= Solution
Take $\alpha>0$, $t\ne0$, and the negative-norm convention for a <timelike curve> used by the printed metric. Writing $v=dx/dt$ and $ds^2=-c^2d\tau^2$ gives
$$
\frac{d\tau}{dt}=\frac{|t|}{\sqrt\alpha}\sqrt{1-\frac{v^2}{c^2}}.
$$
Spatial translation is a <Killing vector field>. Equivalently, the <geodesic Lagrangian> has the <cyclic coordinate> $x$, so its conserved momentum is
$$
P=\frac{t^2}{\alpha}\frac{dx}{d\tau}
=\frac{|t|}{\sqrt\alpha}\frac{v}{\sqrt{1-v^2/c^2}}.
$$
Therefore the correct <first integral> for this metric is
$$
\boxed{\frac{|t|\,v}{\sqrt{1-v^2/c^2}}=\text{constant}.}
$$
\b[The first-integral assertion in the PDF is false.] Its dimensionless square root requires $v/c$, and even after setting $c=1$ it is missing the factor $|t|$. For example, the nonvertical geodesics below have $v/\sqrt{1-v^2/c^2}$ proportional to $1/|t|$.
Set $b=\sqrt\alpha P/c$. Solving for the velocity, with increasing $t$ as the time orientation, gives $v=cb/\sqrt{t^2+b^2}$. The complete unparametrized family of <timelike geodesics of a quadratic-time conformal metric> is
$$
\boxed{x(t)=x_c+cb\,\operatorname{arsinh}\!\left(\frac{t}{|b|}\right)\quad(b\ne0),\qquad
x(t)=x_c\quad(b=0).}
$$
Each curve is restricted to a connected positive- or negative-$t$ patch; $t=0$ is excluded because the metric becomes degenerate. For $t\ne0$, $|v|<c$, and the two independent constants are $x_c$ and $b$. If an explicit affine parametrization is desired,
$$
\tau-\tau_c=\frac{t\sqrt{t^2+b^2}-b^2\operatorname{arsinh}(t/|b|)}{2\sqrt\alpha}\quad(b\ne0),
$$
while the vertical family has $d\tau=|t|dt/\sqrt\alpha$. A reversal of the time orientation reverses the parameter without changing these paths.
Back to article page
