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Evaluate the metric inner products of the proposed orthonormal tetrad to first order in the vector cosmological perturbation:The orthonormal tetrad therefore has metric signature $(-,+,+,+)$ to the required order. Put $q=ϵ/a^2$ and $b=\mathbf{e}\cdot \mathbf{B}$. The photon momentum components areThey satisfy the null vector condition to first order, with $|\mathbf{e}|=1$.

Use $\eta$ as parameter in the time component of the geodesic equation:The derivative of the direction is first order, since the background photon moves on a straight comoving line; $B_{i}d{e}^i/d\eta$ is consequently second order. With $d\mathbf{x}/d\eta =\mathbf{e}+O(B)$,The supplied Levi-Civita connection coefficients giveThe spatial gradients and all background expansion terms cancel. Since ${ϵ}^{\prime }/ϵ$ vanishes in the background, its product with $b$ is also second order. The photon energy redshift from a vector metric perturbation is thereforeAn independent check uses the covariant component $p_0=g_{0\mu }{p}^{\mu }=-ϵ+O(B^2)$. The covariant geodesic equation gives $d{p}_0/d\eta =(2p^0{)}^{-1}{\partial }_{\eta }{g}_{\alpha \beta }{p}^{\alpha }{p}^{\beta }=ϵe^i{\dot{B}}_i$; the conformal expansion term vanishes by the null vector condition and reproduces the same result.

The Euler-Lagrange equations for the kinetic Lagrangian areExpanding and multiplying by the inverse metric gives the geodesic equationThe kinetic variational principle uses an affine parameter; the energy $g_{ab}{\dot{x}}^a{\dot{x}}^b/2$ is constant along its solutions. Constant curves are also geodesics.

For the Poincare half-plane model, translation in $x$ gives the conserved quantity $p=\dot{x}/y^2$, and the energy is $E=({\dot{x}}^2+{\dot{y}}^2)/(2y^2)$. If $p=0$, the nonconstant geodesics are vertical lines, with $y=y_0{e}^{\pm \sqrt{2E}t}$. If $p\ne 0$,Writing $R^2=2E/p^2$ and integrating $dx/dy=\pm y/\sqrt{R^2-y^2}$ yields $(x-x_0{)}^2+y^2=R^2$. Thus nonconstant geodesic images are vertical lines or upper semicircles meeting the real axis orthogonally.

For the two specified points, the circle centre is $(6,0)$ and the radius is five:For example, an affine unit-speed parametrization is $x=6+5\tanh t$, $y=5\mathrm{sech}t$. The endpoints on the real axis are $(1,0)$ and $(11,0)$, outside the hyperbolic surface.

Let $D_{abc}={\nabla }_a{\nabla }_b{K}_c$. Differentiating the Killing equation gives $D_{abc}=-D_{acb}$. For a covector, the Ricci identity isCall this difference $C_{abc}$. The differentiated Killing equation then givesTaking the first minus the second plus the third yieldsThe pair symmetries of the Riemann curvature tensor and the first Bianchi identity reduce the coefficient to $2R_{cbad}$. Therefore the second covariant derivative of a Killing vector is

For an affine parameter $\lambda$ on a geodesic, its tangent obeys $V^b{\nabla }_b{V}^a=0$. ThusThe second term vanishes by the geodesic equation; the first contracts a symmetric product $V^a{V}^b$ with the antisymmetric derivative from the Killing equation. This establishes the geodesic conserved quantity from a Killing vector:

For the Expanding flat slicing of de Sitter spacetime, take the geodesic LagrangianThe Euler-Lagrange equation for $t$ iswhile each spatial coordinate $x^i$ obeysComparing these equations with the geodesic equation gives the complete list of nonzero Christoffel symbols:

For a slowly moving particle in the weak-field approximation, $dt/d\tau =1+O(\Phi ,v^2)$ and terms containing two spatial velocities may be neglected. The spatial geodesic equation reduces toFor $g_{00}=-(1+2\Phi )$, the Levi-Civita connection isHence the Newtonian limit of general relativity is

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 $\pi a$.

Use the geodesic Lagrangianwhere here the overdot on a coordinate means differentiation with respect to the affine parameter. Applying the Euler-Lagrange equation to each coordinate and comparing with the geodesic equationdetermines the Christoffel symbols. To display derivatives of the metric coefficients unambiguously, write ${\lambda }_t={\partial }_t\lambda$, ${\lambda }_r={\partial }_r\lambda$, and similarly for $\mu$. The nonzero symbols, together with those obtained by symmetry in the two lower indices, areThis is the Christoffel symbols of a diagonal spherical spacetime metric.

Differentiate $B_{\alpha \beta }={\nabla }_{\beta }{X}_{\alpha }$ along $X$ and commute the two covariant derivatives:where the sign is the covector form of the Ricci identity. Apply the product rule to the first term:The first term vanishes by the geodesic equation, and the second isFinally the pair symmetries and antisymmetries of the Riemann curvature tensor giveConsequentlyThis is the Riccati equation for a timelike geodesic congruence; taking its spatial trace is the starting point for the timelike Raychaudhuri equation.

For a smooth curve $\gamma (t)$ on a Riemannian surface, its energy of a curve isChoose a local parameterization $\mathbf{X}(u,v)$ and write $\gamma (t)=\mathbf{X}(u(t),v(t))$. With coefficients of the first fundamental formthe Lagrangian isThe Euler-Lagrange equations are equivalentlyAfter multiplying by the inverse metric these become the geodesic equationwhere the Christoffel symbols are determined by $E,F,G$.

If a straight line segment lies in the surface, parameterize it byThen $\ddot{\gamma }=0$, so its acceleration has zero tangential component. The two displayed equations hold directly, and the segment is a geodesic.

For the one-sheeted hyperboloidputTwo distinct ruling lines through $P$ areIndeed, the direction $d_{\pm }$ satisfiesso substitution shows that every point of the line lies in $H$. These are geodesics by the straight-line argument.

A third geodesic is the meridian through $P$. Choose $s_0$ with $\sinh s_0=z_0$; thenIts acceleration is ${\gamma }_m$, which is normal to $H$, so it is a geodesic. If $z_0\ne 0$, these give the required three distinct subsets.

If $z_0=0$, there is also the equatorial circle. Writing $P=(\cos {\theta }_0,\sin {\theta }_0,0)$,Its acceleration is normal to $H$ along $z=0$, so it is a fourth geodesic distinct from the meridian and the two rulings.

Finally, write $\rho =\sqrt{x^2+y^2}=\sqrt{1+z^2}$. Clairaut's relation gives the conserved quantityChoose initial data in $z>0$ withSince $|\sin \psi |\le 1$, every point of the resulting geodesic satisfies $\rho \ge c$, and thereforeContinuity keeps the geodesic in the component $z>0$, and the stated completeness assumption defines it for every real time. The continuum of choices of $c$ supplies infinitely many such geodesics.

Assume the test body moves slowly, $d{x}^i/dt=O(ϵ)$, and that the field is quasistatic, so time derivatives are of higher order. The spatial geodesic equation, parametrized by coordinate time, then reduces at leading order toDefining the Newtonian potential bygivesthe Newtonian equation of motion. Terms involving $h_{0i}$, spatial velocity, or time derivatives enter beyond the stated order.

Choose at one transverse cross-section a null vector $N^a$ satisfying $N^a{N}_a=0$ and $U^a{N}_a=-1$, then parallel transport it along every generator:Metric compatibility and the affine geodesic equation imply that both $N^2$ and $U\cdot N$ are constant along each generator, so the required normalization persists.

A null geodesic congruence is a smooth family of nonintersecting null geodesics filling an open spacetime region. For its affinely parametrized tangent $U^a$,Writing $B_{ab}={\nabla }_b{U}_a$, differentiation of the null norm and the geodesic equation give respectively

In the absence of scalar anisotropic stress, the traceless spatial Einstein field equations giveThe photon mass-shell condition $g_{\mu \nu }{P}^{\mu }{P}^{\nu }=0$, together with $p^2=g_{ij}{P}^i{P}^j$ and $ϵ=ap$, gives to first orderUse the time component of the geodesic equation, divide it by $d\eta /d\lambda =P^0$, and substitute the stated Christoffel symbols. Keeping the perturbed ratios $P^i/P^0=(1+\Phi +\Psi ){\hat{p}}^i$ where they multiply the background $\mathcal{H}$ terms makes those terms cancel. Comparing the result with the total derivative of $P^0=ϵa^{-2}(1-\Psi )$ yieldsThe first term is the time-dependent gravitational redshift and the second is the gravitational frequency shift along the photon direction.

Forthe nonzero Christoffel symbols needed here areOn a line of constant longitude, $\dot{\varphi }=0$, and the geodesic equation reduces to $\ddot{\theta }=0$ after choosing an affine parameter. Thus every meridian is a geodesic.

For a nonconstant line of constant latitude, $\dot{\theta }=0$ and the $\theta$ equation requiresAway from the coordinate-degenerate poles, this holds only whenso only the equator is a constant-latitude geodesic.

LetThe Euler-Lagrange equation givesSince ${\dot{x}}^{\mu }=Ld{x}^{\mu }/ds$ and $d/d\lambda =Ld/ds$, this becomesExpanding the derivative, multiplying by the inverse metric, and using the Christoffel symbolgives the geodesic equation

Write $q^1=u$, $q^2=v$, ${\sigma }_i={\partial }_i\sigma$, and let $g_{ij}={\sigma }_i\cdot {\sigma }_j$ be the first fundamental form. Sincedifferentiating ${\sigma }_u\cdot \dot{\gamma }$ and ${\sigma }_v\cdot \dot{\gamma }$, or equivalently taking the two inner products of the second formula with ${\sigma }_k$, givesThe matrix $(g_{k\ell })$ is invertible because $\sigma$ is a regular embedded surface parametrization. Consequently the two geodesic equations hold exactly when $\ddot{\gamma }$ is orthogonal to both tangent vectors ${\sigma }_u$ and ${\sigma }_v$, which says precisely that $\ddot{\gamma }$ is a normal vector. This is the ambient acceleration criterion for a surface geodesic.

Because $\dot{\gamma }$ is tangent and $\ddot{\gamma }$ is normal,Thus an affinely parametrized geodesic has constant speed, as recorded by constant speed of an affinely parametrized geodesic.

For a surface of revolution, use profile arc length $s$ and azimuth $\varphi$. Its Riemannian metric isThe azimuth is an ignorable coordinate, so the corresponding geodesic equation has the first integralIf $v=|\dot{\gamma }|$ and $\theta$ is the oriented angle with the parallel, then the component of velocity along the parallel isSince $v$ is constant,This is the Clairaut first integral for a surface of revolution.

The Poincare disc model isIts geodesics through the origin are the Euclidean diameters. To see directly that the radial segment from $0$ to $r{e}^{i{\theta }_0}$ minimizes length, write an arbitrary joining curve as $z(t)=r(t)e^{i\theta (t)}$. Its hyperbolic length satisfiesThe radial segment has constant $\theta$ and monotone $r$, so equality holds. Henceand every radial diameter is length minimizing. Rotational symmetry and uniqueness for the geodesic equation show that these are all geodesics through $0$.

Given any geodesic and a point $p$ on it, a disc-preserving maptakes $p$ to the origin. By part (a), $T$ is a hyperbolic isometry and commutes with $J(z)=1/\overline{z}$. The transformed geodesic is a diameter, hence a generalized circle invariant under $J$. Its inverse image is therefore also a Generalized circle under a Möbius transformation invariant under $J$. Thus every hyperbolic geodesic is the part in $\mathbb{D}$ of a Euclidean line or circle preserved by reflection in the unit circle; equivalently, it is a diameter or a circle orthogonal to the unit circle. This is the Geodesics of the Poincare disc description.

Locally write the null hypersurface as a level set $u=0$ and its normal as $n_a={\nabla }_{a}u$. Since a null normal is also tangent, $n^a$ lies within the hypersurface. The symmetry of the Levi-Civita connection givesThe scalar $n^2$ vanishes on the hypersurface, so its gradient there is normal and hence proportional to $n_a$:This is the nonaffine geodesic equation. A rescaling of $n$ removes $\kappa$, proving that the null hypersurface normal generates null geodesics lying in the hypersurface.