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Let a one-parameter infinitesimal transformation be $\delta q_i=ϵX_i(q,t)$ and suppose its change in the Lagrangian is a total derivative,With canonical momentum $p_i=\partial \mathcal{L}/\partial {\dot{q}}_i$, the chain rule and the Euler-Lagrange equations give, along a motion,Comparison with the assumed total derivative proves Noether theorem:

The given Lagrangian has no explicit time dependence, so its energy function is conserved:It is also invariant under the dilation $(x,y)\mapsto (e^{ϵ}x,e^{ϵ}y)$: the velocities and $y$ all scale by the same factor, while $x/y$ is unchanged. The generator is $(X_x,X_y)=(x,y)$ and $F=0$. SinceNoether's theorem supplies the second first integralThe integrals $E$ and $D$ are independent on an open dense subset of phase space.

The Lagrangian has no explicit time dependence, so its energy function is constant. With total speed $v^2={\dot{r}}^2+r^2{\dot{\varphi }}^2$, division by $M_p$ givesConsequently, between the initial point $(r_i,v_i)$ and any other point,This energy relation holds for nonradial motion too. Radially outward motion is the stipulated approximation for the subsequent probe calculation; angular momentum is then zero.

Define the energy function $\mathcal{E}={\sum }_i{\dot{q}}_i{p}_i-L$. Along solutions of the Euler-Lagrange equations,ConsequentlyFor an ordinary isolated mechanical system with $L=T-U$, a time-independent quadratic kinetic energy and a velocity-independent potential energy, ${\sum }_i{\dot{q}}_i{p}_i=2T$, so $\mathcal{E}=T+U$. For a more general Lagrangian, the conserved quantity is the displayed energy function; identifying it with $T+U$ requires these assumptions.

Use the $+---$ metric signature and abbreviate $f(r)=1-2\mu /r$, with $\mu =GM/c^2$ a length. Dots initially mean differentiation with respect to an affine parameter $\lambda$. The geodesic Lagrangian in the equatorial plane isThe planarity of geodesics in spherical symmetry permits this plane to be chosen. More explicitly, the full angular Euler-Lagrange equation is $\ddot{\theta }+2\dot{r}\dot{\theta }/r-\sin \theta \cos \theta {\dot{\varphi }}^2=0$, so initial data $\theta =\pi /2$, $\dot{\theta }=0$ preserve it.

Both $t$ and $\varphi$ are cyclic coordinates. Their Euler-Lagrange equations give conserved conjugate momenta $p_t=c^{2}f\dot{t}$ and $p_{\varphi }=-r^2\dot{\varphi }$. Defining constants $k,h$ therefore givesThe corresponding second-order geodesic equations areand variation of $r$ givesThe geodesic Lagrangian has no explicit $\lambda$ dependence. Its conserved Lagrangian energy function is ${\dot{x}}^{\alpha }\partial \mathcal{L}/\partial {\dot{x}}^{\alpha }-\mathcal{L}=\mathcal{L}$, since it is quadratic in the tangent components. Thus twice this function is a constant:This is the constant squared tangent norm of an affinely parametrized geodesic.

For a massive particle choose $\lambda =\tau$, its proper time, and obtain $\kappa =c^2$. For a photon, the tangent is null and $\kappa =0$. A general rescaling of a massive particle's affine parameter rescales a positive $\kappa$; the numerical value $c^2$ requires the proper time choice. With that choice, $h=L_z/m$ is signed specific angular momentum, of units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$, and $k=E/(m{c}^2)$ is dimensionless conserved specific Killing energy. The time coordinate is normalized to proper time at infinity, so this $E$ is the particle's conserved energy referred to infinity. A photon's constants scale with the normalization of its null tangent; it has no rest frame or proper time parametrization.

Substitution of $\dot{t}=k/f$ and $\dot{\varphi }=h/r^2$ into the norm gives the general radial first integralFor a massive particle this becomesAdopting the usual half-normalization for the effective potential for timelike Schwarzschild geodesics, defineAll terms have units ${\mathrm{m}}^2{\mathrm{s}}^{-2}$, or energy per unit mass. The first two terms are the Newtonian gravitational and centrifugal contributions; the last is the relativistic correction.

A circular orbit requires both $\dot{r}=0$ and $V_{\mathrm{eff}}^{\prime }(R)=0$. Differentiating at fixed $h$ givesThe radial first integral also givesFor the future-directed circular orbit at $R=4\mu$, these constants are $|h|=4\mu c$, $k=1$, and $f=1/2$. ThereforeSince Schwarzschild time agrees with the clock time of a static observer at infinity, the intended infinity-referred tangential orbital speed isThis uses the circumferential areal radius $R$ with the clock at infinity; it is not a local measurement made at the orbit. A local static observer's clock instead has $d{\tau }_{\mathrm{stat}}=\sqrt{f}dt$ and measures $v_{\mathrm{local}}=v_{\infty }/\sqrt{f}=\boxed{c/\sqrt{2}}$. These two operationally different speeds explain the significance of the phrase “at infinity”.

The effective-potential stability of a circular orbit follows fromConsequently the $R=4\mu$ orbit exists but is unstable to radial perturbations. The original plot below shows its position at a maximum, for $h=4\mu c$.

For a weak-field elliptic orbit, the relativistic correction makes the direction of closest approach advance after each radial cycle. The derivations under i and ii below giveper orbit to first order, where $a_{\mathrm{orb}}$ is the semi-major axis and $e$ the orbital eccentricity. For Mercury, the additional Schwarzschild perihelion precession is approximately 43 arcseconds per century, over and above the much larger Newtonian precession produced primarily by other planets. It is the unexplained residual after those effects have been included that provides the classic quantitative test of general relativity; one should not identify the whole observed precession with this relativistic correction.^[ref]

The labels i and ii in the original paper introduce optional orbit-equation aids, rather than separate mandatory subquestions. They are retained here for derivations of both aids and of the perihelion result.

Define the canonical momentum $p_i=\partial L/\partial {\dot{q}}_i$ and the energy function $E={\sum }_i{p}_i{\dot{q}}_i-L$. Along an Euler-Lagrange equation solution,Thus an autonomous Lagrangian has conserved $E$. For a simple harmonic oscillator, $L=m{\dot{x}}^2/2-k{x}^2/2$ and $\boxed{E=m{\dot{x}}^2/2+k{x}^2/2}$ is constant. For ordinary time-independent mechanical coordinates this is the total mechanical energy.