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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.

The coordinate expression for the Lie bracket of vector fields isFor the Levi-Civita connection,The final term vanishes because the Christoffel symbols are symmetric in $\mu ,\nu$ while the vector product in parentheses is antisymmetric. Hence

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.

The Levi-Civita connection is a torsion-free connection. Its torsion form isSince $T=0$,In coordinates this is the symmetry ${\Gamma }_{jk}^i={\Gamma }_{kj}^i$ of the Christoffel symbols.

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.

The Schwarzschild exterior is vacuum, soand consequently $C_{\alpha \beta \gamma \delta }=R_{\alpha \beta \gamma \delta }$. PutThe relevant Christoffel symbols areWith the conventionone findsSince $f^{\prime \prime }=-4M/r^3$,This is the leading Radial Weyl curvature of Schwarzschild spacetime. The opposite convention for the Riemann tensor reverses the sign but not the physical tidal magnitude.

For the Friedmann-Lemaître-Robertson-Walker metric in Cartesian spatial coordinates, the only time-dependent components are $g_{ij}=a^2{\delta }_{ij}$. The Christoffel symbol formula therefore givesSubstituting these into the definition of the Ricci tensor givesThus, with no summation over $i$, the requested components are ${\Gamma }^0{}_{ii}=a\dot{a}$, ${\Gamma }^i{}_{0i}={\Gamma }^i{}_{i0}=\dot{a}/a$, and $R_{ii}=a\ddot{a}+2{\dot{a}}^2$.

Writing $f=1-2M/R$, orthogonality from part a givesThe needed Christoffel symbols areThe radial and azimuthal transport equations reduce toHenceWith the stated initial 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

At an arbitrary point choose normal coordinates, so the Christoffel symbols vanish there. Torsion freedom and commutation of partial derivatives then give the algebraic first Bianchi identityDifferentiate the coordinate curvature expression and cyclically antisymmetrize. Third derivatives cancel, giving the second Bianchi identityBoth statements are tensorial and hence hold in every coordinate system. Contracting the differential identity, using the curvature symmetries and metric compatibility, yieldsthe contracted Bianchi identity.

Apply the definition to the coordinate basis, for which $[{\partial }_{\mu },{\partial }_{\nu }]=0$ and ${\nabla }_{\mu }{\partial }_{\tau }={\Gamma }^{\rho }{}_{\tau \mu }{\partial }_{\rho }$. Comparing coefficients givesequivalent to the paper's ordering after commuting scalar factors. Although the individual Christoffel symbols are not tensors, the preceding intrinsic definition proves that this complete combination is tensorial.

The background spatial metric is $g_{ij}=a^2{\delta }_{ij}$. For the large spatial diffeomorphismhomogeneity and isotropy imply that the purely spatial background Christoffel symbols vanish. HenceThe shift is constant, transverse and traceless, and is therefore the zero-momentum adiabatic tensor mode.

A scalar transforms by its Lie derivative:Fourier transformation and integration by parts in momentum space giveThe term proportional to ${\omega }_i{}^i$ vanishes because the deformation is traceless.

For coordinates $x^{\alpha }$, the Christoffel symbols of the Levi-Civita connection arewhere $g^{\alpha \delta }$ is the inverse metric.

For the two-dimensional hyperbolic metric in polar coordinates,Only $g_{\varphi \varphi }$ has a nonzero derivative, namelyIt follows that the nonzero connection coefficients are

An affinely parameterized geodesic $x^{\mu }(\lambda )$ satisfiesFor the conformally flat metric $g_{\mu \nu }={\phi }^2{\eta }_{\mu \nu }$, its Christoffel symbols areLet $u^{\mu }=d{x}^{\mu }/d\lambda$. Along a null geodesic, ${\eta }_{\alpha \beta }{u}^{\alpha }{u}^{\beta }=0$, so the equation reduces toThe acceleration is parallel to the tangent. A change of parameter removes it, leaving a straight null line of the Minkowski metric. Thus conformal rescaling changes the affine parameter but not the unparametrized light ray.

Consequently Nordstrom theory of gravitation predictsfor light passing a star. This contradicts the observed nonzero deflection and the successful general-relativistic value $4GM/(b{c}^2)$.

For local coordinates $x^1,\dots ,x^n$, a curve is an affinely parametrized geodesic exactly when it satisfies the geodesic equationwhere the Christoffel symbols of the Levi-Civita connection are

Write ${\dot{x}}^{\mu }=d{x}^{\mu }/ds$ and $L=\sqrt{-g_{\mu \nu }{\dot{x}}^{\mu }{\dot{x}}^{\nu }}$. Varying $T=\int Lds$ with fixed endpoints and then choosing proper time as parameter, for which $L=1$, gives the Euler-Lagrange equationExpanding the derivative, raising the free index with the inverse metric, and using the symmetry of ${\dot{x}}^{\alpha }{\dot{x}}^{\beta }$ gives the geodesic equationThese are the Christoffel symbols of the Levi-Civita connection.

Put $h_{ab}=\delta g_{ab}$. Varying $g^{ac}{g}_{cb}={\delta }_b^a$ givesThe determinant identity $\delta \log |\det g|=g^{ab}{h}_{ab}$ givesFinally, varying the formula for the Christoffel symbols and rewriting partial derivatives covariantly gives the tensorThese are the basic metric variation identities.

Using the stated Christoffel symbols in the definition of the Riemann curvature tensor, the two nontrivial contractions areHenceand the vacuum Einstein field equations reduce toThis is the diagonal Rosen coordinates for a plane gravitational wave equation.

For the plus-polarized wave, setComparison of the transverse metric components gives, to linear order,orThereforeThe Vacuum Einstein equations are consequently satisfied at linear order. A real gravitational wave is obtained by taking the real part of the complex plane-wave notation.

Affinely parametrized geodesics are the critical curves of the geodesic LagrangianFor $w=1+r^2$ and $w^{\prime }=2r$, the nonzero Christoffel symbols, up to symmetry in the lower indices, areThey follow either from the Euler-Lagrange equations or directly from the Levi-Civita connection formula.