After some calculations, we find that the geodesic equation becomes
$$\frac{d^2t}{d\lambda^2} = 0, \quad \frac{d^2x^i}{d\lambda^2} = 0$$
The geodesic equation is given by
$$\frac{d^2r}{d\lambda^2} = -\frac{GM}{r^2} + \frac{L^2}{r^3}$$ moore general relativity workbook solutions
The equation of motion for a radial geodesic can be derived from the geodesic equation. After some algebra, we find
where $\lambda$ is a parameter along the geodesic, and $\Gamma^\mu_{\alpha\beta}$ are the Christoffel symbols.
The gravitational time dilation factor is given by After some calculations, we find that the geodesic
Consider the Schwarzschild metric
Derive the geodesic equation for this metric.
Using the conservation of energy, we can simplify this equation to Using the conservation of energy, we can simplify
$$\Gamma^0_{00} = 0, \quad \Gamma^i_{00} = 0, \quad \Gamma^i_{jk} = \eta^{im} \partial_m g_{jk}$$
$$\frac{d^2x^\mu}{d\lambda^2} + \Gamma^\mu_{\alpha\beta} \frac{dx^\alpha}{d\lambda} \frac{dx^\beta}{d\lambda} = 0$$
Derive the equation of motion for a radial geodesic.
where $L$ is the conserved angular momentum.
$$\frac{t_{\text{proper}}}{t_{\text{coordinate}}} = \sqrt{1 - \frac{2GM}{r}}$$