Deriving the geodesic equation

Question

I having been reading a general relativity book, but when in comes to the geodesic equation, it is not derived. How does one go about doing this?

Answer 1

The geodesic equation can be derived by extremizing the length ("proper time" in the case of general relativity) of a path connecting two fixed points.

One requires that, after introducing a parameter $\lambda$ so that for the geodesic $x^{\mu} = x^{\mu}(\lambda)$ connecting points $A$ and $B$: $$\delta\int_{A}^{B}\sqrt{g_{\mu\nu}\frac{dx^{\mu}}{d\lambda}\frac{dx^{\nu}}{d\lambda}} d\lambda = 0$$ under the variation $x^{\mu}(\lambda) \rightarrow x^{\mu}(\lambda) + \delta{x^{\mu}}(\lambda)$. This is a straightforward problem in the calculus of variations; the square root can be taken as the Lagrangian: $$L(x, \frac{dx}{d\lambda}) = \sqrt{g_{\mu\nu} \frac{dx^{\mu}}{d\lambda} \frac{dx^{\nu}}{d\lambda}}$$.

The solution is given by the Euler-Lagrange equations: $$\frac{\partial L}{\partial x^{\mu}} - \frac{d}{d\lambda}\left(\frac{\partial L}{\partial \left( \frac{dx^{\mu}}{d\lambda}\right)}\right) = 0$$ Plugging in the above Lagrangian, and remembering that there is a position dependence in the metric, leads directly to the geodesic equation (for convenience, one normally prefers to choose the proper time itself as a parameter, and extremize the square of this Lagrangian instead).

Answer 2

Assume a path $X:\mathbb{R}\to\mathcal{M}$ in your manifold $\mathcal{M}$ parameterized by $s\in\mathbb{R}$ linking points $X(0),\,X(1)\in\mathcal{M}$. Write the length of this path as the functional $\mathcal{L}(X)=\int_0^1\sqrt{g(\mathrm{d}_s\,X,\,\mathrm{d}_s\,X)}\,\mathrm{d} s$. Now use the calculus of variations, i.e. apply Euler-Lagrange equation to this functional.

Alternatively, define the geodesic as a curve such that parallel transport along this curve preserves the tangent to the curve. Informally, grab a vector by its tail and shove it so that it moves always in the direction that it "wants" to go in (this is called finding the exponential map - see the Exponential Map (Riemannian Geometry) Wiki page). This definition tells us that the geodesic fulfills the equation $\nabla_\dot{X}\,\dot{X}=0$, where $\nabla$ is the covariant derivative or connection. If your connection is the Levi-Civita connexion (as it always is in GR) then this is exactly the same equation that you end up with if you work through the calculus of variations method.