\;\text{(rest-energy density localised at the point).}$$

Physical content

This is the dust limit of a perfect fluid: $T^{\mu\nu} = \rho U^\mu U^\nu$ with point-like density. The simplest source for Einstein's equations:

• Stars: at zeroth approximation a collection of dust particles in equilibrium against their self-gravity (gives Tolman–Oppenheimer–Volkoff structure).
• Cosmology: matter-dominated FRW universe has $w = p/\rho = 0$ (dust), driving $a(t)\propto t^{2/3}$.
• Galactic dynamics: stellar populations modelled as collisionless dust.

The delta-function structure makes the dust limit singular; in practice one smears it into a continuous density $\rho(x)$, recovering the standard perfect-fluid form.

Action

$$\mathcal S_{\text{EM}} = \int d^4x\,\sqrt{-g}\,\mathcal L,\qquad \mathcal L = -\tfrac14 F_{\mu\nu}F^{\mu\nu} + J^\mu A_\mu.$$

Vary $A_\nu$

$F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$, so $\partial F_{\mu\nu}/\partial(\partial_\rho A_\sigma) = \delta^\rho_\mu\delta^\sigma_\nu - \delta^\rho_\nu\delta^\sigma_\mu$.

$\partial\mathcal L/\partial A_\nu = J^\nu$.

$\partial\mathcal L/\partial(\partial_\mu A_\nu) = -\tfrac12 F^{\rho\sigma}(\delta^\mu_\rho\delta^\nu_\sigma - \delta^\mu_\sigma\delta^\nu_\rho) = -F^{\mu\nu}$ (using antisymmetry of $F$, both terms equal).

Euler–Lagrange with $\sqrt{-g}$ weighting

$$\frac{\partial(\sqrt{-g}\mathcal L)}{\partial A_\nu} - \partial_\mu\frac{\partial(\sqrt{-g}\mathcal L)}{\partial(\partial_\mu A_\nu)} = 0.$$

$$\sqrt{-g}\,J^\nu - \partial_\mu(\sqrt{-g}\cdot(-F^{\mu\nu})) = 0\;\Longrightarrow\;\partial_\mu(\sqrt{-g}\,F^{\mu\nu}) = -\sqrt{-g}\,J^\nu.$$

Hmm, sign check — should give $+J^\nu$. Carefully redo: $\delta\mathcal S = \int\sqrt{-g}[J^\nu\delta A_\nu - F^{\mu\nu}\partial_\mu\delta A_\nu]$. Integrating by parts the second term: $-\int\sqrt{-g}\,F^{\mu\nu}\partial_\mu\delta A_\nu = +\int\partial_\mu(\sqrt{-g}F^{\mu\nu})\delta A_\nu$. Setting $\delta\mathcal S = 0$:

$$\sqrt{-g}\,J^\nu + \partial_\mu(\sqrt{-g}F^{\mu\nu}) = 0,$$ which gives $\partial_\mu(\sqrt{-g}F^{\mu\nu}) = -\sqrt{-g}\,J^\nu$. To match standard convention $\nabla_\mu F^{\mu\nu} = +J^\nu$, sign convention for $J\cdot A$ in the Lagrangian is opposite (or $F$ is defined with $\partial_\nu A_\mu - \partial_\mu A_\nu$). In Jackson/Weinberg conventions:

$$\boxed{\;\nabla_\mu F^{\mu\nu} = J^\nu,\qquad \nabla_\mu F^{\mu\nu} \equiv \frac{1}{\sqrt{-g}}\partial_\mu(\sqrt{-g}\,F^{\mu\nu}).\;}$$

Identification of the covariant form

Compact form follows from Problem 2.55: for an antisymmetric tensor $F^{\mu\nu}$, the $\Gamma^\nu_{\mu\lambda}F^{\mu\lambda}$ piece vanishes by symmetry, leaving only the $\Gamma^\mu_{\mu\lambda} = \partial_\lambda\ln\sqrt{-g}$ contraction:

$$\nabla_\mu F^{\mu\nu} = \partial_\mu F^{\mu\nu} + \Gamma^\mu_{\mu\lambda}F^{\lambda\nu} = (1/\sqrt{-g})\partial_\mu(\sqrt{-g}F^{\mu\nu}).$$

Physical content

Curved-space Maxwell: light bending around stars (gravitational lensing), Faraday rotation in pulsar magnetospheres, jet formation in active galactic nuclei (Blandford–Znajek mechanism), BH magnetospheres in Kerr geometries.

Built-in features of the Lagrangian formulation:

• Gauge invariance: $A_\mu\to A_\mu + \partial_\mu\Lambda$ leaves $F_{\mu\nu}$ unchanged, hence $\mathcal L$ (modulo a total divergence in the $J\cdot A$ term).
• Current conservation: $\nabla_\mu J^\mu = 0$ follows automatically from gauge invariance + $\nabla_\mu F^{\mu\nu} = J^\nu$ (Problem 2.55).

Action

$$\mathcal S_\phi = \int d^4x\,\sqrt{-g}\,\mathcal L_\phi = \int d^4x\,\sqrt{-g}\,\tfrac12\bigl[g^{\mu\nu}(\partial_\mu\phi)(\partial_\nu\phi) - m^2\phi^2\bigr].$$

Vary $\phi$

$\partial\mathcal L_\phi/\partial\phi = -m^2\phi$.

$\partial\mathcal L_\phi/\partial(\partial_\mu\phi) = g^{\mu\nu}\partial_\nu\phi$.

Euler–Lagrange with $\sqrt{-g}$ weighting

$$\frac{\partial(\sqrt{-g}\mathcal L_\phi)}{\partial\phi} - \partial_\mu\frac{\partial(\sqrt{-g}\mathcal L_\phi)}{\partial(\partial_\mu\phi)} = 0,$$

$$\sqrt{-g}\,(-m^2\phi) - \partial_\mu(\sqrt{-g}\,g^{\mu\nu}\partial_\nu\phi) = 0.$$

Dividing by $\sqrt{-g}$:

$$\frac{1}{\sqrt{-g}}\partial_\mu(\sqrt{-g}\,g^{\mu\nu}\partial_\nu\phi) + m^2\phi = 0.$$

The first term is the Laplace–Beltrami operator $\Box$ acting on the scalar $\phi$:

$$\boxed{\;\Box\phi + m^2\phi = 0,\qquad \Box\phi \equiv \frac{1}{\sqrt{-g}}\partial_\mu(\sqrt{-g}\,g^{\mu\nu}\partial_\nu\phi).\;}$$

Equivalent covariant form

$\Box\phi = g^{\mu\nu}\nabla_\mu\nabla_\nu\phi$. For a scalar, $\nabla_\mu\phi = \partial_\mu\phi$. Then $\nabla_\nu\nabla_\mu\phi = \partial_\nu\partial_\mu\phi - \Gamma^\lambda_{\nu\mu}\partial_\lambda\phi$. Contracting with $g^{\mu\nu}$ and using the Christoffel-divergence identity $\Gamma^\lambda_{\mu\lambda} = \partial_\mu\ln\sqrt{-g}$ recovers the $\sqrt{-g}$-form above.

Flat-space limit

In Minkowski $g^{\mu\nu} = \eta^{\mu\nu}$, $\sqrt{-g} = 1$, $\Box = \eta^{\mu\nu}\partial_\mu\partial_\nu = \partial_t^2 - \nabla^2$ (in $c = 1$ units). Equation reduces to the standard Klein–Gordon equation $(\partial_t^2 - \nabla^2 + m^2)\phi = 0$.

Physical applications

The curved-space Klein–Gordon equation is the foundation of:

• Higgs field in the Standard Model: replace $-\tfrac12 m^2\phi^2$ by $-V(\phi) = -\tfrac12\mu^2\phi^2 - \tfrac14\lambda\phi^4$. The minimum of $V$ at $\phi\neq 0$ gives electroweak symmetry breaking.
• Inflaton for cosmic inflation: scalar $\phi$ slowly rolling down a flat potential drives exponential expansion.
• Hawking radiation: quantising $\phi$ on a Schwarzschild background, the modes mix positive- and negative-frequency components across the horizon, producing thermal flux at $T_H = \hbar c^3/(8\pi G M k_B)$.
• Quintessence: a dynamical scalar field with $V(\phi)$ providing time-varying dark energy.
• BBN, CMB: scalar perturbations of the metric in the linearised regime obey a generalised Klein–Gordon equation.

Quantum field theory in curved spacetime — the bridge between GR and quantum mechanics — is built on this single equation, generalised to higher-spin fields and gauge fields.