Nil Trujillo
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General relativity 1.01

I have recently started a course on general relativity, following some classical textbooks, where I am learning how to describe the curvature of spacetime in a nutshell. I came across an interesting exercise proposed by a professor, where one must compute the geodesics for a given 2×22 \times 2 metric. In this document, I will solve this exercise as part of a new series of posts where I document my progress in general relativity.

Geodesics equations

We consider the coordinates:

x1=xx2=y\begin{aligned} x^1 = x \\ x^2 = y \end{aligned}

The metric tensor is given by:

gij=(−1−by0011+by)g_{ij} = \begin{pmatrix} -1 -\frac{b}{y} & 0 \\ 0 & \frac{1}{1+\frac{b}{y}} \end{pmatrix}

Our goal is to derive the geodesic equations for these coordinates using variational principles:

dumds+12gmi(∂kgip+∂pgik−∂igkp)ukup=0\frac{du^m}{ds} + \frac{1}{2} g^{mi} \left( \partial_k g_{ip} + \partial_p g_{ik} - \partial_i g_{kp} \right) u^k u^p = 0

Where:

u1=dx1ds=dxdsu2=dx2ds=dyds\begin{aligned} u^1 &= \frac{dx^1}{ds} = \frac{dx}{ds} \\ u^2 &= \frac{dx^2}{ds} = \frac{dy}{ds} \end{aligned}

Since gmig^{mi} is the inverse of gijg_{ij}, it satisfies:

gijgmi=δjmg_{ij} g^{mi} = \delta^m_j

The inverse metric tensor is computed as:

gmi=(−11+by001+by)g^{mi} = \begin{pmatrix} -\frac{1}{1+\frac{b}{y}} & 0 \\ 0 & 1+\frac{b}{y} \end{pmatrix}

Geodesic equation for xx

For the coordinate x1=xx^1 = x, the only relevant term in the geodesic equation is g11g^{11}:

d2xds2+12g11(∂kg1p+∂pg1k−∂1gkp)ukup=0.\frac{d^2x}{ds^2} + \frac{1}{2} g^{11} \left( \partial_k g_{1p} + \partial_p g_{1k} - \partial_1 g_{kp} \right) u^k u^p = 0.

The only nonzero Christoffel term arises when k=2k=2 and p=1p=1:

∂2g11+∂1g12−∂1g21=by2\partial_2 g_{11} + \partial_1 g_{12} - \partial_1 g_{21} = \frac{b}{y^2}

Thus, the geodesic equation for xx simplifies to:

d2xds2−12b(1+by)y2dydsdxds=0\frac{d^2x}{ds^2} - \frac{1}{2} \frac{b}{(1+\frac{b}{y}) y^2} \frac{dy}{ds} \frac{dx}{ds} = 0

Geodesic equation for yy

Similarly, for the coordinate yy, only the g22g^{22} term contributes, and the relevant partial derivatives are:

p=1,k=1⇒−∂2g11=−by2,p=2,k=2⇒∂2g22+∂2g22=2∂2g22=2b(1+by)2y2.\begin{aligned} p=1, k=1 & \Rightarrow -\partial_2 g_{11} = -\frac{b}{y^2}, \\ p=2, k=2 & \Rightarrow \partial_2 g_{22} + \partial_2 g_{22} = 2 \partial_2 g_{22} = \frac{2b}{(1+\frac{b}{y})^2 y^2}. \end{aligned}

Thus, the geodesic equation for yy takes the form:

d2yds2−12by2(1+by)(dxds)2+b(1+by)y2(dyds)2=0\frac{d^2y}{ds^2} - \frac{1}{2} \frac{b}{y^2} \left( 1+\frac{b}{y} \right) \left( \frac{dx}{ds} \right)^2 + \frac{b}{(1+\frac{b}{y}) y^2} \left( \frac{dy}{ds} \right)^2 = 0