The Geodesic Equation & Parallel Transport
The Geodesic Equation & Parallel Transport
In curved spacetime, freely falling particles follow geodesics — curves that parallel-transport their own tangent vectors. The geodesic equation replaces Newton's second law in GR. Parallel transport around a closed loop reveals spacetime curvature via holonomy.
Geodesic Equation
A geodesic satisfies: \(\frac{d^2 x^\mu}{d\lambda^2} + \Gamma^\mu_{\nu\sigma}\frac{dx^\nu}{d\lambda}\frac{dx^\sigma}{d\lambda} = 0\). For timelike geodesics, \(\lambda = \tau\) (proper time). Parallel transport of \(V^\mu\) along \(x^\mu(\lambda)\): \(\frac{DV^\mu}{d\lambda} = \dot x^\nu \nabla_\nu V^\mu = 0\). A geodesic parallel-transports its own tangent: \(\frac{Du^\mu}{d\lambda}=0\).
Variational Derivation
Varying \(S = \int g_{\mu\nu}\dot x^\mu \dot x^\nu\,d\lambda\) gives the Euler-Lagrange equations. Since \(g_{\mu\nu}\) depends on \(x^\alpha\), the \(\partial_\alpha g_{\mu\nu}\) terms produce the Christoffel symbols automatically: \(\ddot x^\mu + \Gamma^\mu_{\nu\sigma}\dot x^\nu \dot x^\sigma = 0\).
Example 1: Geodesics on \(S^2\)
Find geodesics on a unit sphere using \(\Gamma^\theta_{\phi\phi}=-\sin\theta\cos\theta\), \(\Gamma^\phi_{\theta\phi}=\cot\theta\).
Solution: The geodesic equations yield \(\ddot\theta - \sin\theta\cos\theta\,\dot\phi^2 = 0\) and \(\ddot\phi + 2\cot\theta\,\dot\theta\dot\phi = 0\). Solutions are great circles, confirming geodesics are shortest paths on \(S^2\).
Example 2: Parallel Transport Holonomy
Parallel-transport a vector around latitude circle \(\theta = \theta_0\) on \(S^2\). Find the rotation angle.
Solution: The holonomy angle is \(\Delta\phi_{\text{rot}} = 2\pi\cos\theta_0\). At the equator (\(\theta_0=\pi/2\)) there is no rotation; approaching the pole the vector rotates by \(2\pi\).
Practice
- Derive the geodesic equation from \(\delta\int g_{\mu\nu}\dot x^\mu \dot x^\nu d\lambda = 0\).
- Show that along a geodesic, \(g_{\mu\nu}\dot x^\mu \dot x^\nu\) is constant.
- Find geodesics in 2D flat polar coordinates \((r,\phi)\) and verify they are straight lines.
- Explain the geometric meaning of holonomy and its relation to enclosed area on \(S^2\).
Show Answer Key
1. Action: $S = \int g_{\mu\nu}\dot{x}^\mu\dot{x}^\nu d\lambda$. Euler-Lagrange: $\frac{d}{d\lambda}(2g_{\mu\nu}\dot{x}^\nu) - \partial_\mu g_{\alpha\beta}\dot{x}^\alpha\dot{x}^\beta = 0$. Expanding $\frac{d}{d\lambda}(g_{\mu\nu}\dot{x}^\nu) = g_{\mu\nu}\ddot{x}^\nu + \partial_\sigma g_{\mu\nu}\dot{x}^\sigma\dot{x}^\nu$ and rearranging with the Christoffel symbol definition: $\ddot{x}^\mu + \Gamma^\mu_{\alpha\beta}\dot{x}^\alpha\dot{x}^\beta = 0$.
2. $\frac{d}{d\lambda}(g_{\mu\nu}\dot{x}^\mu\dot{x}^\nu) = 2g_{\mu\nu}\dot{x}^\mu\ddot{x}^\nu + \partial_\sigma g_{\mu\nu}\dot{x}^\sigma\dot{x}^\mu\dot{x}^\nu$. Using the geodesic equation $\ddot{x}^\nu = -\Gamma^\nu_{\alpha\beta}\dot{x}^\alpha\dot{x}^\beta$ and the definition of $\Gamma$: the expression equals $2(-g_{\mu\nu}\Gamma^\nu_{\alpha\beta}+\frac{1}{2}\partial_\alpha g_{\mu\beta}+\frac{1}{2}\partial_\beta g_{\mu\alpha}-\frac{1}{2}\partial_\mu g_{\alpha\beta})\dot{x}^\mu\dot{x}^\alpha\dot{x}^\beta$... which vanishes identically by the Christoffel definition. So $g_{\mu\nu}\dot{x}^\mu\dot{x}^\nu = \text{const}$.
3. In 2D polar $(r,\phi)$: $ds^2 = dr^2+r^2d\phi^2$. Geodesic equations: $\ddot{r}-r\dot{\phi}^2=0$ and $\ddot{\phi}+\frac{2}{r}\dot{r}\dot{\phi}=0$. The second gives $r^2\dot{\phi}=L$ (conserved angular momentum). Solving: these produce straight lines $r\sin(\phi-\phi_0)=d$ (verified by converting to Cartesian). ✓
4. Parallel-transport a vector around a closed loop on $S^2$. The vector rotates by an angle $\Delta\theta = \int_A R\,dA = A/R^2$ (Gaussian curvature $K=1/R^2$ times enclosed area $A$). Holonomy is the net rotation: it detects curvature. For a triangle with angles summing to $\pi + \epsilon$: holonomy angle $= \epsilon = A/R^2$ (angular excess equals solid angle).