Moore General Relativity Workbook Solutions Apr 2026

$$\frac{d^2t}{d\lambda^2} = 0, \quad \frac{d^2x^i}{d\lambda^2} = 0$$

Derive the equation of motion for a radial geodesic.

$$\frac{d^2r}{d\lambda^2} = -\frac{GM}{r^2} \left(1 - \frac{2GM}{r}\right) \left(\frac{dt}{d\lambda}\right)^2 + \frac{GM}{r^2} \left(1 - \frac{2GM}{r}\right)^{-1} \left(\frac{dr}{d\lambda}\right)^2$$

$$\frac{d^2x^\mu}{d\lambda^2} + \Gamma^\mu_{\alpha\beta} \frac{dx^\alpha}{d\lambda} \frac{dx^\beta}{d\lambda} = 0$$ moore general relativity workbook solutions

where $\lambda$ is a parameter along the geodesic, and $\Gamma^\mu_{\alpha\beta}$ are the Christoffel symbols.

This factor describes the difference in time measured by the two clocks.

where $\eta^{im}$ is the Minkowski metric. $$\frac{d^2t}{d\lambda^2} = 0

$$\frac{t_{\text{proper}}}{t_{\text{coordinate}}} = \sqrt{1 - \frac{2GM}{r}}$$

The gravitational time dilation factor is given by

The geodesic equation is given by

$$\Gamma^0_{00} = 0, \quad \Gamma^i_{00} = 0, \quad \Gamma^i_{jk} = \eta^{im} \partial_m g_{jk}$$

After some calculations, we find that the geodesic equation becomes