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The $x$ Euler-Lagrange equation has the first integralwhere the constant follows from the initial data. Combining this with proper-time normalization gives the Timelike geodesic in the expanding flat slicing of de Sitter spacetime:ConsequentlyIntegrating from $t=0$, where $\tau =0$, givesAs $t\to -\infty$, the first ratio tends to $1$, soThus $t$ tends to $-\infty$ at the finite proper time ${\tau }_{-}$.

For an affine parameter $\lambda$, the geodesic Lagrangian isVarying $S=\int \mathcal{L}d\lambda$ gives the geodesic equations

The rotational Killing vector fields of the spherical symmetry conserve the angular-momentum vector. Its direction is fixed, so every nonradial orbit lies in the plane through the symmetry centre orthogonal to that vector. A spatial rotation can make this the equatorial plane. Equivalently, the initial conditions $\theta =\pi /2$ and $\dot{\theta }=0$ solve the $\theta$ equation uniquely. This is the planarity of geodesics in spherical symmetry; a radial geodesic has zero angular momentum and may be assigned any plane.

On $\theta =\pi /2$, the cyclic coordinates $t$ and $\varphi$ produce geodesic conserved quantities from Killing vectors,For a timelike geodesic parametrized by proper time, the proper-time normalization is $g_{ab}{\dot{x}}^a{\dot{x}}^b=-1$. Substitution of $E$ and $L$ gives the first-order radial equationThis is the timelike geodesic effective potential for static Anti-de Sitter spacetime with curvature radius one.

A geodesic passing through $r=0$ must have $L=0$. If it moves away from the origin, $E>1$, andTaking proper time $s=0$ at departure givesIt turns around at $s=\pi /2$ and first returns to the origin atThis energy-independent refocusing is the characteristic radial timelike geodesic in anti-de Sitter spacetime result; for curvature radius $a$, the answer is $\pi a$.

The initial data and $\ddot{u}=0$ give $u=\tau$. To first order in $ϵ$, replace $x$ and $y$ on the right-hand sides of the transverse equations by $x_0$ and $y_0$. Twice integrating with the initial rest conditions givesFor $\tau >U$, define $A_0={\int }_{-\infty }^{\infty }a(s)ds$ and $A_1={\int }_{-\infty }^{\infty }s,a(s)ds$. ThenHenceThese permanent changes are forms of displacement memory and velocity memory.

There is a discrepancy in the question's final instruction. The longitudinal equation actually givesand thereforeThus $z$ does not vanish to first order for a general allowed profile. It vanishes after the pulse under the additional hypothesis $A_0=0$ used in part (iv), but it need not vanish while that pulse is passing. The proper-time normalization independently gives the same relation $\dot{z}=xyA+O({ϵ}^2)$.

Interpret the orbit in the Schwarzschild metric as an unforced circular timelike geodesic and choose its plane as $\theta =\pi /2$. Put $f=1-2\mu /r$. The radial geodesic equation at constant radius uses the Christoffel symbolsand gives the Angular velocity of a circular Schwarzschild geodesicThe proper-time normalization then yieldsOver one revolution, the Proper period of a circular Schwarzschild geodesic is thereforeBoth the gravitational and orbital kinematic contributions to time dilation are included. Circular timelike geodesics exist for $r>3\mu$; they are stable for $r>6\mu$ and unstable for $3\mu <r<6\mu$, with marginal stability at $6\mu$.

If “circular orbit” instead means a forced trajectory with an externally chosen angular velocity, the radius alone does not fix the answer. Its elapsed time is $\Delta \tau =\Delta t\sqrt{1-2\mu /r-r^2{\Omega }^2/c^2}$, or $2\pi \sqrt{1-2\mu /r-r^2{\Omega }^2/c^2}/|\Omega |$ per revolution. The geodesic interpretation supplies the missing angular velocity.

A freely moving massive test particle in an FLRW metric has physical momentum $p=m\gamma v$ relative to a comoving observer, where $v$ is its peculiar velocity. For a radial geodesic, the geodesic Lagrangian gives $C=a^{2}d\chi /d\tau$ constant, and the local radial momentum is $p=mad\chi /d\tau =mC/a$. The proper-time normalization gives $dt/d\tau =\sqrt{1+C^2/(a^2{c}^2)}$, soConsequently nonrelativistic peculiar velocity scales as $a^{-1}$, while the exact relativistic law keeps $|v|<c$. The momentum magnitude of a free photon also scales as $a^{-1}$, giving cosmological redshift through $E=cp$. These statements concern force-free propagation; a bound system need not follow the cosmological expansion.