# Geodesics VS Projectile

by Black Integra
Tags: geodesics, projectile
 P: 56 Yes, I want to make sure that geodesics of a particle moving in curved space time is the same thing of projectile trajectories. I start from assuming that $1-\frac{2GM}{r}\approx1-2gr$ and then calculate the schwarzschild metric in this form $\Sigma_{\mu\nu}=\begin{bmatrix}\sigma & 0\\ 0 & -\sigma^{-1}\end{bmatrix}$ where $\sigma = 1-2gr$ and I calculated for the Christoffel symbols for this metric: $\Gamma^0_{\mu\nu}=-\sigma g\begin{bmatrix}0 & 1\\ 1 & 0\end{bmatrix}$ $\Gamma^1_{\mu\nu}=-\frac{g}{\sigma^2}\Sigma_{\mu\nu}$ I plugged them to a geodesics equation $\partial^2_\tau x^\mu = -\Gamma^\mu_{\alpha\beta}\partial_\tau x^\alpha\partial_\tau x^\beta$ where $d\tau^2 = dx^\mu dx^\nu\Sigma_{\mu\nu}$ and I got these ugly conditions: $\partial^2_\tau t = \sigma\partial_\tau t\partial_\tau \sigma$ $\partial^2_\tau \sigma = \frac{2g^2}{\sigma^2}$ what I expect is just something like $x=-\frac{g}{2}t^2$ I havn't finished these differential equations yet. But I want to know that I'm going through the right track, right? Any suggestion?
 Emeritus Sci Advisor PF Gold P: 5,585 In the level of approximation you're using, you might as well make your life easier and approximate $\sigma^{-1}$ as $1+2gr$. I would also define a new coordinate $\rho=gr$ to avoid having to write all the factors of g.
 P: 56 I can't see why we can assume that $\sigma^{-1} = 1+2gr$, they're not quitely equal.(at least at the earth's surface) But i think i can apporximate σ to be -2gr because 1 is very small comparing with -2gr. But I still can't find a way to prove this. Please, any can help me?
Emeritus
P: 7,599
Geodesics VS Projectile

 Quote by Black Integra I can't see why we can assume that $\sigma^{-1} = 1+2gr$, they're not quitely equal.(at least at the earth's surface)
It IS an approximation. The taylor series expansion of 1/(1-x) is 1+x+x^2 + o(x^3), basically. So if x is small, it's a good approximation.
 PF Gold P: 4,087 I don't think this approach can give the Newtonian result because you are using kinematic equations. The full equation for r is $$\ddot{r}=-\frac{\left( m\,{r}^{2}-4\,{m}^{2}\,r+4\,{m}^{3}\right) \,{\dot{t}}^{2}-m\,{r}^{2}\,{\dot{r}}^{2}}{{r}^{4}-2\,m\,{r}^{3}}$$ setting $\dot{t}=1$ and doing a Maclaurin-Taylor expansion of the RHS in m $$\ddot{r}=-\frac{\left( 1-{\dot{r}}^{2}\right) \,m}{{r}^{2}}+\frac{2\,\left(1+{\dot{r}}^{2}\right) \,{m}^{2}}{{r}^{3}}+\frac{4\,{\dot{r}}^{2}\,{m}^{3}}{{r}^{4}}+ ...$$ assuming m << r we get $$\ddot{r}=-\frac{\left( 1-{\dot{r}}^{2}\right) \,m}{{r}^{2}}$$ which does not have a closed form solution. In this m=GM/c2. However it is possible to deduce Newton's law of gravitation from GR by another approach.
P: 56
 Quote by pervect It IS an approximation. The taylor series expansion of 1/(1-x) is 1+x+x^2 + o(x^3), basically. So if x is small, it's a good approximation.
That's the point. I use 1-2GM/r = 1-2gr because I calculate in case where g=9.8 and r is around the earth's radius (not small, is it?)

 Quote by Mentz114 I don't think this approach can give the Newtonian result because you are using kinematic equations.
Oh. I have never heard something like this before, it's new for me. What's the name of the other way, other than kinematic equation?

 Quote by Mentz114 However it is possible to deduce Newton's law of gravitation from GR by another approach.
What is the other approach to deduce the classical mechanics? Mainly, I just want to find out that Euler-Lagrange Equation and Geodesics Equation are the same concept.
 PF Gold P: 4,087 The geodesic equation is found by extremizing the action for a free particle which is $$\int_{\lambda_1}^{\lambda_2}\frac{ds}{d\lambda}d \lambda = \int_{\lambda_1}^{\lambda_2}\sqrt{g_{\mu\nu}\frac{dx^\mu}{d\lambda}\fra c{dx^\nu}{d\lambda}}d\lambda$$ where s is the proper length. Look up 'weak field theory' in the context of GR to see how Newton's law can be inferred from GR. It's too involved for me to reproduce here.
 PF Gold P: 4,087 After some reading I found the correct procedure. From my post #5 $$\ddot{r}=-\frac{\left( 1-{\dot{r}}^{2}\right) \,m}{{r}^{2}}+\frac{2\,\left(1+{\dot{r}}^{2}\right ) \,{m}^{2}}{{r}^{3}}+\frac{4\,{\dot{r}}^{2}\,{m}^{3 }}{{r}^{4}}+ ...$$ Now for static rest particle $\dot{r}=0$ so the leading term gives the Newtonian value. $$\ddot{r}=-\frac{m}{{r}^{2}}=-\frac{GM}{{r}^{2}}$$ A longer way is to start with $$g_{\mu\nu}=\eta_{\mu\nu}+f_{\mu\nu}$$ and $$\frac{d^2 x^a}{d\tau^2}= -\Gamma^a_{bc}\frac{dx^b}{d\tau}\frac{dx^c}{d\tau}$$ throwing away lots of stuff and setting the 4-velocities to (1,0,0,0) getting $$\frac{d^2 x^a}{dt^2}= -\Gamma^a_{00}= \frac{1}{2}\eta^{ab}g_{00,b}= \frac{1}{2}\eta^{ab}f_{00,b}$$ which works for $f_{00}=2m/r$
 P: 56 thx, that's clear :)

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