Lorentzian geometry

In Euclidean geometry the straight line is the shortest path because moving sideways adds $+dx^2$ to the length; in Minkowski spacetime the spatial displacement enters with the opposite sign from $-c^2dt^2$, so any spatial detour subtracts from $-ds^2$ and therefore shrinks the integral of $\sqrt{-ds^2}/c$. That is the entire resolution of the twin paradox: the twin who stays in an inertial frame follows the straight worldline in spacetime and ages the most; the twin who flies away and turns back takes a bent worldline and ages less.

**Step 1 (adapt the inertial frame to $\gamma_0$).** Because $\gamma_0$ is a straight timelike line, we can choose an inertial coordinate system $(t,x,y,z)$ in which $\gamma_0$ sits at rest at the spatial origin: $p=(0,\mathbf 0)$, $q=(T,\mathbf 0)$ with $T>0$, and $\gamma_0(t)=(t,\mathbf 0)$ for $t\in[0,T]$. Since the metric $ds^2 = -c^2 dt^2 + dx^2 + dy^2 + dz^2$ is invariant under Lorentz transformations, the proper time of any curve is the same in every inertial frame.

**Step 2 (compute $\tau(\gamma_0)$).** Along $\gamma_0$ the spatial velocity is $\mathbf v(t)=\mathbf 0$, so $$\tau(\gamma_0) = \int_0^T \sqrt{1 - 0/c^2}\,dt = T.$$

**Step 3 (parameterize the competing curve $\gamma$).** Any future-directed timelike curve $\gamma$ from $p$ to $q$ has $dt/d\lambda > 0$ everywhere (since $c^2 dt^2 > |d\mathbf x|^2 \ge 0$), so we may parameterize it by the coordinate time $t\in[0,T]$ as $\gamma(t)=(t,\mathbf x(t))$ with $\mathbf x(0)=\mathbf x(T)=\mathbf 0$ and velocity $\mathbf v(t)=d\mathbf x/dt$ satisfying $|\mathbf v(t)|<c$. Its elapsed proper time is $\tau = \int \sqrt{1 - v(t)^2/c^2}\,dt$ over $[0,T]$.

Step 4 (pointwise bound and equality case). For every $t\in[0,T]$, $|\mathbf v(t)|^2\ge0$ implies $0 < \sqrt{1 - |\mathbf v(t)|^2/c^2} \le 1$, with equality at a given $t$ if and only if $\mathbf v(t)=\mathbf 0$. Integrating over $[0,T]$, $$\tau(\gamma) = \int_0^T \sqrt{1 - \frac{|\mathbf v(t)|^2}{c^2}}\,dt \le \int_0^T 1\,dt = T = \tau(\gamma_0).$$ Because the integrand is continuous and $\le 1$, equality $\tau(\gamma)=T$ holds if and only if $\mathbf v(t)=\mathbf 0$ for all $t\in[0,T]$, which together with $\mathbf x(0)=\mathbf 0$ forces $\mathbf x(t)\equiv\mathbf 0$, i.e. $\gamma=\gamma_0$.

Just as Euler–Lagrange turns stationary action into a differential equation for a particle path, varying the scalar curvature integrated over spacetime turns stationary action into a partial differential equation for the metric itself. The $-\tfrac12 Rg_{\mu\nu}$ term comes specifically from varying the volume factor $\sqrt{-g}$, and it is the exact term needed to make the curvature side divergence-free via the contracted Bianchi identity — matching the conservation of energy and momentum on the matter side.

Step 1 (decompose the variation of the gravitational integrand). Write $R=g^{\mu\nu}R_{\mu\nu}$. By the product rule, $$\delta\big((R-2\Lambda)\sqrt{-g}\big) = R_{\mu\nu}\,\sqrt{-g}\,\delta g^{\mu\nu} + g^{\mu\nu}(\delta R_{\mu\nu})\sqrt{-g} + (R-2\Lambda)\,\delta\sqrt{-g}.$$

**Step 2 (Jacobi's formula for $\delta\sqrt{-g}$ and the Palatini identity).** Using Jacobi's identity $\delta(\det g)=(\det g)\,g^{\mu\nu}\delta g_{\mu\nu}=-(\det g)\,g_{\mu\nu}\delta g^{\mu\nu}$, we get $\delta\sqrt{-g}=-\tfrac12\sqrt{-g}\,g_{\mu\nu}\delta g^{\mu\nu}$. Meanwhile, the Palatini identity expresses the Ricci variation as a covariant divergence, $g^{\mu\nu}\delta R_{\mu\nu}=\nabla_\alpha\big(g^{\mu\nu}\delta\Gamma^\alpha_{\mu\nu}-g^{\alpha\nu}\delta\Gamma^\mu_{\mu\nu}\big)$, which integrates to zero on $M$ by Stokes's theorem for variations $\delta g^{\mu\nu}$ supported away from the boundary.

**Step 3 (assemble $\delta S=0$).** Combining Steps 1 and 2 and the definition $T_{\mu\nu}\equiv-\dfrac{2}{\sqrt{-g}}\dfrac{\delta S_{\mathrm{matter}}}{\delta g^{\mu\nu}}$, $$\delta S = \int_M \left[\frac{c^4}{16\pi G}\left(R_{\mu\nu}-\tfrac12 Rg_{\mu\nu}+\Lambda g_{\mu\nu}\right) - \frac12 T_{\mu\nu}\right]\delta g^{\mu\nu}\,\sqrt{-g}\,d^4x = 0.$$ Since $\delta g^{\mu\nu}$ is an arbitrary symmetric tensor variation, the bracket must vanish pointwise, giving $R_{\mu\nu} - \tfrac{1}{2}R g_{\mu\nu} + \Lambda g_{\mu\nu} = \dfrac{8\pi G}{c^4} T_{\mu\nu}$.

Step 4 (contracted Bianchi identity and conservation). The second Bianchi identity $\nabla_\lambda R_{\rho\sigma\mu\nu}+\nabla_\rho R_{\sigma\lambda\mu\nu}+\nabla_\sigma R_{\lambda\rho\mu\nu}=0$, contracted twice with $g^{\lambda\mu}g^{\rho\nu}$, yields $\nabla^\mu R_{\mu\nu}-\tfrac12\nabla_\nu R=0$, i.e. $\nabla^\mu G_{\mu\nu} = 0$ for $G_{\mu\nu}=R_{\mu\nu}-\tfrac12 Rg_{\mu\nu}$. Because $\nabla^\mu g_{\mu\nu}=0$ (metric compatibility), taking $\nabla^\mu$ of $R_{\mu\nu} - \tfrac{1}{2}R g_{\mu\nu} + \Lambda g_{\mu\nu} = \dfrac{8\pi G}{c^4} T_{\mu\nu}$ forces $\nabla^\mu T_{\mu\nu}=0$.