Geodesic

Definition

If a tangent vector at a point of the curve is parallel transported along the curve and remains as its tangent vector, $$\nabla_U\vec{U}=0,$$ then the curve is called a geodesic. $$\begin{align} \nabla_U\vec{U} &= U^\beta U^\alpha_{;\beta} = 0\\ U^\beta U^\alpha_{,\beta} + \Gamma^\alpha_{\mu\beta} U^\mu U^\beta &= 0.\\ \end{align}$$ Let $\lambda$ be the parameter of the curve, then $U^\alpha=dx^\alpha/d\lambda$ and $U^\beta\partial/\partial x^\beta=d/d\lambda$. Substitute back, we have $$\bbox[5px, border: 2px solid #fff]{ \frac{d}{d\lambda}\left(\frac{dx^\alpha}{d\lambda}\right) + \Gamma^\alpha_{\mu\beta}\frac{dx^\mu}{d\lambda}\frac{dx^\beta}{d\lambda} = 0. }$$ This is known as the geodesic equation. For a parameter defined as $$\phi = a\lambda + b,$$ for some constant $a$ and $b$, the parameter $\phi$ also satisfies the geodesic equation. Parameters of the geodesic equation is called affine parameters.

Extremal Length

Distance of a curve is given by $$l = \int\sqrt{-g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}}d\lambda,$$ where $\lambda$ is a parameter of the curve. The extremal length is occured when $\delta l=0$, in which $L$ satisfies the Euler-Lagrange equation, where $L=\sqrt{-g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}}$. First prove that when $L^2$ satisfies the Euler-Lagrange equation, $L$ also satisfies the Euler-Lagrange equation. Then, the Euler-Lagrange equation will give the geodesic equation. This means that geodesic is the shortest distance between two points on a manifold. Thus, it is equivalent to the straight line in Euclidean space.

Euclidean Space

For flat Euclidean spacetime, the Christoffel symobl is zero, i.e. $\Gamma^\alpha_{\mu\beta}=0$. Then, $$\begin{align} \frac{d}{d\lambda}\left(\frac{dx^\alpha}{d\lambda}\right) &= 0\\ x^\alpha &= a\lambda + b, \end{align}$$ for some real constant $a$ and $b$. Thus, the geodesic in Euclidean flat spacetime is a straight line.

Lagrangian and Killing Vector

As the distance is defined as integration of the square root of the line element, and that distance is an extrema, the Lagrangian is given by $$\mathcal{L} = \left[-g_{\alpha\beta}\frac{dx^\alpha}{d\sigma}\frac{dx^\beta}{d\sigma}\right]^{1/2}$$ The Euler-Lagrange equation is $$\frac{d}{d\sigma}\left(\frac{\partial \mathcal{L}}{\partial dx^\mu/d\sigma}\right) = \frac{\partial \mathcal{L}}{\partial x^\mu}$$ When the Lagrangian is independent of $x^\mu$, according to the Euler-Lagrange equation, $$\frac{d}{d\sigma}\left[\frac{\partial \mathcal{L}}{\partial(dx^\mu/d\sigma)}\right] = 0$$ Let $\bf{\xi}$ be a unit vector in the direction of $x^\mu$. $$\begin{align} \frac{\partial \mathcal{L}}{\partial(dx^\mu/d\sigma)} &= \frac{1}{2}\left(-g_{\alpha\beta}\frac{dx^\alpha}{d\sigma}\frac{dx^\beta}{d\sigma}\right)^{-1/2}\left(-g_{\mu\beta}\frac{dx^\beta}{d\sigma} - g_{\alpha\mu}\frac{dx^\alpha}{d\sigma}\right) \\ &= -g_{\mu\beta}\frac{1}{\mathcal{L}}\frac{dx^\beta}{d\sigma} \\ &= -g_{\mu\beta}\frac{dx^\beta}{d\tau} \\ &= -g_{\alpha\beta}\xi^{\alpha}u^{\beta} \\ &= -\bf{\xi}\cdot\bf{u}. \end{align}$$ $\bf{\xi}$ is called a Killing vector. The dot product of a Killing vector, $\bf{\xi}$, and a four-velocity, $\bf{u}$, is a constant.