GR: Conservation Of Energy

Date:

26 Sep 2019

Derivation

Step 1

\[\begin{split}\begin{align*} & \text{I. Earlier we had}\\ & ds^2 = -(1-{r_s\over r}) dt^2+(1-{r_s\over r})^{-1} dr^2 +r^2 d\theta^2+ r^2 \sin^2 \theta d\phi^2\\ & u^2=u\cdot u = g_{ab}{dx^{(a)} \over d\tau}{dx^{(b)} \over d\tau} = {ds^2\over -ds^2} =-1\\ & (1-{r_s\over r}) {dt\over d\tau}=\text{const}\equiv e \;\;\;\;\;\;\;\; r^2 \sin^2 \theta {d\phi \over d\tau}=\text{const}\equiv l \\ & \text{II. Consider the case }\theta ={\pi\over 2} \text{ (Plane Orbit)}\\ & \Rightarrow g_{00} ({dt \over d\tau})^2 + g_{11} ({dr \over d\tau})^2 + g_{33} ({d\phi \over d\tau})^2 = -1 \\ & \Rightarrow -(1-{r_s\over r}) ({dt \over d\tau})^2 + (1-{r_s\over r})^{-1} ({dr \over d\tau})^2 + r^2 ({d\phi \over d\tau})^2 = -1 \\ & \Rightarrow -(1-{r_s\over r})^{-1} e^2 + (1-{r_s\over r})^{-1} ({dr \over d\tau})^2 + {l^2\over r^2} = -1 \\ & \Rightarrow - e^2 + ({dr \over d\tau})^2 + (1-{r_s\over r}){l^2\over r^2} = -(1-{r_s\over r}) \\ & \Rightarrow \underbrace{ {1\over 2}({dr \over d\tau})^2 }_{\text{Kinetic}} + \underbrace{ (-{1\over 2}){r_s\over r}+{1\over 2}{l^2\over r^2}\boxed{-{1\over 2}{r_s l^2\over r^3}} }_{\text{Potential } \tilde{V}}= \underbrace{ {1\over 2}(e^2-1)\equiv \tilde{E} }_{\text{Total}} \\ & \;\;\;\; {d\tilde{E}\over d\tau}=0\Rightarrow {d^2r \over d\tau^2} =-{M\over r^2}+{l^2\over r^3}-{3Ml^2\over r^4}\\ & \;\;\;\; \text{Recall when }r\gg r_s \Rightarrow r_s\approx 2M \\ & \;\;\;\; \text{In Newtonian: }\underbrace{ {1\over 2}({dr \over d\tau})^2 }_{\text{Kinetic}} + \underbrace{ (-){M\over r}+{1\over 2}{l^2\over r^2} }_{\text{Potential}}= \underbrace{ E }_{\text{Total}} \\ & \text{III. Applications}\\ & (1)\text{ Possible + Stable circular orbits:}\\ & \;\;\;\;{d\tilde{V}\over dr}=0 \Rightarrow r={l^2\pm \sqrt{ l^4-12M^2l^2 }\over 2M} ={6M\over 1\pm \sqrt{ 1-12M^2/l^2 }}\\ & \;\;\;\;{d^2\tilde{V}\over dr^2}>0 \Rightarrow {2\over r}\underbrace{(-{M\over r^2}+{l^2\over r^3}-{3Ml^2\over r^4})}_{=-d\tilde{V}/ dr=0} +{l^2\over r^4}-{6Ml^2\over r^5}>0 \\ & \;\;\;\; \;\;\;\; \Rightarrow l>\sqrt{12}M, r>6M \\ & (2)\text{ Perihelion Shift:}\\ & \;\;\;\; r^2 {d\phi \over d\tau}= l \Rightarrow {dr\over d\tau}={l\over r^2}{dr\over d\phi} \;\;\;\; \;\;\;\; u\equiv {1\over r} \\ & \;\;\;\; \Rightarrow \tilde{E}={1\over 2}(-l{du\over d\phi})^2 -Mu+{1\over 2}l^2u^2-Ml^2u^3\\ & \;\;\;\; {d\tilde{E}\over d\phi }\overset{!}{=}0 \Rightarrow {d^2 u\over d\phi^2}-{M\over l^2}+ u-M 3u^2=0 \\ & \;\;\;\; \text{Perturbation theory: } u=u_0+c u_1 \;\;\;\; |u_1|\ll |u_0| \\ & \;\;\;\; {d^2 u_0\over d\phi^2}-{M\over l^2}+ u_0=0 \;\; [1]\Rightarrow u_0= {M\over l^2}(1+\epsilon\cos\phi) \\ & \;\;\;\; {d^2 cu_1\over d\phi^2}+ cu_1-c M 3u_0^2=0 \;\; [2] \\ & \;\;\;\; \text{Note that we will let }c=1 \text{ at the end, you can} \\ & \;\;\;\; \text{sum up [1] and [2] to check consistency}\\ & \;\;\;\; \text{Assume } u_1(\phi)=A+B\phi\sin\phi+C\phi\cos\phi+D\sin2\phi+E\cos2\phi\\ & \;\;\;\; \Rightarrow A= {3M^3\over l^4}(1+{\epsilon^2\over 2}), B={3M^3\epsilon\over l^4}, E=-{M^3 \epsilon^2\over 2l^4} \\ & \;\;\;\; \text{Finally }u\approx {M\over l^2}(1+\epsilon\cos\phi+ {3M^2\epsilon\over l^2}\phi\sin\phi) \\ & \;\;\;\; \approx {M\over l^2}(1+\epsilon\cos[(1-{3M^2\over l^2})\phi]) \Rightarrow \Delta \phi\approx 2\pi(1+{3M^2\over l^2}) \\ & \;\;\;\; \text{Semi-major Axis of elliptical orbit: } a={l^2\over M(1-\epsilon^2)}\\ \end{align*}\end{split}\]