Conserved quantities and equatorial radial reduction of timelike geodesics of the Schwarzschild exterior model

(The fields ${\mathrm{\partial }}_0$ and ${\mathrm{\partial }}_{\phi }$ are Killing.) For $\xi ={\mathrm{\partial }}_0$ the lowered field is ${\xi }_b=g_{0b}=-f{\delta }_b^0$, and the Killing equation reads ${\mathrm{\partial }}_a{g}_{0b}+{\mathrm{\partial }}_b{g}_{0a}-2{\mathrm{\Gamma }}^c{}_{ab}{g}_{0c}=0$; substituting [L1], the only potentially nonzero components are $(a,b)=(0,1)$: ${\mathrm{\partial }}_1{g}_{00}+2f{\mathrm{\Gamma }}^0{}_{01}=-f^{\mathrm{\prime }}+f^{\mathrm{\prime }}=0$, and $(a,b)=(0,0)$: $2{\mathrm{\partial }}_0{g}_{00}=0$; all other components involve derivatives of vanishing components or vanishing symbols, so they vanish. For $\xi ={\mathrm{\partial }}_{\phi }$ the lowered field is ${\xi }_b=r^2{\sin }^2\vartheta {\delta }_b^3$, and the equation is ${\mathrm{\partial }}_a{g}_{3b}+{\mathrm{\partial }}_b{g}_{3a}-2r^2{\sin }^2\vartheta {\mathrm{\Gamma }}^3{}_{ab}=0$; its potentially nonzero components are $(a,b)=(1,3)$: $2r{\sin }^2\vartheta -2r^2{\sin }^2\vartheta \cdot (1\mathrm{/}r)=0$, $(a,b)=(2,3)$: $2r^2\sin \vartheta \cos \vartheta -2r^2{\sin }^2\vartheta \cdot \cot \vartheta =0$, and $(a,b)=(3,3)$: $2{\mathrm{\partial }}_3{g}_{33}=0$; all other components vanish. Hence both fields are Killing by [L2], and $e=-g(U,{\mathrm{\partial }}_0)$ and $\mathrm{\ell }=g(U,{\mathrm{\partial }}_{\phi })$ are constant along $\gamma$. For positivity: $n=f^{-1\mathrm{/}2}{\mathrm{\partial }}_0$ is a future unit timelike field by [L5], and $-g(n,U)>0$; since $-g(n,U)=f^{-1\mathrm{/}2}e$, we get $e>0$. This proves (a).

(Equatorial reduction.) The coordinate geodesic system [L3] and [L1] has the equations ${\ddot{x}}^0+\frac{f^{\mathrm{\prime }}}{f}{\dot{x}}^0\dot{r}=0,\ddot{r}+\frac{f{f}^{\mathrm{\prime }}}{2}({\dot{x}}^0{)}^2-\frac{f^{\mathrm{\prime }}}{2f}{\dot{r}}^2-rf{\dot{\vartheta }}^2-rf{\sin }^2\vartheta {\dot{\phi }}^2=0,$ $\ddot{\vartheta }+\frac{2}{r}\dot{r}\dot{\vartheta }-\sin \vartheta \cos \vartheta {\dot{\phi }}^2=0,\ddot{\phi }+\frac{2}{r}\dot{r}\dot{\phi }+2\cot \vartheta \dot{\vartheta }\dot{\phi }=0.$ Set $\vartheta =\pi \mathrm{/}2$ and $\dot{\vartheta }=0$ in the $x^0,r,\phi$ equations. The resulting smooth three-equation subsystem, written as a first-order system for positions and velocities, has an open state domain containing the initial data because $r({\tau }_0)>s$ and the chosen product chart is open. Its right-hand side is smooth and hence locally Lipschitz, so [L7] gives a local solution with the same positions and velocities as $\gamma$ at ${\tau }_0$. Adjoining the constant coordinate $\vartheta =\pi \mathrm{/}2$ gives a solution of the full four equations, because the $\vartheta$ equation vanishes and the other equations reduce to that subsystem. This constructed geodesic has the same event and tangent as $\gamma$ at ${\tau }_0$, so uniqueness in [L3] makes them agree on a neighborhood of ${\tau }_0$. Let $E:=\{\tau \in I:\vartheta (\tau )=\pi \mathrm{/}2\text{ and }\dot{\vartheta }(\tau )=0\}\mathrm{.}$ It is nonempty, closed by continuity, and open by the same local construction at each of its points. Since $I$ is an interval, $E=I$. Thus $\vartheta \equiv \pi \mathrm{/}2$ on $I$, proving (b); in particular ${\sin }^2\vartheta =1$ there.

(Normalization identity.) By [L4], $-c^2=g(U,U)=-f({\dot{x}}^0{)}^2+f^{-1}{\dot{r}}^2+r^2{\dot{\phi }}^2$ on the equatorial geodesic. Substituting ${\dot{x}}^0=e\mathrm{/}f$ and $\dot{\phi }=\mathrm{\ell }\mathrm{/}{r}^2$ from the definitions of $e,\mathrm{\ell }$ and multiplying by $f$ gives $-c^{2}f=-e^2+{\dot{r}}^2+f{\mathrm{\ell }}^2\mathrm{/}{r}^2$, that is, ${\dot{r}}^2=e^2-f(c^2+{\mathrm{\ell }}^2\mathrm{/}{r}^2)=e^2-W_{\mathrm{\ell }}(r)$. This is the normalization identity of (c).

(Radial equation.) The $r$-component of the geodesic equation [L3] is, by [L1], $\ddot{r}+\frac{f{f}^{\mathrm{\prime }}}{2}({\dot{x}}^0{)}^2-\frac{f^{\mathrm{\prime }}}{2f}{\dot{r}}^2-rf{\dot{\phi }}^2=0$. Substituting ${\dot{x}}^0=e\mathrm{/}f$ and $\dot{\phi }=\mathrm{\ell }\mathrm{/}{r}^2$ turns this into $\ddot{r}+\frac{f^{\mathrm{\prime }}{e}^2}{2f}-\frac{f^{\mathrm{\prime }}}{2f}{\dot{r}}^2-\frac{f{\mathrm{\ell }}^2}{r^3}=0$. The claimed equation reads, after substituting $W_{\mathrm{\ell }}^{\mathrm{\prime }}$, $\ddot{r}+\frac{f^{\mathrm{\prime }}}{2}(c^2+\frac{{\mathrm{\ell }}^2}{r^2})-\frac{f{\mathrm{\ell }}^2}{r^3}=0$. The difference between the two left-hand sides is $\frac{f^{\mathrm{\prime }}}{2f}(e^2-{\dot{r}}^2-f(c^2+{\mathrm{\ell }}^2\mathrm{/}{r}^2))=\frac{f^{\mathrm{\prime }}}{2f}(e^2-{\dot{r}}^2-W_{\mathrm{\ell }}(r))$, which vanishes precisely by the normalization identity of step 1.3; the substitution of $W_{\mathrm{\ell }}^{\mathrm{\prime }}$ and the rearrangement use the derivative rules of [L6]. Hence, given that identity, the $r$-equation and $\ddot{r}=-\frac{1}{2}{W}_{\mathrm{\ell }}^{\mathrm{\prime }}$ are equivalent. This proves (c).

(Conclusion.) Steps 1.1, 1.2, 1.3 and 2.1 prove clauses (a)-(c) for the given geodesic on its chart interval; clause (d) records that the reduction is local and asserts nothing about completeness, extension or null geodesics.