1

I am supposed to derive the newtonian equation for gravitational acceleration ($-GM/r^2$) from the Schwarzschild metric $$ds=-(1-\frac{r_s}{r})c^2dt+(1+\frac{r_s}{r})^{-1}dr+0$$ where $$r_s=\frac{2GM}{c^2}$$ as we are only on a radial path (falling inward towards the planet) we are allowed to set $d\theta$ and $d\phi$ to $0$ so I didn't bother writing them. From here we are supposed to extremize $\int \frac{\sqrt{-ds}}{c}$ in order to somehow get acceleration = $$\frac{d^2r}{dt^2}=\frac{GM}{r^2}.$$ We are allowed to approximate $r_s<<r$ and $\frac{dr}{dt}<<c$ but I can't find a way to use either of these without totally breaking the equation

Attempt at a solution:

$$ds^2=-(1-\frac{r_s}{r})c^2dt+dr^2(1-\frac{r_s}{r})^{-1}\therefore\frac{\sqrt{-ds^2}}{c}=\sqrt{(1-\frac{r_s}{r})dt^2-dr^2((1-\frac{r_s}{r})c^2)^{-1}}=L$$ therefore $$L=dt\sqrt{(1-\frac{r_s}{r})-\frac{\dot{r}^2}{c^2-\frac{2MG}{r}}}$$

I should somehow use $$\frac{d}{dt}(\frac{dL}{d\dot{r}})-\frac{dL}{dr}=0$$ but the derivatives are so messy I don't know how to continue. I've seen other "solutions" like the Lagrangian approach on wikipedia but none seem to derive $GM/r^2$.

Qmechanic
  • 201,751
Big_Tubbz
  • 11
  • see how it is done here https://en.wikipedia.org/wiki/Schwarzschild_geodesics#Lagrangian_approach – atarasenko Jan 28 '19 at 10:53
  • They don't seem to actually derive GM/R^2 in there, I am also not supposed to have a phi factor, whereas they do have a factor of phi. – Big_Tubbz Jan 28 '19 at 19:13
  • Possible duplicates: https://physics.stackexchange.com/q/211930/2451 and links therein. – Qmechanic Jan 28 '19 at 20:21
  • Usually you solve the geodesic equation to derive $\ddot{r}$, therefore you need to solve the Christoffel symbols first, see http://f.yukterez.net/einstein.equations/files/2.html – Yukterez Feb 22 '20 at 16:36

1 Answers1

1

Action of a particle in the Schwarzschild metric is: $$S=-mc\int \sqrt{-ds^2}=-mc^2\int\sqrt{1-\frac{r_s}{r}-\frac{1}{1-r_s/r}\frac{\dot r^2}{c^2}}dt$$ The integrand is the Lagrange function: $$L=-mc^2\sqrt{1-\frac{r_s}{r}-\frac{1}{1-r_s/r}\frac{\dot r^2}{c^2}}$$ Because $|\dot r|\ll c$ and $r_s\ll r$, and using Tailor series $\sqrt{1+x}\approx1+x/2$ it can be approximated as: $$L\approx-mc^2\sqrt{1-\frac{r_s}{r}-\frac{\dot r^2}{c^2}}\approx-mc^2\left(1-\frac{r_s}{2r}-\frac{\dot r^2}{2c^2}\right)$$ Largange equations are: $$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot r}\right)-\frac{\partial L}{\partial r}=0 \implies m\ddot r=-\frac{mc^2r_s}{2r^2}=-\frac{GMm}{r^2}$$ In the last equation formula $r_s=2GM/c^2$ was used.

atarasenko
  • 1,009