2 added 307 characters in body edited Oct 9 '12 at 14:21 Jerry Schirmer 32.3k22 gold badges5858 silver badges111111 bronze badges The second equation doesn't reduce to the first. There's no term proportional to $$x$$ and without derivatives of $$x$$ anywhere in the second equation. You could couple GR to some sort of potential and get something like the spring force, but that's rarely done. If you did, you'd have some sort of answer like $$k\,x^{a} = {\ddot x}^{a} + \Gamma_{bc}{}^{a}{\dot x}^{b}{\dot x}^{c}$$ Models like this don't end up being very relevant in the regime where general relativity is important. You could use this to analyze the simple harmonic oscillator in curvilinear coordinates, though. (As an added aside, since the Newtonian "force" term arises from GR due to the $$\Gamma_{tt}{}^{r}$$ term in the Schwarzshild solution in standard coordinates, I suppose you could play the same game and have $$\Gamma_{tt}{}^{r} = kr$$ for some metric, and have your second equation reduce to the first one. The second equation doesn't reduce to the first. There's no term proportional to $$x$$ and without derivatives of $$x$$ anywhere in the second equation. You could couple GR to some sort of potential and get something like the spring force, but that's rarely done. If you did, you'd have some sort of answer like $$k\,x^{a} = {\ddot x}^{a} + \Gamma_{bc}{}^{a}{\dot x}^{b}{\dot x}^{c}$$ Models like this don't end up being very relevant in the regime where general relativity is important. You could use this to analyze the simple harmonic oscillator in curvilinear coordinates, though. The second equation doesn't reduce to the first. There's no term proportional to $$x$$ and without derivatives of $$x$$ anywhere in the second equation. You could couple GR to some sort of potential and get something like the spring force, but that's rarely done. If you did, you'd have some sort of answer like $$k\,x^{a} = {\ddot x}^{a} + \Gamma_{bc}{}^{a}{\dot x}^{b}{\dot x}^{c}$$ Models like this don't end up being very relevant in the regime where general relativity is important. You could use this to analyze the simple harmonic oscillator in curvilinear coordinates, though. (As an added aside, since the Newtonian "force" term arises from GR due to the $$\Gamma_{tt}{}^{r}$$ term in the Schwarzshild solution in standard coordinates, I suppose you could play the same game and have $$\Gamma_{tt}{}^{r} = kr$$ for some metric, and have your second equation reduce to the first one. 1 answered Oct 9 '12 at 2:07 Jerry Schirmer 32.3k22 gold badges5858 silver badges111111 bronze badges The second equation doesn't reduce to the first. There's no term proportional to $$x$$ and without derivatives of $$x$$ anywhere in the second equation. You could couple GR to some sort of potential and get something like the spring force, but that's rarely done. If you did, you'd have some sort of answer like $$k\,x^{a} = {\ddot x}^{a} + \Gamma_{bc}{}^{a}{\dot x}^{b}{\dot x}^{c}$$ Models like this don't end up being very relevant in the regime where general relativity is important. You could use this to analyze the simple harmonic oscillator in curvilinear coordinates, though.