Timelike geodesic (${\nabla }_{u}u=0$)

A timelike geodesic has timelike tangent and vanishing covariant acceleration in an affine parameter. Choosing proper time gives $g(\dot{\gamma },\dot{\gamma })=-1$.

Physically, an ideal massive test particle in free fall follows a timelike geodesic when nongravitational forces and appreciable self-gravity can be neglected. The normalization above uses $c=1$; in SI units it is $g(u,u)=-c^2$. Zero covariant acceleration does not mean constant coordinate velocity: a dropped object accelerates relative to the ground. Conversely, a person standing on the ground is pushed away from free fall by the supporting surface. A photon follows a null geodesic, whose parameter cannot be proper time. See Tong's geodesics in spacetime.

For $\alpha >0$ and $t\ne 0$, the conformal factor $t^2/\alpha$ makes spatial translation a Killing vector field. Using proper time, its conserved momentum is $P=(t^2/\alpha )dx/d\tau$. With $v=dx/dt$ and $d{s}^2=-c^{2}d{\tau }^2$ this becomesThus $v/\sqrt{1-v^2/c^2}$ alone is not constant. Set $b=\sqrt{\alpha }P/c$. Every unparametrized timelike geodesic in a connected nonzero-$t$ patch has the formThe nonvertical paths have $v=cb/\sqrt{t^2+b^2}$ and proper-time parametrizationThe chart excludes $t=0$, where its metric is degenerate; a vertical family must not be lost by dividing by the conserved momentum.

For $d{s}^2=(1+f(x){)}^{2}d{x}^2/(1+x^2)-(1+x^2)d{t}^2$, with smooth odd $f$ and $|f|<1$, the conserved energy gives $dt/d\tau =E/(1+x^2)$ and $(dx/d\tau {)}^2=(E^2-1-x^2)/(1+f{)}^2$. Both half-cycle integrals over $[-\sqrt{E^2-1},\sqrt{E^2-1}]$ contain an even weight times $1+f$. The odd contribution vanishes, leaving the unperturbed half-period $\pi$ in each clock. The waveforms need not be sinusoidal, but both periods are unchanged for $E>1$.

In a positive-time patch, proper-time first integrals give $\dot{x}=p{t}^2$ and ${\dot{t}}^2=t^2+p^2{t}^4$. If $p\ne 0$, the unparametrized timelike geodesics are $t^2=(x-x_c{)}^2-p^{-2}$ on an appropriate branch; in the expanded quadratic form their discriminant is positive. If $p=0$, there is the separate family $x=x_c$, $t=t_{\ast }{e}^{\pm \tau }$. A description using only the nonvertical quadratic family omits genuine timelike geodesics.