Chapter 6: Geodesics
Geodesics are the straightest possible paths in curved spacetime. Free-falling particles follow timelike geodesics; light rays follow null geodesics. This is the essence of Einstein's equivalence principle: gravity is geometry.
The Geodesic Equation
A geodesic is a curve whose tangent vector is parallel transported along itself:
\(\frac{d^2x^\mu}{d\lambda^2} + \Gamma^\mu_{\nu\rho} \frac{dx^\nu}{d\lambda} \frac{dx^\rho}{d\lambda} = 0\)
Second-order ODE for the worldline xΞΌ(Ξ»)
Timelike Geodesics
ds^2 < 0: massive particles, Ξ» = proper time Ο
Null Geodesics
ds^2 = 0: light rays, Ξ» is affine parameter
Spacelike Geodesics
ds^2 > 0: tachyons (hypothetical), Ξ» = proper length
Extremal Property
Geodesics extremize the action \(S = \int ds\)
Constants of Motion
Killing vectors generate symmetries that yield conserved quantities along geodesics:
Energy Conservation
Time translation symmetry βt gives \(E = -g_{t\mu} \frac{dx^\mu}{d\tau} = \left(1 - \frac{2M}{r}\right)\frac{dt}{d\tau}\)
Angular Momentum Conservation
Axial symmetry βΟ gives \(L = g_{\phi\mu} \frac{dx^\mu}{d\tau} = r^2 \frac{d\phi}{d\tau}\)
Normalization
\(g_{\mu\nu} \frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau} = -1\) for timelike geodesics
Effective Potential Analysis
Using conserved quantities, radial motion reduces to a 1D problem with effective potential:
\(\frac{1}{2}\left(\frac{dr}{d\tau}\right)^2 + V_{eff}(r) = \frac{1}{2}(E^2 - 1)\)
\(where V_{eff} = -\frac{M}{r} + \frac{L^2}{2r^2} - \frac{ML^2}{r^3}\)
The last term (-MLΒ²/rΒ³) is the relativistic correction, responsible for:
- Perihelion precession of Mercury
- Unstable circular orbits at r = 3M (photon sphere)
- Innermost stable circular orbit (ISCO) at r = 6M
Python: Geodesic Integration in Schwarzschild
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Fortran: Null Geodesics (Light Rays)
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Practice Problems
Problem 1:Write the geodesic equation for the radial coordinate $r$ in the Schwarzschild metric and identify the effective potential for massive particles.
Solution:
Step 1: The Schwarzschild metric gives the Lagrangian $2\mathcal{L} = (1-2M/r)\dot{t}^2 - (1-2M/r)^{-1}\dot{r}^2 - r^2\dot{\phi}^2$ (equatorial plane $\theta = \pi/2$).
Step 2: Conserved quantities: energy $E = (1-2M/r)\dot{t}$ and angular momentum $L = r^2\dot{\phi}$.
Step 3: The normalization $g_{\mu\nu}\dot{x}^\mu\dot{x}^\nu = 1$ for massive particles gives:
$\dot{r}^2 = E^2 - \left(1 - \frac{2M}{r}\right)\left(1 + \frac{L^2}{r^2}\right)$
Step 4: This has the form $\dot{r}^2 + V_{\text{eff}}(r) = E^2$ with:
$V_{\text{eff}}(r) = \left(1 - \frac{2M}{r}\right)\left(1 + \frac{L^2}{r^2}\right) = 1 - \frac{2M}{r} + \frac{L^2}{r^2} - \frac{2ML^2}{r^3}$
Answer: The effective potential is $V_{\text{eff}}(r) = 1 - \frac{2M}{r} + \frac{L^2}{r^2} - \frac{2ML^2}{r^3}$. The last term $-2ML^2/r^3$ is the GR correction absent in Newtonian gravity.
Problem 2:Find the radius of circular orbits in the Schwarzschild geometry by solving $V_{\text{eff}}'(r) = 0$.
Solution:
Step 1: Differentiate $V_{\text{eff}}$: $V_{\text{eff}}'(r) = \frac{2M}{r^2} - \frac{2L^2}{r^3} + \frac{6ML^2}{r^4} = 0$.
Step 2: Multiply through by $r^4/(2)$: $Mr^2 - L^2 r + 3ML^2 = 0$.
Step 3: Solve the quadratic in $r$:
$r = \frac{L^2 \pm \sqrt{L^4 - 12M^2L^2}}{2M} = \frac{L^2}{2M}\left(1 \pm \sqrt{1 - 12M^2/L^2}\right)$
Step 4: Circular orbits exist only when $L^2 \geq 12M^2$, i.e., $L \geq 2\sqrt{3}\,M$.
Answer: $r_{\pm} = \frac{L^2}{2M}\left(1 \pm \sqrt{1 - 12M^2/L^2}\right)$. The $+$ root is stable, the $-$ root is unstable. They merge at $L = 2\sqrt{3}\,M$.
Problem 3:Derive the innermost stable circular orbit (ISCO) radius for the Schwarzschild black hole.
Solution:
Step 1: The ISCO occurs when both $V_{\text{eff}}'(r) = 0$ (circular orbit) and $V_{\text{eff}}''(r) = 0$ (marginally stable) hold simultaneously.
Step 2: From Problem 2, circular orbits satisfy $Mr^2 - L^2r + 3ML^2 = 0$, giving $L^2 = Mr^2/(r - 3M)$.
Step 3: Compute $V_{\text{eff}}''(r) = -\frac{4M}{r^3} + \frac{6L^2}{r^4} - \frac{24ML^2}{r^5} = 0$. Substitute $L^2$: $-4M + \frac{6Mr}{r-3M} - \frac{24M^2r}{(r-3M)r} = 0$.
Step 4: After simplification: $r^2 - 6Mr + 9M^2 - 9M^2 + ... $ reduces to $r(r - 6M) = 0$.
Answer: $r_{\text{ISCO}} = 6M = 3r_s$ where $r_s = 2M$ is the Schwarzschild radius. The corresponding angular momentum is $L_{\text{ISCO}} = 2\sqrt{3}\,M$ and orbital energy is $E_{\text{ISCO}} = \sqrt{8/9} \approx 0.943$.
Problem 4:Derive the deflection angle of light passing a mass $M$ at impact parameter $b$ in the weak-field limit.
Solution:
Step 1: For null geodesics ($ds^2 = 0$), the orbit equation using $u = 1/r$ is: $\frac{d^2u}{d\phi^2} + u = 3Mu^2$.
Step 2: The zeroth-order solution (straight line) is $u_0 = \sin\phi/b$.
Step 3: First-order correction: substitute into the RHS: $\frac{d^2u_1}{d\phi^2} + u_1 = \frac{3M}{b^2}\sin^2\phi = \frac{3M}{2b^2}(1 - \cos 2\phi)$.
Step 4: The particular solution gives a correction. The total deflection angle integrates to:
$\delta\phi = \frac{4GM}{c^2 b}$
Answer: $\delta\phi = \frac{4GM}{c^2 b}$. For the Sun ($M = M_\odot$, $b = R_\odot$): $\delta\phi = 1.75''$, famously confirmed in the 1919 eclipse expedition.
Problem 5:A photon is emitted radially outward from $r = 3M$ in Schwarzschild spacetime. Find the coordinate velocity $dr/dt$ and compare it to the speed of light.
Solution:
Step 1: For a radial null geodesic ($ds^2 = 0$, $d\theta = d\phi = 0$): $(1 - 2M/r)\,dt^2 = (1-2M/r)^{-1}dr^2$.
Step 2: Solve for coordinate velocity: $\frac{dr}{dt} = \pm(1 - 2M/r)$.
Step 3: At $r = 3M$: $\frac{dr}{dt} = 1 - \frac{2M}{3M} = 1 - \frac{2}{3} = \frac{1}{3}$ (in units where $c = 1$).
Step 4: Note that $dr/dt \to 0$ as $r \to 2M$ (event horizon), even though the local speed of light is always $c$.
Answer: $dr/dt = c/3$ in coordinate time. This is a coordinate effect; the photon's locally measured speed is always $c$. The coordinate velocity vanishes at the horizon due to the extreme gravitational time dilation.