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## EXPLICIT LIGHT CONE GEODESICS IN A LIGHT CONE METRIC

I took Differential Geometry at Princeton in my sophomore year in 1992 and have become incredibly rusty so it is time to do some computations in General Relativity to get back in shape.  An elementary problem: find the light-cone geodesics for the metric

$g = dt^2 - e^{-2Ht} dx^2$

which satisfy $g_{ab} \dot{x}^a \dot{x}^b=0$.  The Christoffel symbols are computed from

$\Gamma_{abc} = \frac{1}{2}( g_{ca,b} + g_{cb,a} - g_{ab,c})$

There are too many of these to compute exhaustively so we note first that

$g_{ab,c} = 0$ unless $a=b=2, c=1$

Therefore the only nonzero $\Gamma_{abc}$ are those with two 2 and one 1 in the indices.  For example,

$\Gamma_{212} = \frac{1}{2} ( g_{21,2} + g_{22,1} - g_{12,2}) = \frac{1}{2} g_{22,1} = -H e^{-2Ht}$

and

$\Gamma_{122}=\Gamma_{212}=\Gamma_{221} = -H e^{-2Ht}$

Then we have to raise the first index which has the solution

$\Gamma^1_{ab} = g^{11} \Gamma_{1ab} + g^{21} \Gamma_{2ab} = \Gamma_{1ab} = -He^{-2Ht}$

$\Gamma^2_{ab} = g^{12}\Gamma_{2ab} + g^{22}\Gamma_{2ab} = e^{+2Ht} \Gamma_{2ab} = - e^{+2Ht} H e^{-2Ht} = -H$

Now the geodesic equation is

$\frac{ d^2{x}^c}{ds^2} + \Gamma^c_{ab} \frac{dx^a}{ds} \frac{dx^b}{ds} = 0$

For coordinates $t,x$ we have

$\ddot{t} + (-He^{-2Ht}\dot{x}^2 = 0$

$\ddot{x} - 2H\dot{t}\dot{x} = 0$

I don’t know how to solve this system of equations in general.  But the light cone condition $g_{ab}\dot{x}^a\dot{x}^b =0$ in this case is

$\dot{t}^2 = e^{2Ht} \dot{x}^2$

which we plug into the first equation $\ddot{t} - H\dot{t}^2=0$.  This is $\dot{z} = Hz^2$ solved by integrating $\dot{z}{z^2} = H$, so $z=-\frac{1}{Hs}$ and $t = -\frac{1}{H} \log(s)$.

The second equation now simplifies when we plug in $\dot{t} = -\frac{1}{Hs}$.

$\ddot{x} +2/s \dot{x} = 0$

Again $\dot{z} = -(2/s)z$ and $\log(z) = -\log(s^2) +C$ and $x = \int^{s} \exp(-r^2) dr$.